Non-harmonic tones

This'll just be a quick post, on music theory this time rather than the historical stuff.

Non-harmonic tones are basically the notes that aren't part of a chord per se, but are there to make the music more pretty and melodious. After all, most music isn't exactly comprised of entirely block chords, unless the composer's using them for a certain effect.

For example, in the example below, the A in the soprano note is a non-harmonic tone because it isn't a part of chord IVb.

The different kinds of non-harmonic tones also have different names. In the above example, the A is a passing note because the soprano line "passes" through there to get from the G to the B. This particular passing note is an unaccented passing note because it doesn't fall on the same beat as the rest of the chord. If it did- for example, if the G and the A were switched around- then it would be an accented passing note.

Now, let's look at the other kinds of non-harmonic tones!

In the above example, the A and the F# are neighbouring tones, so called because they are the "neighbour" of the notes on either side. Bear in mind that the two notes on either side must be the SAME note. If the note before is a step higher than the note in the middle, and the note after is a step lower than the note in the middle, or vice versa, then the middle note is a passing tone, not a neighbouring tone.

In the above example, the A is an anticipation tone. That is because it comes in earlier than it really needs to. The first chord is a IV chord... but wait! There's an A in the soprano! There's no A in C major chords... but there is one in D Major chords, and that's precisely what the second chord is! The A has simply come in too early, "anticipating" the next chord. That's why it's an anticipation tone.

(By the way, I've noticed that my cadences are pretty dodgy in some places. Apologies for that.)

The note after an anticipation tone can be either higher or lower than the anticipation tone. In the above example, we move from an A (anticipation tone in this case) to B.

Retardations and suspensions kind of work in the opposite way. Instead of being the same as the note in the next chord, retardations and suspensions carry over a note from a previous chord:

In the above examples, G is a note in the IV chord, but not in the ii chord. In the first example, G is a retardation tone because the note afterwards is higher. At the risk of being somewhat politically incorrect, you could say that the note is "retarded" because it has to wait longer until it can finally move up in pitch. In the second example, G is a suspension tone because the next note is lower. You could say that the G is "suspended in mid-air" until it finally falls on the F#.

An appogiatura is where you have a larger gap leading to the non-harmonic tone (normally an interval of a 3rd) and then a step downwards. Like this:

Except C is in an A minor chord, so I'm not entirely sure if it qualifies as a non-harmonic tone.

Escape tones are the opposite- a step and then a larger gap. Like it's trying to "escape" by stepping away but then realises that it can't.

(I just reversed the order of the chords from the last example. Original, aren't I?)

Those are pretty much all of the non-harmonic tones that you need to know for now. Isn't music theory just so incredibly fun? w00t.

Lit notes - The Turning by Tim Winton

I have a Lit exam on Wednesday next week and I sure as hell don't feel prepared. I've done a couple of close readings (or rather, I've done a couple of half close readings- an intro and a body paragraph on poetry and an intro and a body paragraph and a bit of another body paragraph on prose) and I started trying to organise my notes on the texts that we've done this year. (Note- this was at the time that I started writing this post, which I think was on Thursday.) My organisation is pretty shoddy though, just like my room where everything kind of has its place but really doesn't.

Anyway. Here are my notes for The Turning (the only reason I'm typing them up here is because I'm procrastinating over writing practise essays...). Bear in mind that I'm not that great a Lit student, so don't take my word as gospel. Or actually, maybe you can, since half my notes are taken from handouts and class powerpoints and whatnot.

ANYWAY...

The Turning by Tim Winton

Context
The Turning is set in Australia.

• Long, wide spaces- Australia is supposed to be one of the most sparsely populated countries so we get lots of space which is nice. There's a lot of times during the novel (or during the 17 short stories, if you prefer) where characters have to cross these long, wide spaces, like when Vic goes to find his dad in "Commission," when Vic is coming home after his cadet camp in "Immunity," or when Gail and Vic are going to visit Daisy and Fenn in "Defender." And then there's a chapter titled "Long, Clear View" which kind of adds to this because you wouldn't really be having a long, clear view if it wasn't for the long, wide spaces available. And then I guess you could also kind of argue that there's a long, wide space between Vic and Gail in terms of their relationship. Especially since Gail's continually visiting Angelus and even going out with some other guy there to boot.
• The outback- Well, it's pretty wild and interesting and unique to Australia, which is why there's always been lots of poems and stories and whatnot dedicated to it (bush rangers, anyone?). Sure enough, there's a few stories in here in which the wildness of Australia's outback is important- stories like "Aquifer," for instance, where Alan Mannering drowns in a swamp, as well as "Fog" where Bob Lang and Marie are stuck in the bush overnight.
• Connection between people and place- particularly important to Aboriginal culture, but it can still be seen in this book even though there aren't that many Aboriginals. For example, the swamp is a big part of the children's life in "Aquifer" and then there's the whole idea of returning to one's hometown to search for answers just as Frank does in "Family" and Gail kind of does in "Damaged Goods" (but not really since Gail's trying to find answers about Vic's life).
• Mainly secular society, but there are still many people who practise a religion of some description. The Turning has several characters who are Christians, such as Sherry and Dan in "The Turning" and Agnes' family. Christianity certainly helps Raelene in her time of need.
• There are a lot of immigrants but sometimes there's still a sense of who's "Australian" and who's an immigrant. Agnes' family is British, and Agnes refers to herself as "fresh off the boat."
• Notion of the "underdog" eventually gaining superiority: Frank was the "underdog" in his younger years- bullied by Max and never quite able to walk among the men just as Max could. However, he eventually becomes a professional footballer (though not for long).
• Archetypal Australian men and women: apparently men are meant to be all tough and independent and  masculine and whatnot and women are meant to be "stoic," which (according to Google) means "a person who can endure pain and hardship without showing their feelings or complaining." Now, there are characters that naturalise these ideas, as well as other characters who challenge them. There are whole bunch of characters who are, well, not quite as tough and independent as the "ideal" male figure. Max, for instance, is a violent alcoholic and we're pretty much set up to hate him through the point of view (we mainly see him from the point of view of the characters who he victimises- more on this later), Frank walks off the football pitch after a rather short career, Vic is so absorbed in his past that nothing in the present holds his attention, and so on. On the other hand, though, there are other characters that might be considered as upholding the ideals of dependability and so on such as Dan who recovers from alcoholism and I think Dyson from "Small Mercies" as well for being able to be relied on by Fay and her parents (if I remember correctly- it's been a while since I read that story. In fact it's been a while since I read the book). There are then also characters that represent the "stoic woman," such as Carol Lang, who doesn't buckle under that other woman's accusation that Carol stole her earrings, and Raelene, who puts up with Max's abuse for a while without complaining. I'm not so sure about which characters challenge this ideal. Maybe Fay? I don't know. As I said, I haven't read this book for a while, and I've only read this particular chapter/story once.

Genre

• The Turning can be considered to be Realist fiction, which is a less ornate style which uses verisimilitude (a very cool term which I think means that the story is made to be as realistic as possible) to illuminate truths about political and social elements of society, such as class differences. Realist fiction also often has individuals raging against the world rather than any massed effort to change the system. You can see a lot of individuals raging against society across the stories- Bob Lang leaves Angelus because there is something about the police force that makes him feel uncomfortable (potential corruption?), Rae is locked in an abusive relationship but undergoes spiritual growth to become internally stronger than Max, Frank is still struggling with being victimised by Max which results in him walking off the football pitch, and so on. You can also kind of see class differences being illuminated, such as in "On Her Knees" in the contrast between the unnamed home owner and Carol and Vic Lang who are left to clean it (after receiving empty accusations about stealing the home owner's earrings), though I don't recall seeing any huge differences.
• The Turning can also be considered to be Romantic fiction, using Romantic themes such as the power of nature as restorative, a lone man against society and freedom of the individual. (I wrote a little bit about the Romantic period on my post about music and society during the Romantic period.) As I said above, there are quite a few times where there are lone people against society. There are also quite a few instances throughout the novel where there are people alone in nature, using nature's restorative power to heal themselves. Raelene in "The Turning" often uses her walks along the beach to think, which leads her to come to revelations which eventually lead to her finding her own internal freedom from Max. Complementing this is language that revels in the beauty of nature through various kinds of imagery, such as in "Fresh-mown grass felt good beneath her feet and over the green smell of it you could almost taste the sea." Nature is green and fresh and feels good, contrasting with the horrors of the abusive relationship that Raelene is locked into. Another character who can be said to heal in nature is Bob Lang, because in his new home tucked away in the bush, he becomes the trustworthy, honest character known as "Honest Bob"- a character who he wasn't quite able to be when in society. As well as nature having the power to heal, nature is also powerful and dangerous as can be seen through Alan Mannering drowning in the swamp and Vic becoming injured on a hook when their fishing boat is overtaken by a wave.
• Finally, The Turning can be considered to be a series of parables- stories that are meant to teach us moral and/or spiritual lessons. For example, Max is a pretty mean character who eventually gets eaten by a shark. Also, before his death, Raelene became stronger than him through finding God. And then of course there's Carol Lang upholding the virtue of dignity by her actions in "On Her Knees." (I don't really have many more notes on how this book is a series of parables though.)

Language and Generic Conventions

I'm just going to put a few points down here because, well, a novel has lots of words and thus lots of conventions and stuff to write about. It really depends on how close you want to look. (For me- not very.)

• Possibly the one generic convention that will hit you the hardest in this book is the constantly changing point of view. The point of view moves between 1st person, 2nd person and 3rd person, the character in focus changing between stories. "Big World" is told through the 1st person perspective of an unnamed narrator who's just graduated from high school, "Abbreviation" is told through 3rd person limited focusing on Vic Lang, and so on. So what's so important about these points of view? Well, point of view in a novel really affects the way that we see characters and events in the story. It's just like how in class someone else might get a mark that doesn't seem so bad to you but it might seem devastating to the person in question. Now, let's look at Max's character. We're pretty much set up to hate Max through the point of view. The three stories in which Max is involved- "The Turning," "Sand" and "Family"- are through 3rd person limited perspective focusing on Raelene and Frank, two characters who are badly treated by Max, so we sympathise more with Raelene and Frank and the struggles that they have to endure against a seemingly tyrannical character. If one of the stories was told with a greater focus on Max, though, perhaps a slightly different story might be told. I'm not sure that a focus on Max would make us like him given all the other stuff that he's done, but if, say, Max actually felt remorseful for what he does, and the story was told from his point of view so that we could see his remorse, we might hate him less. Vic Lang is another character who is described from multiple points of view, but unlike Max, we can see both sides of the story. "Long, Clear View" allows us to see Vic's side of the story, but in 2nd person, which perhaps makes us closer to the raw emotion within the story. On the other hand, you have Gail talking about Vic's obsession with the past in "Damaged Goods," as well as 3rd person limited focusing on Gail in "Defender" in which Gail outright tells Vic about her frustration.
• Just like pretty much every other book, there are objects that can be read as symbols. Unfortunately I don't have many notes on the symbols in this book and I'm pretty bad at telling what can be considered as a symbol and what's best just left alone. Anyway, fire, water and sharks are all (possibly) symbols that recur throughout the novel. Fire is seen as damaging but cleansing, like the destruction of an old life and the start of a new one... oh wait! The novel title is, after all, "The Turning," "turning" also suggesting such radical changes! And I don't know about the cover of your copy, but the cover on mine has a burning car... okay, maybe I'm getting ahead of myself, because Lit doesn't really have visual analysis in it. But still. Now, where do we see fire in The Turning? Well, there's Agnes, who may or may not have burnt her "dead inside" house down at the end, causing the family to have to start anew. Then, at the end of "Big World," the narrator tells about all the changes and stuff that happens in the future after he talks about how the car "[smouldered] and [hissed]." As for water, well, pretty much all the people live near beaches, like a lot of people in Australia. Quite a few significant events happen in the water too, like Vic's injury in "Abbreviation," Alan Mannering's death in "Aquifer," Boner McPharlin is found with broken legs at Thunder Beach, and so on. And, where there are beaches, there are sharks. Boner McPharlin must have a thing about them because he catches and kills a shark at the bonfire party at Massacre Point and he gives a shark sticker to Jackie (well, according to my notes he does anyway). Then there are symbols only relevant to one story, like Raelene's snow globe in "The Turning" which reminds her of Jesus and gives her the strength to keep fighting Max.
• And, of course, like pretty much every other book under the sun (except for maybe some crazy postmodern book out there) has characters in it. These characters can then be read as representations of class, cultures, gender and so on. Like I guess you could say that Max is a representation of a violent drunk and that Carol Lang is a representation of an archetypal Australian stoic woman. Vic Lang's characterisation can be read as examining the effects of the past on the present. Etcetera. Oh, and don't pay too much attention to what I just said because I have no idea what the hell I'm doing.

Intertextuality

• The epigraph is part of a verse from T.S. Eliot's poem "Ash Wednesday." According to my notes, Ash Wednesday is the beginning of the Lent season, so it symbolises repentance, loss and waiting for salvation. And, sure enough, you see a hell of a lot of characters repenting stuff they've done before (like Bob Lang), going through some kind of loss (like Dyson in "Small Mercies) or waiting for salvation (like Raelene in "The Turning). In the epigraph, however, Winton removed the first and last line of the verse- the first line being "And pray to God to have mercy upon us" and the last line being "May the judgement not be too heavy upon us." Removing these lines means that Winton has, in essence, secularised the poem a bit, possibly to reflect that Australian society is considered to be a relatively secular society. He has also secularised concepts such as redemption and forgiveness suggesting that these concepts are universal to all, not just those who practise a religion. Another point of interest is the word "hope" in the line "Because I do not hope to turn again." "Hope" suggests a lack of control, because if you had control over doing something, then you would make it happen, rather than crossing your fingers and hoping that it might happen.
• Although not really intertextuality per se, the short stories that The Turning is comprised of are all interconnected. Obviously, there are characters that appear in multiple stories: Vic Lang, for instance, appears in "Abbreviation," "Damaged Goods," "On Her Knees," "Long, Clear View," "Reunion," "Commission," "Immunity" and "Defender." Max appears in the middle three stories: "The Turning," "Family" and "Sand." Apart from just characters, though, there are other themes that crop up constantly. Symbols such as fire, water and sharks are mentioned in multiple stories. Although the society is mainly secular, there are religious messages in some stories, like "Cockleshell" and "The Turning." The power of nature is also alluded to in many stories throughout the novel. There are also similarities between different characters in different stories, just like how Vic's infatuation with Strawberry Alison in "Damaged Goods" is somewhat mirrored by an unnamed girl's infatuation with Vic. Finally, there are the ideas of being stuck in the past- like Vic Lang, or like the unnamed narrator of "Aquifer"- and of needing to escape, like the many characters who seek to get out of Angelus and the several others who seek escape through alcohol like Max does.
• There are several other allusions to other texts. Biblical stories get mentioned a couple of times, as do some other books, such as Catcher in the Rye (Jackie briefly mentions it in "Boner McPharlin's Moll"). Unfortunately, even though I've actually read Catcher in the Rye, that was a while ago and to be honest I had a hard time getting my head around what was going on. (I'm a pretty poor reader. Get over it. Or maybe I'm not that bad a reader but I have to read something a few times for the words to sink in? I dunno.)
• Finally, the chapter "Family" starts with a poem by Chuang Tzu. I guess the main idea in this poem- that the need to win a prize drains an archer of his ability to shoot- is reflected in the story by how Frank's desire to prove people wrong eventually leads to him walking off the football pitch, which really doesn't help matters much.

Ideologies and Discourses

This is probably my shakiest point. Never mind the fact that Lit is already a shaky subject for me. Recipe for disaster much?

• Romantic ideas about escape via nature and the confinement of society are all throughout the novel. Although I've mentioned quite a few Romantic ideas back in the section about genre, I've got a few more to add on here. First of all, being in nature means that you don't have to worry about all the complex and cultural values that you have to encounter in society so you can act a bit more freely, like Max and Frank do in "Sand." Also, although this isn't really glorifying nature per se, small towns (which are a bit closer to nature due to their smaller size) are favoured over big cities, as can be seen through the way that the small towns like Angelus and White Point are named but the city is not. Society is also seen as confining through setting such as Raelene and Max's claustrophobic caravan home. Vic's dad also breaks down as a result of stuff happening in the police force, and his breakdown causes his relationship with his other family members to become more strained until at last he can't handle it any longer and he leaves.
• There are some ideas about masculinity and patriarchy. In "Sand," Max and Frank are becoming men, or rather, Max walks among the men while Frank hasn't quite become a man yet. In "Cockleshell," Brakey gets stung by a catfish but he refuses to tell his mother, believing that he is man enough to take it on by himself. Also, male figures are important to Raelene, so much so that in order to reject one man (Max), she has to accept another in his place (Jesus). Of course, some ideas about masculinity are challenged by some of the other characters, like Bob Lang who resorts to drinking for a while.
• Finally there are lots of ideas about family. Although there are lots of dysfunctional families in this story, like the Langs after Bob left, Raelene's family, and Frank and Max, there is still some sort of importance attached to family. For example, Frank and Max depend on each other.

I might expand this bit later. But I'm tired and I should probably get around to practising clarinet before my sister comes home (she's not a huge fan of Poulenc). TTFN.

Chemical Equilibrium

This is one topic which has quite a lot of new stuff to learn, so hold onto your hats!

The first two dot points, regarding enthalpy diagrams and collision theory, are explained in my post on Reaction Rates: http://year11misadventures.blogspot.com.au/2012/10/reaction-rates.html.

Now for the new stuff! Half this stuff probably needs diagrams but I can't be bothered doing them just yet, so I might get back to this post later (key word here is "might." Great blog writer I am).

Describe and explain the characteristics of a system in dynamic chemical and physical equilibrium

A lot of the reactions that we have dealt with thus far go to completion: that is, all of the reactants are fully consumed to form products (unless there is a limiting reagent, in which case only one of the reactants is fully consumed). However, there are a lot of reactions that do not go to completion. One such reaction which you would have probably encountered is the ionisation of acetic acid. Acetic acid only partially ionises into acetate/ethanoate ions and hydrogen ions: the rest remains as acetic acid molecules. Well, actually, that isn't totally true, because at any one time some acetic acid is dissociating into ions and some acetate and hydrogen ions are combining to form acetic acid molecules. The catch is, both of these processes are happening at the same rate, so the system is therefore at equilibrium.

Since the two reactions are proceeding at the same rate, the decrease in concentration of any one particular substance is counterbalanced by an equal increase due to the opposite reaction, and vice versa. Therefore, when a system is at equilibrium, the concentrations of the substances in the system are constant. This also means that other macroscopic properties ("macroscopic" being a word I only learned just then by glancing in my textbook), such as colour, also remain constant.

Write equilibrium law expressions for homogeneous and heterogeneous systems

Writing equilibrium law expressions is the easiest part of learning about equilibrium. Basically, on the top line, you have the concentrations of each of the products raised to the power of their coefficients, and on the bottom line you do the same thing for the reactants. For example, in the Haber process of producing ammonia, N₂ + 3H₂ ßà 2NH<sub>3, the equilibrium law expression is:

K = ([</sub>N₂] + [H₂]^3)/ ([NH₃]^2)

(K is the symbol used for equilibrium constants. Don't ask.)

Remember that solids and liquids do not get included in this expression, as they don't have concentrations. Only aqueous solutions and gases get included.

If your products are all solids and/or liquids, put 1 on the top line. One way of thinking about that is that the concentration of a solid or a liquid is equal to 1 (though obviously not strictly true as, as I just said, solids and liquids don't have concentrations).

Use K and equilibrium law expressions to explain the relative proportions of products and reactants in a system of dynamic chemical equilibrium

Since the products are the numerators and the reactants are the denominators, when there's a relatively large quantity of products formed, the equilibrium constant is relatively big. Similarly, a small equilibrium constant is indicative of the formation of a relatively small amount of products.

Explain, using the collision theory, the effect on the position of equilibrium when the following changes are made to a system initially at chemical equilibrium.

Changes in solution concentration: When the concentration of one of the aqueous solutions in the system gets changed, the system is no longer at equilibrium because the equilibrium law expression no longer equals whatever equilibrium constant it was equal to before (and the only time that an equilibrium constant changes for a solution is when the temperature is changed). Therefore, equilibrium shifts in favour of the reaction that will help to bring the concentration of said solution back to what it was before: if the concentration was increased, then the reaction that consumes that solution will be favoured, and if the concentration was decreased, then the reaction that forms that solution will be favoured.

Oh, wait, I was meant to use the collision theory. Oops. Well, increasing the concentration of an aqueous solution means that there are more particles of that solution, so that the chances of them colliding with other reactant particles are increased and, therefore, that reaction proceeds faster. If the concentration is decreased, then they collide less often and, therefore, that reaction proceeds slower.

Changes in partial pressures of gases: Partial pressures are kind of the same as solution concentrations (as far as I know, anyway). So just see what I've written for solution concentrations above. However, if the overall system pressure is changed, other stuff happens too, which I'll talk about later.

Addition of a catalyst: Catalysts do not affect equilibrium- they only affect the rate at which a system reaches equilibrium.

Predict, using Le Châtelier's principle, the impact of certain changes to a system initially at chemical equilibrium

Before I dive into all the prediction stuff, I first need to explain what Le Châtelier's principle is. Basically, Le Châtelier's principle states that if a change is made to a system at equilibrium, for example the concentration of one of the substances has been altered, or the temperature has been changed, then the system will do something to at least partially counteract this change. For example, if the pressure is increased, the rate of the reaction that produces fewer particles and thus partially counteracts the change in pressure will be favoured. And so on, and so forth. Let's take a closer look:

Addition of a catalyst- I'm starting with this one, because it's easy. As I said before, catalysts do not affect equilibrium- only the rate at which a system reaches equilibrium.

Changes in solution concentration/ partial pressure of gases: I accidentally touched on this in the previous section (the section where I was meant to talk about collision theory).

Change in pressure: When pressure is raised, all gases have a higher concentration, so both forward and reverse reactions proceed faster. However, one reaction will initially proceed faster than the other. The reaction favoured in this case is the one that produces fewer particles. For example, in the Haber process, N₂ + 3H₂ ßà 2NH₃, the production of ammonia would be increased if pressure was to increase as the forward reaction produces fewer particles than the reverse reaction (2 particles of ammonia for the forward reaction and 1 particle of nitrogen gas and 3 of hydrogen gas for the reverse reaction). The inverse is also true- if pressure is decreased, the reaction producing more particles is favoured.

Change in temperature: When temperature is raised, both reactions proceed more quickly (see my previous post on Reaction Rates). However, one will proceed more quickly than the other. To cool down the system, the endothermic reaction is favoured. The inverse is also true- if temperature is decreased, the exothermic reaction is favoured in order to heat up the system.

By the way, changing the temperature is the only way to change the equilibrium constant (K).

Interpret changes, such as colour changes, of physical and chemical systems at equilibrium

For changes like these, simply work out what colour (or whatever) each "side" of the reaction is meant to have- for example, the reactant ions might be yellow while the product ions are orange. As you do stuff to the system, work out whether the changes move more in favour of the product side or the reactant side, for example, is the solution becoming more yellow or more orange?

Conditions of Industrial Processes

Sometimes you get given some random industrial process and you have to predict the ideal conditions under which to perform these processes.

First of all, you should pretty much always suggest using a catalyst, as catalysts increase reaction rate without having any effect on the yield.

Secondly, use Le Châtelier's principle to work out whether an increase or decrease in temperature will favour the products. Then do the same for pressure. If an increase is required, that's great, since increasing temperature and pressure also increases reaction rate. If a decrease is required, then say that a "medium" temperature or pressure is required, since having a really low temperature or pressure will just make the reaction proceed at a snail's pace.

However, high temperatures and pressures cannot be indefinitely high, as there's economic and safety limits on stuff. Plus our technology is not infinitely good either.

Describe and explain the conjugate nature of buffer solutions

Buffer solutions? Never heard of these before... Oh, wait, my textbook has them in the next chapter. So maybe that's why I haven't heard of them before. Maybe this dot point got put in the wrong place or something. Or maybe we're meant to apply Le Châtelier's principle to buffer solutions once we learn about them. Stay tuned...

Exponentials and Logs

I'm going to very very briefly go over stuff that was covered last year, and then explain the new stuff.

Differentiating exponential equations: Multiply by the derivative of the power of e, keep everything else the same. (e.g. e^(2x) becomes 2e^(2x))
Integrating exponential equations: Divide by the derivative of the power of e, keep everything else the same (e.g. e^(2x) becomes (1/2)e^(2x))

ln (natural log) is a log to base e
A log with no base specified is a log to base 10

Converting logs to exponentials:

Remember, the bottom number (the a in this case) is the base.
According to Wikipedia, the b is called the argument in the logarithmic function (top) and the answer in the exponential function (bottom).
Also according to Wikipedia, the c is called the answer in the logarithmic function and the power in the exponential function.
Though Wikipedia also suggests that the terms "argument" and "power" are interchangeable.

Adding and subtracting logs: You know how when you multiply numbers with powers, you add the powers, and when you divide numbers with powers, you subtract the powers? Well, logs are the opposite. You multiply when you need to add and divide when you need to subtract. I know, I'm terrible at explaining. And there's the power law too, which involves bringing the power around to the front- also a pretty terrible explanation. So here's some formulae, courtesy of Microsoft Word:

Differentiating logarithmic equations: This method ONLY works for natural logs- logs with base e, or ln.

Derivative of the function ln (f(x)) is given by (f '(x))/ (f(x)) - that is, the derivative of the argument divided by the argument (see "converting logs to exponentials," two sections above).

If you don't have a natural log, you need to convert whatever you have into a natural log. To convert an exponential into a natural log, take a natural log of both sides. For example, 2^x = y can be rearranged to ln (2^x) = ln y and then x ln 2 = ln y and finally x = (ln y) / (ln 2). From there you can find dx/dy and then dy/dx.

To convert a log of another base into a natural log, first convert to exponential form and then use the above method.

Integrating to give logarithmic equations: The new stuff! Finally!

Remember how back in 3AB we were told that you can't find the integral of x^(-1) using the old "raise the power by one and divide by the new power" rule because then you'd end up dividing by 0? Well, fear no more, because now we have a new secret weapon! Natural logs!

You see, x^(-1) = 1/x. And the derivative of ln x = 1/x! Therefore, the integral of x^(-1) must be ln x! (EDIT: Actually, it's ln x + c.)

Well, actually, it's slightly more complicated than that. You see, if you want to find the area under the curve of y = x^(-1) you'd want to use integration, which would lead you to using ln x. The thing is, though, the curve of y = x^(-1) exists for negative values as well as positive values, while ln x is undefined for negative values. To get around this, we actually have to write the integral of x^(-1) as ln |x|. No, ln |x| + c. Oops.

Now for some more fancy stuff! Clearly, there are other logarithmic functions out there than ln x, and therefore lots more equations with logs as their integrals. Let's see how we can work this one out.

First of all, put the bit at the bottom into a natural log (probably not really correct terminology, but whatever). That is to say, if you have 2x + 3 at the bottom, make it ln (2x + 3) and move it to the top. Now, hopefully you also have a multiple of the derivative- in this case 2- sitting at the top with it. (If not, you can't integrate.) Then divide this multiple by the derivative.

Wow, today must be a day for terrible explanations. So here's an example that might shed some light on the matter.

(integral sign) (2 + cos 2x)/(4x + sin 2x) dx

Step 1: Put the bit at the bottom into a natural log and move it on top.

(2 + cos 2x) ln (4x + sin 2x) dx

Step 2: Divide by the derivative of the stuff that was originally at the bottom.

((2 + cos 2x) ln (4x + sin 2x))/(4 + 2 cos 2x)
= ((2 + cos 2x) ln (4x + sin 2x))/(2 (2 + cos 2x))
= (1/2) ln(4x + sin 2x)

Step 3: DON'T FORGET THE PLUS C!!!!

= (1/2) ln(4x + sin 2x) + c

(Please note: I don't know for sure if the above method is mathematically correct or whatever... but hey, it works. Just make sure you've got the derivative of the stuff at the bottom at the top, and you should be fine. Otherwise, use the substitution method as I outlined here: http://year11misadventures.blogspot.com.au/2013/05/integrating-by-substitution.html.)

Differential Equations

So what is a differential equation you might ask? Well, it's an equation that involves derivatives. For example,  dy/dx = 2 is a differential equation. To solve an equation like this, we just use integration, and integrate both sides with respect to x, giving us y = 2x + c. (Don't forget the +c!!!)

Solutions to differential equations involving a + c are called general solutions to the equation. If you're given more information, though, like what y is at a certain point, then you can find a particular solution.

My textbook's also reminding me that, technically, when you integrate both sides, you get a +c on both sides. The integral of (dy/dx) is actually y + c and the integral of 2 is 2x + c. You only need to leave in one of the cs, and this c is technically equal to the difference between the two cs.

Now, if you have something like y(dy/dx) = 2x + 3, you still integrate both sides with respect to x... (and yes, I'm sorry that I still don't know what to type in as a good substitute for the integral sign)

(integral sign) y(dy/dx) dx = (integral sign) 2x + 3 dx
(integral sign) y dy = (integral sign) 2x + 3 dx
(y^2)/2 + c = x^2 + 3x + c
(y^2)/2 = x^2 + 3x + c

Now, look closely at the second line. See how the y is with the dy and the x is with the dx? This is what you should aim to do when you rearrange the equation. Although dy/dx isn't really a fraction- it's the limit of a fraction- thinking of it as one might help you to rearrange the equation so that the ys are with the dy and the xs are with the dx.

Wow, that was a relatively short post (for me, anyway). I can't really be bothered doing area under a curve now because you've probably already done it before and I have other stuff to write about like integrating logarithmic functions, 3D vectors, the shapes of molecules, and Shakespeare's Hamlet. I might write about area under a curve once I've revised everything else that I need to (or maybe while I'm procrastinating over revising those things).

Oh, and by the way...

DON'T FORGET THE + C!!!!!!!!!!!!!!!!!!

Integrating by Substitution

This (hopefully) shouldn't take too long to explain. Basically, you generally use integrating by substitution when you've got something in brackets raised to a power and something else sitting outside those brackets that isn't a multiple of the derivative of the stuff in the brackets.

When you integrate by substitution, you set the stuff in the brackets equal to u. For example, if you want to find the integral of 20x(2x + 3)^3, you might then make 2x + 3 = u.

You then need to play with this equation to find two things: an equation for x and an equation for u.

Rearranging the equation, x = (u - 3)/2.
Also, du/dx = 2. Therefore, dx/du = (1/2).

I don't know how to type in integral signs on Blogger, and I don't know what the shorthand is either. So bear with me here.

(integral sign) 20x (2x + 3)^3 dx
= (integral sign) 20((u - 3)/2)(u)^3(dx/du) du
= (integral sign) 10 (u - 3) (u)^3(1/2) du
= (integral sign) (5u - 15)(u)^3 du
= (integral sign) 5u^4 - 15u^3 du
= u^5 - (15/4)u^4 + c
= u^4(u - (15/4)) + c
= (1/4)(u^4)(4u - 15) + c
= (1/4)(2x + 3)^4(8x + 12 - 15) + c
= (1/4)((2x + 3)^4)(8x - 3) + c

You have to make sure that you pop in your substitution for x (in this case x = (u - 3)/2) otherwise you'll end up trying to integrate two variables at once, which I'm told isn't allowed. This is also why you need to have a multiple of the derivative "sitting outside the brackets" in order to integrate. You don't really have to use the substitution method when you do have a multiple of the derivative present, but you still can.

(Well, I'm currently looking at another example which is to find the integral of 8 cos (2x) (sin 2x)^5... but I'm not really sure how to find x in terms of u using the substitution u = sin 2x. At least in this question a multiple of the derivative is present...)

Integrating Trig Functions

Okay. I HATE INTEGRATING TRIG FUNCTIONS. Absolutely hate doing it. But I have to for Spec. So I'm going to explain how to do it.

The Basics

Probably one of the main reasons why I hate integrating trig functions is because I keep mixing up the derivatives of the trig functions with their integrals. Plus you might get some that seem tricky to work out and then you realise that the trig function in question is simply a multiple of one of the three derivatives of tan x.

Here are the integrals that you need to know:

Integral of sin x = -cos x
Integral of cos x = sin x
Integral of (1/(cos x)^2), (sec x)^2 or 1 + (tan x)^2 = tan x

All the other rules work as before. If the trig function's raised to a power, raise the power by one and divide by the new power and the derivative of the trig function- but make sure that a multiple of said derivative is also present. So you can't differentiate (sin x)^3 directly because sin x's derivative, -cos x, isn't "sitting outside the brackets." (There are, however, other ways to differentiate this, which I will talk about later.) This is where I always screw up because I always divide by the integral by accident, which means that I get the positive/negative signs all wrong.

Oh, and don't forget the...

+ c

If you don't have the derivative sitting outside the brackets... (for trig functions only)

If you have sin x or cos x raised to an odd power, you can do a dirty trick as follows...

If you have, say, (sin x)^3, you can split that up into (sin x)(sin x)^2. Now, (sin x)^2 is also equal to 1 - (cos x)^2. This gives you (sin x)(1 - (cos x)^2), which then gives you sin x - sin x(cos x)^2 which can be integrated to -cos x + (1/3)(cos x)^3 + c!

(And yes, while I was working that one out I made my old mistakes of dividing by the integral instead of the derivative and forgetting the plus c... hopefully I don't make those mistakes in tests...)

If you have sin x or cos x raised to an even power, well, there's a dirty trick for that too, but it's not as convenient.

Here are two more trig identities to memorise (I hate memorisation, so that's just another reason why I hate integrating these things). (cos x)^2 = ((1 + cos 2x)/2) and (sin x)^2 = ((1 - cos 2x)/2). You can go ahead and prove them if you want to feel confident that they work. These functions can be integrated.

If you have (cos x)^6 or whatever, (cos x)^6 = ((cos x)^2)^3 = ((1 + cos 2x)/2)^3. Then you have to do the fun, fun work of expanding that and integrating it. And so on. Argh. I hate these things. I've pretty much said everything about them that I have to, though. Now for some more posts on integration by substitution, differential equations and maybe a quickie on area under a curve as well...

Implicit Differentiation and Differentiating Parametric Equations

More differentiation!

I'm going to do parametric equations first, because they're easier.

Differentiating Parametric Equations

Parametric equations are where you have separate equations for x and y in terms of a third variable, like t. For example, you could have x = 2t and y = t^2 + 1. Sometimes you can combine the two functions to make one single function in terms of either x or y. In this example, you could rearrange x = 2t to become t = (x/2), and then substitute that into the y equation to find an equation for y in terms of x. You could then differentiate that way. Sometimes, though, it's rather tiresome and annoying to do all this substitution and whatnot, which is why there's a method for directly differentiating parametric equations. It's a pretty simple method too, which is nice.

First, you find dx/dt (or, if x isn't given in terms of t, just replace the t with whatever variable x is given in terms of). Then you find dy/dt. Now, by the chain rule, dy/dx = (dy/dt)(dt/dx). But wait! you might say. We have dx/dt, but not dt/dx! Well, the handy thing is, dt/dx is simply equal to 1/(dx/dt).

For the above equations, x = 2t and y = t^2 + 1, dx/dt = 2 and dy/dt = 2t. dy/dx = (dy/dt)(dt/dx) = (2t)(1/2). Therefore, dy/dx = t. If you want to find this in terms of x, just use one of your original equations to find what t is. x = 2t can be arranged to t = x/2. Therefore, dy/dx = t = x/2. You can get the same answer using the substitution method that I described above.

Differentiating Implicitly Defined Functions

Most of the time, when you see equations, they're in the form of y in terms of x or x in terms of y. But sometimes, you might get an equation where x and y are on the same side, like in x^2 + y^2 = 16 (which, by the way, is the graph of a circle, centre (0, 0) and radius 4). One way of differentiating such equations is to rearrange them and then differentiate, but sometimes this is hard or even impossible to do. That's where implicit differentiation comes in!

Implicit differentiation, however, is pretty long-winded, so if you're doing a calculator assumed part of a test and you see an implicit differentiation question only worth 1 or 2 marks, use your calculator instead! It's not worth the effort!

Basically, in implicit differentiation, you have to differentiate each term separately with respect to x (or one of the two variables in the equation). Then you have to factor out dy/dx (or dy/dt or whatever the variables are) to get your derivative.

Here's how to differentiate the different terms. Let x = whatever variable you're differentiating with respect to and y = the other variable.

Number on its own: Derivative is 0 (like always)
x, or powers of x: Differentiate as normal
y: Derivative of y is dy/dx. (dy/dx does mean "derivative of y with respect to x"...)
Powers of y: Use the chain rule as well as the above fact. y^2 becomes (2y)(dy/dx)
Mix of x and y: Separate into two groups, one with the x and one with the y. For example, 6xy could become (6x)(y). Then use the product rule. Derivative of 6xy with respect to x would end up being 6y + 6x(dy/dx).

Here's a pretty simple example for implicit differentiation: finding the derivative of x^2 + y^2 = 16.

(d/dx)(x^2) + (d/dx)(y^2) = (d/dx)(16)
2x + 2y(dy/dx) = 0
(dy/dx)(2y) = -2x
(dy/dx) = (-2x)/(2y)
(dy/dx) = -(x/y)

I'm too lazy to find and type up a harder example... but hopefully you get the idea. You have to differentiate each term separately using the rules that I stated above (well, I kind of made them up as I went along so I hope they work all the time), and then rearrange the equation to get an equation for dy/dx in terms of x and y.

If you have any more questions, just ask!

Differentiating Trig Functions

I'm going to skip ahead in the book a bit and go straight to the differentiation and integration bits, because I don't really have to draw diagrams to explain them (and drawing diagrams on Paint is annoying).

Differentiating Trig Functions- First Principles

First principles- the random long-winded way of differentiation which you only do when you're trying to learn the ideas and concepts behind it, but then drop when you learn quicker, easier methods.

I'm not sure whether or not I've explained first principles before on my blog, so I'll try and briefly explain the concepts behind it here. Differentiation is all about finding the gradient, and gradient can be calculated by rise over run (i.e. difference in y-values divided by difference in x-values). Therefore, one way of finding the gradient at any given point along a curve is to find the gradient between that point and another point very very close to it. You can keep adjusting the distances between the two points until there's only a very infinitesimally small difference between the points, and then you'll get the gradient.

Let's say that you're looking for the gradient at the point where x = x on the curve y = f(x). When x = x, the y-coordinate of the curve is given by f(x), so our first point is (x, f(x)). Our next point is h more units along the x-axis, so its x-coordinate is x+h, and its y-coordinate is f(x+h), giving us the point (x + h, f(x+h)). Now, gradient is rise over run, so the gradient formula is (f(x+h)-f(x))/((x+h)-x), which simplifies down to (f(x+h)-f(x))/h. Now, to get the gradient at point x, you want the distance between the two points (h) to equal zero. But wait! If h = 0, then you're dividing by 0, and you can't do that!

That's why we use a limiting equation, as represented by that "lim h->0" thingy.

(I love Word's "insert equation" feature...)

Now, let's put this into action. Chances are, you know that the derivative of y = x^2 is simply 2x. How do we go about proving this with first principles?

Well, on the top line we first substitute f(x + h) for (x + h)^2 (since the function, f(x), is x^2) and f(x) for x^2. Then we expand and simplify this so we have (2hx + h^2)/h, h approaching zero. Now we factor out h to get (h(2x + h))/h and then cancel out the hs so we get 2x + h. Now we don't have to bother about h approaching zero. Instead, we can allow h to equal zero, since we're no longer dividing by h and therefore making h zero won't make the universe explode. That leaves us with 2x as the derivative of x^2.

We can use a similar approach to find out the derivatives of sin x and cos x, but there's a couple of things that you have to know first.

As x approaches zero, (sin x)/x = 1 and (1-cos x)/x = 0. These can be determined by using tables of values (i.e. working out the values of these two functions by calculating for x = 0.01, x =0.001 etc.) or by looking at the graphs of the functions.

Here's how the derivative of sin x can be calculated using first principles:

As for the derivative of cos x... I'll leave that one for you to do as practise! (Actually I'm only saying that because I can't be stuffed typing that one up.)

The derivative of sin x is cos x, and the derivative of cos x is -sin x. REMEMBER THAT. It's helpful knowledge to have on hand.

tan x has a retarded derivative, or rather derivatives (yes, it has more than one acceptable derivative. One way to differentiate tan x is by using the quotient rule, since tan x = (sin x) / (cos x).

When y = (sin x) / (cos x), dy/dx = (cos x cos x - sin x (-sin x)) / (cos x)^2
= ((cos x)^2 + (sin x) ^2) / (cos x)^2
But (cos x)^2 + (sin x)^2 = 1 (Pythagorean identity, which I realise I haven't mentioned yet on this blog)
So dy/dx = 1 / (cos x)^2.

So one possible derivative of tan x is 1 / (cos x)^2.

Now, apart from sine, cosine and tangent, there are three more trig functions (which you don't really need to know very well for now) which are the reciprocals of the first three functions. They are secant (sec), cosecant (cosec) and cotangent (cot). Secant is 1 / cos x, cosecant is 1 / sin x and cotangent is 1 / tan x.

Since sec = 1 / cos x, (sec x)^2 = 1 / (cos x)^2, which is the derivative of tan x. Therefore, another possible derivative of tan x is (sec x)^2.

Also, there's an alternative way of simplifying my second line of working for the derivative of tan x using the quotient rule. This alternative simplification leads to a third acceptable derivative of tan x.

((cos x)^2 + (sin x) ^2) / (cos x)^2 = ((cos x)^2) / (cos x)^2 + ((sin x)^2) / (cos x)^2))
= 1 + (tan x)^2

Now I'm going to summarise the derivatives for the three main functions, because it's handy having everything in one place.

Derivative of sin x = cos x
Derivative of cos x = -sin x
Derivative of tan x = (1 / (cos x)^2), (sec x)^2 or 1 - (tan x)^2, depending on your mood and the question that you're trying to solve

Rules of Differentiating Trig Functions

Differentiating trig functions is annoying (though nowhere near as annoying as integrating them) because you have to make sure not to get mixed up between the derivatives and stuff (my main problem is getting the positive/negative signs the right way around). At least they go by most of the same rules as pretty much every other function with regards to differentiation: you can still use the good ol' product, chain and quotient rules.

I'll just do a quick example with the chain rule because I believe that that's where it's easiest to get mixed up with regards to what order to do the steps in.

Let's use y = (cos 5x)^3 as an example.

Let cos 5x = u, and 5x = s. Therefore ds / dx = 5. u = cos 5x = cos s, so du/ds = - sin s. Finally y = u^3, so dy/du = 3u^2.

Now, due to the chain rule, dy/dx = (dy/du)(du/ds)(ds/dx).
Therefore, dy/dx = (3u^2)(-sin s)(5).
Substituting to get an answer in terms of x, dy/dx = 3((cos 5x)^2)(- sin 5x)(5) = (-15 sin 5x)(cos 5x)^2.

Wow, explaining that in full was more complicated than I was expecting since cos 5x warranted an extra chain as well (you need to multiply by the derivative of 5x). Just remember that the power takes precedence, then the stuff in the brackets. And take care in working out the derivative of the stuff in the brackets too.

That's all from me on differentiating trig functions for now, but if you want more clarification, feel free to ask!

Ways to get some stuff done

This is a bit of a stupid post. You know those lazy days when you have to get some stuff done but you can't be bothered doing anything? Well, here's some really stupid ways that you may or may not be motivated to do something (maybe not a lot, but something).

Method #1: I Really Want to Play Video Games but I Have to Get Some Stuff Done

I actually finished off the Chapter 5 and Chapter 6 end-of-chapter questions in my book while using this method. It took a while, but at least I got my gaming fix done at the same time.

You have to pick a game that can be played in short bursts. For example, in something like Pokémon Mystery Dungeon, you could divide it up into floor at a time. Then, pick an assignment or task that can be divided up into short bursts. For example, with my chemistry questions, I just did a question at a time. (However, make sure that the assignment chunks take at least roughly the same amount of time as the gaming chunks so that you are getting stuff done.) Then alternate between the two activities until you've gotten everything done.

Method #2: Last Minute Warrior

Some people do better under pressure. This method is for them. I'm sure you all know how it works without me having to explain.

WARNING: Don't do this if you can't handle stress, or if you actually have something important to do the next day which might require you to have had a good night's rest the night before.

Method #3: The List

Write a list of everything that you have to do. Break some stuff down into smaller chunks. For example, a 2000 word essay could be broken down into "write 500 words," "write next 500 words" and so on. If writing 500 words still looks like a massive hurdle, break it down further into 250 words or something.

Method #4: Go With the Flow

This only really works for nerdy people who sometimes find studying fun. It works better when there isn't anything more interesting to do.

Basically the idea is that you study when you feel like it and you don't study when you don't feel like it. It works if you have a genuine interest in the subjects that you're studying because if you're like me, once you get started you can study for hours (though sometimes the trouble is getting started in the first place). However, you might be better advised not to use this method when it comes to studying for something that you hate and you tend to avoid at all costs (like me and Lit).

Method #5: Race Your Friends

It works for some people. I'm not that competitive though so I don't use this method. But hey, I guess Facebook has its study advantages...

Method #6: Do Non-Computer Stuff First

The reason that I say this is that I know that I tend to procrastinate and play random games like Bush Whacker 2 when I'm on the computer. If I do non-computer stuff first, then I get more stuff done.

Method #7: Create a Non-Admin Account on your Computer

Never done this, but I have considered it. If you create a non-admin account and then put on parental controls so that you can't play games and do all your normal procrastination stuff, then you might get stuff done. Still, make sure to reward yourself afterwards.

How to make the most out of Language Perfect

Okay, just a quick diversion here to talk about Language Perfect!

Language Perfect (www.languageperfect.co.nz) is a website where you can revise vocabulary words in whatever second language you're studying. Some schools use this website quite a bit (like mine, for instance).  This is just a little guide to get the most out of using this program.

Okay, first of all, do not make Language Perfect your be-all-and-end-all for language studying. Language Perfect is great for studying vocab, but you need a range of study materials to have a well-rounded knowledge of the language you're studying. Here's some stuff that you should try and get your hands on (apart from a dictionary, because that's kind of obvious):

• Books in whatever language you're studying. I suggest getting non-fiction books geared at kids because then the vocab is simple but the topics are still interesting. Another alternative is getting translated versions of books that you've read before in English. I have 小屁孩日记 (Diary of a Wimpy Kid) and some 杰罗尼莫·斯蒂顿 (Geronimo Stilton) books in Chinese. Reading books allows you to see how sentences work in your target language as well as being a source for new words.
• Some textbooks have good example sentences. I find the content of many textbooks to be rather stilted, though. I'm currently working through a pretty decent Chinese textbook called 中文, which I think is targeted at kids learning Chinese in the weekend Chinese schools here (well, I know at least one of the weekend Chinese schools uses this book). Some of the stories are kind of annoying because they have an overly cheery atmosphere, but there are other interesting biographical stories as well on famous people like Newton and Beethoven. The example sentences in the books are also really good.
• Movies in whatever language you're studying so that you can get some listening practise in.
• And, of course, if you can find people that you can communicate with using your target language, that just makes the experience more rewarding!

Now that's out of the way, how do we make the most out of Language Perfect? Well...

First of all, I recommend nailing all your school vocab just to get it out of the way. My teacher uploaded lists for the class, but if you don't have a teacher that does that, add words in yourself, pulling words from your textbook or any other word lists that you may have received in class. (To create a list, click on the "Control Panel" link on the login screen, or click on "Create a List" at the top of the list of available lists for whatever "module" you're doing.) (Wow, a list of lists.)

If there are any entries in any of these lists that are giving you trouble for anything other than you are simply having trouble learning that particular word, for example if the English definition is spelled incorrectly or the Chinese characters cannot be inputted, don't waste time with that word. Instead, click on "edit list" and remove those words. You can also re-add the word with correct spelling or whatever if need be. If you are using a school-created list, Language Perfect will keep the original list with all words in it and then create a copy of the list with your changes, under the "My Lists" tab.

Once you've mastered your school vocab, you can also add lists for the new words that you're encountering in books and movies. I think that doing this is more useful than just going through other random Language Perfect lists that might not be relevant to you. You're not really going to remember a word if you're not seeing it in books and movies and in life.

Recently, Language Perfect has also included support for other subjects such as chemistry and maths. I haven't really experimented with these yet, but they might come in handy. Language Perfect's system would be great for learning those ions, for example. Oh, and those compound names that you're just meant to know, such as sulfuric acid, sulfurous acid and so on.

3AB Reactions, Equations and Stoichiometry Index plus a bit more, including absolute zero

Okay, well, I'm pretty sure I've already covered most of the stoichiometry stuff in my year 11 posts, but again, there's a little bit of new stuff to add. So first I'll put an index of everything that's carried over from last year and then add the new bits.

List of useful stoichiometry posts from Year 11:

• Formulae (naming ionic, metallic and covalent compounds, as well as some fixed names that you should know)
• Simple Chemistry Calculations (molar mass and percentage composition)
• Chemical Equations (how to write them, including balancing equations)
• Simple Calculations Involving Moles (mass to mole, mole to mole, mole to mass)
• Reactions and Equations part deux (standard reactions, ionic equations)
• Empirical Formulae and Limiting Reagents and Some Other Stuff (the "other stuff" consists of mass to volume and volume to moles calculations, as well as working out molecular formulae from empirical formulae and molar mass)

New stuff:

There's two main new things that you have to learn, and that is all the gas stuff and percentage purity.

The gas stuff

Back in 2AB, we calculated the moles of a gas at STP by using n = V/22.71. Well, you can still use that if the question says STP, but now we're going to find new questions that don't use STP.

Some geniuses used the facts that pressure is inversely proportional to volume, volume is directly proportional to temperature and pressure is inversely proportional to temperature to form the following VERY USEFUL equation:

PV = nRT

where P = pressure in kPa

V = volume in L

n = number of moles

R = 8.314 (a constant- it's on the data sheet)

and T = temperature in Kelvin

The nice thing about R is that it never changes, though some textbooks and websites might have slightly different values due to rounding (use your data sheet in the exam). Then you just need 3 of the other 4 things to work out what the 4th thing is.

If you don't already have the pressure, volume, or temperature in the units given above, then you'll have to convert. I won't talk about volume conversions because by now you'll probably feel like I'm taking you for an idiot if I do that, but I will briefly mention the conversion ratios for the other two.

I'll do temperature first, because that's easiest (well, unless you throw Fahrenheit in there, but I've never seen a question with Fahrenheit in it yet). Also, I get to cover another dot point which got put on its own under another heading, and I don't want to do a whole post for that one dot point.

I *think* that the way that zero degrees Kelvin got calculated was that someone found graphs of the average kinetic energy for different particles at different temperatures and extrapolated them all the way down to where the kinetic energy would theoretically be zero and the particles would theoretically stop moving. All of the graphs stopped at the same point, at a point estimated to be -273.15 degrees Celsius. Therefore, 0 degrees Kelvin = -273.15 degrees Celsius.

The great thing about the Kelvin scale is that it increases at the same rate as the Celsius scale. Therefore, to change from Celsius to Kelvin, just add 273.15 degrees, and vice versa.

Now for pressure! The bad news is that the pressure conversions don't appear to be on the data sheet, so if they do have a pressure conversion question then I might be stuffed because I don't remember them off the top of my head, but the good news is that in light of that, they might decide not to put one in.

From my textbook, 1 atm (atmosphere) = 101.3 kPa = 760 mmHg (millimetres of mercury- I think this measurement had something to do with some old system of measuring air pressure).

(I think that 101.3 kPa is atmospheric pressure, which is why it's 1 atmosphere.)

The PV = nRT equation is very useful for when you encounter gases in stoichiometry problems. Use and abuse it at will.

Percentage Yield

This is a bit different from percentage composition, but if you understand percentages in general, this isn't too hard to pick up. (Yes, percentage composition is still in the course. If you need a refresher, it's in my simple chemistry calculations post.)

When we calculate how much is going to be formed as a result of a reaction by using our amazing Stoichiometry skills, what we're calculating is the theoretical value. Unfortunately, life being life, some reactions form a smaller mass of products than the theoretical value, that is to say, their percentage yield is less than 100%.

If you're given the reactants and the percentage yield of the reaction and you're asked to find out what mass of one of the products is produced, you need to do two things:

1. Work out the theoretical yield of the product in question using everything else that you've learned in stoichiometry. Remember to watch for limiting reagents and other tricks that they might throw at you (but at the same time be wary of questions that provide too much information).
2. Multiply the theoretical yield of the product by the percentage yield (e.g. if the percentage yield is 91%, multiply the theoretical yield by 0.91).

(You could probably do the two steps in the reverse order as well, i.e. you could probably multiply the amount of reactant by the percentage yield and then work out the yield of products.)

If you're given the products and the percentage yield of the reaction and you're asked to find out what mass of one or more the reactants is required, you need to follow the steps below:

1. Work out how much product you were "theoretically" meant to have by first dividing by the percentage and then multiplying by 100 (e.g. if the percentage yield is 91%, divide the mass by 91 and then multiply by 100. Or you could divide by 0.91. It's the same thing really, since 0.91 = 91/100 and when you're dividing by 91/100, you're actually multiplying by 100/91.) 
2. Calculate the amount of reactants from there.

Alternatively, you could calculate the amount of reactant using the amount of product that was actually yielded, and then divide by 91 and multiply by 100.

A bit more on solutions

The Solutions part of this course outline is pretty similar to the stuff we did in Year 11, so you can visit my old post on solutions for a refresher.

There are a couple of things that I didn't explicitly mention there, however, so I will mention them here.

Concentration Calculations

Concentration of substances is represented using a variety of measurements, such as moles per litre, grams per litre, parts per million etc. Here's how to calculate the concentration:

mol/L: Divide number of moles by number of litres. (Makes perfect sense when you consider that a forward slash is also considered to be a divide sign on some computers/calculators.)
g/L: Divide number of grams by number of litres.
ppm (parts per million): Divide number of milligrams of solute by the number of kilograms of solvent.
Percentage composition by mass: See my post on simple chemistry calculations.

Concentration of ions in solution for strong electrolytes

This is easy. Since strong electrolytes completely ionise, the concentration of the ions in solution is the same as the concentration of the strong electrolyte.

Calculation of concentration and volume in dilution and addition of solutions

For this we use a nifty formula: C₁V₁ = C₂V₂, where C₁ = concentration of initial solution, V₁ = volume of initial solution, C₂ = concentration of final solution and V₂ = volume of final solution.

This works because the number of moles of the solute isn't changing when you dilute a solution: it's only the concentration that is. You see, the good ol' concentration formula C = n/V can be rearranged to n = CV. Now, C and V might be different for the initial solution and the final solution, but n is the same.

Intermolecular Bonding

Okay, now I'm not going to completely adhere to the dot point guide. That's because some of the dot points here are to do with molecular shape, which involves drawing diagrams, which I really can't be bothered doing right now. So I'm just going to do intermolecular bonding because I can kind of get away with doing that without making diagrams.

Previously, I've talked about 3 kinds of strong bonding: ionic bonding, metallic bonding, and covalent bonding, and in the covalent bonding post, I wrote "there are other random types like hydrogen bonds, but they're not important for now." Well, hydrogen bonds are now important, and so are other "random types" of bonds like dispersion and dipole-dipole bonds.

I said that I wasn't going to completely adhere to the dot points, but I didn't mean that I wouldn't use them at all. I'm still going to do all the dot-points that are relevant to intermolecular bonds.

Explain polar and non-polar covalent bonds in terms of the electronegativity of the atoms involved in the bond formation

Wow, that sounds so high-brow and sciencey. Not to worry though. Or maybe you should, because my explanation for this is probably going to be a little shoddy? I don't know.

Now, you know how electrons are shared between atoms in a covalent bond? (If not, revisit my post on covalent bonding.) Well, here's the catch: from what I understand, they aren't always shared equally. They're shared equally enough so that both atoms get to have a stable full outer shell configuration, but sometimes those pesky electrons decide that they like one atom more than another. That's due to electronegativity differences: the more electronegative element attracts the electrons more and therefore the electrons spend more time near that atom than the other atom. This results in a polar covalent bond, where one atom is more positively charged than the other. How "positive" or "negative" each end is depends on the difference in electronegativities between the two atoms (the periodic table on my phone- yes, I'm so nerdy that I have a periodic table on my phone- gives information on the electronegativities of atoms of each element). When there's a super big difference in electronegativities, the electrons spend pretty much all their time around the negative end, which results in an ionic bond. (My Chem teacher was saying how it's like one atom is "stealing" electrons from the other, which is like "extreme sharing." Pretty damn good joke, if you ask me.)

Non-polar covalent bonds occur when both atoms on either side of the bond are of the same element because then there's absolutely zero difference between the electronegativities of each atom.

Use the relationship between molecule shape and bond polarity to predict and explain the polarity of a molecule

I'll explain this better later, after I've discussed molecule shape, but basically, if the atoms on one side of the molecule tend to be on the "negative" side of polar covalent bonds and the atoms on the other side tend to be on the "positive" side of polar covalent bonds, then the molecule is probably polar. If there's no general trend like this, then the molecule is probably non-polar. Yeah, I really need a diagram to explain this properly.

Explain the differences between intermolecular and intramolecular forces.

This is easy (unless there's more to it than I'm aware of). Intermolecular forces are forces between molecules. Intramolecular forces are forces between the atoms that make up the molecules. In covalent molecular substances, the intramolecular covalent bonds are strong while the intermolecular Van der Waals forces tend to be weaker.

Describe and explain the origin and relative strength of dispersion forces, dipole-dipole attractions, hydrogen bonds and ion-dipole interactions for molecules of a similar size.

Dipole-dipole attractions are seen only in polar molecules. This is because one side of a polar molecule is more positively charged while the other side is more negatively charged. The more positively-charged side of one molecule can attract the more negatively-charged side of another molecule. Dipole-dipole forces get stronger as the difference in electronegativity between atoms within the molecule increases (because that would cause the ends to be "more positive" and "more negative").

Hydrogen bonds are kind of like dipole-dipole attractions, but with a catch: they only exist when there's an O-H, N-H or F-H bond somewhere in the molecule. Hydrogen bonds are much stronger than dipole-dipole attractions (though still not as strong as ionic, covalent or metallic bonds).

Dispersion forces are seen in all kinds of molecules. In a covalent bond, the electrons do not always remain in the space between the two atoms. Instead, the electrons move around (though with a net movement of zero). Therefore, there are times when there are temporarily more electrons on one side of the bond (and the molecule) than the other. This causes a temporary "positive" dipole and a temporary "negative" dipole. Oppositely-charged temporary dipoles of different molecules can attract each other, much like in dipole-dipole forces, but in the case of dispersion forces, these attractions are not as strong since the dipoles are only temporary. Dispersion forces get stronger as the number of electrons in a molecule increase.

Ion-dipole interactions occur when soluble ionic solids are dissolved in aqueous solution, particularly when said solution is polar. The positive ions of the solid can be attracted by the negative dipole of the molecules in the solution, and vice versa. This causes the ionic solid to dissolve. Complex ions (not really sure how to explain these) are formed by such ion-dipole interactions between ion(s) and whatever other molecules make up the complex ion.

Explain the relationships between physical properties including melting and boiling point, and the types of intermolecular forces present in substances with molecules of similar size.

Melting and boiling involves the breaking of intermolecular bonds between molecules, so stronger bonds result in more heat energy required which in turn results in higher melting and boiling points, and vice versa. Of the types of intermolecular bonds discussed above (except I'm not going to talk about ion-dipole for now since that type of bond doesn't occur between molecules of a covalent molecular substance), hydrogen bonding is the strongest. Dipole-dipole is stronger than dispersion if the molecules are small and don't have that many electrons, but in bigger molecules it might be the other way around. In fact, often it will be the dispersion forces that have a greater effect on melting and boiling points rather than dipole-dipole forces.

Apply an understanding of intermolecular interactions to explain the trends in melting and boiling points of hydrides of groups 15, 16 and 17 accounting for the anomalous behaviour of ammonia, water and HF.

Yay, now I can provide an example for my previous point! Let's use the group 15 hydrides as an example. In group 15 the first four hydrides are ammonia, PH₃, AsH₃ and SbH_{3. Ammonia has the highest boiling point, then} PH<sub>3 has the lowest boiling point. After that, the boiling points gradually increase.

Ammonia has the highest boiling point because it contains hydrogen bonds between molecules due to the N-H bonds within ammonia molecules. The other three do not contain hydrogen bonds so their boiling points are not as high, but as the size of the molecules and their numbers of electrons increases, their dispersion forces also become stronger and their boiling points gradually get higher.

Similar trends are observed in groups 16 and 17.

Explain and describe the interaction between solute and solvent particles in a solution.</sub>

When a solute dissolves in a solvent, the bonds between solute particles and the bonds between solvent particles break. Then, new bonds are formed between solute and solvent particles.

Now, of course, breaking bonds requires energy, so in order to make up for that, the amount of energy that is released when bonds are reformed must be close to the amount of energy that was required in the first place. Therefore, dissolving generally only happens when the strength of the bonds within solute and solvent are reasonably similar.

Use the nature of the interactions, including the formation of ion-dipole and hydrogen bonds to explain water's ability to dissolve ionic, polar and non-polar solutes.

Now I have an excuse to provide an example for my last point!

Water contains hydrogen bonds. If you want to dissolve something non-polar in water that only has dispersion forces between molecules (since non-polar covalent molecular substances can't contain dipole-dipole or hydrogen bonds), we have a problem. You see, hydrogen bonds between water molecules would have to be broken (as well as dispersion forces between the non-polar molecules), but when bonds form between the non-polar substance and water, only dispersion forces would technically be formed as the non-polar substance wouldn't be able to bond to the water using dipole-dipole and/or hydrogen bonds. Therefore, a lot of energy would technically be required while very little would be released: a rather selfish reaction, which is why it generally doesn't happen.

The dipole-dipole bonds in polar substances are closer to the strength of hydrogen bonds, so polar substances are generally able to dissolve in water.

Finally, many ionic substances can dissolve as a result of ion-dipole reactions, which I mentioned earlier but I'll say it again. And, because I'm really lazy, I'm just going to copy-paste it from above, so you can skip over this paragraph if you've already read it.

Ion-dipole interactions occur when soluble ionic solids are dissolved in aqueous solution, particularly when said solution is polar. The positive ions of the solid can be attracted by the negative dipole of the molecules in the solution, and vice versa. This causes the ionic solid to dissolve. Complex ions (not really sure how to explain these) are formed by such ion-dipole interactions between ion(s) and whatever other molecules make up the complex ion.

Describe the variation of gas solubility in aqueous solution with temperature

Okay, this is a bit that I'm going to have to revise because I don't really remember. I know that gas solubility tends to decrease with temperature, and I think that is because the added temperature would just cause the gas to evaporate out of the solution. I don't know for sure though, so I guess I'll have to look at my textbook again.

And, of course, this topic seems to be pretty damn elusive in my textbook, so I guess I'll probably have to Google it or search a bit harder. I'll do that later. (Yeah, I am pretty lazy.)

Atomic Structure and Periodic Table, take 2

Time for a post on Chem now... again I'm going to work off the good ol' dot points under the first heading, or rather the dot-points under the first subheading ("Atomic structure and Periodic Table" is under the heading "Atomic structure and bonding").

Back on topic: if you need to review the basics of atomic structure, go to my post on Atomic Structure and the Periodic Table (oh my, it has the word "the" in it as opposed to the subheading written here on my course outline! Okay, I'll stop being stupid now).

Explain the structure of the atom in terms of protons, neutrons and electrons.

See my first post on Atomic Structure and the Periodic Table.

Write the electron configuration using the shell model for the first 20 elements.

Also see my first post on Atomic Structure and the Periodic Table.

Explain trends in first ionisation energy, atomic radius and electronegativity across periods and down groups (for main group elements) in the Periodic Table.

Yay! Something new!

I'm going to start with atomic radius, because I think it's the easiest to explain, and then I can explain the other two from there.

When you go down groups, atomic radius generally increases. This is because you're adding more shells and therefore the atom is getting bigger. Also, when you go up groups, atomic radius generally decreases.

When you go across periods (from left to right), atomic radius generally decreases. This is because the number of shells remains consistent across a period, but the number of protons increases, making the nucleus "more positive" and therefore pulls the electrons in the shells closer to it (if that made any sense). Therefore, the radius decreases. Again, the inverse is true: as you move left across a period, atomic radius generally increases.

I'm saying "generally" because occasionally you get some retarded element that doesn't want to follow the rules exactly. All these things are trends only. Sort of like how there might be a fashion trend where a lot of people are dressing according to that fashion trend but then there'd be someone like me who doesn't give a damn about it.

Now for electronegativity! Electronegativity is an atom's ability to attract electrons and, in doing so, become more electronegative. Now, since it's the positively-charged protons in the nucleus that are doing most of the attracting (opposites attract, remember?) electronegativity is largely based on two things: how many protons are in the nucleus, and how many shells of electrons are surrounding that nucleus (remember, likes repel, so the more shielding shells there are, the harder it is for an atom to attract an electron). Therefore, electronegativity decreases as you go down groups as there are more "shielding" shells of electrons, and increases as you go across periods since there are more protons for the same number of "shielding" shells. (The inverse is also true.)

Now, what is this "first ionisation energy" thing you might ask? (I've started like 3 sentences with the word "now" in the past few minutes... argh.) Well, "first ionisation energy" is the energy required to remove one electron from an atom. As electronegativity of an atom increases, the electrons are held more tightly to the atom and thus require more energy to remove, and vice versa. Thus trends in first ionisation energy are the same as trends in electronegativity: first ionisation energy increases as you go right across a period, or up a group.

Explain the trend in successive ionisation energies

Apart from first ionisation energy, there's also second ionisation energy, third ionisation energy, and so on, as you take more and more electrons away from an atom. Second ionisation energy is higher than the first, third is higher than the second, and so on. This is because it is harder to take an electron away from an atom which is turning into a more and more positive ion.

Ionisation energy jumps significantly when you've finished taking away all the electrons from one shell and have to start on the next. This is because this next shell is closer to the nucleus and therefore is bound more strongly by the protons in the nucleus of the atom. Analysing ionisation energy numbers, therefore, allows you to get a pretty good idea of how many electrons are in the outer shell. For example, if there's a big difference between the 3rd and 4th ionisation energies, you can assume that there's 3 electrons in the outer shell, and that the big jump is due to the 4th electron having to be taken from the shell second furthest from the nucleus.

Describe and explain the relationship between the number of valence electrons and an element's bonding capacity, position on Periodic Table and physical and chemical properties.

I've explained everything apart from position on Periodic Table back at my first post on Atomic Structure and the Periodic Table.

Position on Periodic Table is pretty simple though. Elements are arranged according to the number of electrons that they have. They are then arranged into periods according to how many shells they have, and groups according to how many valence electrons they have (1 valence electron- group 1, 2 valence electrons- group 2, 3 valence electrons- group 13, 4- group 14, 5- group 15 until 8- groups 18). Then there are the transition metals in the middle which I can't be bothered explaining, but they're also listed in order of number of protons.