Chemical equations and states (1 Viewer)

Kaido · Jan 2, 2015

Sometimes, I forget which states to use.
Apparently (from a lecturer at UNSW) if I mess up just a single state, my entire equation becomes invalid and thus losing all the marks allocated to a correct equation

So, should I just leave the states out if i'm unsure or should i go back and rote the **** out of every equation. (Or is this lecturer just making bs up)

photastic · Jan 2, 2015

Kaido said:
Sometimes, I forget which states to use.
Apparently (from a lecturer at UNSW) if I mess up just a single state, my entire equation becomes invalid and thus losing all the marks allocated to a correct equation

So, should I just leave the states out if i'm unsure or should i go back and rote the **** out of every equation. (Or is this lecturer just making bs up)

For the HSC, an equation is worth 1 mark. You will lose an entire mark if there are no states or incorrect states so this UNSW lecturer is correct. Remember this is science and marking is very serious.

SuchSmallHands · Jan 3, 2015

Kaido said:
Sometimes, I forget which states to use.
Apparently (from a lecturer at UNSW) if I mess up just a single state, my entire equation becomes invalid and thus losing all the marks allocated to a correct equation

So, should I just leave the states out if i'm unsure or should i go back and rote the **** out of every equation. (Or is this lecturer just making bs up)

Hmm, depends on the way the question is put. If you add the equation in a written response, depending on the marking criteria, you may get full marks awarded even with a wrong state. If the question just asks you to write an equation and you get a state wrong you will lose a mark though, which as already explained, will be all the marks for that question in all likelihood.

If you know your solubility rules well, and you know the conditions under which a reaction is taking place, you seriously shouldn't need to rote learn states though. Just use your knowledge of chemistry and you'll be fine.

dan964 · Jan 3, 2015

yes and no
some questions in hsc do not require states but most do
e.g. precipitation, combustion, long response where needed to understand concept
and also the one mark allocated to it
usually a "write a balanced equation"
safest bet write states.

InteGrand · Jan 3, 2015

Kaido said:
Sometimes, I forget which states to use.
Apparently (from a lecturer at UNSW) if I mess up just a single state, my entire equation becomes invalid and thus losing all the marks allocated to a correct equation

So, should I just leave the states out if i'm unsure or should i go back and rote the **** out of every equation. (Or is this lecturer just making bs up)

This can be analysed using probabilities. Sometimes the marking criteria requires states, in which case, you lose the mark either if you leave off states, or if you get state(s) wrong. Other times, the marking criteria does NOT require states, in which case you get the marks even if you put in no states, but I'm not sure if you lose marks if you put wrong states and states aren't required (don't think you do). The problem is, they don't tell you if they're marking for states for a given question.

Let's say you're X% sure you've got the states right, and that there's a Y% chance that they're marking for states. i.e. the probability that the states you think go in the equation are correct is $\frac{X}{100}=p_1$ , say. So the probability you've got at least one state wrong is $p_2 = 1- p_1$ . The probability that they're marking for states is $\frac{Y}{100}=q_1$ , say, so the probability that they're not marking for states is $q_2 = 1 - q_1$ .

So, there's four possibilities: you guess the states with the ones you think are right OR leave off states, AND states are being marked OR states aren't being marked.

So, if you do go with your guess for states, and the mark for getting the equation right according to the criteria is 1, and mark for wrong equation is 0, your expected mark is:

E_guess = P(guess is right)×(mark for correct answer) + P(guess is wrong)×P(states are being marked)×(mark for wrong answer) + P(guess is wrong)×P(states not being marked)×(mark for correct answer)

$\Rightarrow E_{\text{guess}}=p_1 \cdot 1+p_2\cdot q_1 \cdot 0 + p_2 \cdot q_2\cdot1$

i.e. $E_{\text{guess}}=p_1 + p_2 q_2 = p_1 + (1-p_1)q_2$ .

Your expected mark if you decide not to guess states is

E_{no guess} = P(states not being marked)×(mark for correct answer) + P(states are being marked)×(mark for wrong answer)

$\Rightarrow E_{\text{no guess}}=q_2\cdot1 + q_1 \cdot 0$

$\Rightarrow E_{\text{no guess}}=q_2$

Expanding and rearranging the formula for E_guess,

$E_{\text{guess}}= p_1 + (1-p_1)q_2=p_1 + q_2 -p_1 q_2$

$\Rightarrow E_{\text{guess}}= p_1 (1-q_2)+q_2 > q_2 = E_{\text{no guess}$ for any p₁ > 0 and q₂ ≠ 1.

So what we've shown is that if it's not certain that states are not going to be marked (i.e. q₂ ≠ 1), you should always guess.

So, if you're unsure, always guess, unless you're told that states aren't required (in which case it's obvious you shouldn't guess).

Note. This was done assuming that, when states aren't required, incorrect states don't lose you marks. The calculations change if incorrect states still lose you marks even if states aren't required.

InteGrand · Jan 3, 2015

SuchSmallHands said:
If you know your solubility rules well, and you know the conditions under which a reaction is taking place, you seriously shouldn't need to rote learn states though. Just use your knowledge of chemistry and you'll be fine.

But then you'll need to rote learn the solubility rules, so you still need to end up roting, don't you?

Kaido · Jan 3, 2015

(Integrand has wayyyyyyyyyyyyyyyyyyyyy to much time on his hands + he's a freak)

Yeah, I guess I'll go rote the rules + past papers to familiarise.

Cheers guys

SuchSmallHands · Jan 3, 2015

InteGrand said:
But then you'll need to rote learn the solubility rules, so you still need to end up roting, don't you?

Yeah but everyone has to do that for the course anyway. Besides it's a lot easier/less rote intensive to remember the solubility rules than remembering states for every singe equation.

Librah · Jan 3, 2015

SuchSmallHands said:
Yeah but everyone has to do that for the course anyway. Besides it's a lot easier/less rote intensive to remember the solubility rules than remembering states for every singe equation.

Solubility table was the hardest part of the chemistry course. I've only remembered that all nitrates are soluble lol gg.

Librah · Jan 3, 2015

InteGrand said:
This can be analysed using probabilities. Sometimes the marking criteria requires states, in which case, you lose the mark either if you leave off states, or if you get state(s) wrong. Other times, the marking criteria does NOT require states, in which case you get the marks even if you put in no states, but I'm not sure if you lose marks if you put wrong states and states aren't required (don't think you do). The problem is, they don't tell you if they're marking for states for a given question.

Let's say you're X% sure you've got the states right, and that there's a Y% chance that they're marking for states. i.e. the probability that the states you think go in the equation are correct is $\frac{X}{100}=p_1$ , say. So the probability you've got at least one state wrong is $p_2 = 1- p_1$ . The probability that they're marking for states is $\frac{Y}{100}=q_1$ , say, so the probability that they're not marking for states is $q_2 = 1 - q_1$ .

So, there's four possibilities: you guess the states with the ones you think are right OR leave off states, AND states are being marked OR states aren't being marked.

So, if you do go with your guess for states, and the mark for getting the equation right according to the criteria is 1, and mark for wrong equation is 0, your expected mark is:

E_guess = P(guess is right)×(mark for correct answer) + P(guess is wrong)×P(states are being marked)×(mark for wrong answer) + P(guess is wrong)×P(states not being marked)×(mark for correct answer)

$\Rightarrow E_{\text{guess}}=p_1 \cdot 1+p_2\cdot q_1 \cdot 0 + p_2 \cdot q_2\cdot1$

i.e. $E_{\text{guess}}=p_1 + p_2 q_2 = p_1 + (1-p_1)q_2$ .

Your expected mark if you decide not to guess states is

E_{no guess} = P(states not being marked)×(mark for correct answer) + P(states are being marked)×(mark for wrong answer)

$\Rightarrow E_{\text{no guess}}=q_2\cdot1 + q_1 \cdot 0$

$\Rightarrow E_{\text{no guess}}=q_2$

Expanding and rearranging the formula for E_guess,

$E_{\text{guess}}= p_1 + (1-p_1)q_2=p_1 + q_2 -p_1 q_2$

$\Rightarrow E_{\text{guess}}= p_1 (1-q_2)+q_2 > q_2 = E_{\text{no guess}$ for any p₁ > 0 and q₂ ≠ 1.

So what we've shown is that if it's not certain that states are not going to be marked (i.e. q₂ ≠ 1), you should always guess.

So, if you're unsure, always guess, unless you're told that states aren't required (in which case it's obvious you shouldn't guess).

Note. This was done assuming that, when states aren't required, incorrect states don't lose you marks. The calculations change if incorrect states still lose you marks even if states aren't required.

I can't believe i actually read that whole thing...

Chemical equations and states (1 Viewer)

Kaido

be.

photastic

Well-Known Member

SuchSmallHands

Well-Known Member

dan964

what

InteGrand

Well-Known Member

InteGrand

Well-Known Member

Kaido

be.

SuchSmallHands

Well-Known Member

Librah

Not_the_pad

Librah

Not_the_pad

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