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March 29, 2024, 11:40:18 pm

Author Topic: HSC Chemistry Question Thread  (Read 1040817 times)  Share 

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winstondarmawan

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Re: Chemistry Question Thread
« Reply #2340 on: June 25, 2017, 02:17:23 pm »
0
I'd say the answer is A.
The carbon in methane has an oxidation state of -4 and hydrogen always has a state of +1 unless bonded as a metal hydride; and molecules like methane must have a total oxidation state of 0.

The carbon on the reactant side (methane) gets converted into carbon dioxide on the product side. Carbon dioxide has 2 oxygens of -2, thus carbon must be +4 to bring the charge down to 0.
Therefore, the carbon has changed oxidation states from -4 to +4, which is being oxidised by losing electrons to become a positive species.

The O2 oxygen also has a change in oxidation state by going from 0 (diatomic) to -2 (carbon dioxide). However, the options only had oxidised oxygen, which is incorrect.

Hope this helps :) Look out for changes in oxidation state!!

Yes!! Thank you heaps.

Piza

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Re: Chemistry Question Thread
« Reply #2341 on: June 25, 2017, 04:01:32 pm »
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Hey! Good answer; here's how I would have structured it, to engage a bit more in the question, and the additional information I would add/stuff I would leave out.

...

Again, it is all about making every word of every sentence count. Just load your sentences with information, and make sure to really engage with the question.

Thank you, this is super helpful!!

beau77bro

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Re: Chemistry Question Thread
« Reply #2342 on: June 25, 2017, 04:46:27 pm »
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hey so what would a processing skills test for chem look like? like what can i expect it to include?
thanks

Kekemato_BAP

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Re: Chemistry Question Thread
« Reply #2343 on: June 26, 2017, 12:58:05 pm »
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Hey! The reaction QUOTIENT a value used to describe a reaction at a particular point in time. However, the equilibrium constant is a value used to describe a reaction when it reaches equilibrium. They are calculated in exactly the same way (ie. same formula). However, you can use the reaction quotient (Q), when compared with the equilibrium constant (k) in order to determine which direction the reaction will shift.

If Q>k

Where the reaction quotient is greater than the given equilibrium value, the reaction will favour the reactants. Thus, it will move to the left until it reaches equilibrium.

If Q<k

Where the reaction quotient is less than the given equilibrium value, the reaction will favour the products. Thus, it will move to the right until it reaches equilibrium.

If Q=k

Where there reaction quotient is equal to the given equilibrium value, the reaction is at equilibrium!

In a question, you would likely be given an equilibrium value (eg. 2.4), and then a number of concentrations from which you can calculate the current reaction quotient. From that, you can decide in which direction the reaction will move!

Could you explain the things about excluding solids and liquids in the equilibrium? And also the thing about temperature being the only affecting variable.
Hello

beau77bro

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Re: Chemistry Question Thread
« Reply #2344 on: June 26, 2017, 05:05:08 pm »
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Could you explain the things about excluding solids and liquids in the equilibrium? And also the thing about temperature being the only affecting variable.

ok so the temperature being the only effecting variable, is basically the fact that at a stable temperature - when you add more of a substance (concentration) or decrease the volume the equilibrium is taking place it will change concentrations on both sides accordingly to the reaction. so for the extra products it would balance out essentially. similarly in the change of volume the pressure is proportional to concentration, so it will move to the side with less gaseous moles increasing the concentration on the other side. but since the pressure has changed and there are now more gaseous molecules squeezed in a smaller space (and there are more moles on the side which the equilibrium has moved away) so it essentially balances out. the more moles has decreased in overall molecules - but there are more in a small space so increased concentration, the less moles side has gained more overall molecules and these take up a smaller space increasing the concentration but since there are less moles overall it's concentration increases more, but equals out in the equation.

now temperature is different because it doesn't balance out like the other changes do. it makes the reaction tend more to one side full stop. so for an endothermic if you increase the temperature, you push it to the right increasing the products. so in the equation the numerator increases drastically and the denominator shrinks - changing the K value. that's why they are for given temperatures.

now im not super sure on the solids and liquids part - i think its to do with the fact gas solubility increases as it cools. or not. but either way gaseous-solid or liquid reactions will favour the side with less gaseous moles, so if you increase the pressure they will just dissolve or make solids in a way that the equation doesnt allow for. i hope a moderator can shine some more light on this.

kiwiberry

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Re: Chemistry Question Thread
« Reply #2345 on: June 26, 2017, 05:50:07 pm »
+4
Could you explain the things about excluding solids and liquids in the equilibrium? And also the thing about temperature being the only affecting variable.

Solids and liquids have a fixed concentration - we exclude them from the equilibrium constant because their concentrations do not change over time

hey so what would a processing skills test for chem look like? like what can i expect it to include?
thanks

In my processing skills test we got asked to consolidate information from a passage into a table and answer questions about it, draw graphs and do some basic calculations, so make sure you remember how to draw proper graphs/tables (line of best fit, title, etc.) :)
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kylesara

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Re: Chemistry Question Thread
« Reply #2346 on: June 28, 2017, 03:16:49 pm »
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Hi, I was just going through the 2016 Chem past paper and don't understand why the answer for question 11 in Multiple choice is D and not A.
Thanks.

kiwiberry

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Re: Chemistry Question Thread
« Reply #2347 on: June 28, 2017, 03:45:24 pm »
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Hi, I was just going through the 2016 Chem past paper and don't understand why the answer for question 11 in Multiple choice is D and not A.
Thanks.

The numbers in D add up to 7 whereas the numbers in A add up to 8 - we pick the one with the lowest sum, so D is the answer :)
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beau77bro

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Re: Chemistry Question Thread
« Reply #2348 on: June 28, 2017, 11:25:54 pm »
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Please help Q26b - how do u do percentage (w/v)?

Thanks - beau

Shadowxo

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Re: Chemistry Question Thread
« Reply #2349 on: June 30, 2017, 03:40:40 pm »
+1
(Image removed from quote.)

Please help Q26b - how do u do percentage (w/v)?

Thanks - beau

So if you have the molarity of it, you have the number of mols per L = number of mols per 1000mL
1 mol of acetic acid = 60.0 g (rounding to one decimal place using the periodic table, your one may have more decimal places and you can use this)
So convert the number of mols of acetic acid to the weight of it (g) (multiply the number of mols by 60.0 to get the weight in g per 1000mL)
%w/v would be weight in grams / volume in mL * 100%
So divide the grams of acetic acid in each litre by 1000 (the mLs ie volume of vinegar) then multiply by 100 to get the percentage

Hope this helps :) bit rusty on my chem though
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winstondarmawan

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Re: Chemistry Question Thread
« Reply #2350 on: June 30, 2017, 07:49:52 pm »
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Hey can someone please help with Q1, all parts except (iii) and (v) and an explanation would help alot! TIA

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MisterNeo

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Re: Chemistry Question Thread
« Reply #2351 on: June 30, 2017, 09:27:59 pm »
+3
Hey can someone please help with Q1, all parts except (iii) and (v) and an explanation would help alot! TIA

https://scontent-syd2-1.xx.fbcdn.net/v/t34.0-12/19679939_1257428117716042_1386574065_n.jpg?oh=a0a1bdfe654c145b7f2552a18f855b8d&oe=5957A740

I

Neutralisation reactions are NOT redox reactions. There is no change in oxidation state or exchange of electrons as hydrogen in HCl is already in a +1 state and ends as +1.
II

Oxidation

Reduction

This combustion one is a bit complex because there are a crapload of electrons involved for longer alkanes. Anyhow, carbon can have oxidation states from -4 to +4, and ethane carbon exists as -3 because of the 3 +1 hydrogens bonded to each; and molecules must have a total state of 0. Therefore, carbon changes from a -3 to a +4 from ethane to carbon dioxide. And, oxygen goes from 0 in diatomic to -2 in carbon dioxide and water.
IV
Esterification is a redox reaction.
The alcohol is oxidised, and the alkanoic acid is reduced.
This guide explains it.
VI

This decomposition reaction is not a redox reaction and has no oxidation/reduction involved because the calcium, oxygen, and carbon all start and end with the same oxidation states of 2+, 2-, and 4+ respectively.
VII

This is the same with question I because it is also a neutralisation reaction. There is an exchange in proton to form ammonium, which then bonds with the already negative chlorine. Hence, there are no changes in oxidation states.
VIII
This one is a redox reaction because aluminium 3+ ions are reduced into solid aluminium by gaining 3 electrons each.
The oxygen is oxidised because it starts as a 2- ion and gets gains the electrons from aluminium ions to form oxygen gas.
This explains it nicely with equations.
IX

Oxidation

Reduction

This is a redox reaction because hydrogen goes from +1 in water to 0 in diatomic hydrogen, and oxygen goes from -2 in water to 0 in diatomic oxygen.
Basically, you have to watch out for changes in oxidation states where electrons are being exchanged.
Hope this helps ;)


winstondarmawan

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Re: Chemistry Question Thread
« Reply #2352 on: July 01, 2017, 01:32:57 pm »
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I

Neutralisation reactions are NOT redox reactions. There is no change in oxidation state or exchange of electrons as hydrogen in HCl is already in a +1 state and ends as +1.
II

Oxidation

Reduction

This combustion one is a bit complex because there are a crapload of electrons involved for longer alkanes. Anyhow, carbon can have oxidation states from -4 to +4, and ethane carbon exists as -3 because of the 3 +1 hydrogens bonded to each; and molecules must have a total state of 0. Therefore, carbon changes from a -3 to a +4 from ethane to carbon dioxide. And, oxygen goes from 0 in diatomic to -2 in carbon dioxide and water.
IV
Esterification is a redox reaction.
The alcohol is oxidised, and the alkanoic acid is reduced.
This guide explains it.
VI

This decomposition reaction is not a redox reaction and has no oxidation/reduction involved because the calcium, oxygen, and carbon all start and end with the same oxidation states of 2+, 2-, and 4+ respectively.
VII

This is the same with question I because it is also a neutralisation reaction. There is an exchange in proton to form ammonium, which then bonds with the already negative chlorine. Hence, there are no changes in oxidation states.
VIII
This one is a redox reaction because aluminium 3+ ions are reduced into solid aluminium by gaining 3 electrons each.
The oxygen is oxidised because it starts as a 2- ion and gets gains the electrons from aluminium ions to form oxygen gas.
This explains it nicely with equations.
IX

Oxidation

Reduction

This is a redox reaction because hydrogen goes from +1 in water to 0 in diatomic hydrogen, and oxygen goes from -2 in water to 0 in diatomic oxygen.
Basically, you have to watch out for changes in oxidation states where electrons are being exchanged.
Hope this helps ;)



Thank you for clearing this up! My school doesn't teach oxidation states which is why I struggle with these questions.

jakesilove

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Re: Chemistry Question Thread
« Reply #2353 on: July 02, 2017, 04:26:40 pm »
+4
Just wanted to say thank you so much to all the incredible users answering questions at the moment. Responses have been comprehensive, clear, and correct, and it is so so impressive to watch this community grow. I'm on a bit of a hiatus, as I'm out of the state for the next few days, but will still try to check in every now and again. I expected heaps of Chem questions but, alas, they have all already been answered! Again, thank you to everyone who is a part is this beautiful beautiful community; we love you all, we need you all, and we appreciate you all.

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Annie657

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Re: Chemistry Question Thread
« Reply #2354 on: July 03, 2017, 12:05:17 pm »
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Hi guys,
So I'm just writing my notes for steps in titration, and was wondering is the end point when the solution first changes colour or when it sustains a colour for 10 seconds? If so when should you record your measurement?

Thanks! Annabelle
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