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March 30, 2024, 02:30:34 am

Author Topic: HSC Chemistry Question Thread  (Read 1040854 times)  Share 

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RuiAce

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Re: Chemistry Question Thread
« Reply #405 on: July 13, 2016, 09:12:37 pm »
+1
I'm having trouble with a following two questions in one of my school's past trial exams:

Spoiler
"A student used this equipment [experimental setup for alkanol's heat of combustion prac: i.e. retort stand, clamp, spirit burner, water, thermometer, conical flask] to heat 250g of water. The mass of the spirit burner, which contained ethanol, decreased from 296.52g to 295.95g.

Given the heat of combustion of ethanol is 29.7kJ/g, calculate the maximum possible change in the temperature of the water."


--> this question I'm confused about the kJ/g...seeing ∆H = Q/n for kJ/mol, i just thought of doing ∆H=Q/m as it was in kJ/g (kilojule per gram), and solving for Q, as: 29.7 = Q/0.57 (0.57 is mass of ethanol burned)...yeah i got lost :(

Spoiler
"Ozone concentrations are measured in Dobson units (DU). DU are the standard way to express ozone concentration in the stratosphere. A concentration of one DU means there would be 2.7x1020 ozone molecules in a layer of air that was one square metre in area and 0.01mm thick.

A baseline value of 220 DU is chosen as the starting point for an ozone hole in the stratosphere since total ozone values of less than 200 DU were not found in historic obsevations over Antarctica over 1979.
Which of the following concentrations, in moles per cubic metre (molm'3), is most nearly equivalent to 220 DU?"

A) 0.05 molm-3
B) 10 molm-3
C) 5000 molm-3
D) 10 000 molm-3

--> this one i am just lost in.
If they give you the enthalpy change per gram, then ∆H=-Q/m (don't forget the negative) is correct.

Q = mc∆T = 16.929 kJ

Continue.
_________________________________
This question tests how well you know your units.








Tell me if I am wrong.

wesadora

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Re: Chemistry Question Thread
« Reply #406 on: July 13, 2016, 09:17:38 pm »
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That's correct, answer was D. I'll have a look at the working and redo the first question and see what I get.

Thanks Rui, also 1 more quick question:
I just came across a question using crystalline oxalic acid (aka oxalic acid dihydrate) as a primary standard. This comes in the form (COOH)2.2H2O

I'm confused because in class we used it and my teacher said to include the 2H2O in the molar mass, while i SWEAR my chem tutor said not to. And this makes a BIG difference in the calculation (obviously) ._.
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RuiAce

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Re: Chemistry Question Thread
« Reply #407 on: July 13, 2016, 09:18:53 pm »
+1
That's correct, answer was D. I'll have a look at the working and redo the first question and see what I get.

Thanks Rui, also 1 more quick question:
I just came across a question using crystalline oxalic acid (aka oxalic acid dihydrate) as a primary standard. This comes in the form (COOH)2.2H2O

I'm confused because in class we used it and my teacher said to include the 2H2O in the molar mass, while i SWEAR my chem tutor said not to. And this makes a BIG difference in the calculation (obviously) ._.
I'd say to include the molar mass of the dihydrate. I will ask around because I haven't done this type of question in ages.

This post will be deleted.
________

A response I got

Well I guess if crystalline oxalic acid (just (COOH)2 without the hydrates) was used, adding it to water to get those extra dihydrates shouldn't impact the experimental results, so I think he shouldn't include it in his calculations. Because, according to me, adding the oxalic acid to the water just causes the acid to essentially dissolve in the water and pick up that dihydrate (similar to changing from a solid to aqueous subscript) and so it should be treated as just a change of state.

But I'm not completely sure.
« Last Edit: July 13, 2016, 10:57:30 pm by RuiAce »

ProfLayton2000

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Re: Chemistry Question Thread
« Reply #408 on: July 13, 2016, 10:51:39 pm »
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Hi there,

So on the HSC Chem 2011 paper q29b: "Why does the neutralisation of any strong acid in an aqueous solution by any strong base always result in a heat of reaction approximately -57kJ mol^-1

Could you please explain why it is what it is?

RuiAce

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Re: Chemistry Question Thread
« Reply #409 on: July 13, 2016, 10:58:45 pm »
+1
Hi there,

So on the HSC Chem 2011 paper q29b: "Why does the neutralisation of any strong acid in an aqueous solution by any strong base always result in a heat of reaction approximately -57kJ mol^-1

Could you please explain why it is what it is?
Because:
a) The reaction goes to completion between two completely ionised substances
b) The reaction H3O+ + OH- -> 2 H2O(l) has ∆H=-57 kJ mol-1 (approximately).

That's all you need. See if you can realise how that just builds up to an answer.

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Re: Chemistry Question Thread
« Reply #410 on: July 14, 2016, 12:44:28 am »
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I was wondering for chemcial equations how do you know the state of each substance/compound whether it is aqueoud, liquid, gas or solid?

jakesilove

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Re: Chemistry Question Thread
« Reply #411 on: July 14, 2016, 10:16:29 am »
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I was wondering for chemcial equations how do you know the state of each substance/compound whether it is aqueoud, liquid, gas or solid?

Hey!

To some extent, it's just intuition after you've done a hell of a lot of practice. The only liquid that you really deal with is water: Assume water is liquid, unless it is clearly steam (in which case it is a gas). Anything else, in like a 'liquid' form, but containing metal ions etc. will always be aqueous. That means any acidic solution, any basic solution, any ions in solution, will be aqueous. Gas, again, is usually obvious from the question or from having done heaps of past papers. Things like hydrogen gas, oxygen, nitrogen (all diatomic) will be in a gaseous form. I can't give much more advice than that, except to say DO HEAPS OF PAST PAPERS and this will all because clear. Think about what is actually happening in the reaction, and you'll be sweet :)

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RuiAce

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Re: Chemistry Question Thread
« Reply #412 on: July 14, 2016, 12:09:06 pm »
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I was wondering for chemcial equations how do you know the state of each substance/compound whether it is aqueoud, liquid, gas or solid?
On top of what Jake said, some are trivial whilst others are logic.

E.g. The water you produce in neutralisation between an acid and base has to be a liquid. You're mixing two liquids together.

As opposed to the combustion of ethanol where you're achieving such high temperatures and water must have been boiled already.


An example of what I call a special case I know is that aluminium refuses to react with liquid water, but rather steam.

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Re: Chemistry Question Thread
« Reply #413 on: July 15, 2016, 08:22:01 pm »
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can you check my response attached below thanks

when talking about conditions for certain process involving equilibrium  do i have to talk about what happens if (temperature, pressure) is too low or too high   or do i just need to talk about the compromise temperature (the temperature is not too high or low thus favours the forward reaction)?

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Re: Chemistry Question Thread
« Reply #414 on: July 15, 2016, 08:37:44 pm »
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When going down a group in the periodic table, a new electron shell is added to the atom. Why is this the case, is this "by definition"

Also this would increase (what I said above) the atomic radius as the outermost electrons are placed further from the nucleus. Why is this the case?

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Re: Chemistry Question Thread
« Reply #415 on: July 15, 2016, 08:43:28 pm »
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When going down a group in the periodic table, a new electron shell is added to the atom. Why is this the case, is this "by definition"

Also this would increase (what I said above) the atomic radius as the outermost electrons are placed further from the nucleus. Why is this the case?

The periodic table was constructed in such a way that patterns in the elements are made obvious and systematic (to a limited degree of course, trends break down as you go down a group).
Every time a new electron shell is added, the number of valence electrons can be thought of as being 'reset', hence elements in the same group have the same number of valence electrons and have similar (although by no means identical) properties.

As more electrons are added, they need to occupy higher energy levels which are thus more distant from the nucleus. This is a consequence of orbital theory which restricts the number of electrons in a particular energy state. The lowest energy state in an element (1s) can hold 2e-, once this is filled, it goes to the next orbital (2s), and so on, with the electrons becoming progressively more distant from the nucleus.

Across a period, as you gain more electrons (and consequently more protons), the atomic radius actually decreases because the pull on each valence electron becomes greater (due to more protons)
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Re: Chemistry Question Thread
« Reply #416 on: July 15, 2016, 08:46:09 pm »
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Another question but why did Mendeleev leave gaps in his periodic table?

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Re: Chemistry Question Thread
« Reply #417 on: July 15, 2016, 08:50:08 pm »
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Another question but why did Mendeleev leave gaps in his periodic table?

That was purely due to the fact that the 'next' element actually skipped a property, and Mendeleev believed that there was a missing element which hadn't been discovered which belonged there. Later on, that element was discovered.
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Happy Physics Land

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Re: Chemistry Question Thread
« Reply #418 on: July 16, 2016, 06:44:57 pm »
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Another question but why did Mendeleev leave gaps in his periodic table?

Because not all the elements can be found in the order as we see them now - some are more difficult to synthesise or extract than other due to the technological deficiency of the time.
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Happy Physics Land

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Re: Chemistry Question Thread
« Reply #419 on: July 16, 2016, 07:00:06 pm »
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can you check my response attached below thanks

when talking about conditions for certain process involving equilibrium  do i have to talk about what happens if (temperature, pressure) is too low or too high   or do i just need to talk about the compromise temperature (the temperature is not too high or low thus favours the forward reaction)?

Hey Katherine,

Your response is certainly very detailed and encompasses the essential things you need to know. When you get asked about equilibrium questions involving ammonia, you usually talk about why there is a compromise. I see you have already included those into your response so thats very good.

A couple of things that would be beneficial for you to add:
- You want to be more specific with the type of catalyst you are using. Iron oxide is ok, but l think its better to say Fe3O4 (magnetite). This shows the teacher you really know your stuff.
- With temperature, you dont want very high temperature also because it can damage the catalyst. Since catalysts are quite expensive to replace and relatively vulnerable, you want to preserve them in good conditions by using lower temperatures.
- You must also add the liquidification of ammonia as a part of your conditions. By this l mean constantly turning ammonia gas into liquid to remove gaseous ammonia from the equilibrium and hence the equilibrium will shift right, favouring the production of more ammonia. You touched on this in your last condition but its very important to mention LIQUIDIFICATION of ammonia.

Other than those points, I think this is a very good table to study for ammonia section of chemical monitoring and management. Kudos to you, you have done a very good job!

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