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March 28, 2024, 08:13:24 pm

Author Topic: HSC Chemistry Question Thread  (Read 1040508 times)  Share 

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david.wang28

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Re: Chemistry Question Thread
« Reply #3675 on: December 27, 2018, 09:34:11 pm »
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Hello,
I'm having trouble with Q 17, Q 19 b) and Q 20, but I have shown my working out in the attachments. May I get a detailed answer please? Thanks :)
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fun_jirachi

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Re: Chemistry Question Thread
« Reply #3676 on: December 27, 2018, 11:21:51 pm »
+3
For question 17, I think you've understood quite well what you're doing. Just stick a zero in for the initial concentration of nitrogen gas, since it doesn't say anything about nitrogen at the beginning of the question. The assumption is that if it isn't mentioned, it doesnt exist at the start. Because what's given is added to an empty vessel, nothing was in the vessel to begin with and nothing more or less than what is added is added. Crunch the numbers and you should get that


For 19, I think you should use an ICE table. What you've done there is add some ammonia that doesnt exist by subbing directly into the equilibrium constant equation, and that doesnt work. Using an ICE table, you eventually get to an expression like this:

I'm going to trust your maths skills here, but if you need any help with this part ask again :). Solve for x, which is roughly 8.28 by 10 to the -3. Since the equilibrium constant is zero at the start (since ammonia has zero concentration at the start), and now it has a value, then it follows that equilibrium shifts right.

For 20, a) is pretty much spot on. At the start, Q is 150, so at equilibrium, there will be more reactants and less products. So that part's great :D With b) you have to notice what's implied by this. As the temperature increases, there are more products (implied by increase in constant). This tells you that when you increase the temperature, by LCP the system shifts right to minimise the change, which would be to absorb it. Therefore the forward reaction is endothermic, and therefore the enthalpy change is positive.

Hope this helps :)
« Last Edit: December 27, 2018, 11:27:08 pm by fun_jirachi »
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david.wang28

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Re: Chemistry Question Thread
« Reply #3677 on: December 28, 2018, 09:56:11 pm »
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For question 17, I think you've understood quite well what you're doing. Just stick a zero in for the initial concentration of nitrogen gas, since it doesn't say anything about nitrogen at the beginning of the question. The assumption is that if it isn't mentioned, it doesnt exist at the start. Because what's given is added to an empty vessel, nothing was in the vessel to begin with and nothing more or less than what is added is added. Crunch the numbers and you should get that


For 19, I think you should use an ICE table. What you've done there is add some ammonia that doesnt exist by subbing directly into the equilibrium constant equation, and that doesnt work. Using an ICE table, you eventually get to an expression like this:

I'm going to trust your maths skills here, but if you need any help with this part ask again :). Solve for x, which is roughly 8.28 by 10 to the -3. Since the equilibrium constant is zero at the start (since ammonia has zero concentration at the start), and now it has a value, then it follows that equilibrium shifts right.

For 20, a) is pretty much spot on. At the start, Q is 150, so at equilibrium, there will be more reactants and less products. So that part's great :D With b) you have to notice what's implied by this. As the temperature increases, there are more products (implied by increase in constant). This tells you that when you increase the temperature, by LCP the system shifts right to minimise the change, which would be to absorb it. Therefore the forward reaction is endothermic, and therefore the enthalpy change is positive.

Hope this helps :)
For Q19, I seemed to have trouble finding x. I've tried to expand the equation using binomial theorem (it seems surreal, I know), but that has led me to nowhere. Can you please give me guidance on how to find x please? Thanks. :)
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fun_jirachi

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Re: Chemistry Question Thread
« Reply #3678 on: January 09, 2019, 09:39:12 pm »
+1
Sorry I didn't reply, I was away for quite a while!

Firstly please note my answer of 8.28 x 10^-3 is erroneous, ignore that please! made a calculation error. It's more like 7.75x10^-3. This question is a bit of a tough one because you're given a quartic to solve. There might be an easier method, but I found that Newton's method of finding a root approximation was relatively okay. I subbed in values for f(x) which happens to be 1.404x^4-0.7722x^3-3.8479x^2-0.01235+0.000325 (best expanded with binomial theorem, saves a lot of work). I just looked for a value that had 5 zeroes after the decimal point before trying newton's method (for reference i started with x1 being 0.0078, going down in small increments). Finding f'(x), you use this method quite a few times till I guess you get a good enough approximation for our standards (i chose something pretty small when subbing into f(x)). I guess with this method it's up to your own discretion where you stop, but as for this, I'm not too sure. That's how I did it anyway, you may have a different better method!

Hope this helps :D
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david.wang28

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Re: Chemistry Question Thread
« Reply #3679 on: January 11, 2019, 11:56:06 am »
+1
Sorry I didn't reply, I was away for quite a while!

Firstly please note my answer of 8.28 x 10^-3 is erroneous, ignore that please! made a calculation error. It's more like 7.75x10^-3. This question is a bit of a tough one because you're given a quartic to solve. There might be an easier method, but I found that Newton's method of finding a root approximation was relatively okay. I subbed in values for f(x) which happens to be 1.404x^4-0.7722x^3-3.8479x^2-0.01235+0.000325 (best expanded with binomial theorem, saves a lot of work). I just looked for a value that had 5 zeroes after the decimal point before trying newton's method (for reference i started with x1 being 0.0078, going down in small increments). Finding f'(x), you use this method quite a few times till I guess you get a good enough approximation for our standards (i chose something pretty small when subbing into f(x)). I guess with this method it's up to your own discretion where you stop, but as for this, I'm not too sure. That's how I did it anyway, you may have a different better method!

Hope this helps :D
This is what I did for Q19.
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fun_jirachi

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Re: Chemistry Question Thread
« Reply #3680 on: January 11, 2019, 12:29:05 pm »
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Well, should've thought of that one. Looking at that now, it seems pretty obvious :)  (should really stop overthinking things)

Good job! :D
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david.wang28

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Re: Chemistry Question Thread
« Reply #3681 on: January 11, 2019, 12:39:15 pm »
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Well, should've thought of that one. Looking at that now, it seems pretty obvious :)  (should really stop overthinking things)

Good job! :D
Thanks man! Usually chemistry questions shouldn't be too hard, or at least not to a 3u level. Though the knowledge you gain in 3u and 4u should be beneficial to u
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louisaaa01

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Re: Chemistry Question Thread
« Reply #3682 on: January 14, 2019, 10:09:52 am »
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I came across this question and I was pretty confused:

The question is asking to write an equation for the reaction between propyl methanamine and hydrochloric acid and name all organic products.

I said the products would be N-chloromethanamine and propane (really more a guess than anything) but I wasn't sure whether this was right or not. What are the products of this reaction?

« Last Edit: January 14, 2019, 10:14:51 am by louisaaa01 »
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louisaaa01

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Re: Chemistry Question Thread
« Reply #3683 on: January 14, 2019, 10:13:00 am »
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Can someone please help with the attached question? Thank you!
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janeaustin

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Re: Chemistry Question Thread
« Reply #3684 on: January 14, 2019, 08:10:12 pm »
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For Acid + Metal Carbonate reactions, is the state of the metal carbonate reactant always a solid?

For example, the reaction between phosphoric acid and sodium hydrogen carbonate:
H3PO4 (aq) + 3NaHCO3 (s) →  Na3PO4 (aq) + 3CO2 (g) + 3H2O (l)
Why is NaHCO3 a solid? According to the solubility rules, all Group 1 metals (in this case, Na) are soluble, so isn't this carbonate meant to exist in aqueous form?

Jessthemess

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Re: Chemistry Question Thread
« Reply #3685 on: January 14, 2019, 10:55:24 pm »
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So, I thought this dot point was simple but now I'm confused.

analyse examples of non-equilibrium systems in terms of the effect of entropy and enthalpy, for example:
– combustion reactions
– photosynthesis


I thought combustion was a static equilibrium reaction rather than a non-equilibrium one.

myopic_owl22

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Re: Chemistry Question Thread
« Reply #3686 on: January 16, 2019, 03:54:25 pm »
+1
For Acid + Metal Carbonate reactions, is the state of the metal carbonate reactant always a solid?

For example, the reaction between phosphoric acid and sodium hydrogen carbonate:
H3PO4 (aq) + 3NaHCO3 (s) →  Na3PO4 (aq) + 3CO2 (g) + 3H2O (l)
Why is NaHCO3 a solid? According to the solubility rules, all Group 1 metals (in this case, Na) are soluble, so isn't this carbonate meant to exist in aqueous form?

Here, sodium bicarbonate (NaHCO3) is added in its solid form - it can be in aqueous form if you add water and dissolve it. Solubility rules only apply when the ions inside dissociate into its ions in the presence of water!
In an exam or somewhere you're asked to write down the states of an equation, they should give you the states of the reactants as you wouldn't know otherwise. But as there's water involved (phosphoric acid is aqueous and therefore dilute), any soluble products will be aqueous to some degree (remember that precipitates can form as according to each compound's ksp values!) 
hope this helps :)

janeaustin

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Re: Chemistry Question Thread
« Reply #3687 on: January 16, 2019, 04:08:05 pm »
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When writing ionic equations for the dissociation of a compound, are we to always use a reversible arrow?

For example, for the reaction FeCl2 (s) ⇌ Fe2+ (aq) + 2Cl- (aq)
Why is an equilibrium arrow used and not just a forward arrow? How do we know when to use which arrow?

janeaustin

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Re: Chemistry Question Thread
« Reply #3688 on: January 16, 2019, 04:09:51 pm »
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Here, sodium bicarbonate (NaHCO3) is added in its solid form - it can be in aqueous form if you add water and dissolve it. Solubility rules only apply when the ions inside dissociate into its ions in the presence of water!
In an exam or somewhere you're asked to write down the states of an equation, they should give you the states of the reactants as you wouldn't know otherwise. But as there's water involved (phosphoric acid is aqueous and therefore dilute), any soluble products will be aqueous to some degree (remember that precipitates can form as according to each compound's ksp values!) 
hope this helps :)

Yes this helped, thank you so much!

myopic_owl22

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Re: Chemistry Question Thread
« Reply #3689 on: January 16, 2019, 05:20:20 pm »
+1
When writing ionic equations for the dissociation of a compound, are we to always use a reversible arrow?

For example, for the reaction FeCl2 (s) ⇌ Fe2+ (aq) + 2Cl- (aq)
Why is an equilibrium arrow used and not just a forward arrow? How do we know when to use which arrow?


Depends on whether the reaction is reversible or not. If the reaction is reversible, the arrow goes both forwards and backwards. Reactions like combustion or many precipitation reactions will not be reversible, as the enthalpy/ entropy is too varied for ΔG to = 0 (which happens when a reaction is in equilirbium). I'm assuming that everything is in a closed system, by the way.
Sometimes, the single forwards arrow also denotes an equilibrium reaction that has gone to completion, but it isn't used too often. Let me explain what I mean...

Iron (II) chloride, like most compounds, are soluble to some degree. At one point, though, the solution will become saturated (no more water molecules can attach to the dissociated ions) and the stuff will precipitate as solid. How much you need to add in to achieve this effect is determined by each compunds' solubility/ ksp value. FeCl2 is pretty soluble given it's a halide salt. In a large amount of water, a small amount of it will be easily dissolved and the equation above will achieve completion. Of course, we can't guarantee that this situation will always be the case, so the double arrow covers all possibilities of amounts of water/ compound, etc. This also includes other factors like increasing the temperature. In short, all dissolution reactions will be reversible and should be written with a double arrow :)