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March 29, 2024, 12:47:00 am

Author Topic: HSC Chemistry Question Thread  (Read 1040560 times)  Share 

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IkeaandOfficeworks

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Re: Chemistry Question Thread
« Reply #60 on: February 24, 2016, 08:40:37 pm »
+1
I just wanted to ask about how to answer an ASSESS question. Like the structure (i.e how you start the response, what to include etc.) in answering these questions. Thank you!

jakesilove

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Re: Chemistry Question Thread
« Reply #61 on: February 24, 2016, 08:46:54 pm »
+2
I just wanted to ask about how to answer an ASSESS question. Like the structure (i.e how you start the response, what to include etc.) in answering these questions. Thank you!

Hey!

To answer an assess question, you essentially need to include an extra step with is a "judgement" or "reasoning" step.

You won't be required to conclude with some sort of actual assessment. For instance, at the end of an "assess the use of Ethanol as an alternate fuel source" you don't need to say "Ethanol is definitely good as an alternative fuel source". Rather, at the end you should say something like "Ethanol has many benefits as an alternate fuel source, however further research is required to increase the efficiency" etc.

The difference between "Explain" and "Assess" is the "reasoning" step. For instance, with ethanol, if you need to assess advantages and disadvantages it is not enough to just state that Ethanol has a lower heat of combustion. Rather, you need to say "As Ethanol has a lower heat of combustion, it is less productive than regular fuel when used. This is a negative result, as it means an individual needs to purchase more fuel to travel the same distance, as compared to Octane". Notice I've explained WHY it is a disadvantage. That is the main extra component of an "assess" question!

Hope this helps!

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IkeaandOfficeworks

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Re: Chemistry Question Thread
« Reply #62 on: February 24, 2016, 10:27:53 pm »
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Thanks Jake! One more question, how different is an Evaluate question from an Assess question? Thanks again!

Maz

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Re: Chemistry Question Thread
« Reply #63 on: February 25, 2016, 07:47:01 pm »
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Hey
i have a test on equilibrium tomorrow and i was wandering if someone could please help me?
The general idea is that if a system is in equilibrium the foreword and reverse reactions are occurring at the same rate...
if a reactant is added the system will move to reduce and counter-react this added reactant by producing more products.
but what happens if the newly added reactant is so much that the other reactant doesn't even have enough particles available to
react with it? does the excess just exist in a non-reacted state in the system? and how do you determine that, just by being given a
equilibrium reaction?
Would really appreciate and help with this...
thankyou so so much in advance  :)
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jakesilove

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Re: Chemistry Question Thread
« Reply #64 on: February 25, 2016, 08:08:24 pm »
0
Hey
i have a test on equilibrium tomorrow and i was wandering if someone could please help me?
The general idea is that if a system is in equilibrium the foreword and reverse reactions are occurring at the same rate...
if a reactant is added the system will move to reduce and counter-react this added reactant by producing more products.
but what happens if the newly added reactant is so much that the other reactant doesn't even have enough particles available to
react with it? does the excess just exist in a non-reacted state in the system? and how do you determine that, just by being given a
equilibrium reaction?
Would really appreciate and help with this...
thankyou so so much in advance  :)

You're absolutely right! Any "excess" particles just sit there in an unreacted state. You can only figure this out if you are given molar values. For instance, in the reaction between Hydrochloric acid and Sodium Hydroxide, the ratio of the two reactants is 1:1. Therefore, if there are more moles of one than the other, the "excess" will just sit there gathering dust (not literally, obviously!).

In a generic "Extra reactant x is added, what will happen", you can assume that this "excess" principle is totally irrelevant. It is only when you are expected to do molar calculations that you should even think about principles like excess reagents etc.

Hope this helps!

Jake
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Re: Chemistry Question Thread
« Reply #65 on: February 25, 2016, 09:41:21 pm »
0
You're absolutely right! Any "excess" particles just sit there in an unreacted state. You can only figure this out if you are given molar values. For instance, in the reaction between Hydrochloric acid and Sodium Hydroxide, the ratio of the two reactants is 1:1. Therefore, if there are more moles of one than the other, the "excess" will just sit there gathering dust (not literally, obviously!).

In a generic "Extra reactant x is added, what will happen", you can assume that this "excess" principle is totally irrelevant. It is only when you are expected to do molar calculations that you should even think about principles like excess reagents etc.

Hope this helps!

Jake
it helps a lot
thankyou!  :)
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amandali

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Re: Chemistry Question Thread
« Reply #66 on: February 26, 2016, 03:53:46 pm »
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why does redox reaction occur when the cathode and anode are not separated?

jakesilove

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Re: Chemistry Question Thread
« Reply #67 on: February 26, 2016, 03:59:26 pm »
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why does redox reaction occur when the cathode and anode are not separated?

Hey Amandali!

Redox reactions like the one in the picture you've attached actually don't need to be separated! As long as there are two dissimilar metals and a way for ions to travel back and forth (in this case, the solution connecting the two is that method) then a reaction will occur. The separation of the typical Galvanic cell is not actually required. Electrons still travel from one substance to the other, and ions travel in the opposite direction in the solution.

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RuiAce

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Re: Chemistry Question Thread
« Reply #68 on: February 26, 2016, 07:25:43 pm »
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Hey Katherine!

I actually get a different answer (12.5) and have looked through my working a few times and am not sure where I've gone wrong. Perhaps your answers are wrong? Otherwise, hopefully someone in the community can help me out!

(Image removed from quote.)

Jake

Hey Jake, with this question I found out your mistake.

When I typed 0.12/Ans into the calculator, where Ans=Molar Mass of Ca(OH)2, I found the answer to be 1.619...x10-3

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Re: Chemistry Question Thread
« Reply #69 on: February 26, 2016, 07:27:05 pm »
+1
Thanks Jake! One more question, how different is an Evaluate question from an Assess question? Thanks again!
To answer this question, it doesn't.

RuiAce

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Re: Chemistry Question Thread
« Reply #70 on: February 26, 2016, 07:33:30 pm »
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why does redox reaction occur when the cathode and anode are not separated?
Hey Amandali!

Redox reactions like the one in the picture you've attached actually don't need to be separated! As long as there are two dissimilar metals and a way for ions to travel back and forth (in this case, the solution connecting the two is that method) then a reaction will occur. The separation of the typical Galvanic cell is not actually required. Electrons still travel from one substance to the other, and ions travel in the opposite direction in the solution.

Jake
To reinforce what Jake said here, note that for the seperated galvanic cell, the salt bridge is there only for a migration of ions. Note that in the single unit, the migration of ions is equally possible!

amandali

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Re: Chemistry Question Thread
« Reply #71 on: February 26, 2016, 10:41:50 pm »
+1
0.300g of solid NaOH was added to 1.00L of 5.00*10^-3 mol/L HNO3
Assuming no volume change, what is the pH of final solution
ans: 11.4

what i did was:
-show that HNO3 is limiting reagent with moles (5*10^-3)
-n(H20) = (5*10^-3)  since 1 mol of HNO3 produce 1 mole of H20
-find n(H+)= n(H20)
-[H+]= n/v
-  pH=-log(H+)


Happy Physics Land

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Re: Chemistry Question Thread
« Reply #72 on: February 26, 2016, 11:08:42 pm »
+2
0.300g of solid NaOH was added to 1.00L of 5.00*10^-3 mol/L HNO3
Assuming no volume change, what is the pH of final solution
ans: 11.4

what i did was:
-show that HNO3 is limiting reagent with moles (5*10^-3)
-n(H20) = (5*10^-3)  since 1 mol of HNO3 produce 1 mole of H20
-find n(H+)= n(H20)
-[H+]= n/v
-  pH=-log(H+)

Hey Amanda:

Good question. Ok so for me firstly what I would have done is to establish a balanced neutralisation equation.

NaOH (s) + HNO3 (aq) --> NaNO3 (aq) + H2O (l)

So clearly with this equation no further balancing is required. Now we need to work out the moles of HNO3 and moles of NaOH to determine the limiting reagent.

n of NaOH = 0.3/(22.9 + 16 + 1.008) = 0.0075173 moles
n of HNO3 = CV = 5 x 10^-3 x 1 = 0.005 moles

Ok so quite clearly since the molar ratio in the balanced neutralisation equation is 1:1:1:1, HNO3 is the limiting reagent, as you have correctly identified, well done!

Now theres an extra step here we need to be careful of, because the H+ ions and OH- ions that are dissociated from HNO3 and NaOH would neutralise one another and become H2O, we need to calculate the amount of moles of NaOH that is in excess of:

Excess n(NaOH) = 0.0075173 - 0.005 = 0.0025173 moles
Assuming theres no volume change, the concentration of NaOH would be C = 0.0025173/1 = 0.0025173 mol/L

So now we know that there are 0.0025173 moles per litre of OH- ions in the solution. In order to figure out the concentration of H+ ions and the pH values, we need to use the water constant (Kw = 1.0 x 10^-14).

Kw = [H+] [OH-]
[H+] = 1.0x10^-14/0.0025173 = 3.97251 x 10^-12

Now we can calculate the pH:

pH = -log(3.97251 x 10^-12)
Hence pH = 11.4

Hope you understood my reasoning there amanda, a very good question indeed! If you have any further queries please dont hesitate to ask! :)

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Re: Chemistry Question Thread
« Reply #73 on: February 27, 2016, 08:43:01 am »
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A student is given 25ml of 0.50mol/L of HCl(aq).  He was asked to prepare a solution with pH=1.0 . What will be the final volume of the solution? Ans: 125ml

I dont know what went wrong with my working

I found:
 -  final [H+]=10^-1
since HCl --> H+ + Cl-
 -  [HCl] = 10^-1
 - initial v(HCl) = 25*10^-3 L
 - initial [HCl] = 25*10^-3*0.5 = 12.5*10^-3

using CiVi=CfVf

Vf=0.003L

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Re: Chemistry Question Thread
« Reply #74 on: February 27, 2016, 08:54:57 am »
+4
A student is given 25ml of 0.50mol/L of HCl(aq).  He was asked to prepare a solution with pH=1.0 . What will be the final volume of the solution? Ans: 125ml

I dont know what went wrong with my working

I found:
 -  final [H+]=10^-1
since HCl --> H+ + Cl-
 -  [HCl] = 10^-1
 - initial v(HCl) = 25*10^-3 L
 - initial [HCl] = 25*10^-3*0.5 = 12.5*10^-3

using CiVi=CfVf

Vf=0.003L

[H+]=10-pH=10-1mol L-1
Strong acid - Assume the acid fully ionises (Degree of ionisation = 100%)
Note: Degree of ionisation is why this happens. NOT because HCl → H+ + Cl- because what if you had acetic acid: CH3COOH ⇄ H+ + CH3COO-
Therefore [HCl] = [H+] = 10-1mol L-1
In the dilutions formula, we now have Vf=10-1mol L-1 ----> This you already have

Vi=2.5*10-2 L   ----> This you already have
Ci=0.50 mol L-1   ---->This is what you did that confused me. I am unsure as to why you chose to do V * C when the concentration was given already. The formula n=CV does allow us to find the moles of acid present, but the concentration was already given to us.

0.50 * 2.5*10-2 = 0.1 * Vf
Vf = 0.125 L