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March 28, 2024, 09:06:35 pm

Author Topic: VCE Chemistry Question Thread  (Read 2313028 times)  Share 

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rani_b

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Re: VCE Chemistry Question Thread
« Reply #8025 on: June 11, 2019, 10:32:07 am »
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For the above reaction, which set of factors would produce the greatest yield of SrO(s)? Would it be high temperature/low pressure, or high temperature/high pressure?

Decreasing the pressure will yield more SrO as the system will favour the side will more particles ( aka the right ) to compensate for the low pressure

Doesn't changing the pressure only affect gaseous equilibrium?
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rani_b

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Re: VCE Chemistry Question Thread
« Reply #8026 on: June 11, 2019, 10:35:33 am »
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Hey guys!

So the data booklet now specifies that 1mL of water is 0.997 grams. I was watching one of the AtarNotes chem videos and the lecturer mentioned we should now always use that conversion, but it also says density of water at 25 degrees Celsius. Shouldn't this mean I only use this conversion when the question specifies the reaction is taking place in a room at 25 degrees?
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leonm19

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Re: VCE Chemistry Question Thread
« Reply #8027 on: June 11, 2019, 11:38:55 am »
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Hey guys!

So the data booklet now specifies that 1mL of water is 0.997 grams. I was watching one of the AtarNotes chem videos and the lecturer mentioned we should now always use that conversion, but it also says density of water at 25 degrees Celsius. Shouldn't this mean I only use this conversion when the question specifies the reaction is taking place in a room at 25 degrees?

You're right, it does say at 25 degrees on the data booklet, but from my experience I think it's okay to use it even if the temperature isn't 25 degrees. For example, I use it when calculating q = mc delta t if I need to find the mass of water for that equation, even though the temperature isn't constant. You might wanna double check with someone else though, I could be wrong.

briv01

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Re: VCE Chemistry Question Thread
« Reply #8028 on: June 12, 2019, 08:55:45 pm »
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Doesn't changing the pressure only affect gaseous equilibrium?

You’re right, it’s better to refer to it as concentration instead of pressure I’m pretty sure ( both mean the Same but pressure is used for gases ). But the answer I said should still be right ( once you replace pressure with concentration )

Ionic Doc

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Re: VCE Chemistry Question Thread
« Reply #8029 on: June 12, 2019, 09:03:31 pm »
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have an exam coming and still struggling with the whole mole concept  :-[

anyways the question reads
A sample of Sodium Carbonate Na2CO3 has a mass of 16.6 g

a)Calculate the amount in mol of sodium carbonate present
I got 6.75 mol

b)) Calculate the amount in mol of sodium atoms present
I know I use the n=m/M formula but what would I sub in for those values

thnx
« Last Edit: June 12, 2019, 09:05:55 pm by Ionic Doc »
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Bri MT

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Re: VCE Chemistry Question Thread
« Reply #8030 on: June 12, 2019, 09:13:46 pm »
+1
have an exam coming and still struggling with the whole mole concept  :-[

anyways the question reads A same of Sodium Carbonate ( Na2CO3) has a mass of 16.6g

a) Calculate the amount in mol, of sodium atoms present
answer = 6.75mol (pretty sure that's correct)

b) Calculate the amount in mol, of sodium atoms present
I know I would use the n=m/M formula but what do I substitute

thnx
I'm not sure what you're asking since you've listed the same question twice - hopefully this helps anyway.
For M you use the periodic table and add up the atomic masses   (I'm recalling the numbers off my head here so they won't be perfect but..)

M = 23*2 + 12*1 + 16*3  g/mol
m =  16.6  g



Ionic Doc

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Re: VCE Chemistry Question Thread
« Reply #8031 on: June 12, 2019, 09:16:14 pm »
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I'm not sure what you're asking since you've listed the same question twice - hopefully this helps anyway.
For M you use the periodic table and add up the atomic masses   (I'm recalling the numbers off my head here so they won't be perfect but..)

M = 23*2 + 12*1 + 16*3  g/mol
m =  16.6  g

for question b im asking for the amount of sodium atoms present in sodium carbonate but in mol
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Matthew_Whelan

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Re: VCE Chemistry Question Thread
« Reply #8032 on: June 12, 2019, 09:22:35 pm »
+3
have an exam coming and still struggling with the whole mole concept  :-[

anyways the question reads
A sample of Sodium Carbonate Na2CO3 has a mass of 16.6 g

a)Calculate the amount in mol of sodium carbonate present
I got 6.75 mol

b)) Calculate the amount in mol of sodium atoms present
I know I use the n=m/M formula but what would I sub in for those values

thnx
It should be 16.6(m) / 106(M)
= 0.1566 mol

for question b im asking for the amount of sodium atoms present in sodium carbonate but in mol
If you want the number of atoms you use avogadros constant.
To find mol the good two to remember are n=m/M and n=cv


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Bri MT

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Re: VCE Chemistry Question Thread
« Reply #8033 on: June 12, 2019, 09:28:45 pm »
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for question b im asking for the amount of sodium atoms present in sodium carbonate but in mol


Right ok. So for every amount of sodium carbon you have, you have 2 amounts of sodium atoms.
This means if you have x mols of sodium carbonate, you have 2x mols of sodium atoms.

a) You do n = m/M   (make sure you do m/M rather than M/m) to obtain the number of mols of sodium carbonate
b) you multiply your answer for a by 2


Ionic Doc

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Re: VCE Chemistry Question Thread
« Reply #8034 on: June 12, 2019, 09:32:13 pm »
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Right ok. So for every amount of sodium carbon you have, you have 2 amounts of sodium atoms.
This means if you have x mols of sodium carbonate, you have 2x mols of sodium atoms.

a) You do n = m/M   (make sure you do m/M rather than M/m) to obtain the number of mols of sodium carbonate
b) you multiply your answer for a by 2

ahh so whatever I got from part A I would multiply by 2 because there are 2 sodium atoms in every sample of sodium carbonate

thnx both of u for your help  :)
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briv01

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Re: VCE Chemistry Question Thread
« Reply #8035 on: June 12, 2019, 09:33:53 pm »
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for question b im asking for the amount of sodium atoms present in sodium carbonate but in mol

For part a. You’re tryna find the mols, which is mass/ molar mass. Therefore, it’s 16.6/106. Try to always link things to the formula

For part b, you look at the ratio between the number if molecules of silver carbonate vs the number of atoms of silver. Which is 1:2. Therefore, you multiply the mols of silver carbonate by 2

EnslaveAMollusc

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Re: VCE Chemistry Question Thread
« Reply #8036 on: June 13, 2019, 10:24:30 pm »
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Hello,
I don’t think I’ve ever asked a question before so this is my first time, and I’m a bit hesitant! Hopefully somebody will be able to help!!

My question is what ester is formed when you combine Pentanol with Decanoic Acid?

C5H11OH+C10H20O2

Matthew_Whelan

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Re: VCE Chemistry Question Thread
« Reply #8037 on: June 13, 2019, 10:39:39 pm »
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Hello,
I don’t think I’ve ever asked a question before so this is my first time, and I’m a bit hesitant! Hopefully somebody will be able to help!!

My question is what ester is formed when you combine Pentanol with Decanoic Acid?

C5H11OH+C10H20O2
Hi there,
I believe it would form pentyl decanoate (or something like that), i think of it by omitting the H2O as esterification is a condensation reaction.
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EllingtonFeint

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Re: VCE Chemistry Question Thread
« Reply #8038 on: June 14, 2019, 08:51:39 pm »
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Hello,
I'm a bit confused with the following question  ???
I got part a but I'm a lil stuck with part b (and c).
For a, I got the charge will equal 4+ because according to the electrochemical series  Sn ^4+ (aq) + 2 e- (those double sided arrows) Sn ^2+ (aq) so I just took that 4+ charge. Is that alright?
For part b, I attempted using Q= I x t
and I got...
25 * 18000 = 450 000 coulombs.

Then I would have to use Q= n (e-) x F... right??
 So would I rearrange to find n(e-) = Q/F,
but then if I did that, what would I do about the 4:2 ratio?? would I divide by 2 or times by 2 or ...?

I'm not quite sure how to do part c either

I just started this topic today so I'm still a bit unsure about it.

Thank you :)

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Tatlidil

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Re: VCE Chemistry Question Thread
« Reply #8039 on: June 14, 2019, 09:20:53 pm »
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Hey AN!
So a little bit of background information on me, last year (yr 11) I did English/spesh/methods/physics/chem, but this year (and I'm not afraid to say it) because of the teacher I dropped chem for psych. Now, I really loved chemistry and I am eager to learn everything taught in yr 12 and maybe more, how would I go about doing that? Is there a class somewhere I can go to learn?I doubt ill be able to learn it in uni since it requires the foundation provided in VCE but let me know please!