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November 12, 2019, 10:51:45 pm

Author Topic: VCE Chemistry Question Thread  (Read 987275 times)  Share 

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briv01

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Re: VCE Chemistry Question Thread
« Reply #8040 on: June 12, 2019, 08:55:45 pm »
+1
Doesn't changing the pressure only affect gaseous equilibrium?

Youíre right, itís better to refer to it as concentration instead of pressure Iím 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 #8041 on: June 12, 2019, 09:03:31 pm »
0
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 #8042 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


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Ionic Doc

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Re: VCE Chemistry Question Thread
« Reply #8043 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 #8044 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 #8045 on: June 12, 2019, 09:28:45 pm »
+2
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

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Ionic Doc

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Re: VCE Chemistry Question Thread
« Reply #8046 on: June 12, 2019, 09:32:13 pm »
0

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 #8047 on: June 12, 2019, 09:33:53 pm »
+3
for question b im asking for the amount of sodium atoms present in sodium carbonate but in mol

For part a. Youíre tryna find the mols, which is mass/ molar mass. Therefore, itís 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 #8048 on: June 13, 2019, 10:24:30 pm »
+2
Hello,
I donít think Iíve ever asked a question before so this is my first time, and Iím 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

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Re: VCE Chemistry Question Thread
« Reply #8049 on: June 13, 2019, 10:39:39 pm »
+3
Hello,
I donít think Iíve ever asked a question before so this is my first time, and Iím 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 #8050 on: June 14, 2019, 08:51:39 pm »
0
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 #8051 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!
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Matthew_Whelan

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Re: VCE Chemistry Question Thread
« Reply #8052 on: June 14, 2019, 09:39:25 pm »
+1
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!
I don't know about any classes but you can certainly learn the content via resources such as the internet, khan academy and maybe online lectures. The ATAR Notes summary book for chem would be a decent option for looking at the year 12 course specifically. Also Psych is great so its a fair trade. There is also Edrolo but I access it through my school, I'm not sure how you'd go about accessing it.
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EllingtonFeint

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Re: VCE Chemistry Question Thread
« Reply #8053 on: June 14, 2019, 10:41:19 pm »
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Heyy,
I'm also a bit unsure about what to do with this question...
I think it's mainly the 250ml of 1.0M which confuses me.
How would I convert that to moles??
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Re: VCE Chemistry Question Thread
« Reply #8054 on: June 15, 2019, 01:43:08 pm »
+5
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 :)



Multiplying by 2 would be for if it takes half a mole of electrons to make 1 mole of tin
Dividing by 2 is for if it takes 2 moles of electrons to make 1 mole of tin

Heyy,
I'm also a bit unsure about what to do with this question...
I think it's mainly the 250ml of 1.0M which confuses me.
How would I convert that to moles??
n = c v
c= = 1.0 M
v = 0.250 L
n = 1.0*0.250 = 0.250 mol

This is the amount of mol copper sulfate in solution before you run the current, in other questions you might want to compare that to how many mols of electrons are provided by running the current to see which is the limiting factor but as briv01 picked up the anode is made of copper so you don't need to do this..


Hope this helps :)

Edited for clarity in regards to the anode being copper.
« Last Edit: June 16, 2019, 12:42:27 pm by Bri MT »
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