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October 18, 2019, 12:22:49 am

Author Topic: VCE Chemistry Question Thread  (Read 953561 times)  Share 

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f0od

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Re: VCE Chemistry Question Thread
« Reply #8010 on: May 11, 2019, 11:48:19 am »
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Hey, I just wanted to check my answers for an electrolytic cell as I'm not the best at chemistry

For the electrolysis of 1M copper(II) bromide using inert electrodes, would
the reduction reaction be Cu2+(aq) + 2e- > Cu(s)
the oxidation reaction be 2Br-(aq) > Br2(l) + 2e-
the overall reaction be Cu2+(aq + 2Br-(aq) >  Cu(s) + Br2(l) ?
I wasn't sure if the copper(II) bromide solution was supposed to be CuBr2(aq) or not because using the electrochemical series the solution should be copper ions with bromide ions but you can't make them into a solution without bromine becoming Br2? Sorry if the question is confusing!

Also, for the 'source of electric current' in this reaction/any electrolytic cell, what would it be?

Thanks so much!
class of 2019

jasheel

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Re: VCE Chemistry Question Thread
« Reply #8011 on: May 11, 2019, 04:12:58 pm »
+3
Hey, I just wanted to check my answers for an electrolytic cell as I'm not the best at chemistry

For the electrolysis of 1M copper(II) bromide using inert electrodes, would
the reduction reaction be Cu2+(aq) + 2e- > Cu(s)
the oxidation reaction be 2Br-(aq) > Br2(l) + 2e-
the overall reaction be Cu2+(aq + 2Br-(aq) >  Cu(s) + Br2(l) ?
I wasn't sure if the copper(II) bromide solution was supposed to be CuBr2(aq) or not because using the electrochemical series the solution should be copper ions with bromide ions but you can't make them into a solution without bromine becoming Br2? Sorry if the question is confusing!

Also, for the 'source of electric current' in this reaction/any electrolytic cell, what would it be?

Thanks so much!

Hey, your half-equations look good.

The potential  at the anode and cathode would be: \(\ce{Cu^2+}\), \(\ce{Br-}\) and \(\ce{H2O}\).

At the cathode the strongest oxidant reacts, so that would be \(\ce{Cu^2+}\), so that's right.

At the anode the strongest reductant reacts, so that would be \(\ce{Br-}\), so that's right. You don't have to worry about \(\ce{Br2(l)}\) being a liquid it just means it won't dissolve into the water (as it is a non-polar molecule). You can search up pictures of bromine in water to see an example.

The source of electric current would just be your power supply (usually a battery).
 


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EllingtonFeint

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Re: VCE Chemistry Question Thread
« Reply #8012 on: May 15, 2019, 09:18:40 pm »
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Hey,
So I am really very confused about this question and my teacher cant explain it to me either, so Im really hoping someone can help me make sense of it...

1. The equilibrium constant for the decomposition of ammonia is 100 M^2 at 255C for the equation
\(\ce{2NH3(g) <=> N2(g) + 3H2(g)}\)

a. Write an expression for the equilibrium constant for the equation:
\(\ce{N2(g) + 3H2(g) <=> 2NH3(g)}\)

Soooo, I did this and I got it correct (I would upload a pic but I suck at uploading pictures on this website... please someone direct me to a detailed explanation of how to do so from an iPhone please 🙏🏼)

But I dont understand part b.

b. Calculate the equilibrium constant for the equation given in part a.

THIS is where I got stuck! I dont understand how ~spoiler  ::) ~ the answer is 0.01M^-2. I understand where the M^-2 comes from but HOW do they get 0.01M?? My teacher said something about a square root or something but I do not get how or why the answer is that!  :P

Maybe Im just missing out on something super obvious, idk  :-[ someone help please. 

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sweetcheeks

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Re: VCE Chemistry Question Thread
« Reply #8013 on: May 15, 2019, 10:20:02 pm »
+1
Hey,
So I am really very confused about this question and my teacher cant explain it to me either, so Im really hoping someone can help me make sense of it...

1. The equilibrium constant for the decomposition of ammonia is 100 M^2 at 255C for the equation
\(\ce{2NH3(g) <=> N2(g) + 3H2(g)}\)

a. Write an expression for the equilibrium constant for the equation:
\(\ce{N2(g) + 3H2(g) <=> 2NH3(g)}\)

Soooo, I did this and I got it correct (I would upload a pic but I suck at uploading pictures on this website... please someone direct me to a detailed explanation of how to do so from an iPhone please 🙏🏼)

But I dont understand part b.

b. Calculate the equilibrium constant for the equation given in part a.

THIS is where I got stuck! I dont understand how ~spoiler  ::) ~ the answer is 0.01M^-2. I understand where the M^-2 comes from but HOW do they get 0.01M?? My teacher said something about a square root or something but I do not get how or why the answer is that!  :P

Maybe Im just missing out on something super obvious, idk  :-[ someone help please. 



How would you convert the equilibrium expression from the top equation to the bottom equation? You inverse it (raise to the power of -1). What you do to one side of the equation, you must do to the other. So if you inverse the equilibrium expression, you must also inverse the equilibrium constant.

EllingtonFeint

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Re: VCE Chemistry Question Thread
« Reply #8014 on: May 16, 2019, 09:57:42 am »
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How would you convert the equilibrium expression from the top equation to the bottom equation? You inverse it (raise to the power of -1). What you do to one side of the equation, you must do to the other. So if you inverse the equilibrium expression, you must also inverse the equilibrium constant.

I just don't understand why you would raise it to -1... Where does -1 come from?? :-[
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EllingtonFeint

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Re: VCE Chemistry Question Thread
« Reply #8015 on: May 16, 2019, 01:52:17 pm »
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Okay, so I think I understand why it's to the negative 1 power but now I'm stuck on part 3.  :(

1. The equilibrium constant for the decomposition of ammonia is 100 M^2 at 255C for the equation
\(\ce{2NH3(g) <=> N2(g) + 3H2(g)}\)

c. Write an expression for the equilibrium constant for the equation:
\(\ce{NH3(g) <=> 1/2N2(g) + 3/2H2(g)}\)
(so I wrote the equation, but now I don't know how to calculate the equilibrium constant... Surely it should be 100/2... so 50 M to the power of 1??

The answer is actually 10 M.

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jasheel

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Re: VCE Chemistry Question Thread
« Reply #8016 on: May 16, 2019, 06:01:56 pm »
+3
Okay, so I think I understand why it's to the negative 1 power but now I'm stuck on part 3.  :(

1. The equilibrium constant for the decomposition of ammonia is 100 M^2 at 255C for the equation
\(\ce{2NH3(g) <=> N2(g) + 3H2(g)}\)

c. Write an expression for the equilibrium constant for the equation:
\(\ce{NH3(g) <=> 1/2N2(g) + 3/2H2(g)}\)
(so I wrote the equation, but now I don't know how to calculate the equilibrium constant... Surely it should be 100/2... so 50 M to the power of 1??

The answer is actually 10 M.

The change that you made to the equation is halfing all the coefficients.

Remember the formula for K is Pcoefficients/Rcoefficients.

That means if you half all the coefficients, you're powering the whole thing by 1/2 (which is the same as square rooting). So K becomes K1/2.

Therefore, it will be 1001/2, which equals 10.

The same thing applies to the units, which will be (M2)1/2, which equals M.

Hence the answer 10 M.

Hope that helped!
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EllingtonFeint

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Re: VCE Chemistry Question Thread
« Reply #8017 on: May 16, 2019, 08:54:53 pm »
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The change that you made to the equation is halfing all the coefficients.

Remember the formula for K is Pcoefficients/Rcoefficients.

That means if you half all the coefficients, you're powering the whole thing by 1/2 (which is the same as square rooting). So K becomes K1/2.

Therefore, it will be 1001/2, which equals 10.

The same thing applies to the units, which will be (M2)1/2, which equals M.

Hence the answer 10 M.

Hope that helped!


Ohmigosh yesss! Thank you!  :)
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Donut

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Re: VCE Chemistry Question Thread
« Reply #8018 on: May 18, 2019, 08:11:23 pm »
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Is there a chemical equation for the anaerobic decomposition of trash into methane?

whoopybird

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Re: VCE Chemistry Question Thread
« Reply #8019 on: May 19, 2019, 12:52:26 am »
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what are the physical properties of biodiesel?
thank you

Lear

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Re: VCE Chemistry Question Thread
« Reply #8020 on: May 19, 2019, 01:14:41 am »
+1

what are the physical properties of biodiesel?
thank you

Have you tried checking your textbook first?
I remember there being some excellent information on this in the textbooks :)
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whoopybird

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Re: VCE Chemistry Question Thread
« Reply #8021 on: May 19, 2019, 03:22:13 am »
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Have you tried checking your textbook first?
I remember there being some excellent information on this in the textbooks :)

is it under the properties of ester?
i thought biodiesel was non-polar so it has only dispersion force."the result of the very long non-polar hydrocarbon chains is to reduce the overall polarity of the molecule so that it can be considered to be non-polar. With only weak Van der Waal's forces (dispersion or London forces) acting between the biodiesel molecules" by website ausetute

but the textbook by Taylor 1n2 state "Esters have relatively low boiling points because the intermolecular attraction is dipole-dipole (and dispersion forces)"under reaction on pg 165.

so yea,i don't really know

Ribozyme

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Re: VCE Chemistry Question Thread
« Reply #8022 on: May 19, 2019, 09:08:19 am »
+2
is it under the properties of ester?
i thought biodiesel was non-polar so it has only dispersion force."the result of the very long non-polar hydrocarbon chains is to reduce the overall polarity of the molecule so that it can be considered to be non-polar. With only weak Van der Waal's forces (dispersion or London forces) acting between the biodiesel molecules" by website ausetute

but the textbook by Taylor 1n2 state "Esters have relatively low boiling points because the intermolecular attraction is dipole-dipole (and dispersion forces)"under reaction on pg 165.

so yea,i don't really know

Hey!

I'm pretty sure that a biodiesel molecule can have dipole moments (and hence dipole-dipole interactions) due to the presence of highly electronegative oxygen atoms in the ester functional groups. That being said, the overall impact of this could be minimal if the biodiesel molecule itself has large segments of non-polar regions. But yeah its probably best to have a read of your textbook on this as well.

:)

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Re: VCE Chemistry Question Thread
« Reply #8023 on: May 21, 2019, 05:18:29 pm »
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Why is temperature the only thing that changes an equilibrium constant?
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EllyJayne

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Re: VCE Chemistry Question Thread
« Reply #8024 on: May 23, 2019, 02:24:30 pm »
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Does anyone have any ideas or websites for the practical investigation sac thats done in unit 4? our school is doing it in unit 3 and I have no idea what to do.
Thanks !