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Author Topic: VCE Chemistry Question Thread  (Read 2313330 times)  Share 

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rocksonchan

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Re: VCE Chemistry Question Thread
« Reply #6375 on: June 22, 2017, 04:53:11 pm »
+1
Hi I need some help!

We did an electrolysis of copper (II) sulfate, with a stainless steel cathode and a copper metal anode.

One of our questions is: If an inert carbon electrode was used instead of a strip of copper metal, would the experiment have worked?
And: Calculate the current efficiency of the cell, how do I do this? It's supposedly is the actual mass of copper deposited / theoretical amount of copper deposited. How do I calculate the theoretical amount?

Thanks!

So if an inert carbon electrode was used (as the anode) instead of a strip of copper metal, the experiment would most likely still work because the Copper (II) Sulfate supplies the electrolytic cell with copper ions which can be reduced to deposit copper metal onto the stainless steel cathode. The only way that the experiment would not work is if you had very little or very low concentrations of Copper (II) Sulfate so that the copper ions run out and cannot be reduced. What the strip of copper metal does is oxidise to produce additional copper ions that can readily be reduced.

The theoretical amount of copper deposited would most likely be calculated using Faraday's laws (formulas: Q = It and n(e-) = Q/F), where you hopefully recorded both the current (I) and the time (t) that you operated the electrolytic cell for. After finding Q, you can subsequently find n(e-) and then use the molar ratios in the half-equations to determine the amount of moles that should be theoretically deposited. Lastly, use n = m/M to find out the theoretical mass.

Hope this helps!! :) :) :) :)

miguel_1

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Re: VCE Chemistry Question Thread
« Reply #6376 on: June 22, 2017, 04:57:42 pm »
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Tbh, wouldn't use ferric oxide. Only geologists will still use weird terminology like that, it's going out of phase for iron (III) which is a lot more straightforward.
You may be getting stumped by the context - the trick with chemistry is, if they ask you something and it looks like you haven't learned it, you need to strip away all the scary stuff and just work with what you have.

Okay, thank you
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ardria

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Re: VCE Chemistry Question Thread
« Reply #6377 on: June 22, 2017, 08:58:19 pm »
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Why are smaller alcohols more water-soluble than larger alcohols?

miguel_1

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Re: VCE Chemistry Question Thread
« Reply #6378 on: June 22, 2017, 09:40:04 pm »
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Why are smaller alcohols more water-soluble than larger alcohols?

Alcohols with a smaller hydrocarbon chain are more water-soluble because as the length of the hydrocarbon chain increases, the solubility in water decreases. With four carbon in the hydrocarbon chain and higher, the decrease in solubility becomes visible as the mixture forms two immiscible layers of liquid. The reason why the solubility decreases as the length of hydrocarbon chain increases is because it is requires more energy to overcome the hydrogen bonds between the alcohol molecules as the molecules are more tightly packed together as the size and mass increases.
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MisterNeo

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Re: VCE Chemistry Question Thread
« Reply #6379 on: June 22, 2017, 09:45:20 pm »
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Why are smaller alcohols more water-soluble than larger alcohols?

Smaller alkanols are more soluble because they have a shorter carbon chain.
An alkanol has that polar -OH group which hydrogen bonds with water.
BUT
They also have that non-polar carbon chain which forms dispersion forces that repel the polar nature of water.

Thus, as alkanols become bigger, the carbon chain's dispersion forces become stronger. This means that as you go from methanol to something big like octanol, the dispersion forces repel so much that it makes the alkanol insoluble.
You see what I mean? Basically, dispersion force repulsion overcome hydrogen bonding as alkanols become longer, which will cause it to not be soluble.
And also, the longer molecule prevents water from touching the -OH group, but instead they touch the non-polar carbons that are not soluble.
 :D
« Last Edit: June 22, 2017, 09:47:54 pm by MisterNeo »

miguel_1

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Re: VCE Chemistry Question Thread
« Reply #6380 on: June 22, 2017, 09:58:29 pm »
+1
Smaller alkanols are more soluble because they have a shorter carbon chain.
An alkanol has that polar -OH group which hydrogen bonds with water.
BUT
They also have that non-polar carbon chain which forms dispersion forces that repel the polar nature of water.

Thus, as alkanols become bigger, the carbon chain's dispersion forces become stronger. This means that as you go from methanol to something big like octanol, the dispersion forces repel so much that it makes the alkanol insoluble.
You see what I mean? Basically, dispersion force repulsion overcome hydrogen bonding as alkanols become longer, which will cause it to not be soluble.
And also, the longer molecule prevents water from touching the -OH group, but instead they touch the non-polar carbons that are not soluble.
 :D

Way better explanation than mine  :o
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lzxnl

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Re: VCE Chemistry Question Thread
« Reply #6381 on: June 22, 2017, 11:00:23 pm »
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Smaller alkanols are more soluble because they have a shorter carbon chain.
An alkanol has that polar -OH group which hydrogen bonds with water.
BUT
They also have that non-polar carbon chain which forms dispersion forces that repel the polar nature of water.

Thus, as alkanols become bigger, the carbon chain's dispersion forces become stronger. This means that as you go from methanol to something big like octanol, the dispersion forces repel so much that it makes the alkanol insoluble.
You see what I mean? Basically, dispersion force repulsion overcome hydrogen bonding as alkanols become longer, which will cause it to not be soluble.
And also, the longer molecule prevents water from touching the -OH group, but instead they touch the non-polar carbons that are not soluble.
 :D

Most explanations about this are generally a bit murky. It's not that the dispersion forces are getting stronger, so it becomes less soluble. The most important part is actually what you said in the end; the molecule is getting larger, so for a molecule to dissolve (aka get surrounded by water), you need to bond more water molecules to it, which means you're breaking more hydrogen bonds between water molecules, which becomes less worth it.

Dispersion forces technically don't repel. What they do is provide a much, much less attractive bonding option, so in effect they're repelling water molecules, but it's not the same as true repulsion.
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ham0019

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Re: VCE Chemistry Question Thread
« Reply #6382 on: June 23, 2017, 12:22:28 am »
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um ive done really bad on my chemistry sac's for both of them I got a C does anyone know if there is still a chance I could get a study score of 40 if I try harder on my other 2 sacs

Sine

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Re: VCE Chemistry Question Thread
« Reply #6383 on: June 23, 2017, 09:37:15 am »
+1
um ive done really bad on my chemistry sac's for both of them I got a C does anyone know if there is still a chance I could get a study score of 40 if I try harder on my other 2 sacs
To answer your question directly yes there is a chance but whether it is probable is upto you and what you do between now and the end of year chemistry exam.

It really depends on whether your C for Unit 3 scales up - if you get C A+ A+ (very high A+'s/full marks) your study score would still be in the 30's + whether you get A+'s for Unit 4 and the exam.

Don't waste too much time obsessing over stuff (as hard as it is) like this just try your best to put yourself in a position to succeed.

Gogo14

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Re: VCE Chemistry Question Thread
« Reply #6384 on: June 23, 2017, 05:28:42 pm »
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I don't understand what the textbook means here, or why it works. Thanks
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keltingmeith

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Re: VCE Chemistry Question Thread
« Reply #6385 on: June 23, 2017, 05:42:16 pm »
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I don't understand what the textbook means here, or why it works. Thanks

Tbh, it's not super important - this is more just motivation as to why we care about chirality in molecules. That is, it has this control over the way light bends, and the "opposite" (in some sense of the word) chirality will cause it to bend the other way.

Essentially, I'm going to shine some light at a wall. We know that light is a wave (like a sine wave), but it can travel in lots of different directions. If you put it through a polariser, all the light disappears except for the light travelling in one specific direction, which is controlled by the polariser. We call this the polarised light.

If you take this polarised light, and put a chiral molecule in front of it, then you can do some experiments that will tell you that the light is now travelling in a different direction. That's what the books trying to get across.

What might be confusing you is when I keep saying the waves of light are travelling in different directions (with the polarised light only travelling in one direction), but we only have one beam of light - it's only travelling forwards. So even though the light is travelling forwards, the waves in the light could be oscillating in lots of different directions. It's this oscillating that we keep track of when polarisers become involved.


But as I said - it's not super important you know this stuff, it's more important that you just know about it, and give some sort of motivation as to why we care about chirality/some of the properties of chiral molecules

zhen

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Re: VCE Chemistry Question Thread
« Reply #6386 on: June 23, 2017, 11:11:33 pm »
+1
I was wondering why increasing the concentration of my electrolyte solution increased the mass of electroplating given that I kept the power supply on a fixed voltage setting (this is what I obtained experimentally). I link it to collision theory and how an increase in concentration of these reactants will increase the rate of reaction. Also relating it to I=V/R, I get a lower resistance with the higher concentration, increasing the current and thus increasing the eclectroplating rate. The electrolytic cell used is in the attached image below. I was hoping that someone could confirm or fix my reasoning behind these results. Thanks in advanced.

MisterNeo

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Re: VCE Chemistry Question Thread
« Reply #6387 on: June 23, 2017, 11:47:35 pm »
+3
I was wondering why increasing the concentration of my electrolyte solution increased the mass of electroplating given that I kept the power supply on a fixed voltage setting (this is what I obtained experimentally). I link it to collision theory and how an increase in concentration of these reactants will increase the rate of reaction. Also relating it to I=V/R, I get a lower resistance with the higher concentration, increasing the current and thus increasing the eclectroplating rate. The electrolytic cell used is in the attached image below. I was hoping that someone could confirm or fix my reasoning behind these results. Thanks in advanced.

That's correct!!  ;D
The increased concentration does increase the rate of reaction according to collision theory.
With the mass thing, I think it's because the higher concentration electroplated faster, thus the mass would be higher after x-amount of time.
Imagine a car and a snail racing from A to B, starting at the same point at the same time. After 1 minute, the car (higher concentration) would have made it much further (mass) because of its higher speed (rate of reaction).

This explains the effects on electroplating when factors change.
This explains the collision theory with higher concentrations.

zhen

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Re: VCE Chemistry Question Thread
« Reply #6388 on: June 24, 2017, 12:09:49 am »
+1
That's correct!!  ;D
The increased concentration does increase the rate of reaction according to collision theory.
With the mass thing, I think it's because the higher concentration electroplated faster, thus the mass would be higher after x-amount of time.
Imagine a car and a snail racing from A to B, starting at the same point at the same time. After 1 minute, the car (higher concentration) would have made it much further (mass) because of its higher speed (rate of reaction).

This explains the effects on electroplating when factors change.
This explains the collision theory with higher concentrations.

That's what I got too relating to the mass. Thought it was obvious that a higher rate for a fixed time will give a higher mass. But the thing is I'm not too sure if that explanation for the higher rate is sufficient, since I'm not sure if collision theory is applicable, as for these redox reactions they don't actually collide, instead they merely transfer electrons.
« Last Edit: June 24, 2017, 12:19:18 am by zhen »

zsteve

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Re: VCE Chemistry Question Thread
« Reply #6389 on: June 25, 2017, 03:51:24 pm »
+4
That's what I got too relating to the mass. Thought it was obvious that a higher rate for a fixed time will give a higher mass. But the thing is I'm not too sure if that explanation for the higher rate is sufficient, since I'm not sure if collision theory is applicable, as for these redox reactions they don't actually collide, instead they merely transfer electrons.

Increasing the concentration of your electrolyte would decrease its resistance, yes. Since the electrolyte is a crucial part of the circuit, this will result in a larger current flowing through -> more electrons -> more mass at the cathode.
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