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July 06, 2020, 02:58:43 pm

Author Topic: VCE Chemistry Question Thread  (Read 1208904 times)  Share 

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Gogurt

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Re: VCE Chemistry Question Thread
« Reply #8640 on: June 13, 2020, 01:59:36 pm »
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Can someone explain why the mole ratios between lead and the electrons is 1:1? I genuinely don't understand the logic behind it.
Thanks! (for d)

Chocolatepistachio

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Re: VCE Chemistry Question Thread
« Reply #8641 on: June 13, 2020, 05:14:57 pm »
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Do you think itís better to actually write your own set of notes or just use the atar notes/ other peopleís notes.

keltingmeith

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Re: VCE Chemistry Question Thread
« Reply #8642 on: June 14, 2020, 04:37:38 am »
+7
Can someone explain why the mole ratios between lead and the electrons is 1:1? I genuinely don't understand the logic behind it.
Thanks! (for d)

Might help if you think of it as 2:2 instead of 1:1. Mathematically, the ratio is the exact same, of course - but remember that the half-reaction that the lead goes through is:

Pb -> Pb2+ + 2e-

Except, there's not one lead atom in that equation - there's two of them. "But hang on", you might say. "There's two lead atoms, so don't both of them require 2 electrons???"

Well, you'd be exactly right - except, the reduction half reaction is this:

Pb4+ + 2e- -> Pb2+

So you see, the two electrons move from one lead molecule, to the OTHER lead molecule. So while both need 2 electrons, if those 2 electrons are the same, then you don't need 4 electrons for everything to happen! Hence, per equation, there are 2 lead molecules, but only 2 electrons moving around. This is more obvious if you go through the process of building the full equation yourself:

Pb -> Pb2+ + 2e-
Pb4+ + 2e- -> Pb2+

Next, add the sulfuric acid:
Pb + H2SO4 -> PbSO4 + 2e- + 2H+
Pb4+ + H2SO4 + 2e- -> PbSO4 + 2H+

Now, turn the Pb4+ into PbO2, predicting where the oxygens will likely move:
Pb + H2SO4 -> PbSO4 + 2e- + 2H+
PbO2 + H2SO4 + 2e- -> PbSO4 + 2H2O

Finally, balance out any discrepancies and add them both together:
Pb + H2SO4 -> PbSO4 + 2e- + 2H+
PbO2 + H2SO4 + 2H+ + 2e- -> PbSO4 + 2H2O
--------------------------------------------------------
Pb + PbO2 + H2SO4 -> 2PbSO4 + 2H2O

Which matches what they've given us. Notice that in the final step (which shows the full half-reactions), there are only two electrons that move, but also 2 lead atoms in the final reaction.

Do you think itís better to actually write your own set of notes or just use the atar notes/ other peopleís notes.

Depends entirely on the person - for me, I like to write my own notes, because the process of writing my notes helps me understand what's going on and commit it to memory.
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thatdumbstudent

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Re: VCE Chemistry Question Thread
« Reply #8643 on: June 16, 2020, 09:59:49 pm »
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how do you name this hydrocarbon?
would this be 2,3,4-trimethylbutamine? We haven't really gone through amines in class so i'm a little confused. I also don't know if it should be butan- or pentan- ? do you need a line connecting to an amine or does that count as carbon (?)

Erutepa

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Re: VCE Chemistry Question Thread
« Reply #8644 on: June 17, 2020, 06:39:10 pm »
+4
how do you name this hydrocarbon?
would this be 2,3,4-trimethylbutamine? We haven't really gone through amines in class so i'm a little confused. I also don't know if it should be butan- or pentan- ? do you need a line connecting to an amine or does that count as carbon (?)
To answer your last question first, the line connecting to the amine group does not indicate an additional carbon (as can be seen in the drawing below).
When naming molecules its important we identify and name based off the longest carbon chain - in this case it would be 5 carbon's long. We then number the carbons along this longest chain prioritizing the most important functional groups; in this case we have alkyl groups and an amine group. The Amine is more important so we number a 5 carbon chain in such a way to give the amine group the lowest number possible (in this case this is carbon number 2). This makes the molecule 3,4-dimethyl pentan-2-amine.

Hopefully this clarifies any confusions but feel free to point out anything I might not have explained clearly enough!
« Last Edit: June 17, 2020, 07:11:08 pm by Erutepa »
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thatdumbstudent

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Re: VCE Chemistry Question Thread
« Reply #8645 on: June 17, 2020, 06:49:35 pm »
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To answer your last question first, the line connecting to the amine group does not indicate an additional carbon (as can be seen in the drawing below).
When naming molecules its important we identify and name based off the longest carbon chain - in this case it would be 5 carbon's long. We then number the carbons along this longest chain prioritizing the most important functional groups; in this case we have alkyl groups and an amine group. The Amine is more important so we number a 5 carbon chain in such a way to give the amine group the lowest number possible (in this case this is carbon number 2). This makes the molecule 3,4-dimethyl pentan-2-amine.

Hopefully this clarifies any confusions but feel free to point out anything I might not have explained clearly enough!


hi yep i've recently found out the correct answer but your explanation did clarify more stuff so thank you! :-)
also, just another question, if i was given a name (4-ethyl-3-methylpent-2-ene) and i had i drew the structural formula out but lets say the double bond was located at the right. For the semi-structural formula, do i have to write the structure from left to right or does that not matter? if that makes any sense

so like CH3CH(CH2CH3)C(CH3)=CHCH3

btw i dont know if that's right in the first place haha but what i mean is you can see the double bond is towards the end however you'd still read it as 4-ethyl-3-methylpent-2-ene right? or do i need to draw it backwards to the double bond is located at the front?

whys

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Re: VCE Chemistry Question Thread
« Reply #8646 on: June 17, 2020, 07:06:58 pm »
+4
hi yep i've recently found out the correct answer but your explanation did clarify more stuff so thank you! :-)
also, just another question, if i was given a name (4-ethyl-3-methylpent-2-ene) and i had i drew the structural formula out but lets say the double bond was located at the right. For the semi-structural formula, do i have to write the structure from left to right or does that not matter? if that makes any sense

so like CH3CH(CH2CH3)C(CH3)=CHCH3

btw i dont know if that's right in the first place haha but what i mean is you can see the double bond is towards the end however you'd still read it as 4-ethyl-3-methylpent-2-ene right? or do i need to draw it backwards to the double bond is located at the front?
The structural formula is read left to right, but the double bond could be at the start or end. If you decide to put the double bond at the end of the structural formula, then all other functional groups must be numbered from right to left, and vice versa. With 4-ethyl-3-methylpent-2-ene, putting the double bond at the end means that the methyl will be on the carbon to the left of that, as you've shown, with the ethyl on the next carbon (as you've also shown). As long as it represents the molecule, it will be correct. You can always check this by drawing it out and seeing if it still matches the name. Personally, it is much easier to write the structural formula from left to right. It is also not a requirement to show double/triple bonds in the structural formula, but that is a personal choice and they won't mark you down if you decide to include it/not include it.
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Azila2004

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Re: VCE Chemistry Question Thread
« Reply #8647 on: June 17, 2020, 07:40:23 pm »
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Hello,

I am a little confused about this question: 4.15 grams of tungsten was burned in chlorine and 8.95 grams of tungsten chloride (WCl6) was formed. Find the relative atomic mass of tungsten.

Isnít the relative atomic mass just the number on the periodic table? Iím not really used to practising this exact type of question, so help would be appreciated :)
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IThinkIFailed

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Re: VCE Chemistry Question Thread
« Reply #8648 on: June 19, 2020, 04:30:47 pm »
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Hey, Iím struggling to identify the variables in these experimental set ups:

Experiment 1:
Place two drops of cyclohexane on one watch glass and two drops of cyclohexene on another. Light the two liquids with a match and record the appearance of each flame

Experiment 2:
Using a fume cupboard, place cyclohexane to a depth of 1 cm in one test tube and cyclohexene to the same depth in another tube. Add two drops of a solution of iodine in hexane to each liquid. If there is no immediate reaction, stopper the tube and place it in sunlight for 5 minutes. Record your observations

Experiment 3:

Shake five drops of cyclohexane with 10 drops of water in a test tube. Note whether the saturated hydrocarbon is soluble in water. Repeat using the unsaturated hydrocarbon, cyclohexene

My attempt at identifying the variables is this:
Experiment 1:
Independent variable: type of hydrocarbon (saturated or unsaturated)
Dependent variable: Completion of combustion/ quality of flame

Experiment 2:
Independent variable: Type of hydrocarbon (saturated or unsaturated)
Dependent variable:
Progression of a reaction with iodine (addition reaction)

Experiment 3:
Independent variable: Type of hydrocarbon (saturated or unsaturated)
Dependent variable: solubility in water

Could someone please verify if Iím right, and if Iím wrong, point me in the right direction?
Thanks
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keltingmeith

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Re: VCE Chemistry Question Thread
« Reply #8649 on: June 19, 2020, 04:57:04 pm »
+5
Hello,

I am a little confused about this question: 4.15 grams of tungsten was burned in chlorine and 8.95 grams of tungsten chloride (WCl6) was formed. Find the relative atomic mass of tungsten.

Isnít the relative atomic mass just the number on the periodic table? Iím not really used to practising this exact type of question, so help would be appreciated :)

I'm gonna be honest, I might need to think harder about this, but I'm not convinced you have enough information for this unless you use a periodic table to get the molecular weight of chlorine.

You are correct, of course - the relative atomic mass is the same as the molecular weight. Which begs the question, if you've got a periodic table to get the molecular weight of chlorine, why not use it to get the amount of chlorine? Is there more to the question you didn't give us? (Like previous parts)

Hey, Iím struggling to identify the variables in these experimental set ups:

Experiment 1:
Place two drops of cyclohexane on one watch glass and two drops of cyclohexene on another. Light the two liquids with a match and record the appearance of each flame

Experiment 2:
Using a fume cupboard, place cyclohexane to a depth of 1 cm in one test tube and cyclohexene to the same depth in another tube. Add two drops of a solution of iodine in hexane to each liquid. If there is no immediate reaction, stopper the tube and place it in sunlight for 5 minutes. Record your observations

Experiment 3:

Shake five drops of cyclohexane with 10 drops of water in a test tube. Note whether the saturated hydrocarbon is soluble in water. Repeat using the unsaturated hydrocarbon, cyclohexene

My attempt at identifying the variables is this:
Experiment 1:
Independent variable: type of hydrocarbon (saturated or unsaturated)
Dependent variable: Completion of combustion/ quality of flame

Experiment 2:
Independent variable: Type of hydrocarbon (saturated or unsaturated)
Dependent variable:
Progression of a reaction with iodine (addition reaction)

Experiment 3:
Independent variable: Type of hydrocarbon (saturated or unsaturated)
Dependent variable: solubility in water

Could someone please verify if Iím right, and if Iím wrong, point me in the right direction?
Thanks


Looks fine to me
« Last Edit: June 19, 2020, 05:02:06 pm by keltingmeith »
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Coolgalbornin03Lo

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Re: VCE Chemistry Question Thread
« Reply #8650 on: June 29, 2020, 01:22:08 pm »
+1
Hi guys! Iím a year 12 revising U3 Chem and was wondering to what extent should we know about energy conversions? I came across a checkpoints question which was asking which is the most efficient. Does anyone know the pathway?

I.e thermalóó> mechanical óó-> electrical?

Or is this more for the galvanic cells topic? (Iím going over fuels atm)
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uwuuu

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Re: VCE Chemistry Question Thread
« Reply #8651 on: June 30, 2020, 04:45:51 pm »
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Hi I was wondering if someone could explain to me in detail how does varying the surface area of electrodes affect the mass of metal formed deposited on the cathode in an electrolytic cell and consequently increase the rate of the chemical reaction? And why can the mass deposited deviate from the theoratical mass deposited using Faraday's Law?

1729

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Re: VCE Chemistry Question Thread
« Reply #8652 on: June 30, 2020, 07:46:29 pm »
+10
Hi I was wondering if someone could explain to me in detail how does varying the surface area of electrodes affect the mass of metal formed deposited on the cathode in an electrolytic cell and consequently increase the rate of the chemical reaction? And why can the mass deposited deviate from the theoratical mass deposited using Faraday's Law?
When you say deviated from theoretical mass, are you speaking of possible errors?
And for the other first part increasing surface area increases number of collisions, hence faster rate of reaction.
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uwuuu

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Re: VCE Chemistry Question Thread
« Reply #8653 on: July 01, 2020, 08:50:37 am »
+1
Yes, the possible errors because I'm doing a practical investigation on the effect of changing surface area on mass of metal deposited on cathode. I was am unsure how to explain in detail in linking electrolysis and rate of reaction with surface area

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Re: VCE Chemistry Question Thread
« Reply #8654 on: July 01, 2020, 11:13:34 am »
+5
Yes, the possible errors because I'm doing a practical investigation on the effect of changing surface area on mass of metal deposited on cathode. I was am unsure how to explain in detail in linking electrolysis and rate of reaction with surface area
Oh well, lots of reactions in chemistry, particularly aqueous ones, are diffusion controlled - that is, it takes time for the proper species to be in just the right place and just the right time to react just the right way. By increasing surface area, you are increasing the likelihood that this event will occur. More spots for the reaction to happen, essentially. Things that promote redox reactions are increasing surface area, increasing electric current, increasing diffusion of products, increasing concentration of reactants, amongst other things.
I also point to you, you should be familar with the cottrell equation to have a deeper understanding (:
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If you are wondering about my avatar, It was inspired by a problem I did which asked me to prove that the graphs of xy = 1 and y^2 = x^2 + 2 intersected at a 90 degree angle. The resulting figure in the middle kinda like a 8-sided square in hyperbolic space. It also resembles the conformal map of the complex square root.

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