VCE Chemistry Question Thread

[sweetcheeks]

Hi guys,
This is a bit of a random question but it's been bugging me for a while. In the DNA's phosphodiester bond, why is the -C-O bond classified as an ester? Don't esters have to also have another double bond with an O?

Where did you learn that? The C-O bond isn't classified as an ester. You are correct that an ester functional group must contain a carbon with a double bond to an oxygen and a single bond to an oxygen that is bonded to another carbon.

[tigerclouds]

Where did you learn that? The C-O bond isn't classified as an ester. You are correct that an ester functional group must contain a carbon with a double bond to an oxygen and a single bond to an oxygen that is bonded to another carbon.

A video explaining phosphodiester bonds stated that. So if that's not the case where's the ester bond in a phosphodiester bond?

[radiant roses]

what do you do if an equilibrium questions gives the initial amounts in mol of both reactants and one of the two products? The ice table is meant to be used for this question. Are both reactants still given - for the change section, and the products still +  ? Is there anything different that should be done since the mol of one of the products is given? Usually only the moles of the reactants are given

[Erutepa]

what do you do if an equilibrium questions gives the initial amounts in mol of both reactants and one of the two products? The ice table is meant to be used for this question. Are both reactants still given - for the change section, and the products still +  ? Is there anything different that should be done since the mol of one of the products is given? Usually only the moles of the reactants are given

Rice (Reactant, initial concentration, change in concentration, equilibirum concentration) Tables are used typically when dealing with finding the equilibrium concentrations of all reactants and products of a reaction where all initial concentrations are known, and the equilibirum concentration is given (often only the reactants concentrations are given as there are no products intially present - however this still means you known the products initial concentraiton).
The question you are dealing with is a bit more of a simpler one which probably doesn't need a RICE table - although you could draw up something similar to help you work through it. I presume you have the initial concentrations for all reactants and products, and are given a final concentration for one of the products (there might potentially be no intial products, thus you might have only been given the initial reactant concentration). From this you can determine the change in the one known product and then apply stoichiometry to determine the quantities of all the other reactants and products at equilibirum.

Lets work through a question as an example:
hydrogen gas and iodine gas will react and form an equilibirum system as described by the equation:

3 mol of iodine and 2 mol of hydrogen are added  to a vessel. Calculate the equilibrium concentrations of all reactants and products if the mol of HI present at equilibrium is 1.2mol?
Note that initially we know the concentration/amounts of all reactants and products since the products are absent.

we know that 1.2 of HI is present at equilibrium. Since the initial concentration of HI is 0, 1.2 mol of HI was produced.
Looking at the above equation, we can use the ratios of the products to HI to determine the amount of that product produced.
For hydrogen gas:
 - 1 mol of gas reacts to produce 2 mol of HI
 - thus 0.5 mol of hydrogen gas reacts for every 1 mol of HI
 - since 1.2 mol of HI is produced, 0.5*1.2 =0.6 mol of hydrogen gas reacts
 - since there is initially 2 mol of hydrogen gas, 2-0.6=1.4 mol remain at equilibirum
For Iodine gas:
 - 1 mol of iodine gas reacts to produce 2 mol of HI
 - thus 0.5 mol of iodine gas reacts for every 1 mol of HI
 - since 1.2 mol of HI is produced, 0.5*1.2 = 0.6 mol of iodine gas reacts
 - since there is initially 3 mol of iodine gas 3-0.6=2.4 mol reamin at equilibrium.

If concentrations of products/reactants are given instead of molar amounts in the question stem, then you can convert these concentrations into molar amounts prior to the calculations performed in the example shown.

Hopefully this helps

[IThinkIFailed]

Why exactly are the electrodes in a fuel cell porous? I’m confused about the reasoning and I’m struggling to visualise it. I’m really scared that there’ll be an application question about this, and I won’t have a good enough understanding to answer it.

[sandywu]

Why exactly are the electrodes in a fuel cell porous? I’m confused about the reasoning and I’m struggling to visualise it. I’m really scared that there’ll be an application question about this, and I won’t have a good enough understanding to answer it.

To increase the surface area of the reactants, allowing the redox reaction to occur at a faster rate.

[Geoo]

Why exactly are the electrodes in a fuel cell porous? I’m confused about the reasoning and I’m struggling to visualise it. I’m really scared that there’ll be an application question about this, and I won’t have a good enough understanding to answer it.

So, since fuel cells use gas's as the sources of fuel, e.g oxygen and hydrogen, think about how they are going to come into contact with the electrolyte? How are they going to get across the electrolyte? They need to be porous (imagine very tiny holes) so that the fuels (gases) can diffuse across into the electrolyte. If they couldn't diffuse across, if we take a hydrogen fuel cell, the H+ ions that split from the H2 gas wouldn't be able to diffuse across the electrolyte to the oxygen to produce water. So, the fuel cell would stop functioning/producing energy.
I hope this helps. I personally haven't seen many questions related to to porous electrodes specifically. As long as you say that it is porous to allow separation of fuels but still allow diffusion to the electrolyte, are coated in a catalyst to speed up the reaction, and are generally expensive because of the metal (think platinum), that's all you really need to know.

[IThinkIFailed]

So, since fuel cells use gas's as the sources of fuel, e.g oxygen and hydrogen, think about how they are going to come into contact with the electrolyte? How are they going to get across the electrolyte? They need to be porous (imagine very tiny holes) so that the fuels (gases) can diffuse across into the electrolyte. If they couldn't diffuse across, if we take a hydrogen fuel cell, the H+ ions that split from the H2 gas wouldn't be able to diffuse across the electrolyte to the oxygen to produce water. So, the fuel cell would stop functioning/producing energy.
I hope this helps. I personally haven't seen many questions related to to porous electrodes specifically. As long as you say that it is porous to allow separation of fuels but still allow diffusion to the electrolyte, are coated in a catalyst to speed up the reaction, and are generally expensive because of the metal (think platinum), that's all you really need to know.

Thanks so much for the help, I understand the design principles behind it now.

[/quote]

To increase the surface area of the reactants, allowing the redox reaction to occur at a faster rate.

I understand how it’d increase the redox reaction’s rate, but I’m confused about how it’d increase the surface area of the reactants... is it that in a porous electrode, there is more surface area exposed for the H2 gas molecules and able to react? Or does the porous electrode itself have a larger surface area, which enables the H2 gas to react faster?

[FrankieDens]

Hi everyone! 

My Chem teacher posted this question on our Google Classroom but I'm kind of confused on how to do it heh. How would you guys solve this?

When 2.45 mol of N₂(g) and 1.83 mol of H₂(g) are mixed in a 3.00L vessel the following reaction takes place:
N₂(g) + 3H₂(g) <=> 2NH₃(g). When equilbirum is reached it was found that there is 0.423mol of NH3. Determine the equilibium constant of the reaction at this temperature and post it, including units with correct significant figures (of course we can't use subscripts and superscripts).

[Geoo]

Hi everyone! 
My Chem teacher posted this question on our Google Classroom but I'm kind of confused on how to do it heh. How would you guys solve this?
When 2.45 mol of N₂(g) and 1.83 mol of H₂(g) are mixed in a 3.00L vessel the following reaction takes place:
N₂(g) + 3H₂(g) <=> 2NH₃(g). When equilbirum is reached it was found that there is 0.423mol of NH3. Determine the equilibium constant of the reaction at this temperature and post it, including units with correct significant figures (of course we can't use subscripts and superscripts).

So this is asking you to the the equailibrium constant K_c. The best way to find this is to set up an ICE table.
I= Inital amount (products are 0)
C= is the change you are going to see and is repersented by x
E= is the amount (in mol) at quailibrium
So lets set it up:
Fill out the table with the infomation you have.


No we need to find the C and E for everything.
The C, is what has changed. Reactants (N2 and H2) will loss mass so it will be -x, but if it has a coefficient like 3H2 does, it becomes -3x. For the products since it will gain mass, it will be +x
Lets add this to the table:


Now we need to work out the mole after equailbrium for N2 and H2.
As we can see, we have all the info in the NH3 colum to work out X.
2X=0.423
X=0.2115
So we can use this to find out info in the other tables.
To work out N2, 2.45-X,
So, 2.45-0.2115 = 2.2385.
To work out H2, 1.83- 3X
1.83- 3(0.2115) = 1.955
So we can no imput this into our table:


No Kc is found from conenctrations, not mol.
So we need to convert the numbers found in the E part of the table.
We use C= n/V
The volume in your question is 3.00 L
So for N2 = 2.2385/3.00 = 0.734
For H2 = 1.955/3.00= 0.652
For NH3 = 0.423/3.00 = 0.141
Now, set up the Kc equation and sub these in+
Kc = [NH2]^2/ [N2] [H2]^3
Sub in, Kc= [0.141]^2 / [0.734] [0.652]^3
Kc = 0.0977
For the unit sub in M to the Kc formula = M^2/ M x M^3 = M^2/ M^4 = M-2
So, Kc = 0.0997 M-2.
(I may be wrong on the units)

[FrankieDens]

So this is asking you to the the equailibrium constant K_c. The best way to find this is to set up an ICE table.

Thank you so much! Our teacher hasn't taught us how to use the ICE table yet only how to solve simple equilibrium constant questions so this was very helpful!!

[Snow Leopard]

How pedantic are VCAA for Chemistry. Is there anything that we need to draw/write a specific way not to get marked down?

For example, when drawing electron dot diagrams, would we get marked down for drawing the electrons for one atom as dots and the electrons for the other atom as crosses?
Thanks in advance

[whys]

How pedantic are VCAA for Chemistry. Is there anything that we need to draw/write a specific way not to get marked down?

For example, when drawing electron dot diagrams, would we get marked down for drawing the electrons for one atom as dots and the electrons for the other atom as crosses?
Thanks in advance

VCAA would not ask you to draw an electron dot diagram in the 3/4 exam. For 1/2, your teachers will be marking your exams, so you should consult with them regarding how pedantic they are and what they would accept. Regarding your specific question, I doubt it would matter if you drew the electrons as dots or crosses as long as it is clear they are electrons.

[ErnieTheBirdi]

Someone please help! I can't even begin. I suck at chemistry, no matter how hard I try or how much work I do but I still screw everything up. I just did a short quiz for school during quarantine and I got 5/11 which should have been 6/11 but I didn't write 12500 as 1.25*10^4. Can someone please help me? I spent all night and most of today studying for it and I could do all the practice questions and everything and it was all correct but when I did the quiz/test I basically failed? I need chem if I want to do medicine but I'm so bad at it. Can someone please help? I don't know what to do

[ksel03]

Hi Ernie.
I am not doing chem this year but I am doing physics and I know they're obviously very different, but I had a similar experience that I will share. It took me ages to 'get' physics. First SACs were really bad. I just couldn't grasp the concepts. Anyway, I want to say, don't be discouraged by bad scores. I would suggest talking to your teachers. This worked for me. Just tell them where you're at, why your not happy with your results and set a goal and tell your teacher what that goal is. Ask them for their opinion. Ask them what you should do differently to get your goal grade for your next SAC. Make sure they give you specific answers. I found that don't think of the teacher as someone trying to catch you out in SACs and ask questions that you don't know. Because this isn't true. They are there to help. So just communicate with them. Ask things like 'What are you looking for in this question?' 'How can I improve my response on this question?' I know sometimes it is hard to accept that your having trouble in a certain subject. I was so used to acing all my subjects that when I did physics it was extremely hard for me to accept that I sucked at it. I was too proud. It took me ages to actually go up to my teacher and ask for help. But just do it. My last two SACs were A+. Trust me, you'll have their support and your grades will improve.
Hope this helps!