* Chemistry * Predictions / discussion / aftermath * (1 thread please)

Iceuil

Quick question on equilibrium. Do we have to remember the formulas for the experiments (the cobalt, chromium and the CNS ones)? Or do we just need to know the colour changes?

Thanks.

Dongl

Gavarn wrote: »

Looking over the bleach experiments that came up in the examcraft mock paper and it asks to show how the 50cm^3 sample of bleach was diluted in 500cm^3.. In our lab there are only 250 volumetric flasks, so is there a flask that does 500 or do you have to split it? Stupid question probably but I want to be sure! Thanks

If it helped answer the required question you could assum you have access to a 5000cm^3 volumetric

DM360

Iceuil wrote: »

Quick question on equilibrium. Do we have to remember the formulas for the experiments (the cobalt, chromium and the CNS ones)? Or do we just need to know the colour changes?

Thanks.

Both I would think

Dongl

Iceuil wrote: »

Quick question on equilibrium. Do we have to remember the formulas for the experiments (the cobalt, chromium and the CNS ones)? Or do we just need to know the colour changes?

Thanks.

formulae and colour changes can be asked.

pinkballetdance

aranciata wrote: »

Would it be alright at this stage to leave out saponification and recrystallisation of benzoic acid? (both came up last year)

Also thinking of leaving out the analysis techniques - spectroscopy, chromatography etc.

I had ever though until this morning that they would ever come up but be warned! On our physics paper this morning an experiment that id totally left out cos it came up last year came up... Hopefully i still did ok cos i did an extra exp q!
The moral of the story is leave ALL the experiments!! Sorry to anyone who doesnt think it can happen!

Also instead of them printing data on the physics paper it said all relevent data is in the formulae and tables available from the super intendant, the chemistry equilivant of this would be the masses at the start and what not so make sure ye can find avagadros constant and the likes in the tables. Talking to people after the exam most people didnt notice the note at the begining and had no idea what to do for the values! :O

Iceuil

formulae and colour changes can be asked.

Arrrghhh.....thanks!

._.

partyatmygaff wrote: »

There are also [LATEX]1000cm^3[/LATEX] and beyond. Standard sizes are 100, 250, 500, 1000, 1500, 2000, 3000, 5000.

Which are never used in LC chemistry... 99% of the time it'll either be 250 or 500

crazylegskane

Any ideas for ordinary experiments to come up?

Sm4shbox

Lads, some questions on FUELS. CAN SOMEONE HELP ME ANSWER?

1) Why does crude oil not dissolve in water?
(Get full marks for saying it's non-polar????)

2) Some of the fractions of the crude oil spill evaporated readily from sea into atmosphere.
NAME the fraction that is LEAST likely to evaporate.
State 1 USE for this fraction.

3) Name process in oil refinery that produces ethene from long-chained molecules.
If a C10 alkane undergoes this process to produce ethene & 1 other product, give the molecular formula of the product.

4) Write a balanced eq. to show how Hydrogen can be produced from natural gas??

Gavarn

A suitable material for the container of the bomb calormeter? Explain you choice?

Exothermic

1. Yes
2. Bitumen. Road surfacing.
3. Hexane (C6H14) Subtract C2H4 from C10H22.
The process is catalytic cracking.
4. Steam reforming of Methane.
CH4 + H2O = 3H2 + CO

partyatmygaff

Exothermic wrote: »

3. Hexane (C6H14) Subtract C2H4 from C10H22.
The process is catalytic cracking.

How is that possible? Shouldn't it be octane?

Exothermic

Whoops. 3 is wrong. Sorry I'm on my iPod :P
It's C8H18. So that's octane.

Exothermic

When we did the bomb calorimeter experiment the bomb was stainless steel. Not 100% why. Either it insulates or doesn't. Or else they know it's heat capacity and can calculate it from there?
Kinda spoofing here :P I should get back to my notes.

Gavarn

Exothermic wrote: »

When we did the bomb calorimeter experiment the bomb was stainless steel. Not 100% why. Either it insulates or doesn't. Or else they know it's heat capacity and can calculate it from there?
Kinda spoofing here :P I should get back to my notes.

Must be because it is a good insulator.. Yeah I'm only on this to stay away from having to study :S god I hope they give us a nice paper!!

Iceuil

I really hope organic is not gonna be 90% of the paper. Planning to do all of the inorganic questions and one question on fuels. *fingers crossed*

red_red_wine

I don't mean to scaremonger, just passing it on, but I was talking to one of my teachers today and his 'gut feeling' is that tomorrow's paper is going to be a difficult paper. Now, there's every chance he's wrong - we'll know this time tomorrow - but just in case it's an incentive for any of ye to study (I know it is for me :P .)

Personally, I prefer the more difficult papers, it's a lot nicer to think that you have done terribly only to be pleasantly surprised with the result than it is to think you've done wonderfully only to be disappointed.

TheFullDuck

gots a equilibrium constant question;


H2 + I2 <---> 2HI

the value of Kc for this reaction is 50 at 721K If 1 mole of hydrogen and 1 mole of iodine were introduced into a sealed vessel at this temp calculate the amount of each substance present when equilibrium is reached?

cxh20y

partyatmygaff wrote: »

There's no need to learn it off. Just check the periodic table and work it out from there.

How do you know by looking at the periodic table?

partyatmygaff

cxh20y wrote: »

How do you know by looking at the periodic table?

Check what group the element belongs to and see how many valence electrons it has.

Group 8 has 0 valence electrons and therefore is monoatomic.
Group 7 has 1 valence electron and therefore has single bonds if diatomic.
Group 6 has 2 valence electrons and therefore has double bonds if diatomic.
Group 5 has 3 valence electrons and therefore has triple bonds if diatomic.

There are of course exceptions but that's the general idea.

MagsK93

TheFullDuck wrote: »

gots a equilibrium constant question;


H2 + I2 <---> 2HI

the value of Kc for this reaction is 50 at 721K If 1 mole of hydrogen and 1 mole of iodine were introduced into a sealed vessel at this temp calculate the amount of each substance present when equilibrium is reached?

You fill in 2x for your value of HI at equilibrium first and sub them into your table for finding the value of kc as normal, to me its kinda like you're working backwards :P

Ratios 1....:.....1...:....2
.......... H2.........I2.......HI
At Start 1..........1.........0
Reacted -x..........-x......+2x
At Equil. 1-x........1-x.......2x


Kc= [HI]^2 / [H2][I2]

= (2x)^2/(1-x)^2 = 50
Get the square root of all of this

2x/1-x= 7.071
2x= 7.071-7.071x
9.071x=7.071
x= .77925

Therefore no. of moles of H2 at eqiul= 1-x
1-.77925= .22075
Rel. molecular mass of H2 is 2, .22075x2= .4415g

Do the same for I2 with the rel. molecular mass of 254= 56.07g

No. of moles of HI=2x= 2(.77925)= 1.5585 moles
Rel. molecular mass= 255, 1.5585x255= 397.42g

Exothermic

If asked about the formation of an ester which is correct?
Condensation reaction
Substitution reaction
Esterification

Confused. :P

jumpguy

Exothermic wrote: »

If asked about the formation of an ester which is correct?
Condensation reaction
Substitution reaction
Esterification

All of them! It's a substituition as the H at the end of the carboxylic acid (eg, HCOOH) is replaced by the alkyl part of the alcohol.

It's condensation as the number of atoms decreases and a small molecule (water) is released.

It's an esterification reaction because an ester has been formed.

Exothermic

I think I had an experience before where I wrote one of those down I was told I should always say a particular one as sometimes one of them isn't on the marking scheme. -.- Of course, I don't remember which one that was :P

Let's hope if that comes up that the marking scheme accepts all answers :P

mpdg

Calculate the concentration of a solution of sulfuric acid given that the pH is 2.3.

My notes my teacher gave us had this example and explained how to do this but I don't get it...

jumpguy

mpdg wrote: »

Calculate the concentration of a solution of sulfuric acid given that the pH is 2.3.

My notes my teacher gave us had this example and explained how to do this but I don't get it...

Is that on? O_O

Anyway, it's basically working backwards (I think).

-log[H+] will give you pH, so you know
-log[H+] = 2.3
log[H+] = -2.3
10^-2.3 = [H+] (laws of logs needed here)

But sulphuric acid is H2SO4, which breaks up into 2H+ and SO4(2-) when in water
H2SO4 -> 2H+ + SO4(2-)
So you've 2 moles of hydrogen ions for every one mole of H2SO4. So, you have the concetration of H+ to get the concentration of H2SO4
10^-2.3/2 = [H2SO4]


That, however, is open to correction, I've never done a question like that before.

Exothermic

mpdg wrote: »

Calculate the concentration of a solution of sulfuric acid given that the pH is 2.3.

My notes my teacher gave us had this example and explained how to do this but I don't get it...

I think this is correct but once again, I'm also open to correction.
If given the pH you use the antilog function on your calculator and make the number a negative. On mine, which is .... "Casio fx-83GT PLUS", it's the 2nd function of the log button.
So what I'd so id antilog(-2.3) = 0.00501187

So [H+] = 0.005 is your answer

So to double check: -log(.005) = pH 2.30

mpdg

Exothermic wrote: »

I think this is correct but once again, I'm also open to correction.
If given the pH you use the antilog function on your calculator and make the number a negative. On mine, which is .... "Casio fx-83GT PLUS", it's the 2nd function of the log button.
So what I'd so id antilog(-2.3) = 0.00501187

So [H+] = 0.005 is your answer

So to double check: -log(.005) = pH 2.30

Thanks to both of you for your help.
What IS this elusive "antilog" function? The notes mention it and I don't know what it is. Part of the solution reads:

2.3= log[H+]
-2.3 = log[H+]
[H+] = antilog - 2.3
[H+] = 0.005

From that, I would imagine antilog means log to the power of -1.
I have both the Casio fx-83GT and Sharp EL-531W Advanced DAL and I can't see the function on either. My second function on the Casio is 10^something. Am I missing something simple?

Bosca

Gah! I'm getting a bad feeling about tomorrow!

macskanadrag

mpdg wrote: »

Thanks to both of you for your help.
What IS this elusive "antilog" function? The notes mention it and I don't know what it is. Part of the solution reads:

2.3= log[H+]
-2.3 = log[H+]
[H+] = antilog - 2.3
[H+] = 0.005

From that, I would imagine antilog means log to the power of -1.
I have both the Casio fx-83GT and Sharp EL-531W Advanced DAL and I can't see the function on either. My second function on the Casio is 10^something. Am I missing something simple?

This is from Revise Wise, isn't it? I came across that too and my calculator is not exactly cutting edge, definitely doesn't have the antilog function. It's ok though because accoring to the Chemistry syllabus:

"1. Use of calculators
Students will be expected to have an electronic
calculator conforming to the examination regulations
for the duration of the course and when answering
the examination paper. It is recommended that
students have available the following keys:
Ordinary level: +, –, x, ÷, 1/x , log10 x, 10^x, EE or EXP,
memory.
Higher level: as above, and x^2, (root)x, x^y.
In carrying out calculations, students should be advised to
show clearly all expressions to be evaluated using a calculator.
The number of significant figures given in the answer to a
numerical problem should match the number of significant
figures given in the question."

Better not see an antilog question on the paper!