Chemistry

munkeehaven · 22-06-2004 6:46pm

i dont mean the alcohol in the ester--methanol--i mean the one that can be ade to a ketone---propanol

penguincakes · 22-06-2004 6:49pm

Originally posted by munkeehaven
i dont mean the alcohol in the ester--methanol--i mean the one that can be ade to a ketone---propanol

There's no propanol, cause for an ester you need a COOH. Which you can't have in the middle of a chain.

CuLT · 22-06-2004 6:51pm

Precarious, are you quite sure the answer to the /_\H was -75.6 and not +75.6 ?

I was under the impression that it was -966 +75.6 to get the -890.4

c0rk3r · 22-06-2004 6:56pm

Originally posted by CuLT
Precarious, are you quite sure the answer to the /_\H was -75.6 and not +75.6 ?

I was under the impression that it was -966 +75.6 to get the -890.4

Thank You! was sweating for a moment there.I got the same as above.

12marks = 3%

munkeehaven · 22-06-2004 6:56pm

NOT THAT remember the qt asked what could the carboxylic acid be derived from--propanoic acid---

PrecariousNuts · 22-06-2004 6:59pm

I thought that heat of formation was nearly all the time negative

BangBeater · 22-06-2004 6:59pm

It HAS to be -75.6 - IT's exothermic

CuLT · 22-06-2004 7:00pm

Originally posted by c0rk3r
Thank You! was sweating for a moment there.I got the same as above.

12marks = 3%

heh np

.

Though too bad I forgot the f*cking ratio for the edta question, I put it as 1:2 . I am BONED and GRILLED for that question now.

munkeehaven · 22-06-2004 7:01pm

I thought that heat of formation was nearly all the time negative

it wont be negative if the reaction is endothermic ie takes in energy

but for methane its exo

penguincakes · 22-06-2004 7:02pm

Originally posted by munkeehaven
NOT THAT remember the qt asked what could the carboxylic acid be derived from--propanoic acid---

Oh right. Well what are you asking? There's still no secondary alcohol. It's propan-1-ol -> propanal -> propanoic acid.

munkeehaven · 22-06-2004 7:04pm

i just thought because you can oxidise a sec alcohol to a ketone that you would be able to do the opp by reduction so that you could do propanol---propanone--propanoic

i did what you did then scribbled it out ****!

CuLT · 22-06-2004 7:05pm

Originally posted by BangBeater
It HAS to be -75.6 - IT's exothermic

Really? Why does does it HAVE to be?

Sure the overall reaction was exothermic, but it can be any-which-way with the individual components, can't it?

penguincakes · 22-06-2004 7:05pm

Originally posted by CuLT
Precarious, are you quite sure the answer to the /_\H was -75.6 and not +75.6 ?

I was under the impression that it was -966 +75.6 to get the -890.4

The CH4 should be on the other side, it's formation, the elements -> the compound, so you have to turn that around to be 966 - 75.6 = 890.4.

penguincakes · 22-06-2004 7:07pm

Originally posted by munkeehaven
i just thought because you can oxidise a sec alcohol to a ketone that you would be able to do the opp by reduction so that you could do propanol---propanone--propanoic

i did what you did then scribbled it out ****!

Err my head hurts I don't understand too much chemistry already today. I hope for your sake your right but for my sake that you're wrong...

BangBeater · 22-06-2004 7:09pm

Q10 (a) anyone?? How do u do dat?

munkeehaven · 22-06-2004 7:09pm

well you know from everyday life that when methane is burned its hot.....endo would be something like

Energy - Examples of Endothermic and Exothermic Reactions.

Most reactions are exothermic.
Examples are

1) Combustion of Methane (Natural Gas).

2) Neutralization of dilute hydrochloric acid and dilute sodium hydroxide.

3) Adding concentrated sulfuric acid to water is highly exothermic
(gives out a lot of heat).

4) Adding water to anhydrous copper(II) sulfate is exothermic.
Anhydrous copper(II) sulfate is white.
Anhydrous is pronounced "an-high-drus"
Anhydrous means "without water".
Anhydrous copper(II) sulfate is copper(II) sulfate which is completely dry.
When water is added to it, it turns into the familiar hydrated blue crystals.
Hydrated is pronounced "high-dray-tid", it means 'with water'.
This is used as a Test for Water.

If blue (hydrated) copper(II) sulfate crystals are heated,
an endothermic reaction occurs, they lose their water,
turn white and become anhydrous copper(II) sulfate crystals again.
The reaction is reversible.

CuLT · 22-06-2004 7:10pm

Originally posted by penguincakes
The CH4 should be on the other side, it's formation, the elements -> the compound, so you have to turn that around to be 966 - 75.6 = 890.4.

Ah nuts, you might indeed be right, but I REALLY don't want to do those questions again.

come on c3...

munkeehaven · 22-06-2004 7:12pm

eh bang what was the qt exactly?

and ps penguin you are prob right im just a desperate girl clutching at straws

BangBeater · 22-06-2004 7:15pm

Right for Q10 (a)

I got that it was 360 g per litre

(i)Then, I put this into Mass/Relativ. Mol. Mass with the RelMolMass being that of HCL

THEN, I divided this by 2.5 to get the mole value required for (i)
Is this right?

BangBeater · 22-06-2004 7:17pm

Actually, where does the 2.5 litres come into this Q at all??

penguincakes · 22-06-2004 7:18pm

Originally posted by BangBeater
Q10 (a) anyone?? How do u do dat?

10 a i

36% of 2500g = 900g.

900/Rmm(HCl) = 900/(35.5+1) <--- I'm sure they meant to put 35 as the value to take for the RAM of Cl, cause it works out nicely as 25M if you do. Otherwise:

= 900/36.5 = 24.65753 moles of HCl spilt.

ii

Ratio = 2:1 HCl : NaCO3

therefore 24.65753/2 = 12.328765 M of NaCO3 needed.

(12.328765)(2(23) + 12 + 3(16)) = (12.328765)(106) = 1306.84909 g of NaCO3 needed.

iii

Volume produced: 1:1 NaCO3 : CO2

therefore 12.328765 moles of CO2 produced. Vol@room temp=24 L/mol (from front)

(24)(12.328765) = 295.89 Litres.

CuLT · 22-06-2004 7:30pm

#LeavingCert on quakenet - "for the joy"

Really. I'll be holding the room if anyone wants to go there, a bit easier than using boards.

c0rk3r · 22-06-2004 7:31pm

Q10 (a)

(i) i got 0.4 moles

hmmm hom hhmmm hom

Seifer · 22-06-2004 7:35pm

Just going to vent my spleen in a random fashion on this. When I came out I felt alright about it, then I looked over some stuff and felt like **** but now I feel better again. I'm pretty sure I lost my A1 not even sure if I'll get my A2. Anyway here's my rundown of the paper:

Q1: perfect! Answers: d (i)8.1x10^-4 (ii)0.081 (iii) 81ppm

Q2:more or less perfect

Q3: Nasty piece of crap that I wasted time on! What the hell are the answers to a(iii) b (ii) (iii). Got 90% as purity.

Q4: Went well. Whats part c answer? I waffled about there being 2 covalent bonds in O2 but only 1 in H2. Part f? Part g: (i) -1 (ii) 5 Part h: 70%

Q5: Perfect

Q7: My stupid brain wanted it to be ethanoic acid so that's what I wrote. I even had methanol so the carbon atoms don't add up! IMPORTANT: will they adjust the answers all the way down as they all depend on you getting the acid? Hope they do cause I got most of it right based on ethanoic acid. Luckily I have an extra question.

Q9: Went nicely. Kc = 0.0813 Did you have to divide all values by 5 because of litre vessel? I did.

Q10: Went well until people here pointed out that you're supposed to use 24L. What a bent thing for them to do! I only looked at the cover for R and the time. Assholes. a (i) 24.658 moles (ii) 8.598 g (iii) 276.1696L (wrong now)
c (iii) 1.204 x 10^-4 m3

Q11: Went fantastically until I released after the exam that my brilliant brain decided to write about alpha particles instead of beta! What a waste of marks on stuff I knew. Luckily I did 3 parts. pH = 2.435

Aaaaaah! So much better now!:)

munkeehaven · 22-06-2004 7:40pm

Q4: Went well. Whats part c answer? I waffled about there being 2 covalent bonds in O2 but only 1 in H2.

one of the few things i got right--van der waals force ---o2 is the bigger molecule so force will be stronger----larger boiling point

for the acid qt is it Ka = to the sq root of the h plus ions times the Mr of the acid?

claire h · 22-06-2004 7:50pm

Originally posted by munkeehaven
one of the few things i got right--van der waals force ---o2 is the bigger molecule so force will be stronger----larger boiling point

for the acid qt is it Ka = to the sq root of the h plus ions times the Mr of the acid?

Isn't it [H+] = sqrt[Ka + Macid]? The Ka is inside the square root, and that's how you get the H+ ions. I hope...

PrecariousNuts · 22-06-2004 7:50pm

Hehe the ads are going crazy, I just saw one for organic chemistry (top right of screen)

munkeehaven · 22-06-2004 7:56pm

im not usually one to fret over such trivial things but i realy did like chemistry and i studied hard for it.i guess you just expect to do really well and then its such a let down when by chance you dont do well......it just wasnt my day.......:(

munkeehaven · 22-06-2004 7:57pm

youre right claire thankfully i actually wrote what you said in the test so phew!!

MatthewVII · 22-06-2004 9:30pm

in relation to question 7 - organic

the methyl propanoate is formed from a substitution reaction (answer) - the H of the propanoic acid is replaced with an alkyl group. it isnt a condensation reaction since the question specifically asks for a type of organic reaction. similarly, the answers to part (iii) are :
propan-1-ol -> propanal : Elimination reaction, H2 eliminated
propanal -> propanoic acid: Addition reaction, O added

did anyone else get that?

Chemistry

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