Chemistry 3/4 2013 Thread

[Edward21]

We did the fuel cell on a small scale as part of our SAC, we electrolysed KOH 1M solution to get H2 and O2, then took the wires out of the power pack, connected it to a voltmeter and watch the gases disappear whilst getting out about 2V pretty cool methinks!!

[barydos]

When forming esters, is it the -OH from alkanol (and -H from carboxylic acid) or the -OH from the carboxylic acid (and -H from alkanol)
that gets donated when producing the H2O molecule?

And what's the difference between amide links and peptide bonds? When do we use which?

Thanks in advance.

[thushan]

When forming esters, is it the -OH from alkanol (and -H from carboxylic acid) or the -OH from the carboxylic acid (and -H from alkanol)
that gets donated when producing the H2O molecule?

And what's the difference between amide links and peptide bonds? When do we use which?

Thanks in advance.

-OH from the carboxylic acid (and -H from alkanol)

But I don't think they expect you to know that.

All peptides are amides. Not all amides are peptides. Peptides are amides within a polypeptide chain (protein).

[Alwin]

-OH from the carboxylic acid (and -H from alkanol)

But I don't think they expect you to know that.

They don't, I asked about it ages ago. Apparently even memorising it is just "too hard" for vce
(the reason why, maybe - but let's not go there ^^)

[lzxnl]

They don't, I asked about it ages ago. Apparently even memorising it is just "too hard" for vce
(the reason why, maybe - but let's not go there ^^)

It sort of is...in acidic solution, you have to remember that the acid OH group acts as a base and abstracts a proton
Then you have to remember that this positive oxygen grabs even more electron density from the carbon, making the carbon more positive
Making it more attractive for attack by a negative oxygen on another alcohol
And that when the nitrogen does attack the carbon, the protonated OH2+ group can leave as it becomes a stable, neutral water molecule, leaving behind an amide.

That's obviously too much detail for any VCE subject.

[Jaswinder]

suppose we have the moles of Cl- in MgCl2.

to find the moles of MgCl2, do we multiply or divide the moles of Cl- by 2? and why?

2) in OCl2, is Cl more electronegative or O? what would be the oxidation number of Cl? why?

[thushan]

MgCl2

The formula itself tells you that for every MgCl2 molecule, there are two Cl atoms.
So you divide the moles of Cl- by 2.

I'm not two sure about number too though, you'll have to rely on nliu for an explanation as to why one of more electronegative than the other haha

O is more electronegative than Cl, so oxidation number of O is -2.

Although O is in group 16 and Cl being in group 17, with core charge in Cl being higher than that in O, O is a significantly smaller atom, so the electron "pulling" power is greater.

[Jaswinder]

how would you figure out how many double bonds C17H31COOH has? is there a formula?

[lzxnl]

MgCl2

The formula itself tells you that for every MgCl2 molecule, there are two Cl atoms.
So you divide the moles of Cl- by 2.

I'm not two sure about number too though, you'll have to rely on nliu for an explanation as to why one of more electronegative than the other haha


Edit: Dayum these scotch college boys got chemistry covered

LOL. Scotch College boys? Nah; thushan did 3/4 chemistry in year 9 or something. He's the exception, not the rule.

Just remember that the order of electronegativity goes fluorine, oxygen, chlorine, nitrogen etc.

how would you figure out how many double bonds C17H31COOH has? is there a formula?

There is a formula, yes, but I'd do this with some common sense.
The COOH bit can be replaced with a hydrogen for simplicity as that is singly bonded to the rest of the molecule and contributes one double bond that you know already from C=O.
So we have C17H32. Recall that alkanes have the form C2H(2n+2). E.g. C2H6, C3H8 for ethane, propane.
C17H36 is your alkane. You've lost four hydrogens, two hydrogens per double bond=>two extra double bonds. Three including C=O.

[clıppy]

Just a quick question in terms of practice exams, how far back do people recommend we go before things start becoming irrelevant?
I know fractional distillation and detailed studies are out (is gel electrophoresis out?) so how far back should we be going? 07/08ish?

[Alwin]

Just a quick question in terms of practice exams, how far back do people recommend we go before things start becoming irrelevant?
I know fractional distillation and detailed studies are out (is gel electrophoresis out?) so how far back should we be going? 07/08ish?

I like how you posted this question on the methods and chem boards

But there's other stuff like enzymes as markers for diseases and electrophoresis that have been taken out of the course for 2013, so watch out for them!

I'm just doing all the relevant ones (2009 onwards NMR, IR and other chem analysis added iirc), then the older ones picking and choosing relevant question only hm

GOOD LUCK

[Jaswinder]

A current of 8 amps is passed through an aqueous solution for 24 000s. In this time, 1 mole of metal is deposited at the cathode and a gas  is collected at the anode. When the gas is cooled to STP conditions, its volume is 11.2L. The aqueous solution could be

ANS: Cu(NO3)2 , could someone explain to me how they got their answer.. I now that q=it and that the n(e) is 2 but what do i do after that? 

[lzxnl]

A current of 8 amps is passed through an aqueous solution for 24 000s. In this time, 1 mole of metal is deposited at the cathode and a gas  is collected at the anode. When the gas is cooled to STP conditions, its volume is 11.2L. The aqueous solution could be

ANS: Cu(NO3)2 , could someone explain to me how they got their answer.. I now that q=it and that the n(e) is 2 but what do i do after that? 

Do NOT just randomly use formulas. Please don't. You'll confuse yourself somewhere.
At STP, 11.2 L of gas is 0.5 moles. You have 0.5 moles of gas and 1 mole of metal.
Presumably, the choices are a metal salt solution, and the metal is being reduced. So we have water being oxidised to form oxygen gas. For every mole of oxygen gas formed, it gave off four moles of electrons. So we have 0.5 moles of gas => 2 moles of electrons, 1 mole of metal. Metal has charge +2.

Or, charge sent through is current times time = 8A * 24000s = 192000 C
Divide by 96500 C/mol to get roughly 2 moles of electrons. 1 mole of metal => two electrons are needed to form one atom of metal, charge on metal is +2.

[clıppy]

Can someone please re-explain ratios to me, I've got no idea what to do when they come up.
Take for example:

The complete combustion of 24.0g of an organic compound produces 1.2mol of CO2 and 1.6mol of H2O. The compound was:
Propane
1-Propanol
Propanoic Acid
Methyl Ethanoate

The answer is 1-Propanol with the working out saying:
n(C) in 24.0 g of the compound = n(CO2) produced
= 1.2 mol
n(H) in 24.0 g of the compound = 2 x n(H2O) produced
= 2 x 1.6
= 3.2 mol
Ratio n(C) : n(H)
= 1.2 : 3.2
= 1 : 2.67
= 3 : 8

n(C3H8) in 24.0 g = m(C3H8) / M(C3H8)
= 24.0 g / 44.0 g mol-1
= 0.545 mol
n(C3H8O) in 24.0 g = m(C3H8O) / M(C3H8O)
= 24.0 g / 60.0 g mol-1
= 0.400 mol

I'm following everything up until the ratio bit, I just can't see how it works, the maths behind it. Any help?

[Jaswinder]

Volume of water: 440 ml
Initial mass of ethyne: 2.88g
Final mass of ethyne: 2.17g
Initial temperature of water: 23.5 C
Final temperature of water 42.6 C

During combustion of ethyne (C2H2), Use the change in temperature of the water to determine the amount of energy in KJ added during the heating.

What I did was E=4.184 x 440 x 19.1 = 35162 = 35.1KJ

but then the solution goes on to find the energy for one mole by dividing it by (0.71/26) giving 1.3 x 10^3 KJ. However isnt that for a mole of ethyne? and the question asks for energy in KJ added during the heating. ?
So shouldn't it be 35.1 KJ?