OCR Salters Chemistry F335 12th June 2013 Exam revision thread

Yeh but whats the actual formula?


Yeah that was weird - i think you just had to use the fact that on c2+ i2 goes to c2i2 (saturated bonds)

Reagents and conditions of Amide formation from ethanol and ethanoyl chloride? Do you have to reflux or add acid or anything?

Reagents and conditions of Amide formation from ethanol and ethanoyl chloride? Do you have to reflux or add acid or anything?

Reagents and conditions of Amide formation from ethanol and ethanoyl chloride? Do you have to reflux or add acid or anything?

Can someone please explain to me how to do Q1 (d) (i) from June 12

http://www.ocr.org.uk/Images/131296-question-paper-unit-f335-chemistry-by-design.pdf

Can someone please explain to me how to do Q1 (d) (i) from June 12

http://www.ocr.org.uk/Images/131296-question-paper-unit-f335-chemistry-by-design.pdf

heres how i did it,
forumula used: Moles = Concentration x Vol
(1 mol of gas occupies 24dm³)

1. i worked out the moles of oxygen, 8.3 x 10^-3 x 24 = 0.1992
2. they told us ratio of nitrogen to oxygen is 4:1, therefore moles of nitrogen is 0.1992 x 4 = 0.7968.
3. since we know the moles of nitrogen, we can work out its concentration using the formula: 0.7968 / 24 = 0.033
Not sure why the mark scheme want it as 0.033 rather than 0.03 though

Right I think I've eventually got it, I reckons its due to the fact that we're told 1/3 of the solution is diluted, and the ratio of [salt]/[acid] HAS to be 2 (from reverse engineering the answer).

So my final calculating came to be:

(0.0045*0.1)/((2/3)*0.05*0.027) which works out to be 1/2 exactly. I think the trick of this question is that the concentration of the acid (HA) has been partially diluted.

part 1. I wouldn'tuse this as a valiud method all the time as while this mark scheme allows it, it doesn't always want you to use the "show that" part of the equation in your working out.

guys

what topics does "Ionisation energy explanations" cover?

This ones just based around Moles = mass/Mr

1% of 1kg is 0.01kg (just divide by 100)

Mr of Sulfur is just 32
Mr of Sulfuric Acid is 98

Find the moles of Sulfur:
0.01/32 = 0.0003125

Find mass of sulfuric acid using the same moles:
0.0003125 x 98 = mass = 0.030625 = 0.031 (2sf)
theres one S atom in the sulfuric acid so I assumed it was 1:1 ratio hence the same moles.

As every value in the question is to 2 significant figures and the question states an appropriate number of sf you get the last mark just for rounding correctly!

I don't know how correct that is but I got the right answer :P I was confused at first thinking that I had to use the Mr of sulphur dioxide but that was earlier in the question and you were only given the mass of sulphur. Also,

ok, but any idea why the mark scheme says 'do not accept 0.03' though?

don't worry haha

Sal Khan actually made me remember this but 'p' infront of anything is just
-log(of something)
[like pH = -log(H⁺)]

K_a is the acidity constant. This usually is a really small number and they cover a large range of values, it's converted to pK_a solely for ease of use and so it can be easily compared to pH - coarse tuning remember!
So if you wanted a pH of 4, you'd pick the acid with the closest pK_a value to 4 and then fine tune with the salt:acid ratio

hope that helps

Did the question ask for an appropriate number of significant figures? If so, then it has to be to two sig figures as that's what the value of oxygen was to (if I remember correctly)


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ok, but any idea why the mark scheme says 'do not accept 0.03' though?