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

partyatmygaff

This may seem stupid but will I lose marks for using X and 1/2X instead of 2X and X in the equilibrium calculation?

KoolAidRelic

rainbowtrout wrote: »

I got the same answer just now.

Did you have:
Kc = (x/12)^2/((5-2x)/12)^2

Because that gives x as .5456... but the [Hydrogen] isn't x, but x/12 which would be .04546

Or did you do it some other way?

M_ark

As far as I remember we were told that it will only be treated as a slip if you show all workings, for example in an Rmm calculation you would have to show all the elements rmms added together for a mistake in the sum to be treated as a slip, otherwise you lose all marks

jumpguy

partyatmygaff wrote: »

This may seem stupid but will I lose marks for using X and 1/2X instead of 2X and X in the equilibrium calculation?

Nah, definitely not, as long as you had them in the right place. You just made life a bit more awkward for yourself. :P

rainbowtrout

partyatmygaff wrote: »

This may seem stupid but will I lose marks for using X and 1/2X instead of 2X and X in the equilibrium calculation?

I have no idea. It would still give you the same answer. Perhaps have a look at the marking scheme for last years equilibrium question to see how it is marked.

http://www.examinations.ie/archive/markingschemes/2010/LC022ALP000EV.pdf

aranciata

Will i lose many marks for having the ratio wrong in the bleach titration? Every figure except for that is correct, just half of what the right answer is. Is that considered a slip?

rainbowtrout

KoolAidRelic wrote: »

Did you have:
Kc = (x/12)^2/((5-2x)/12)^2

Because that gives x as .5456... but the [Hydrogen] isn't x, but x/12 which would be .04546

Or did you do it some other way?

But you've already accounted for the 12L vessel when you divided by 12 in the formula you gave. Why would you divide by 12 a second time?

I did it by square root method , not the -b formula so that could account for the discrepancy between .53 and .54

KoolAidRelic

rainbowtrout wrote: »

But you've already accounted for the 12L vessel when you divided by 12 in the formula you gave. Why would you divide by 12 a second time?

I did it by square root method , not the -b formula so that could account for the discrepancy between .53 and .54

You have to divide by 12 because in your formula, x isn't the concentration of Hydrogen gas, it's the amount.

Look at it like this, Kc is tiny therefore there must be much more HI than Hydrogen but if you're concentration of Hydrogen gas is .53 moles per litre, then there must be 6.36 moles of Hydrogen gas in the entire flask.
Only five moles of Hydrogen Iodide were introduced into the flask at the start so that's obviously wrong.

tl;dr: using x in the formula will give you the concentration of Hydrogen
using x/12 at the start will give you the amount.

rainbowtrout

KoolAidRelic wrote: »

You have to divide by 12 because in your formula, x isn't the concentration of Hydrogen gas, it's the amount.

Look at it like this, Kc is tiny therefore there must be much more HI than Hydrogen but if you're concentration of Hydrogen gas is .53 moles per litre, then there must be 6.36 moles of Hydrogen gas in the entire flask.
Only five moles of Hydrogen Iodide were introduced into the flask at the start so that's obviously wrong.

tl;dr: using x in the formula will give you the concentration of Hydrogen
using x/12 at the start will give you the amount.

Go and have a look at the solution to the 2004 equilibrium question and maybe come back and argue your point. In that one there was a 5L vessel and as I already pointed out you account for the size of the vessel in the third line of your solution 'Concentration at equilibrium' This divides 'x' or whatever value you have by the volume of the vessel. You are then still solving for x.

Nowhere have I ever seen anyone divide twice by the volume of the vessel to get the answer. Concentration of products/reactants is always given in moles so concentration and amount are much the same thing here

tldr; I'm a chemistry teacher

Megan.mcgowan

To be honest, thought the paper was a BALLS.

Im crap at Chem, and have gotten grinds since the mocks. Got 38 and was hoping to get my pass. Question 4 I thought was harder than normal, the organic question was so different to previous years, as were alot of the questions. Got on ok with bleach as it was predicted but definitely failed. Was on the verge of tears in the exam hall!

So unfair it was such a big change from previous years.. And raging I didnt get a chance to revise equilibrium, the mechanisms or the option >:(

100% fail

KoolAidRelic

2004 was different as you had exact amounts, no x.
X is either concentration (moles per litre) in which case solving for x gives you the concentration, or x is the amount of hydrogen gas in which case x/12 is the concentration.

Here's it all written out properly to show what I mean: http://i51.tinypic.com/15ujqx.png

Concentration of products/reactants is always given in moles so concentration and amount are much the same thing here

Concentration and amount are much the same? Only in the trivial case when the flask is one litre.
How do you explain the fact that a concentration (which is always amount per unit volume) of 0.53 moles per litre of Hydrogen could be formed when the initial amount of Hydrogen iodide is in a concentration of .41 moles per litre?

gadgetz

YodaBoy wrote: »

has any1 got the figures for all the calculations please!!!!!!?

1(b) red-brown colour of iodine
excess KI to liberate iodine in the form of iodide (I^- or I>3^-) ions, and then keep it in solution so can titrate the iodine.

1(e) 0.0322 M diluted bleach
dilution factor of 20, so 0.644 M original bleach (12 marks)

1(f) 0.644 moles per litre x 74.5g (mass of NaOCl)
= 47.978 g/l

= 47.978 g / 1000 cm^3
= 4.7979 g / 100 cm^3
= 4.8 % (w/v)

gadgetz

TheFullDuck wrote: »

More of an endothermic man myself

More of the mathsy parts:

4 (e) 2.1 g is H2O 2.1g / 18g = 0.1166 moles

2g MgSO4 = 2g/120g = 0.0166 moles

0.1166 / 0.0166 = 7.02

So x = 7

4 (f) 3 Cu^2+ + 2Al gives 2 Al ^3+ + 3Cu


6(a) reverse 1st and 2nd equations. Leave 3rd one alone.

+ 74.6
+242
- 111
__________________
Delta H = + 205.6 KJ mol^-1 (12 marks)


7(b) pH = -log 10[H+] (6 marks)

= -log10 square root ( 6.3exp-5 x 0.2)
= 2.8

[sulphuric acid] = 2nd log (- 2.8)
= 0.0016 M (14 marks)

gadgetz

9 (a) Kc = [x][x] / (5-2x)^2 = 0.0185

0.0185 = x^2 over
25-10x + 4x^2

0.4625 - 0.185x + 0.074x^2 = x^2

0 = 0.926x^2 + 0.185 x - 0.4625

Use -b formula

x = 0.6 or - 0.8 Disregard the negative answer.

Filling this back in for x gives it close enough to Kc value given, so did something stupid in -b formula maybe!

(I think 12 marks should have been given to this instead of 18 marks, 4.5% of paper)

gadgetz

10 (b) acetaldeyde / ethanal (causes hangover) is the primary metabolite of ethanol
( a bit sly asking this I think, but it was only 3 marks)


C6H12O6 gives 2 C2H5OH + 2 CO2

(i) 2 x 175 ml = 350 ml

12.5% v/v
12.5 cm3 in 100cm3
= 43.75cm3 in 100cm3

D= m/v
0.8g/cm3 = m / 43.75

m = 0.8 x 43.75
m = 35g

35g / 5.3g per hr = 6.6 hours (6 hours 36 mins)


(ii) 35g of ethanol in 28 litres of body fluid
35g in 28,000 ml

0.125g in 100 ml
= 125g / 100ml blood


90 % of this (if necessary !) 112.5mg / 100ml blood

(think I'm right - but a bit much for LC Chem in my opinion)

Pegmin

Damn,just realised that I used 5-x instead of 5-2x for the equilibrium question...I wonder does that mean I lost a whole 18 marks...:(

xdublingirlx

gadgetz wrote: »

10 (b) acetaldeyde / ethanal (causes hangover) is the primary metabolite of ethanol
( a bit sly asking this I think, but it was only 3 marks)


C6H12O6 gives 2 C2H5OH + 2 CO2

(i) 2 x 175 ml = 350 ml

12.5% v/v
12.5 cm3 in 100cm3
= 43.75cm3 in 100cm3

D= m/v
0.8g/cm3 = m / 43.75

m = 0.8 x 43.75
m = 35g

35g / 5.3g per hr = 6.6 hours (6 hours 36 mins)


(ii) 35g of ethanol in 28 litres of body fluid
35g in 28,000 ml

0.125g in 100 ml
= 125g / 100ml blood


90 % of this (if necessary !) 112.5mg / 100ml blood

(think I'm right - but a bit much for LC Chem in my opinion)

I got those answers too! That's a bit reassuring, for some reason i kind of doubted myself on that question after the exam! What question were they expecting you to use the gas constant and molar volume at stp..I couldnt figure that one out because ethanol is liquid?

BrendaN_f

xdublingirlx wrote: »

I got those answers too! That's a bit reassuring, for some reason i kind of doubted myself on that question after the exam! What question were they expecting you to use the gas constant and molar volume at stp..I couldnt figure that one out because ethanol is liquid?

this lol, i was searching the paper thinking i missed a question, usually they only put relevant information on the front of the paper

Gavarn

Pegmin wrote: »

Damn,just realised that I used 5-x instead of 5-2x for the equilibrium question...I wonder does that mean I lost a whole 18 marks...:(

I hope not! Thats what I done, hopefully it's just a slip/error (-3)

._.

It seems I left out a question while calculating up my grade yesterday..
89.25% it is

leaving cert finished 2k11

just a few questions i would like 2 knw if i have the right answers

question 2 part b: wat other substance is distilled accross with the organic substance??

question 5 c: wat do the electron configurations of the elements from scandium to zinc have in common?

question 6 a: wat it the nature of the chemicals of crude oil ??

xdublingirlx

leaving cert finished 2k11 wrote: »

just a few questions i would like 2 knw if i have the right answers

question 2 part b: wat other substance is distilled accross with the organic substance??

question 5 c: wat do the electron configurations of the elements from scandium to zinc have in common?

question 6 a: wat it the nature of the chemicals of crude oil ??

1. Water

2.Their highest energy electron enters a d orbital

3. Hydrocarbons

These were my answers...

._.

I'm pretty proud of myself for correctly predicting that they'd mix Steam Distillation and Ethanoic Acid in Q2

xdublingirlx

._. wrote: »

I'm pretty proud of myself for correctly predicting that they'd mix Ethanal and Ethanoic Acid in Q2

Correct me if i'm wrong but ethanal was not in question 2? It was ethanoic acid and steam distillation....?

._.

xdublingirlx wrote: »

Correct me if i'm wrong but ethanal was not in question 2? It was ethanoic acid and steam distillation....?

Yeah, that's what I said. You must have heard me wrong.. loud in here...

aranciata

I think I got all of Q2 wrong. All of it. I did the preparation of ethanal as a reflux, which it's not.

Didn't attempt the second part of the question, new the first answer was clove oil but we never even did that experiment in school.

aranciata

Also, I got the ratio wrong in the bleach titration and I had the dilution factor as 25 and not 20.

How many marks am I going to lose? Panicking now.

This exam couldn't have gone worse, have been getting an A1 in all my tests and in the mocks.. looking at a B3 now. Not getting my course, not going to college, will be repeating most definitely. FML

JamieDOC

Anyone get 0.04 moles/L for hydrogen in the equilibrium question?

Exothermic

aranciata wrote: »

Also, I got the ratio wrong in the bleach titration and I had the dilution factor as 25 and not 20.

How many marks am I going to lose? Panicking now.

This exam couldn't have gone worse, have been getting an A1 in all my tests and in the mocks.. looking at a B3 now. Not getting my course, not going to college, will be repeating most definitely. FML

I don't think that's too many marks. If you continued your method correctly it's probably only a slip.
Don't be so hard on yourself! A1s all year means your no idiot. The test certainly wasn't what quite a few people were expecting, so you're not the only one. No point over analysing it right now, you don't know how things will turn out! This is the time to revise your second and third choices on your CAO so come August you have realistic options. You could still get your course regardless.

KoolAidRelic

JamieDOC wrote: »

Anyone get 0.04 moles/L for hydrogen in the equilibrium question?

Yep, a few others did a few pages ago.

edit: and me as well