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Chemistry: Post your doubts here!

ahmed abdulla

ahmed abdulla

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AbbbbY said:
Sorry for the delays. Got a bit busy + don't really like opening p5's, personally.
__
2biii:
This is a pretty simple question with a horribly messed up wording. See. Collecting 30dm3 and 24dm3 is not possible with normal lab equipment. Collecting 10cm3 or 20cm3 is.

We know that 1 mol = 123.5g and this produced 30000 cm3 in 2.1 and 24000 in 2.2

So, simply using a ratio, we can find the mass that'd produce x cm3, a measurable value, of gas.

30,000 : 123.5
10 : x

Now you'll simple use this value of x found as the mass of CuCO3 to be used. If it produces 10cm3, it'll be following 2.1. If not, it'll be following 2.2. This can be repeated for several values of gas released and for either of 2.1 or 2.2.
Get it?

3a)

Again. Ratio.

Volume of Sulfuric acid:

10g of H2SO4 : 5.476 cm3
20g = x
x = 10.952

Volume of water:
70g of water : 70.211cm3 of water
20g of water : x
x = 20.060

and so on.

Total volume of 100g of solution:
Simple! Add the volume of Sulfuric acid and Water found.

Density of the solution:

mass / vol
so 100/volume of 100g sol

Easiest question I've seen concerning calculations tbh. If you don't get it let me know and I'll explain it in a more organized manner.
Click to expand...

Thanks abbby . :)
second question , i cant still reach a conclusion as which is the right equation for decomposition
and for the table , if you look at row 5 , you will find Both volume and mass of water not given so how to use the ratio when both not given?
and row 9 has both volume of sulphuric acid and total volume unknown so how to find both
 
AbbbbY

AbbbbY

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Namehere said:
I´ll go in for the rescue!

Lets consider the first row, step 1 is the slowest: This means that the hydrogen peroxide and iodide are involved somehow in the rate determining step. Since they are only involved once (directly or indirectly, you´ll see what I mean later on) and they are in the rate equation, then a=1 and b=1, c=0 because there is no hydrogen ions reacting in step 1.

Lets consider the second row, step 2 is the slowest: This means that the IO- and hydrogen ion are involved somehow in the rate determining step, but the IO- ion doesn´t appear in the rate equation! First, it is clear that the hydrogen ion will give us c=1, since it is only involved once and appears on the rate equation. Secondly, in order for the IO- to be produced, the hydrogen peroxide and hydrogen ion will have to react, and since these are involved in the rate equation, it follows that a=1 and b=1 (since each specie was involved only once), though this time the hydrogen peroxide and hydrogen ion were involved indirectly.

Lets finally consider the third row, step 3 is the slowest: This means that HOI, hydrogen ion and iodide are involved somehow in the rate determining step. Lets analyze HOI. We can see that for HOI to be formed, the IO- and hydrogen ion had to react, but the IO- which was needed for reaction 2 was produced by reaction of hydrogen peroxide with iodide from reaction 1! By looking at the rate equation we see that hydrogen peroxide was involved indirectly once to produce HOI and so we can see a=1. Lets continue with the hydrogen ion. We can see that it is in the rate equation and that it is involved twice, one directly from reaction 3 and one indirectly from reaction 2, since hydrogen ions reacted with IO- to form HOI, one of the reactants somehow involved in the rate determining step in reaction 3. Since it is involved twice, it follows that c=2. Finally with iodide we can also see it is in the rate equation. Here the iodide ion has been involed twice, again once directly from reaction 3 and one indirectly from reaction 1 (since iodide was needed in reaction 1 to produce IO-, which was needed in reaction 2 to make HOI, which was needed in reaction 3 - the rate determining step. Since it is involved twice, it also follows that b=2.

Hope it helps. It is a bit difficult to explain these types of questions by simply writing it. Do ask if you are still unsure about something.
Click to expand...

Oh my! Lovely! Wonder why I didn't see it as individual steps leading up to the slowest! Hah!
 
NaNinG

NaNinG

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AbbbbY said:
Idk I just take somethign that fits on the divisions without awkward calculations, and something that has enough spread to cover 2/3rds of the graph. So, a slight overlap from the max value.
Click to expand...
well i plotted it perfectly.....thanx dear :*
 
AbbbbY

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ahmed abdulla said:
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_w12_qp_53.pdf

Question 2b(v) whats the conclusion
Question 3d & e
Click to expand...

iv-
The decomposition will complete when the plunger stops moving outwards. You'll heat and reheat till the point comes when the plunger stops moving outwards.

v- You've to record the volume of gas collected. You've already calculated how much volume of gas a fixed mass of copper carbonate will give you. You've to simply relate the two here and state that whichever volume is equal, or close enough, to the theoretical value for the given mass (calculated in iii) shows you which equation it is following.

3d
I haven't completed the table so idk if any of the other calculated densities are 1.154. If they are, simply equate the ratio of H2SO4/H2O (or the other way round) in terms of water added to the one which has 1.154.
I mean, (40+X)/60 = Volume of water that makes 1.154 from the table/volume of H2SO4 that makes 1.154 from the table.

If there's no valuee 1.154 in the table, do this:
Calculated Density = Mass of H2O + H2SO4 / Total Volume
Equate this to 1.154


1.154 = (60+X)/32.859+(1.003X)
Solve for X. I simple plugged it into my calculator saving me the time and got X = 140.23
X - volume originally added = volume required
140.23 - 40.00 = 100.23 cm3 required.

Please check the marking scheme and let me know if this is incorrect. The answer.


Explaining my 1.003X,

from the equation, we can see that 40g of H2O = 40.12cm3. So, 1g = 1.003 cm3.
Let me know if this doesn't make sense. I'm not good at explaining. :(

3e
mass error = 0.01/100 * 100 = 0.01%
volume error =
100g of H2SO4 = 54.765cm3
0.25/54.765*100 = 0.456%
 
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