Chemistry: Post your doubts here!

Thought blocker

Thought blocker

XPRS Moderator
Messages
8,545
Reaction score
34,867
Points
718
Zepudee said:
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_w08_qp_4.pdf

Hey guys, how to do Q2(b)?? I dont understand..
Click to expand...
2b)
C40H82 -------> C16H34 + C24H48

a good way of explaining this question is to see how many bonds u have before the reaction and how many bonds u have after the reaction for example...

before the reaction we have 39 C-C bonds and 82 C-H bonds...after the reaction we have 15+22 = 37 C-C bonds and one C=C bond and 48+34 = 82 C-H bonds

so we have broken 39-37 = 2 C-C bonds and formed 1 C=C bond
bond broken = 2 x 350 = +700
bond formed = -610 = -610
enthalpy = + 90 kjmol^-1 !!

thats it..the mark scheme got the answer as +180 because he has used a different equation in cracking instead of forming C24H48 as our alkene he formed 2 moles of C12H24 and in that way we will have 2 C=C bonds instead of one and 4 C-C bonds instead of 2 because one mole of C12H24 contains 1 C=C bond and 10 C-C bonds so the 2 moles will contain 20 c-c bond and 2 C=C bonds 20+15= 35 so 39-35 = 4 C-C bonds so we had broken 4 C-C bonds and formed 2 C=C bond..in that way u will get 180 kj per mole instead of 90..both are correct though depending on the equation u made ;)
 
Kim Shaw

Kim Shaw

Messages
23
Reaction score
15
Points
3
HELP PLEASE IONIC EQUILIBRIA QUESTION:
Lake water contains dissolved sodium carbonate and sodium hydrogen carbonate. The following equilibrium exists:

HCO3- (rev sign) H+ + CO32-

[CO32-]/[HCO3-] = 0.958

When 10cm3 of lake water were titrated with 0.2mol/dm3 HCl, 22cm3 of acid were required to neutralise all the carbonate and hydrogen carbonate ions according to the following equations:

H+ + HCO3- (forward sign) H2O +CO2

2H+ + CO32- (forward sign) H2O +CO2

Calculate the total number of moles of acid used, and thus, by using the ratio quoted, calculate [CO32-] and [HCO3-] in the lake.
 
Metanoia

Metanoia

Messages
603
Reaction score
1,102
Points
153
ShreeyaBeatz said:
Guys help me with q.no 2 and 15!
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_s08_qp_1.pdf
Click to expand...
s08qp1

Just use proportion if its easier to visualise.

mass of N in 100 g of fertiliser = 15 g

mass of N in 14 g of fertiliser = (15/100) x 14= 2.1 g

moles of N in 14 g of fertiliser = 2.1 /14 = 0.15 mol

conc of N in 5 dm^3 of solution = 0.15/5 = 0.03 mol/dm^3

Q15

CaCO3 --> CaO + CO2

start with mass of CaCO3 = 1 200 million tonnes
 
Metanoia

Metanoia

Messages
603
Reaction score
1,102
Points
153
Kim Shaw said:
HELP PLEASE IONIC EQUILIBRIA QUESTION:
Lake water contains dissolved sodium carbonate and sodium hydrogen carbonate. The following equilibrium exists:

HCO3- (rev sign) H+ + CO32-

[CO32-]/[HCO3-] = 0.958

When 10cm3 of lake water were titrated with 0.2mol/dm3 HCl, 22cm3 of acid were required to neutralise all the carbonate and hydrogen carbonate ions according to the following equations:

H+ + HCO3- (forward sign) H2O +CO2

2H+ + CO32- (forward sign) H2O +CO2

Calculate the total number of moles of acid used, and thus, by using the ratio quoted, calculate [CO32-] and [HCO3-] in the lake.
Click to expand...

From [CO32-]/[HCO3-] = 0.958, we have [CO32-] = 0.958[HCO3-]

moles of H+ used = 0.2 x 22/1000 = 0.0044 mol

moles of H+ used
= 1 x moles of HCO3- + 2 x moles of CO32-
= 1 x moles of HCO3- + 2 x 0.985 moles of HCO32-
= 2.97 moles of HCO3-
=0.0044 mol

moles of HCO3- = 0.0044/2.97 = 0.0014815 moles
[HCO3-] = 0.0014815/0.010 =0.148 mol/dm3

[CO32-] =0.14815 x 0.985 = 0.146 mol/dm3
 
You must log in or register to reply here.
Top