Chemistry: Post your doubts here!

[PhyZac]

Nov 2004 MCQs CIE Q7
please help

This must help you, if not, then tell me.
http://www.xtremepapers.com/revision/a-level/chemistry/atoms/bonding/shapes.php

 

[Soldier313]

Aoa wr wb......

Can someone pls explain where the 0.05 and 0.075 in the ms come from??

Thanx a lot

qn:



ms:

 

[PhyZac]

Aoa wr wb......

Can someone pls explain where the 0.05 and 0.075 in the ms come from??

Thanx a lot

qn:

View attachment 22383

ms:

View attachment 22384

See, when you add 10mol/dm3 of NaOH to 10mol/dm3 of the acid...the concentration changes! because the volume change...so now NaOH has 10 mol/2 dm3...since we doubled the volume..thus 0.1/2 is 0.05 concntration!

but why 0.075 ...i am not getting too!

 

[Soldier313]

See, when you add 10mol/dm3 of NaOH to 10mol/dm3 of the acid...the concentration changes! because the volume change...so now NaOH has 10 mol/2 dm3...since we doubled the volume..thus 0.1/2 is 0.05 concntration!

but why 0.075 ...i am not getting too!

thanx bro....
but......look, when you're saying it's because of doubling the volume that the conc of NaOH changes, the ms is saying that 0.05 is the conc of the salt, not of the base (NaOH), (or is it somehow that the conc of the salt and the base is always the same, or something like that?)
that's the bit that i don't get, and the '0.075'.....i too still can't figure out how they get that
Perhaps i'm not making a lot of sense, but well, i'm so confused

 

[PhyZac]

thanx bro....
but......look, when you're saying it's because of doubling the volume that the conc of NaOH changes, the ms is saying that 0.05 is the conc of the salt, not of the base (NaOH), (or is it somehow that the conc of the salt and the base is always the same, or something like that?)
that's the bit that i don't get, and the '0.075'.....i too still can't figure out how they get that
Perhaps i'm not making a lot of sense, but well, i'm so confused

Hmm, good question...!
actually it depend on the equation. and since 1 mole of NaOH gives 1 mole of that salt...so it will be same..!
actually i am not getting the question much, so i am confused bout 'o.o75', i will think of it next morning In Sha Allah...

 

[Soldier313]

Hmm, good question...!
actually it depend on the equation. and since 1 mole of NaOH gives 1 mole of that salt...so it will be same..!
actually i am not getting the question much, so i am confused bout 'o.o75', i will think of it next morning In Sha Allah...

oww okayy thanx, btw i have a test tomorrow morning, that's why i'm desperate :/ I hope i figure out the soln before then inshaAllah.
But thanx a million for your help

 

[PhyZac]

oww okayy thanx, btw i have a test tomorrow morning, that's why i'm desperate :/ I hope i figure out the soln before then inshaAllah.
But thanx a million for your help

Okay, but can you tell which year this question is in..

 

[Soldier313]

Okay, but can you tell which year this question is in..

yes sure, it's on/11, paper 43.

 

[PhyZac]

yes sure, it's on/11, paper 43.

I got it Alhamdulilah...!!
See...the intial acid concentration was 0.250 after adding NaOH it became half...that is 0.125mol/dm3 and then since 0.05 of salt is formed then we need to take 0.05 from salt.....and 0.075 is left ! did u get?

 

[Soldier313]

I got it Alhamdulilah...!!
See...the intial acid concentration was 0.250 after adding NaOH it became half...that is 0.125mol/dm3 and then since 0.05 of salt is formed then we need to take 0.05 from salt.....and 0.075 is left ! did u get?

wooow thanx bro! JazakAllah khair! I understood

PS: the part in red, you meant 0.05 from acid, nah?

 

[PhyZac]

wooow thanx bro! JazakAllah khair! I understood

PS: the part in red, you meant 0.05 from acid, nah?

yea sorry, i meant acid...! Good luck with your test tomorrow, May Allah grant you success Ameen! Wa eyyakum

 

[Soldier313]

yea sorry, i meant acid...! Good luck with your test tomorrow, May Allah grant you success Ameen! Wa eyyakum

Thank you......InshaAllah, Aameen

 

[snowbrood]

why cant we measure the enthalpy change of hydration of anhydrous copper sulfate to copper sulphate crystals other than hess's law

 

[PhyZac]

why cant we measure the enthalpy change of hydration of anhydrous copper sulfate to copper sulphate crystals other than hess's law

The hydration enthalpy is the enthalpy change when 1 mole of gaseous ions dissolve in sufficient water to give an infinitely dilute solution. Hydration enthalpies are always negative.
We can't start with gaseous ion!

 

[Raiyan3]

Please explain question number 9 o/n 10
http://papers.xtremepapers.com/CIE/... AS Level/Chemistry (9701)/9701_w10_qp_11.pdf
and question number 4,8, 10 from this one:
http://papers.xtremepapers.com/CIE/... AS Level/Chemistry (9701)/9701_w10_qp_12.pdf

 

[Umar Farooq]

Anyone know how to do this please?
View attachment 21434
I know the answer has to be either A or C since it's trans, but I have no clue how to calculate the number of double bonds... I thought it would be two as there's a C=C double bond in a cyclohexene ring and the other is given in the question between Carbons 11 and 12 (aldehyde doesn't count as there's no C=C bond). So how would you arrive at a total of 5?

Hey you should read Chemistry AS Muhammad Ayub for this I Dont remember this but its written in that book so that u could understand

 

[daredevil]

Please explain question number 9 o/n 10
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_w10_qp_11.pdf
and question number 4,8, 10 from this one:
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_w10_qp_12.pdf

for q9: find out the total energy released when 1.60 g of fuel is burnt through the enthlpy method then calculate the thermal energy absorbed by water wit Q=mc(delta thita)
sutbtract the amount of energy absorbed by the water from the total energy released and u'll get the thermal energy released in the expreiment. then from there calculate the amount of enrgy per gram of fuel burnt.
i'm sorry i don't have a calculator at hand so i just gave u the steps.... annyways u'll understtand better if u go through with the questions and work it out following the steps urself hope it helped

 

[daredevil]

Please explain question number 9 o/n 10
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_w10_qp_11.pdf
and question number 4,8, 10 from this one:
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_w10_qp_12.pdf

q 4 :

calculate the total energy absorbed in bond breaking of CO and 2H2 :
1077 + 2(436) = 1949

calculate the total energy released in bond making of CH3OH:
3(410) + 360 + 460 = 2050

subtract the energy released from the energy absorbed in the reaction to calculate the enthaply change:
1949-2050 = -101kJ/mol

reply if u don't get anything and i'll xplain... and tell me if its the correct answer

 

[Raiyan3]

for q9: find out the total energy released when 1.60 g of fuel is burnt through the enthlpy method then calculate the thermal energy absorbed by water wit Q=mc(delta thita)
sutbtract the amount of energy absorbed by the water from the total energy released and u'll get the thermal energy released in the expreiment. then from there calculate the amount of enrgy per gram of fuel burnt.
i'm sorry i don't have a calculator at hand so i just gave u the steps.... annyways u'll understtand better if u go through with the questions and work it out following the steps urself hope it helped

Im still struggling a bit :/ i didn't get the "find out the total energy released when 1.60 g of fuel is burnt through the enthlpy method", if you please give a little more explanation i will get the hang of it im sure.

 

[Raiyan3]

q 4 :

calculate the total energy absorbed in bond breaking of CO and 2H2 :
1077 + 2(436) = 1949

calculate the total energy released in bond making of CH3OH:
3(410) + 360 + 460 = 2050

subtract the energy released from the energy absorbed in the reaction to calculate the enthaply change:
1949-2050 = -101kJ/mol

reply if u don't get anything and i'll xplain... and tell me if its the correct answer

And it is the right answerrrr. Thankyou thankyou veryy muchh i understood