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Chemistry: Post your doubts here!

Mairaxo

Mairaxo

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hela said:
PLEASE HELP
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_s08_qp_1.pdf Q6 C Q7 D Q26 C
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_w08_qp_1.pdf Q 26 C
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_w10_qp_12.pdf Q 8 D Q25 D
Click to expand...

Oct 2008 - Q26 - We need to form the reagents C6H5CH2COOH and C2H5OH. In all options the reagents form C6H5CH2OH xept in c
 
Mairaxo

Mairaxo

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hela said:
PLEASE HELP
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_s08_qp_1.pdf Q6 C Q7 D Q26 C
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_w08_qp_1.pdf Q 26 C
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_w10_qp_12.pdf Q 8 D Q25 D
Click to expand...
Oct 2010 - Q8 - Ca(S) --> Ca(g) is +177
then add ionisation energies 1 and 2 wich are 590 and 1150
then Ca+2(g) --> Ca+2(aq) is -1565
so 117+590+1150-1565 = 352

Q25- a is cracking wich needs 500 degree. b needs reflux in ethanol so high temp needed. c also needs heating. only d doesnt need to be heated.
 
Mairaxo

Mairaxo

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iKhaled said:
can someone tell me how to solve these type of questions ? question 4 oct/nov 2009 paper 12

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_w09_qp_12.pdf
Click to expand...
a and d have no dipole. CO2 is linear shaped and so both oxygen atoms cancel out. the same goes for option a. all chlorine molecules cancel eachothers polarity. in b only oxygen is electronegative and it gets -ve charge. in c also there is a dipole but weaker as the chlorine atoms form a -ve charge in the opposite direction so slightly reduce the overall dipole. Hence its b. Just think of it as forces acting at a point. If they act in the same direction the overal force is large. If they act in opposite directions then its less :)
 
Rahma Abdelrahman

Rahma Abdelrahman

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zackle09 said:
can someone pls tell me how to do this?
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_w09_qp_11.pdf

question 2
Click to expand...

You have to try with each one, write the BALANCED chemical equation, then check whether the mole ratios found give 0.2 mol of the hydrocarbon.
First find moles of CO2 and of H2O… à 35.2/44 =0.8 à Carbondioxide
14.4/18 =0.8à Water
Then,
C2H4 + 3O2 à 2CO2 + 2H2O , ratio of the hydrocarbon to CO2 is 1:2 , so this will give 0.4 mol of the hydrocarbon .

But if u try with C4H8 You will get a ratio of 1:4 which will give 0.2 mol of hydrocarbon
 
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zackle09

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Rahma Abdelrahman said:
You have to try with each one, write the BALANCED chemical equation, then check whether the mole ratios found give 0.2 mol of the hydrocarbon.
First find moles of CO2 and of H2O… à 35.2/44 =0.8 à Carbondioxide
14.4/18 =0.8à Water
Then,
C2H4 + 3O2 à 2CO2 + 2H2O , ratio of the hydrocarbon to CO2 is 1:2 , so this will give 0.4 mol of the hydrocarbon .

But if u try with C4H8 You will get a ratio of 1:4 which will give 0.2 mol of hydrocarbon
Click to expand...

ohh i get it thanks alot! :)
 
Mairaxo

Mairaxo

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soso247 said:
can you help me with mine too please , thank you in advance
Click to expand...
Q4- 1st write the combustion equations of all the alkanes. Then calculate how much volume of O2 is needed and how much is remaining. then add the volume of CO2 produced. For example CH4+2O2-->CO2+2H2O
10cm3 of alkane was used so we take that as 1 mole. 2 moles of O2 were needed so 20cm3 of O2 is used up. 70-20=50cm3 remaining. Then 1 mole of CO2 is produced which means 10cm3 so 50+10=60cm3
do it for the other alkanes :)
 
Mairaxo

Mairaxo

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soso247 said:
can you help me with mine too please , thank you in advance
Click to expand...
Q14- Use the Ar of each elements. Then calculate the mole ratio. For example- Mg(NO3)2 is 5g and 3.29 is lost meaning MgO is 1.71g. moles of Mg(NO3)2= 5/148=0.034 moles. then moles of MgO= 1.71/40=0.043moles. mole ratio should be 1:1 so its not magnesium. try it with calcium and u get it :)
Q25- Just substitute n with any integer and youll get the answer. For example take n=1. then u get C3H6 + 4.5O2--> 3CO2 + 3 H2O so it gets balanced :)
Q37- Its not 2 because 2 will form only one ketone and the other will be an acid. D is the only option with no 2 so chose D :D Also 3 is not possible as both the componds are isomers :)
 
Mairaxo

Mairaxo

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ahmed abdulla said:
>? any one
Click to expand...
i just answered the first one right now : Q4- 1st write the combustion equations of all the alkanes. Then calculate how much volume of O2 is needed and how much is remaining. then add the volume of CO2 produced. For example CH4+2O2-->CO2+2H2O
10cm3 of alkane was used so we take that as 1 mole. 2 moles of O2 were needed so 20cm3 of O2 is used up. 70-20=50cm3 remaining. Then 1 mole of CO2 is produced which means 10cm3 so 50+10=60cm3
do it for the other alkanes :)
 
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