Chemistry and Physics AS paper 12 MCQS

[Mairaxo]

snowbrood

 

[Ahmedraza73]

http://papers.xtremepapers.com/CIE/... AS Level/Chemistry (9701)/9701_w12_qp_11.pdf
Can anyone solve Q:11,15,16,17,20

 

[ahmed abdulla]

thanks alot
was very helpful

 

[ahmed abdulla]

?

 

[IGCSEstudent2012]

http://papers.xtremepapers.com/CIE/Cambridge%20International%20A%20and%20AS%20Level/Chemistry%20(9701)/9701_s11_qp_13.pdf
anyone pls tell me how to do ques. 9 and 12
Jazak allah khair!

 

[Mairaxo]

http://papers.xtremepapers.com/CIE/Cambridge%20International%20A%20and%20AS%20Level/Chemistry%20(9701)/9701_s11_qp_13.pdf
anyone pls tell me how to do ques. 9 and 12
Jazak allah khair!

Q9- only bonds are forming here. One X-X bond and 6 X-H bonds. so 6*395=2370
energy to form X-X + energy to form X-H = -2775
X-X - 2370=-2775
X-X=-405
Instead of - we put a + sign because they asked for bond energy and bond energy is always +ve according to its definition. so bond energy= +405
Q12- 4Al + 3O2 --> 2Al2O3
0.02 moles of Al were used so find moles of Al2O3. its in 4:2 ratio so 0.01 moles are formed of Al2O3.
Al2O3 + 6HCl --> 2AlCl3 + 3H2O
0.01 moles of Al2O3 need 0.06 moles of HCl.
moles = V*C
0.06/2 = 0.03dm^3 = 30cm^3

 

[IGCSEstudent2012]

Q9- only bonds are forming here. One X-X bond and 6 X-H bonds. so 6*395=2370
energy to form X-X + energy to form X-H = -2775
X-X - 2370=-2775
X-X=-405
Instead of - we put a + sign because they asked for bond energy and bond energy is always +ve according to its definition. so bond energy= +405
Q12- 4Al + 3O2 --> 2Al2O3
0.02 moles of Al were used so find moles of Al2O3. its in 4:2 ratio so 0.01 moles are formed of Al2O3.
Al2O3 + 6HCl --> 2AlCl3 + 3H2O
0.01 moles of Al2O3 need 0.06 moles of HCl.
moles = V*C
0.06/2 = 0.03dm^3 = 30cm^3

thanksssss a loootttt!!

 

[cute97]

heey i hate chemistry and i dont wanna fail i am doing varient 11 any help ??
thnxx in advance

 

[IGCSEstudent2012]

heey i hate chemistry and i dont wanna fail i am doing varient 11 any help ??
thnxx in advance

try to complete the paper on time!

 

[snowbrood]

snowbrood[/q

snowbrood

seems like u are helped so cheers

 

[snowbrood]

I thought the question would be easy but couldnt solve this please help!
View attachment 28416

in HIO I has oxidation number of +1
in HIO3 I has oxidation number of +5
in I2 I has oxidation number of 0.
look the oxidation number of element in L.H.S=R.H.S
so multiply HIO by 5 to get +5 oxidation number thus balancing oxidation number in LHS and RHS.
as there are 5 iodine ions in left so there should be 5 iodine ions in right.
n=2 p=1 get it now?

 

[h4rriet]

http://papers.xtremepapers.com/CIE/... AS Level/Chemistry (9701)/9701_w12_qp_11.pdf
Can anyone solve Q:11,15,16,17,20

11. A because n is the number of bonds. In B n is the number of bonds too but not all the bonds are broken. In C n is the number of bonds but they're formed and not from separate atoms. In D n is the number of moles.
15. 2-1.32 grams of MO (metal oxide) were formed. Mole ratio of metal nitrate:metal oxide is 1:1. Cross multiple n=m/Mr for the metal nitrate and for the metal oxide.
16. Draw the molecules of NO and SO2 to see if they have lone pairs. Check the oxidation numbers of N in NO2 and S in SO3 and see by how much it increases from NO an SO2.
17. Find the number of moles of O2, then the number of moles of the metal using mole ratios (you'll have to construct a balanced equation for this). Then find the Mr using n=m/Mr.
20. Every single bond is one sigma bond & every double bond is one sigma bond and one pi bond.

 

[Ahmedraza73]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_s12_qp_12.pdf
can anyone solve:Q:7,24,37
plz anyone

 

[h4rriet]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_s12_qp_12.pdf
can anyone solve:Q:7,24,37
plz anyone

7. In the case of B: Initial moles of P, Q & R = 2, 0, 0
Change in moles of each = -x, +2x, +x
Final moles of each = 2-x, 2x, x
Divide all the final moles by 2 and the sum of them will = (2 +x/2).
Do the same for all of them. The sum of the final moles won't equal (2+x/2).
24. All the compounds have less than 8 carbons, so all of them can be obtained from octene.
37. 1 is an alcohol. Dehydration will yield propene. 2 is a polymer of propene. 3 is a halogenoalkane. Elimination will not yield propene entirely, because there'll still be a Br atom to be removed.

 

[IGCSEstudent2012]

Compound X changes the colour of warm acidified sodium dichromate(VI) from orange to green.
1mol of X reacts with 2mol of HCN in the presence of KCN.
What could X be?
A CH3CH2CH2CHO
B CH3COCH2COCH3
C H2C=CHCH2CHO
D OHCCH2CH2CHO

can someone explain me this or how to solve ques like these:/?

 

[haha101]

http://papers.xtremepapers.com/CIE/Cambridge%20International%20A%20and%20AS%20Level/Chemistry%20(9701)/9701_s07_qp_1.pdf

Q no 9 10 12 13 14 16 18 20 22 26 34 37 39 n 40 pleasee

 

[haha101]

http://papers.xtremepapers.com/CIE/Cambridge%20International%20A%20and%20AS%20Level/Chemistry%20(9701)/9701_s07_qp_1.pdf

Q no 9 10 12 13 14 16 18 20 22 26 34 37 39 n 40 pleasee

 

[hope4thebest]

in HIO I has oxidation number of +1
in HIO3 I has oxidation number of +5
in I2 I has oxidation number of 0.
look the oxidation number of element in L.H.S=R.H.S
so multiply HIO by 5 to get +5 oxidation number thus balancing oxidation number in LHS and RHS.
as there are 5 iodine ions in left so there should be 5 iodine ions in right.
n=2 p=1 get it now?

Thanks you made it easy!

 

[mariamh]

37...: the voltmeter is connected across both resistors R and Q hence it shows a reading of 4 V even if the slider position is changed....

Thanks

 

[snowbrood]

Thanks you made it easy!

my pleasure mate