Chemistry: Post your doubts here!

[snowbrood]

can u explain question 14 in more detailed pls ?

what is that troubles u.. any part of this which u cant understand well one.
Step 1
u have to first balance a equation for anhydrous nitrate which has a metal ion from group 2 .
X here is the metal from group 2 as this is anhydrous so no water vapor will be released (2XNO3)2(s) = 2XO(s) + 4N02(g )+O2(g) (where X is a metal from group 2). note that this nitrate equation applies for every group 2 metal nitrate that is anhydrous
Step 2.
u know that substance loses mass after reaction only if there is a product including a gas. in this case that is nitrate and oxygen.
step 3
question says that 5g of this nitrate loses 3.29g(this includes the total mass of both these gases) thus u come to conclude that 5g of that substance will give 1.71(5-3.29) as u know that the total mass of reactants and products is the same.
step 4
1g of nitrate will give (1.71/5)=0.342g of that oxide ok
step5
now make a equation
2X+2(16)/2X+4(14)+12(16)=0.342.
now plug in X calcium magnesium and the other two compounds in the equation and see which satisfies the equation.
if i put calcium 2(40)+2(16)/2(40)+4(14)+12(16) u will see it will equal to 0.342 which means calcium is the right option.
if u put magnesium 2(24)+2(16)/2(24)+4(14)+12(16) this will equal to 0.27 see

 

[snowbrood]

http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_w06_qp_1.pdf

Can anyone explain me question 32? an 24 (Oxygen is a flammable gas I thought :/)
and Q14?

lol its so funny i am ur junior and havent studied anything except organic sec in AS Level yet i am able to answer

 

[snowbrood]

http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_w06_qp_1.pdf

Can anyone explain me question 32? an 24 (Oxygen is a flammable gas I thought :/)
and Q14?

i just figured out question 32 well 1moldm^-3 sulphuric acid is dilute dilute sulphuric acid u will consider an ion as high if there are more moles of that ions formed by the dilute acid than the concentrated one . u see from this equation that 2 moles of H+ ions are formed which are greater than in the concentrated acid where only one is formed. no HS04 ion is formed and the number of moles of SO4 ion is same in both the reactions
H2SO4(l) + aq ==> 2H+(aq) + SO42-(aq)

 

[izzahzainab]

lol its so funny i am ur junior and havent studied anything except organic sec in AS Level yet i am able to answer

lol, thanks buddy! Actually, I got numb and then had to post these questions here was too fed up to use my mind on chemistry

 

[iKhaled]

what is that troubles u.. any part of this which u cant understand well one.
Step 1
u have to first balance a equation for anhydrous nitrate which has a metal ion from group 2 .
X here is the metal from group 2 as this is anhydrous so no water vapor will be released (2XNO3)2(s) = 2XO(s) + 4N02(g )+O2(g) (where X is a metal from group 2). note that this nitrate equation applies for every group 2 metal nitrate that is anhydrous
Step 2.
u know that substance loses mass after reaction only if there is a product including a gas. in this case that is nitrate and oxygen.
step 3
question says that 5g of this nitrate loses 3.29g(this includes the total mass of both these gases) thus u come to conclude that 5g of that substance will give 1.71(5-3.29) as u know that the total mass of reactants and products is the same.
step 4
1g of nitrate will give (1.71/5)=0.342g of that oxide ok
step5
now make a equation
2X+2(16)/2X+4(14)+12(16)=0.342.
now plug in X calcium magnesium and the other two compounds in the equation and see which satisfies the equation.
if i put calcium 2(40)+2(16)/2(40)+4(14)+12(16) u will see it will equal to 0.342 which means calcium is the right option.
if u put magnesium 2(24)+2(16)/2(24)+4(14)+12(16) this will equal to 0.27 see

thanks a lot man..makes more since now!

 

[Beaconite007]

http://www.xtremepapers.com/papers/...d AS Level/Chemistry (9701)/9701_w05_qp_1.pdf
Q. 16 24 38 PLZ HELP

 

[allysaleemally]

Hey guys, how can we find whether a compound has a chiral atom or how many chiral atom are present in the compound. I really cannot understand it.
Q-21: http://www.xtremepapers.com/papers/...d AS Level/Chemistry (9701)/9701_w04_qp_1.pdf

Q-20: http://www.xtremepapers.com/papers/...d AS Level/Chemistry (9701)/9701_w08_qp_1.pdf

Thanks

 

[KurayamiKimmi]

can anyone pls tell me how to un-log a value ?
we know that ph=log[H+]
how do we find concentration of H+ if we know ph

 

[iKhaled]

http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_w05_qp_1.pdf
Q. 16 24 38 PLZ HELP

ok here the the explanation of the answers,

for question 16, in the question it is stated that the compound is decomposed and the solid left is reacted with hydrochloric acid and formed carbon dioxide(Vigorous effervescence).. if we took the first option CaCO3 and we decomposed it calcium oxide and carbon dioxide are formed. now CaO is the solid remained and if we add that to hydrochloric acid a salt and water r formed so no vigorous effervescence..so it cant be A

if we take option C.. you know that a higher temperature is needed to decompose the compound going down group 2 so we have the compound BaCO3.CaCO3 and when it is decomposed ONLY CaCO3 will be decomposed to an oxide and carbon dioxide and Barium carbonate will be left cuz we need a higher energy to decompose it now barium carbonate is the solid left behind and if we took it and added it to hydrochloric acid we will have a vigorous reaction and carbon dioxide will be formed so it must be option C!
here is the equation:

BaCO3.CaCO3 ---> CaO + CO2 + BaCO3(solid left)
BaCO3 + HCl ---> BaCl2 + CO2(effervescence) + H2O

if we take option D..both of the carbonates we have will decompose because they r under each other in the group so definitely its not that option..this option is tricky

i hope you get it now!

for question 24, here is a full explanation of it:
http://www.chemguide.co.uk/mechanisms/elim/dhcomplex.html

for question 38, you can see from the reaction it involved substitution.. it is just like the substitution of halogenoalkane with aqueous NaOH but uk cie examiners they wanna confuse us !

(CH3)3SiCl + C2H5O– → (CH3)3SiOC2H5 + Cl –

you can see from the equation that C2H50- acts as a nucleophilic cuz it donates electrons and it attacks the Si-Cl bond and to form an Si-O-R bond and Cl- so obviously its A!!

if you didn't get somethin from what i explained tell me and i shall explain more i hope i helped!

 

[Shuayb]

Could someone show me how to answer the following question please:
"Apart from peaks associated with solitary nitrogen atoms(at m/e=14) and chlorine atoms(at m/e=35 & m/e=37), the mass spectrum of nitrogen trichloride contains 9 peaks arranged in 3 groups, ranging from m/e=49 to m/e=125. Predict the values of all 9 peaks , and suggest formula for the species responsible for each other one."

 

[iKhaled]

Hey guys, how can we find whether a compound has a chiral atom or how many chiral atom are present in the compound. I really cannot understand it.
Q-21: http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_w04_qp_1.pdf

Q-20: http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_w08_qp_1.pdf

Thanks

for question 21, the answer can be easily found if u understand the definition of a chiral atom.. here in this question the carbon atom should be attached to four other groups so that it can be called a chiral carbon. in A it is chiral cuz u can see the carbon is atom attached to four different groups and same thing goes to B..the trick is in option C and D..see in C the carbon atom is attached to a hydroxyl group and methyl group and 2 different phenols ( see the OH on the carbon in benzene is different ) so it is chiral also but in D it is attached to hydroxyl group and an ester and 2 phenols groups which r the same so it is not chiral because the carbon atom is not attached to 4 different groups!

 

[Shuayb]

can anyone pls tell me how to un-log a value ?
we know that ph=log[H+]
how do we find concentration of H+ if we know ph

Take 10 to the power of the number in the log function.
E.g: log(70)=1.84509804...
10^( 1.84509804...)=70

 

[Samaani]

http://www.xtremepapers.com/papers/... AS Level/Chemistry (9701)/9701_s10_qp_12.pdf

Please explain
Q33-corret answer is D
Q34-correct answer is B
Q35- the correct answer is C . But does SO gets oxidised in air?? and does CO not?

 

[littlecloud11]

http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_s10_qp_12.pdf

Please explain
Q33-corret answer is D
Q34-correct answer is B
Q35- the correct answer is C . But does SO gets oxidised in air?? and does CO not?

Q33) only reaction one is a redox reaction because sulfuric acid is an oxidizing agent and is strong enough to oxidize HBr to form Br2. Alternatively, you can see that only in reaction one there is a change in oxidation state (sulfur from -6 to -4).
In reaction two Phosphoric acid is not an oxidizing agent, there is no change in oxidation state for any element. It is only a reaction to produce hydrogen halide from a halide ion. Reaction 3 is a precipitation reaction.So, D is the correct answer.

q34) Both CaCo3 and CaOH can react with carboxylic acids to neutralize it's acidic properties. The former forms a salt, water and CO2 while the latter forms salt and water. KNO3 does not react with carboxylic acid.

q35) CO is not oxidized by air at room temp. And SO2 is formed initially not SO, both SO2 and NO can be oxidized by air to form SO3 and NO2 respectively (what happens during acid rain)

 

[Samaani]

Q33) only reaction one is a redox reaction because sulfuric acid is an oxidizing agent and is strong enough to oxidize HBr to form Br2. Alternatively, you can see that only in reaction one there is a change in oxidation state (sulfur from -6 to -4).
In reaction two Phosphoric acid is not an oxidizing agent, there is no change in oxidation state for any element. It is only a reaction to produce hydrogen halide from a halide ion. Reaction 3 is a precipitation reaction.So, D is the correct answer.

q34) Both CaCo3 and CaOH can react with carboxylic acids to neutralize it's acidic properties. The former forms a salt, water and CO2 while the latter forms salt and water. KNO3 does not react with carboxylic acid.

q35) CO is not oxidized by air at room temp. And SO2 is formed initially not SO3, both SO2 and NO can be oxidized by air to form SO3 and NO2 respectively (what happens during acid rain)

I guess you meant SO to SO2 in the air thanks

 

[littlecloud11]

I guess you meant SO to SO2 in the air thanks

No actually. SO2 is formed as a result of sulfur combusting in the internal combustion engine. This is oxidized to SO3 in the atmosphere.

 

[trgirl]

Can u help me to solve this question? It is from 2005 q15 on i attached the file here

 

[MrAnon]

Help needed:
http://www.xtremepapers.com/papers/...d AS Level/Chemistry (9701)/9701_w06_qp_1.pdf

Ques 1, 4, 29

 

[autumnsakura]

http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_w03_qp_1.pdf
I don't understand question 28 (answer B) and 40(answer D)
For q40, why can't the second choice be correct as well?

 

[autumnsakura]

Help needed:
http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_w06_qp_1.pdf

Ques 1, 4, 29

For 1, Ca ions are replaced with 2 H ions so H2SO4 is formed. Now it reacts with KOH. But 1 mol of H2SO4 reacts with 2 mol of KOH.
no of mol of KOH=2.5 x 10-4
so no of mol of H2SO4 is half of that= 1.25x10-4
Using m=no of mol x volume, you'll get the answer for concentration of CaSO4 which is A

EDITED:
Question 4,
CH4 + 2O2 = CO2+ 2H2O
10 20 10 <--- volume of gases used and produced.

So 70 - 20 + 10= 60cm3 10cm3 of alkane is not included in the calculation.

Similarly, C3H6 + 7/2 O2 = 2CO2 + 3H20
10 35 20
So, 70-35+20 = 55 cm3
If you do the other 2 you'll find the answer to be D

For Question 29, It must first react with HCN first to form a C with OH and CN functional groups. When H2SO4 is added, the original CN it has will become carboxylic acid. The new CN group it formed from the 1st step will also form carboxylic acid.