Chemistry: Post your doubts here!

[nerdybookworm]

A2 or AS, multi choice or structured paper

mcq

 

[Mustehssun Iqbal]

Assalamu alaikum,
A friend of mine asked this question. I couldn't solve it
pls explain:
http://www.xtremepapers.com/papers/... AS Level/Chemistry (9701)/9701_w09_qp_12.pdf
Answer to the question is D.

 

[étudiante]

Assalamu alaikum,
A friend of mine asked this question. I couldn't solve it
pls explain:
http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_w09_qp_12.pdf
Answer to the question is D.

which question?

 

[Mustehssun Iqbal]

which question?

Apologies. Q.6...

 

[angelgirl:)]

please help me with Q20 with explanation.

it says that there is an aldehyde group and cyclo hexene so C6H10 + CHO = C7H11O
The real moleculer formula is C20H28O...C20 -C 7= C13....
Now,there iz a long aliphatic chain...general formula is CnH2n+2...so C13H28...
for each double bond formation a hydrogen molecule is removed...so if there iz 5 double bond means ...5*2=10 H is removed...
SO OUT OF 28 H in C13H28...28-10=18 is remained...but 1 more H is removed due to attachment...

 

[samwickz]

any1 doing chem ppr 12 2morrow? cuz i am & freaking out down here

 

[bumble26]

View attachment 8865
it says that there is an aldehyde group and cyclo hexene so C6H10 + CHO = C7H11O
The real moleculer formula is C20H28O...C20 -C 7= C13....
Now,there iz a long aliphatic chain...general formula is CnH2n+2...so C13H28...
for each double bond formation a hydrogen molecule is removed...so if there iz 5 double bond means ...5*2=10 H is removed...
SO OUT OF 28 H in C13H28...28-10=18 is remained...but 1 more H is removed due to attachment...

Thanks so much!

 

[étudiante]

Apologies. Q.6...

no problem

calculate the oxidation nos of Cr in all the equations
A) +6 ----> +3
B) +6 ---->+6
C) +6 ----> +6
D) +6 ----> +3

Answer is D because the no of O (per Cr atom) is also reducing along with the ON.. this was the explanation i was given by someone... I'd like it if u guys could confirm please

 

[nerdybookworm]

link please....

http://www.xtremepapers.com/papers/CIE/Cambridge IGCSE/Chemistry (0620)/0620_s03_qp_1.pdf

 

[Student12]

Use of the Data Booklet is relevant to this question.
Lead(IV) chloride will oxidise bromide ions to bromine. The Pb4+ ions are reduced to Pb2+ ions in
this reaction.
If 6.980 g of lead(IV) chloride is added to an excess of sodium bromide solution, what mass of
bromine would be produced?
A 0.799 g B 1.598 g C 3.196 g D 6.392 g

The answer is C ?? explain someone!

 

[Muhammad Asif]

Apologies. Q.6...

the ion which is reduced gain elecrons.....if we compare chromium then:
oxidation state before after
A +6 +6
B +6 +6
C +6 +6
D +6 +3
thus only in D chromium gains electrons to reuce from +6 to +3

any1 doing chem ppr 12 2morrow? cuz i am & freaking out down here

 

[smzimran]

http://www.xtremepapers.com/papers/CIE/Cambridge IGCSE/Chemistry (0620)/0620_s03_qp_1.pdf

This is CIE
not IGCSE section
dear bro

 

[TheMan123]

http://www.xtremepapers.com/papers/CIE/Cambridge IGCSE/Chemistry (0620)/0620_s03_qp_1.pdf

A level questions only!!!
Anyway the answers are
1. C, because cooling a material causes it to lose heat energy, so energy is lost in the cooling tube
2. B, because during change of state there is no temp change, so its B or D, but melting is at a lower temp than boiling, so B

 

[smzimran]

Use of the Data Booklet is relevant to this question.
Lead(IV) chloride will oxidise bromide ions to bromine. The Pb4+ ions are reduced to Pb2+ ions in
this reaction.
If 6.980 g of lead(IV) chloride is added to an excess of sodium bromide solution, what mass of
bromine would be produced?
A 0.799 g B 1.598 g C 3.196 g D 6.392 g

The answer is C ?? explain someone!

AoA,
The key to solving this question is forming the equation
PbCl4 + 2NaBr ------> PbCl2 + Br2 + 2NaCl

Mr(lead chloride) = 207 + 4(35.5) = 349
moles of lead chloride = 6.980 / 349 = 0.02 mol

Since molar ratio b/w lead chloride and Br2 is 1:1
moles of Br2 = 0.02
mass of Br2
= 0.02 * Mr
= 0.02 * (2 * 79.9)
= 3.196

C is the answer!

 

[Muhammad Asif]

Use of the Data Booklet is relevant to this question.
Lead(IV) chloride will oxidise bromide ions to bromine. The Pb4+ ions are reduced to Pb2+ ions in
this reaction.
If 6.980 g of lead(IV) chloride is added to an excess of sodium bromide solution, what mass of
bromine would be produced?
A 0.799 g B 1.598 g C 3.196 g D 6.392 g

The answer is C ?? explain someone!

equation:- PbCl4+2NaBr----->Br2+ other stuff
mol of PbCl4=mol of Br2
6.980 mass
------------------ = -----------------
(207+(35.5*4) 79.9 *2
furthur solve it you will get the answer 3.196

 

[Student12]

Use of the Data Booklet is relevant to this question.
A 5.00 g sample of an anhydrous Group II metal nitrate loses 3.29 g in mass when heated
strongly.
Which metal is present?
A magnesium
B calcium
C strontium
D barium

Ans is B ? There are many such Q's in the past papers. i dont know how to do it !?

 

[TheMan123]

Use of the Data Booklet is relevant to this question.
A 5.00 g sample of an anhydrous Group II metal nitrate loses 3.29 g in mass when heated
strongly.
Which metal is present?
A magnesium
B calcium
C strontium
D barium

Ans is B ? There are many such Q's in the past papers. i dont know how to do it !?

let metal be X

2X(NO3)2 --> 2XO + 4NO2 + O2
5g ............ --> 1.71g +[ 3.29g ]

Mass of 1 mole of (4no2+ o2)=216grams
There are 3.29/216 moles of (4no2+o2)present
Also 3.29/216 moles of 2XO present
Therefore there is (2*3.29)/216 moles of XO
molar mass of XO = g/mol= 1.71/[(2*3.29)/216]=56.1grams
X=56.1-16=40.1 grams permole
Therefore answer is:

B.Calcium

 

[Student12]

http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_w11_qp_13.pdf

QUESTION - ANSWER
15 - B
23 - A
25 - C
36 - A

Someone plz explain ?

 

[Student12]

let metal be X

2X(NO3)2 --> 2XO + 4NO2 + O2
5g ............ --> 1.71g +[ 3.29g ]

Mass of 1 mole of (4no2+ o2)=216grams
There are 3.29/216 moles of (4no2+o2)present
Also 3.29/216 moles of 2XO present
Therefore there is (2*3.29)/216 moles of XO
molar mass of XO = g/mol= 1.71/[(2*3.29)/216]=56.1grams
X=56.1-16=40.1 grams permole
Therefore answer is:

B.Calcium

How did you get the 216g from ?

 

[TheMan123]

How did you get the 216g from ?

216 grams is 4NO2 + O2
N=14grams per mol
O=16 grams per mol