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Chemistry: Post your doubts here!

T

TheMan123

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smzimran

smzimran

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Student12 said:
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!
Click to expand...
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!
 
M

Muhammad Asif

Messages
51
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22
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18
Student12 said:
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!
Click to expand...
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

Student12

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375
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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 !?
 
T

TheMan123

Messages
103
Reaction score
48
Points
38
Student12 said:
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 !?
Click to expand...
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

Student12

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375
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205
Points
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TheMan123 said:
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
Click to expand...

How did you get the 216g from ?
 
S

smartangel

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347
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17
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28
TheMan123 said:
The Cl radical can bond to any CH3 or CH2 carbons by substituting a hydrogen. There are 5 such carbons here, however there are 3 CH3 groups bonded directly to the alpha central carbon. If the Cl radical bonds to any three of this, you get the same molecule. The 3 CH3 groups form only one type of molecule. There are another two carbons which Cl can bond to, which makes 3 Chloroalkanes in total.
Answer: C. 3 (I checked)
Sorry for the messy explanation, not very good at this, ask me to clarify if needed
EDIT: Fixed typo
Click to expand...
no its a great explanation..i got it..thanks :)
 
angelgirl:)

angelgirl:)

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question no32 m/j 2010 ppr 11...some 1 plz help me out ...withthese kind of questions ...coz i can never solve like this kind of question...
 
Student12

Student12

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TheMan123 said:
216 grams is 4NO2 + O2
N=14grams per mol
O=16 grams per mol
Click to expand...

Yea i understood that part.. but i'm still confused. Why did you take 4 for NO2? & the 2 for XO.
We don't take the values of the balancing part for the Molar mass isn't it ?
 
smzimran

smzimran

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S

smartangel

Messages
347
Reaction score
17
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28
35 In a car engine, non-metallic element X forms a pollutant oxide Y.
Further oxidation of Y to Z occurs in the atmosphere. In this further oxidation, 1 mol of Y reacts
with ½ mol of gaseous oxygen.
What can X be?
1 carbon
2 nitrogen
3 sulfur
why cant it be carbon?
 
saudha

saudha

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198
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47
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smzimran said:
AoA,
1 is redox
Br goes from -1 to 0 [oxidation]
S goes from +6 to +4 [reduction]

2 is not redox
P is same at 5
Br is same at -1

As 2 is wrong, you dont need to see 3
Definitely D is the answer but i will solve to clear any doubts!

3 is not redox
N is same at 5
Br is same at -1
Click to expand...
thxx a lott really:)
 
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