Chemistry MCQ thread...

[Vanto1994]

Someone please suggest an explanation for this question ! Would appreciate it !

 

[kewlryan58]

hey can ani 1 help me with mcq 16 ov nov 05...??? i will be thankful to u... :%)

 

[Mobeen]

@shootingstar .
C

 

[abdullah181994]

Someone please suggest an explanation for this question ! Would appreciate it !

answer is D because 1 mole of CS2 reacts with 3 moles of O2 so 30 cm3 of O2 would react and 20 cm3 would remian+20cm3 SO2+10cm3 of CO2=50CM3 .
NaOH will absorb 10 CM3 of CO2 so remianing will be 40 CM3.

 

[abdullah181994]

not sure about SO2 reacting with Naoh @ Vanto1994

 

[Mobeen]

Someone please suggest an explanation for this question ! Would appreciate it !

answer is D because 1 mole of CS2 reacts with 3 moles of O2 so 30 cm3 of O2 would react and 20 cm3 would remian+20cm3 SO2+10cm3 of CO2=50CM3 .
NaOH will absorb 10 CM3 of CO2 so remianing will be 40 CM3.

NaOH will dissolve both CO2 and SO2 .. so the final would be 20cm3

 

[abdullah181994]

please some one explain Q4 and 8of o/N 06 P1
http://www.xtremepapers.me/CIE/Internat ... 6_qp_1.pdf
PLZ PLZ PLZ PLZ PLZ PLZ PLZ ANS ANS ANS ANS IT

 

[abdullah181994]

hey can ani 1 help me with mcq 16 ov nov 05...??? i will be thankful to u... :%)

C IS ANSWER
FOR FIRST ONE 3(1+N-2)=0 N=+1
FOR SECOND ONE 2(1+N)=0 N=-1
FOR THIRD ONE N=-6-1=-7

 

[abdullah181994]

PLEASE SOME ANSWER MY PROBLEMS ALSO WRITTEN ABOVE PLEEEEEEEEEEEEEEEEEEEEEEEEZZZZZZZZZZZZZ

 

[abdullah181994]

@ EVERYONE U PLZ ANSWR MY QUESTIONS

 

[ShootingStar]

Need help with the following question

C is the answer
apply this CxHy + x+y/4...........>xCO2 +y/2 H2O
30/10=3 Carbon atoms
50/10=5 moles of O2 so apply x+y/4=5 x=3 so from here we get y=8 which is number of hydrogen so C3H8

Thanks I get it.

 

[ideggkr]

please some one explain Q4 and 8of o/N 06 P1
http://www.xtremepapers.me/CIE/Internat ... 6_qp_1.pdf
PLZ PLZ PLZ PLZ PLZ PLZ PLZ ANS ANS ANS ANS IT

Wow.. I have NO IDEA after looking at it for 20 minutes.. damn

 

[Mobeen]

please some one explain Q4 and 8of o/N 06 P1
http://www.xtremepapers.me/CIE/Internat ... 6_qp_1.pdf
PLZ PLZ PLZ PLZ PLZ PLZ PLZ ANS ANS ANS ANS IT

Wow.. I have NO IDEA after looking at it for 20 minutes.. damn

its not that hard .. for q8 answer is D because for the enthalpy change of formation the product must be formed from its elements in their standard states .. in other options the reactants have compounds .. but only D has all its reactants as elements in their standard states ..

 

[ideggkr]

please some one explain Q4 and 8of o/N 06 P1
http://www.xtremepapers.me/CIE/Internat ... 6_qp_1.pdf
PLZ PLZ PLZ PLZ PLZ PLZ PLZ ANS ANS ANS ANS IT

Wow.. I have NO IDEA after looking at it for 20 minutes.. damn

its not that hard .. for q8 answer is D because for the enthalpy change of formation the product must be formed from its elements in their standard states .. in other options the reactants have compounds .. but only D has all its reactants as elements in their standard states ..

Ah sorry, didn't even know he asked q8
Yeah q8 is straight forward.
I meant q4

 

[Mobeen]

q4 is easy too.u just form a equation and compare their volume ratios to find the total volume .. it would be only of CO2 and the add the volume of excess oxygen in it ..

 

[ideggkr]

Oh, right. I didn't consider H2O.
Tbh I don't think this question is 'easy' mate.

Well if we set up equation

1. CH4 + 2O2 --> CO2 + 2H2O (20cm^2 of O2 used, 10cm^2 of CO2 formed ) so 70-20+10 = 60
2. C2H6 + 3.5O2 --> 2CO2 + 3H2O (35cm^2 of O2 used, 20cm^2 of CO2 formed) 70-35+20 = 55
3. C3H8 + 5O2 --> 3CO2 + 4H2O (50cm^2 of O2 used, 30cm^2 of CO2 formed) 70-50+30 = 50
4. C4H10 + 6.5O2 --> 4CO2 + 5H2O (65cm^2 of O2 used, 40cm^2 of CO2 formed) 70-65+40 = 45

 

[abdullah181994]

q4 is easy too.u just form a equation and compare their volume ratios to find the total volume .. it would be only of CO2 and the add the volume of excess oxygen in it ..

I DID IT BUT I WAS CALCULATING WRONG VOLUME THANKS

 

[Mobeen]

ur welcome .

 

[Mobeen]

Oh, right. I didn't consider H2O.
Tbh I don't think this question is 'easy' mate.

Well if we set up equation

1. CH4 + 2O2 --> CO2 + 2H2O (20cm^2 of O2 used, 10cm^2 of CO2 formed ) so 70-20+10 = 60
2. C2H6 + 3.5O2 --> 2CO2 + 3H2O (35cm^2 of O2 used, 20cm^2 of CO2 formed) 70-35+20 = 55
3. C3H8 + 5O2 --> 3CO2 + 4H2O (50cm^2 of O2 used, 30cm^2 of CO2 formed) 70-50+30 = 50
4. C4H10 + 6.5O2 --> 4CO2 + 5H2O (65cm^2 of O2 used, 40cm^2 of CO2 formed) 70-65+40 = 45

but i wont call it 'hard' . lets call it a tricky question ..

 

[abdullah181994]

14 A 5.00 g sample of an anhydrous Group II metal nitrate loses 3.29 g in mass on strong heating.
Which metal is present?
A magnesium
B calcium
C strontium
D barium
THIS ONE IS DIFFICULT, ANSWER THIS.