Chemistry MCQ thread...

[Abdulrab]

please do this question from nov 08 Q 30 !!!

PLEASE GIVE THE ANSWER WITH FULL EXPLANATION ! THANX HURRY UP !!!

 

[kshumaila52]

Another Question..

Ams B

According to 1st reaction 1 mole will produce 1.5, and 2nd reaction 1 mole will give 0.1 so total 1.6.

 

[Abdulrab]

PLEASE ANSWER MY QUESTION TOO !!! I WILL BE THANKFUL TO U !!

 

[Mobeen]

please do this question from nov 08 Q 30 !!!

PLEASE GIVE THE ANSWER WITH FULL EXPLANATION ! THANX HURRY UP !!!

u have the mass of ethanol and ethanoic acid so first of all u calculate the moles of both of them and see which one of them is in limiting .. ethanoic acid is limiting and has o.5 moles .
from the mole ratio we can see that the ester would also hav 0.5 moles .. and then multiply 0.5 by the Mr of the ester i.e 88 to get the mass of the ester .. mass is = 0.5*88=44g .. we can find the percentage yield by 22/44 *100% =50% i.e C..
percentage yield = experimental value / theoretical value *100%

 

[kshumaila52]

PLEASE ANSWER MY QUESTION TOO !!! I WILL BE THANKFUL TO U !!

which ???

 

[Mobeen]

the main thing is to check for the limiting reagent .. rest is just simple mole calculations .. hope u understand

 

[sweetiepie]

Q.30 C

%yield= actual mass/calculated massx100
22/44x100
= 50%

I HOPE ITX CLEAR TO U..

 

[beststriker]

hey..i need help in nov10/ variant 12, mcq no 18, 39 and 40.....

 

[ideggkr]

PLEASE ANSWER MY QUESTION TOO !!! I WILL BE THANKFUL TO U !!

It'd be nicer if u can upload the question with screenshot

CH3CH2OH + CH3CO2H ---> CH3CO2CH2CH3 + H2O

30g for both ethanol and ethanoic acid
Mr(CH3CH2OH) = 46
Mr(CH3CO2H) = 60

So 0.65 mole of ethanol, and 0.5 mole of ethanoic acid present
with ethanoic acid being a limiting agent.

That means only 0.5 mole of both reagents will proceed the rxn

Theoretically we r expecting 0.5 mole of ethyl ethanoate to be made.

Mr(CH3CO2CH2CH3) = 88. 0.5 mole of this would be 44g

But only 22g formed. so yield is 50%

 

[Mobeen]

hey..i need help in nov10/ variant 12, mcq no 18, 39 and 40.....

18's D .. the colour is purple of gaseous bromine .

 

[ideggkr]

hey..i need help in nov10/ variant 12, mcq no 18, 39 and 40.....

18's D .. the colour is purple of gaseous bromine .

gaseous bromine is not purple. It's orange/brown

So it's a typo

 

[Abdulrab]

thanks everyone who helped me !! love u guys !! u people rock ! i will post more questions in the morning !! :good:

 

[ideggkr]

Answer with explanation please?
A classic experiment question.

 

[kshumaila52]

hey..i need help in nov10/ variant 12, mcq no 18, 39 and 40.....

Q18 ans B
bromine is non polar so soluble in polar (a is wrong)
it can easily vaporise( c is wrong)
in vapour state it is not purple (d is wrong)

 

[Mobeen]

hey..i need help in nov10/ variant 12, mcq no 18, 39 and 40.....

40 is D because in 1 it has a tertirary compound which makes causes the reaction to go through the Sn1 mechanism .. the Sn1 mechanism has two stages in the reaction and we can see from the diagram that it has two stages ..
all other equations go through a single stage ..

 

[ideggkr]

One more question

Ans with explanation plz

 

[Mobeen]

Answer with explanation please?
A classic experiment question.

the pluger is stationary so the pressure from both inside and from the atmosphere is equal on the plunger .. so 1 is incorrect .. when 1 is incorrect in only leaves C as the option

 

[kshumaila52]

Answer with explanation please?
A classic experiment question.

Ans is C
1 is wrong as pressure will remain same,
and 2 and3 are correct (apply le chatelier principal)

 

[ideggkr]

Answer with explanation please?
A classic experiment question.

the pluger is stationary so the pressure from both inside and from the atmosphere is equal on the plunger .. so 1 is incorrect .. when 1 is incorrect in only leaves C as the option

Cool thanks.

 

[abdullah181994]

11 Swimming pool water can be kept free of harmful bacteria by adding aqueous sodium chlorate(I),
NaOCl. This reacts with water to produce HOCl molecules which kill bacteria.
OCl–(aq) + H2O OH–(aq) + HOCl (aq)
In bright sunshine, the OCl
– ion is broken down by ultra-violet lighT
OCL–(aq) + uv light → Cl–(aq)+ ½O2(g)
Which method would maintain the highest concentration of HOCl (aq)?
A acidify the pool water
B add a solution of chloride ions
C add a solution of hydroxide ions
D bubble air through the water
EXPLAIN PLZ