Chemistry MCQ thread...

[abdullah181994]

CORRECT ANS A

 

[ideggkr]

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

Oh, ok.
If u acidify the pool, the equlibrium would shift to RHS to make more OH-
so conc. of HOCl would be high

 

[ShootingStar]

Another Question..

B

Can you please show the workings aswell

 

[kshumaila52]

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

Oh, ok.
If u acidify the pool, the equlibrium would shift to RHS to make more OH-
its A
By adding acid, OH(hydroxyl) ion will use and reaction 1 will shift in forward direction.
so conc. of HOCl would be high

 

[ideggkr]

I got confused because u didn't upload the screenshot

Whether the equation is reversible or not should be clear.

If
OCL–(aq) + uv light → Cl–(aq)+ ½O2(g)
is not reversible, then any change in conc. of Cl- or O2 would have no effect in equilibrium position of the first equation

 

[ideggkr]

Now can anyone answer my second question (electrolysis)
on previous page please?

 

[abdullah181994]

I got confused because u didn't upload the screenshot

Whether the equation is reversible or not should be clear.

If
OCL–(aq) + uv light → Cl–(aq)+ ½O2(g)
is not reversible, then any change in conc. of Cl- or O2 would have no effect in equilibrium position of the first equation

REVERSIBLE

 

[kshumaila52]

One more question

Ans with explanation plz

Its A
(simple way, all other options are not possible)
in fact , in A2, we ll study( as my teacher said) more positive electrode potential, more easily reduced, means cant oxidised, so silver metal not change into ion, so cant reduce at cathode

 

[Mobeen]

Another Question..

B

Can you please show the workings aswell

yes.
from the first equation we can see that if we have 1 mole NaN3 we get 1.5 moles N2
and from the second equation we can see that 10 moles of Na will give 1 mole of N2 BUT we have only 1 mole of Na(because we only had 1 mole NaN3 and it gives 1 mole Na , u can check from the first eq) .. with 1 mole Na mole ratio would give 0.1 mole N2 ..
so total moles of N2 are 1.5 + 0.1 = 1.6 .. hope it explains ..

 

[abdullah181994]

One more question

Ans with explanation plz

IF ANSWER IS D I CAN EXPLIAN

 

[abdullah181994]

PLEASE TELL IN 2010 M/J AND O/N VARIANTS ARE ALL THE QUESTIONS SAME

 

[Mobeen]

PLEASE TELL IN 2010 M/J AND O/N VARIANTS ARE ALL THE QUESTIONS SAME

i think so .. their just mixed up in the paper ..

 

[abdullah181994]

@Mobeen THANX

 

[kshumaila52]

PLEASE TELL IN 2010 M/J AND O/N VARIANTS ARE ALL THE QUESTIONS SAME

i think so .. their just mixed up in the paper ..

no
nov10/12 and nov10/13 has almost all questions are different.

 

[Mssamgirl15]

PLEASE TELL IN 2010 M/J AND O/N VARIANTS ARE ALL THE QUESTIONS SAME

i think so .. their just mixed up in the paper ..

No guys, the variants of MAY/JUNE are the same but for OCTOBER/NOVEMBER there are two different sets of questions.

 

[Mobeen]

@mssamgirl i only looked at different variants for june papers and saw it was only mixed up .. so assumed the same for the nov papers ..

 

[Hateexams93]

SOME1 PLZZZZZZZZZZZZZZZZZZZZZZZZZZ HELP ME
The density of ice is 1.00 g cm–3.
What is the volume of steam produced when 1.00 cm3 of ice is heated to 323 °C (596 K) at a
pressure of one atmosphere (101 kPa)?
[1 mol of a gas occupies 24.0 dm3 at 25 °C (298 K) and one atmosphere.]
A 0.267 dm3 B 1.33 dm3 C 2.67 dm3 D 48.0 dm3
DO WE TO HAVE USE THIS FORMULA PV=nRT OR WHAT ????? I CAN'T GET THE ANSWER (((((((

 

[libra94]

q.32 plz
http://www.xtremepapers.me/CIE/Internat ... _qp_11.pdf

 

[workinghard]

@hateexams ans is C i used da formula u mentioned n got it
try again or ll post it 2moro
@libra ans is D since oxidation state only changes in 1 as in +6 in SO42- --- +4 in SO2
u can check da otherx dey remain same or u can take ma word cux i hv checked dem!

 

[Mobeen]

SOME1 PLZZZZZZZZZZZZZZZZZZZZZZZZZZ HELP ME
The density of ice is 1.00 g cm–3.
What is the volume of steam produced when 1.00 cm3 of ice is heated to 323 °C (596 K) at a
pressure of one atmosphere (101 kPa)?
[1 mol of a gas occupies 24.0 dm3 at 25 °C (298 K) and one atmosphere.]
A 0.267 dm3 B 1.33 dm3 C 2.67 dm3 D 48.0 dm3
DO WE HAVE USE THIS FORMULA PV=nRT OR WHAT ????? I CAN'T GET THE ANSWER (((((((

we cant use pv=nrt in this .
we use v1/T1 = v2/T2
the initial vol =v1. the mass of ice cube is 1 (mass= density* vol) thus the moles of the water in ice cube would be 1/18.
the initial vol = moles * 24 = 1/18 *24 =1.333
the initial temp is rtp =298
the final temp is 596
so put in the formula
=> 1.333/298 = x/596 ( x = final vol of steam )
x 2.667 C