Chemistry MCQ thread...

[xHazeMx]

Answer this explaining ur choice

 

[Zishi]

Answer this explaining ur choice

The answer is A. Combustion is always exothermic, and atomisation is always endothermic. Imagine a energy profile diagram for graphite and diamond, diamond is higher than graphite in it as the reaction from graphite to diamond is endothermic. Enthalpy change of atomisation is also endothermic, so C atoms will be even more "above" in the diagram than graphite and diamond. Combustion is always exothermic, so CO_2 and H_2 0 will be even lower than graphite in the enthalpy profile diagram. This way, it will help you to understand the answer.

 

[xHazeMx]

zishi, can u explain number 3 more, i dont get it

 

[smartangel]

Please explain this:
In an experiment, 50.0 cm3 of a 0.10 mol dm–3 solution of a metallic salt reacted exactly with
25.0 cm3 of 0.10 mol dm–3 aqueous sodium sulphite.
The half-equation for oxidation of sulphite ion is shown below.
SO3^ − 2 (aq) + H2O(I) → SO4^−2(aq) + 2H+(aq) + 2e–
If the original oxidation number of the metal in the salt was +3, what would be the new oxidation
number of the metal?
A +1 B +2 C +4 D +5

 

[xHazeMx]

The following energy cycle represents the enthalpy changes in the formation of carbon dioxide
from its constituent elements in their standard states.

 

[histephenson007]

The following energy cycle represents the enthalpy changes in the formation of carbon dioxide
from its constituent elements in their standard states.

A?

 

[xHazeMx]

Please explain this:
In an experiment, 50.0 cm3 of a 0.10 mol dm–3 solution of a metallic salt reacted exactly with
25.0 cm3 of 0.10 mol dm–3 aqueous sodium sulphite.
The half-equation for oxidation of sulphite ion is shown below.
SO3^ − 2 (aq) + H2O(I) → SO4^−2(aq) + 2H+(aq) + 2e–
If the original oxidation number of the metal in the salt was +3, what would be the new oxidation
number of the metal?
A +1 B +2 C +4 D +5

the metal could be from group 1,2,3 or 4, having either oxidation number +1 or +2 or +3 or +4. so if u look to the values given
v= 0.05 dm^3
c= 0.10 mol dm^-3

then, n= c x v , 0.10 x 0.05 = 0.005 moles
also,
v= 0.025 dm^3
c= 0.10 mol dm^-3

then, n= c x v , 0.025 x 0.10 = 0.0025 moles

so to know whether this metal is from group 1 or 2
divide the number of moles u got

0.005/0.0025 = 2 .... This shows that the metal is from group 2. Hence, oxidation number is +2
(correct me if i m wrong)

 

[xHazeMx]

The following energy cycle represents the enthalpy changes in the formation of carbon dioxide
from its constituent elements in their standard states.

A?

correct. explain ur answer

 

[histephenson007]

The following energy cycle represents the enthalpy changes in the formation of carbon dioxide
from its constituent elements in their standard states.

A?

correct. explain ur answer

At level Y, the activation energy required for the reaction to take place is achieved. So, all the molecules are ready to react. So, it has to be C(g) and 2O(g).

 

[xHazeMx]

hmm, and why it is not C(g) + O2(g) ?

 

[histephenson007]

Actually, that was the same thing I was unsure about.

But, I thought, if it is in O2 state, another bond needs to be broken for the reaction to occur. So, Activation energy isn't completely achieved.

Correct me if I'm wrong/

 

[xHazeMx]

thats correct ! .. how i didnt think about that

 

[smartangel]

this seems to be an easy question but im not getting the answer.
Use of the Data Booklet is relevant to this question.
When a sports medal with a total surface area of 150 cm2 was evenly coated with silver, using
electrolysis, its mass increased by 0.216 g.
How many atoms of silver were deposited per cm2 on the surface of the medal?
A 8.0 × 10^18
B 1.8 × 10^19
C 1.2 × 10^21
D 4.1 × 10^22

 

[Xthegreat]

answer A

 

[smartangel]

yeah. how?

 

[histephenson007]

Mass of Ag = 0.216 g

n of Ag = 0.216/108 = 0.002

number of atoms of Ag = 0.002 * Avogardo's constant = 0.002*(6.02*10^23) = 1.204 * 10^21

Finally, number of atoms per cm2 = (1.204*10^21) / 150 = 8.0*10^18

Therefore, answer = A

 

[smartangel]

thanks alot

 

[xHazeMx]

this seems to be an easy question but im not getting the answer.
Use of the Data Booklet is relevant to this question.
When a sports medal with a total surface area of 150 cm2 was evenly coated with silver, using
electrolysis, its mass increased by 0.216 g.
How many atoms of silver were deposited per cm2 on the surface of the medal?
A 8.0 × 10^18
B 1.8 × 10^19
C 1.2 × 10^21
D 4.1 × 10^22

its C,

m= 0.216 .. Mr= 108
n= 0.216/108 = 0.002 moles

number of atoms (avogadro's constant = 6.02 x 10^23 )= 0.002 x 6.02 x 10^23 = 1.204 x 10^21 = 1.2 x 10^21
correct me if i m wrong

 

[Xthegreat]

total mass of Ag = 0.216
number of mole of Ag = 0.216/ 108 = 0.002 mole
number of atoms in 0.002 mole = 0.002 x avogadro constant = 0.002 x 6.02 x 10^23 = 1.204 x 10^21
total surface area = 150cm3
number of atoms per cm3 = 1.204 x 10^21 / 150 = 8.02 x 10^18

done.

 

[xHazeMx]

which year is that question, i will check the MS