Chemistry MCQ thread...

Answer this explaining ur choice

xHazeMx said:

Answer this explaining ur choice

Click to expand...

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.

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

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

xHazeMx said:

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

Click to expand...

A?

smartangel said:

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

Click to expand...

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)

histephenson007 said:

xHazeMx said:

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

Click to expand...

A?

Click to expand...

correct. explain ur answer

xHazeMx said:

histephenson007 said:

xHazeMx said:

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

Click to expand...

A?

Click to expand...

correct. explain ur answer

Click to expand...

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).

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

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/

thats correct ! .. how i didnt think about that

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

answer A

yeah. how?

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

thanks alot

smartangel said:

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

Click to expand...

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

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.

which year is that question, i will check the MS