The answer to this item is TRUE. This can be explained through the Graham's law. This law states that the rate at which gases diffuse is inversely proportional to the square root of their densities which is also related to their molecular masses.
It is the boilimg point now aa
Answer:
The entropy change for a real, irreversible process is equal to <u>zero.</u>
The correct option is<u> 'c'.</u>
Explanation:
<u>Lets look around all the given options -:</u>
(a) the entropy change for a theoretical reversible process with the same initial and final states , since the entropy change is equal and opposite in reversible process , thus this option in not correct.
(b) equal to the entropy change for the same process performed reversibly ONLY if the process can be reversed at all. Since , the change is same as well as opposite too . Therefore , this statement is also not true .
(c) zero. This option is true because We generate more entropy in an irreversible process. Because no heat moves into or out of the surroundings during the procedure, the entropy change of the surroundings is zero.
(d) impossible to tell. This option is invalid , thus incorrect .
<u>Hence , the correct option is 'c' that is zero.</u>
Answer:
70.0 %
Explanation:
Step 1: Given data
- Mass of nitrogen (mN): 74.66 g
- Mass of the compound (mNxOy): 250 g
Step 2: Calculate the mass of oxygen (mO) in the compound
The mass of the compound is equal to the sum of the masses of the elements that form it.
mNxOy = mN + mO
mO = mNxOy - mN
mO = 250 g - 74.66 g = 175 g
Step 3: Determine the percent composition of oxygen in the sample
We will use the following expression.
%O = mO / mNxOy × 100%
%O = 175 g / 250 g × 100% = 70.0 %
N⁻²2H4(l) + 2H2O⁻¹2(l) → N⁰2(g) + 4H2O⁻²(g)
N is oxidized and O is reduced
