The answer is A beautiful
Answer:
Explanation:
mass % of C = 0.27/0.45*100 = 60%
mass % of H = 0.02/0.45*100 = 4.4%
mass % of O = 0.16/0.45*100 = 35.6%
Total = 60%+4.4%+ 35.6% = 100%
Answer : The rate of effusion of sulfur dioxide gas is 52 mL/s.
Solution :
According to the Graham's law, the rate of effusion of gas is inversely proportional to the square root of the molar mass of gas.

or,
..........(1)
where,
= rate of effusion of nitrogen gas = 
= rate of effusion of sulfur dioxide gas = ?
= molar mass of nitrogen gas = 28 g/mole
= molar mass of sulfur dioxide gas = 64 g/mole
Now put all the given values in the above formula 1, we get:


Therefore, the rate of effusion of sulfur dioxide gas is 52 mL/s.
Hydrocarbons are compounds formed by only hydrogen atoms and carbon.
Answer (2)
hope this helps!