Answer:
3 : 1
Explanation:
Let the rate of He be R1
Molar Mass of He (M1) = 4g/mol
Let the rate of O2 be R2
Molar Mass of O2 (M2) = 32g/mol
Recall:
R1/R2 = √(M2/M1)
R1/R2 = √(32/4)
R1/R2 = √8
R1/R2 = 3
The ratio of rate of effusion of Helium to oxygen is 3 : 1
Answer: Moles of hydrogen required are 4.57 moles to make 146.6 grams of methane,
.
Explanation:
Given: Mass of methane = 146.6 g
As moles is the mass of a substance divided by its molar mass. So, moles of methane (molar mass = 16.04 g/mol) are calculated as follows.

The given reaction equation is as follows.

This shows that 2 moles of hydrogen gives 1 mole of methane. Hence, moles of hydrogen required to form 9.14 moles of methane is as follows.

Thus, we can conclude that moles of hydrogen required are 4.57 moles to make 146.6 grams of methane,
.
Answer:
Compound B has greater molar mass.
Explanation:
The depression in freezing point is given by ;
..[1]

Where:
i = van't Hoff factor
= Molal depression constant
m = molality of the solution
According to question , solution with 5.00 g of A in 100.0 grams of water froze at at lower temperature than solution with 5.00 g of B in 100.0 grams of water.
The depression in freezing point of solution with A solute: 
Molar mass of A = 
The depression in freezing point of solution with B solute: 
Molar mass of B = 

As we can see in [1] , that depression in freezing point is inversely related to molar mass of the solute.


This means compound B has greater molar mass than compound A,
B, turns red litmus paper to blue