Hi my name is Jaleesa The answer = -1
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.
Answer:

Explanation:
You don't give the reaction, but we can get by just by balancing atoms of Na.
We know we will need the partially balanced equation with masses, moles, and molar masses, so let’s gather all the information in one place.
M_r: 142.04
2NaOH + … ⟶ Na₂SO₄ + …
n/mol: 0.75
1. Use the molar ratio of Na₂SO₄ to NaOH to calculate the moles of NaF.
Moles of Na₂SO₄ = 0.75 mol NaOH × (1 mol Na₂SO₄/2 mol NaOH
= 0.375 mol Na₂SO₄
2. Use the molar mass of Na₂SO₄ to calculate the mass of Na₂SO₄.
Mass of Na₂SO₄ = 0.375 mol Na₂SO₄ × (142.04 g Na₂SO₄/1 mol Na₂SO₄) = 53 g Na₂SO₄
The reaction produces
of Na₂SO₄.
Answer:
2.0x 10-2M
Explanation:
x2/[2.0x10-2]2 [1.0x10-2] =100