The initial volume of a gas at a pressure of 3 atm is 5.2 L. What will the volume be if the pressure is increased to 15 atm?

1 L

Explanation:

Use Boyle's Law, which is defined as:

p1v1 = p2v2

p1 = 3atm

v1 = 5.2L

p2 = 15atm

v2 = ?

We are solving for v2 because all other variables are given.

p1v1 = p2v2

(3atm)(5.2L) = (15atm)(v2)

v2 = (3)(5.2) / (15) = 1.04 L ≈ 1 L

18 g is the mass produced by 4 g of H₂ and 16 g of O₂

Explanation:

The reaction is:

2H₂  +  O₂  →  2H₂O

So, let's find out the limiting reactant as we have both data from the reactants.

Mass / Molar mass = moles

4 g/ 2g/m = 2 moles H₂

16g / 32 g/m = 0.5 moles O₂

2 moles of hydrogen react with 1 mol of oxygen, but I have 0.5, so the O₂ is the limiting.

1 mol of O₂ produces 2 mol of water.

0.5 mol of O₂ produce  (0.5  .2)/1 = 1 mol of water.

1 mol of water weighs 18 grams.