At constant temperature and pressure, 2.05 g of oxygen gas o2 is added to a 1.0 l balloon containing 1.00 g of o2. what is the new volume of the balloon?

1.50 L.

Explanation:

PV = nRT,

Where, P is the pressure if the gas,

V is the volume if the gas container,

n is the no. of gas moles,

R is the general gas constant,

T is the temperature of the gas.

n₁V₂ = n₂V₁.

V₁ = 1.0 L, V₂ = ??? L.

n₁ = mass/molar mass = (2.05 g)/(32.0 g/mol) = 0.064 mol.

n₂ = n₁ + (1.00 g)/(32.0 g/mol) = 0.0953 mol.

∴ V₂ = n₂V₁/n₁ = (0.0953 mol)(1.0 L)/(0.064 mol) = 1.489 L ≅ 1.50 L.

the answer is a and have a great day

the ratio of the conjugate base to the acid required to attain the desired ph.