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.

in scientific notation, the value of the  speed of light in a vacuum

  will be written as 2.995 * 10^8. when this value is represented in the form of

  a * 10^b, then, a = 2.995 and b = 8.

the value of the speed of light in a vacuum given above can also be rounded up to approximately 300000000. in this case, the scientific notation will be written as 3 *10^8.