(0)A mixture of dihydrogen and dioxygen at one bar pressure contains 70% by weight of dioxygen. Calculate the partial pressure of dioxygen. a (ii)Calculate the volume occupied by 8.8 g of CO2 at 31.1°C and 1 bar pressure. R = 0.083 bar L K-1 mol-1

a) 0.13 bar

b) 5.05 L

Explanation:

Let us take the mass of the mixture to be 100 g. Hence, 70% by weight of dioxygen corresponds to 70 g

Mass of dihydrogen = 100g - 70 g = 30g

Number of moles of dioxygen = 70g/32 g/mol = 2.2 moles of dioxygen

Number of moles of dihydrogen = 30g/2g/mol = 15 moles of dihydrogen

Total number of moles = 2.2 + 15 = 17.2 moles

Mole fraction of dioxygen = 2.2/17.2 = 0.13

Partial pressure = mole fraction * total pressure

Partial pressure of dioxygen = 0.13 * 1 = 0.13 bar

ii) number of moles in 8.8 g of CO2 = 8.8g/44g/mol = 0.2 moles

T = 31.1 + 273 = 304.1 K

P = 1 bar

V= ?

R = 0.083 bar L K-1 mol-1

From

PV=nRT

V = nRT/P

V= 0.2 * 0.083 * 304.1/1

V= 5.05 L

8

explanation:

each principal energy level above the first contains one s orbital and three p orbitals. a set of three p orbitals, called the p sublevel, can hold a maximum of six electrons. therefore, the second level can contain a maximum of eight electrons - that is, two in the s orbital and 6 in the three p orbitals.

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