(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

in chemistry, adhesion and cohesion are attractive forces that exist between molecules. they hold the molecules together from each other. they are different in that adhesive force exists between molecules of different substance while cohesive force happens between molecules of same substance.

adhesion between water and glass causes water molecules at the edge of water and glass interface to climb up the wall in a narrow glass tube. this is called capillary action. cohesion is what gives glass its shape and causes water to form a droplet.