What is the density of oxygen gas (O2) at 2.3 atmospheres and 305 Kelvin? Show all of the work used to solve this problem.

2.94g/L

Explanation:

First, let us derive an expression for the density, using the ideal gas equation:

PV = nRT (1)

V = nRT/ P (2)

Recall:

Number of mole(n) = mass(m)/Molar Mass (M)

n = m/M

Substituting the value of n into equation 2:

V = nRT/ P

V = mRT/ MP

Now Divide both side by m

V/m = RT/ MP

Inverting the above equation, we have:

m/V = MP/RT

But Density = m/V

Density = MP/RT

From the question given, we obtained the following data:

P = 2.3atm

T = 305K

Molar Mass of O2 = 16x2 = 32g/mol

R = 0.082atm.L/K /mol

Density = (32x2.3)/(0.082x305)

Density = 2.94g/L

375.3KJ

Explanation:

The following data were obtained from the question:

Mass of water = 166g

Heat of Vaporisation (ΔHv) = 40.7kJ/mol

Heat (Q) =..?

Next, we shall determine the number of mole in 166g of water. This is illustrated below:

Mass of H2O = 166g

Molar mass of H2O = (2x1) + 16 = 18g/mol

Number of mole = Mass/Molar Mass

Number of mole of H2O = 166/18

Number of mole of H2O = 9.22 moles.

Now, we can obtain the heat required to vaporise the water as shown below:

Q = n·ΔHv

Q = 9.22 mol x 40.7kJ/mol

Q = 375.3KJ

Therefore, the heat required to vaporise the water is 375.3KJ.