At standard temperature and pressure, a 0.50 mol sample of H2 gas and a separate 1.0 mol sample of O2 gas have the same (A) average molecular knetic energy

(B) average molecular speed

(C) volume

(D) effusion rate

(E) density

Option E density

Explanation:

This is actually pretty easy to explain. At the beggining it states that at STP which are 1 atm and 273 K, we have two samples of H2 but in different quantities. So using the ideal gas equation, we calculate the volume of each gas:

PV = nRT (1)

Where:

P: pressure in atm

V: volume in L

n: moles

R: gas constant which is 0.082 L atm / K mol

T: temperature in K

So from this equation, we solve for V:

V = nRT/P

Replacing data for both samples we have:

V1 = 0.5 * 0.082 * 273 / 1 = 11.19 L

V2 = 1 * 0.082 * 273 / 1 = 22.38 L

Now, to verify that is option E, let's write the expression for density:

d = m/V (2)

Where:

d: density

m: mass

To calculate the mass, we use the molar weight of hydrogen (2 g/mol) and the moles of the samples so:

m1 = 0.5 * 2 = 1 g

m2 = 1 * 2 = 2 g

Now, replacing in (2):

d1 = 1 / 11.19 = 0.0893 g/L

d2 = 2 / 22.38 = 0.0893 g/L

As d1 = d2 we can conclude that option E is the correct option.