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.

Here, we are required to identify which property is constant for a 0.50 mol sample of H2 gas and a separate 1.0 mol sample of O2 gas at standard temperature and pressure.

According to the question; at Standard pressure and temperature of values 1 atm and 273 K, we have two samples of Hydrogen, H2 in different quantities. So using the ideal gas equation, we calculate the volume of each gas:

PV = nRT…………ideal gas equation.

Where:

P: pressure in atm

V: volume in L

n: moles

R: gas constant which is 0.082 L³ 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 * 0.08205 * 273 / 1 = 11.20 L

V2 = 1 * 0.08205 * 273 / 1 = 22.40 L

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

d = m/V density = mass/volume

The molar mass of hydrogen (2 g/mol) and the no. moles of the samples can be used to evaluate the mass of each specie so:

m1 = n × M = 0.5 * 2 = 1 g

m2 = n× M = 1 * 2 = 2 g

Therefore, by substituting m into d = m/V; we have;

D1 = 1 / 11.20 = 0.0893 g/L

D2 = 2 / 22.40 = 0.0893 g/L

Therefore, since D1 = D2, we can conclude that Density is constant.

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atomic mass = 39

number of neutrons = 20

number of protons= 19

electronic configuration = 1s2 2s2 2p6 3s2 3p6 4s1

the number of electron in first shell = 2

number of electrons in n=3 shell = 8