A particular sample of air contains 3 mol N2 (nitrogen gas),
7 mol O2 (oxygen gas), and 2 mol Ar (argon gas).
What is the mole fraction of oxygen gas in this sample of air?

Question 5 options:

0.20

0.58

0.17

0.42

Mol fraction O2  = 0.583 (option  2) is correct

Explanation:

Step 1: Data given

Number of moles N2 (nitrogen gas) = 3.0 moles

Number of moles O2 (oxygen gas) = 7.0 mol

Number of moles Ar (Argon gas) = 2.0 moles

Step 2: Calculate the total number of moles

Total number of moles = moles of N2 + moles of O2 + moles of Ar

Total number of moles = 3.0 moles + 7.0 moles + 2.0 moles

Total number of moles = 12.0 moles

Step 3: Calculate mole fractions

mol fraction = Number of moles / total number of moles

Mol fraction N2 = 3.0 / 12.0 moles = 0.250

Mol fraction O2 = 7.0 / 12.0 moles = 0.583

Mol fraction Ar = 2.0 / 12.0 moles = 0.167

Mol fraction = 0.250 + 0.583 + 0.167 = 1

the density of gold = 19.3 g/cm3

the relation between density and mass and volume is

density = mass / volume

given :

mass of gold = 13 g

density = 13g / volume = 19.3

volume = 13 /19.3 = 0.67 cm^3

thus volume of gold nugget = 0.67 cm^3

a redox reaction

explanation:

potassium metal is oxidized to potassium ion, and iodine is reduced to iodide ion, and we add (i) and (ii) together to eliminate the electrons.