NET Applied Sciences MCQs

A pure compound has constant composition.
This is also called the law of constant composition.

Percent composition is calculated from formula mass contributions.
The element percentages sum to approximately 100%.

Stoichiometry uses balanced equations to relate amounts.
It supports mass, mole, particle, and gas calculations.

Moles equal mass divided by molar mass.
18 g ÷ 18 g/mol = 1 mol.

Mass equals moles multiplied by molar mass.
2×58.5 = 117 g.

Particles equal moles times Avogadro's number.
0.5×6.022×10^23 = 3.011×10^23 molecules.

Moles equal particles divided by Avogadro's number.
1.2044×10^24 ÷ 6.022×10^23 = 2 mol.

Carbon contributes about 12 and two oxygens contribute 32.
The total molar mass is about 44 g/mol.

Oxygen contributes 16 of the 18 g/mol formula mass.
16/18×100 ≈ 88.9%.

The coefficients of H2 and H2O are equal.
Thus 4 mol H2 produce 4 mol H2O.

The mole ratio H2:NH3 is 3:2.
6 mol H2×2/3 = 4 mol NH3.

Percent yield is actual divided by theoretical times 100.
8/10×100 = 80%.