Answer:
the volume occupied by 3.0 g of the gas is 16.8 L.
Explanation:
Given;
initial reacting mass of the helium gas, m₁ = 4.0 g
volume occupied by the helium gas, V = 22.4 L
pressure of the gas, P = 1 .0 atm
temperature of the gas, T = 0⁰C = 273 K
atomic mass of helium gas, M = 4.0 g/mol
initial number of moles of the gas is calculated as follows;

The number of moles of the gas when the reacting mass is 3.0 g;
m₂ = 3.0 g

The volume of the gas at 0.75 mol is determined using ideal gas law;
PV = nRT

Therefore, the volume occupied by 3.0 g of the gas is 16.8 L.
Answer:
Explanation:
Law of conservation of mass:
According to the law of conservation mass, mass can neither be created nor destroyed in a chemical equation.
Explanation:
This law was given by french chemist Antoine Lavoisier in 1789. According to this law mass of reactant and mass of product must be equal, because masses are not created or destroyed in a chemical reaction.
Chemical equation:
Mg + HCl → H₂ + MgCl₂
24 g + 36.5 g = 2 g+ 95 g
60.5 g = 97 g
The reaction does not hold the law of conservation of mass, because it is not balanced.
Balanced chemical equation:
Mg + 2HCl → H₂ + MgCl₂
24 g + 73 g = 2 g+ 95 g
97 g = 97 g
this equation completely follow the law of conservation of mass.
Answer:
17.5moles
Explanation:
The number of moles in a substance can be calculated by using the formula;
Number of moles (n) = mass (m) ÷ molar mass (MM)
According to this question, mass of ammonia (NH3) = 297g
Molar Mass of NH3 = 14 + 1(3)
= 17g/mol
n = 297/17
n = 17.47
Number of moles of NH3 = 17.5moles