Answer:
0.924 g
Explanation:
The following data were obtained from the question:
Volume of CO2 at RTP = 0.50 dm³
Mass of CO2 =?
Next, we shall determine the number of mole of CO2 that occupied 0.50 dm³ at RTP (room temperature and pressure). This can be obtained as follow:
1 mole of gas = 24 dm³ at RTP
Thus,
1 mole of CO2 occupies 24 dm³ at RTP.
Therefore, Xmol of CO2 will occupy 0.50 dm³ at RTP i.e
Xmol of CO2 = 0.5 /24
Xmol of CO2 = 0.021 mole
Thus, 0.021 mole of CO2 occupied 0.5 dm³ at RTP.
Finally, we shall determine the mass of CO2 as follow:
Mole of CO2 = 0.021 mole
Molar mass of CO2 = 12 + (2×16) = 13 + 32 = 44 g/mol
Mass of CO2 =?
Mole = mass /Molar mass
0.021 = mass of CO2 /44
Cross multiply
Mass of CO2 = 0.021 × 44
Mass of CO2 = 0.924 g.
Answer: (C) conservation of matter
Solution: Law of conservation of matter or mass states that' total mass of the reactants should always be equal to the total mass of the product that is the total mass is remained conserved in a chemical reaction.
A balanced chemical equation always follow this law.
For example:

Mass of hydrogen = 1 g/mol
Mass of Oxygen = 16 g/mol
Total mass on the reactants = 2(2×1)+(2×16)= 36g/mol
Total mass on the product side = 2[(2×1) +16] = 36 g/mol
As,
Mass on reactant side = Mass on the product side
Therefore, a balanced chemical reaction follows Law of Conservation of mass.
This doesn’t have a multiple choice part to answer.
Zn⁰ ----> Zn⁺² + 2e⁻ - oxidation
Hg⁺² + 2e⁻ ----> Hg⁰ - reduction
Zn loses 2 moles of electrons , and Hg gains 2 mole of electrons.
So, number of moles of electrons gained and lost during reaction is equal.