1. an unknown amount of mercury (ii) oxide was decomposed in the lab. mercury metal was formed and 4.50 l of oxygen gas was released at a pressure of 0.970 atm and 390.0 k. what was the initial weight of mercury oxide in the sample? a.) 27.3 grams b.) 38.3 grams c.) 46.2 grams d.) 60.6 grams 2. read the chemical equation. mg + hcl → mgcl2 + h2 how many liters of hydrogen gas is produced at 298 k and 0.960 atm if 3.00 moles of hydrochloric acid react with an excess of magnesium metal? a.) 38.2 liters b.) 42.6 liters c.) 66.7 liters d.) 76.4 liters 3. the pressure and temperature of a gas are held constant. which of the following is true for the volume of the gas? a.) it is inversely proportional to the number of moles of the gas. b.) it is directly proportional to the number of moles of the gas. c.) it is equal to the number of moles of the gas. d.) it is half of the number of moles of the gas.

1. 59.1 g HgO

2. A.) 38.2 L

3. B.) It is directly proportional to the number of moles of the gas.

Step-by-step explanation:

1. Mass of HgO

We know we will need a chemical equation with masses and molar masses, so let's gather all the information in one place.

M_r:   216.59

          2HgO ⟶ 2Hg + O₂

(a) Calculate the moles of O₂  

pV = nRT            Divide each side by RT  

 n = (pV)/(RT)  

Data:

p = 0.970 atm  

V = 4.50 L  

R = 0.082 06 L·atm·K⁻¹mol⁻¹  

T = 390.0 K  

Calculation:

n = (0.970 × 4.500)/(0.082 06 × 390.0)  

n = 0.1364 mol O₂

(b) Calculate the moles of HgO

The molar ratio is 1 mol O₂/2 mol HgO.

Moles of HgO = 0.1364 mol O₂ × (2 mol Hg/1 mol O₂)

Moles of HgO = 0.2728 mol HgO

(c) Calculate the mass of HgO

Mass of HgO = 0.2728 mol HgO × (216.59 g HgO/1 mol HgO)

Mass of HgO = 59.1 g HgO

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2. Volume of Hydrogen

              Mg + 2HCl ⟶ MgCl₂ + H₂

n/mol:              3.00

(a) Calculate the moles of H₂

The molar ratio is (1 mol H₂/2 mol HCl).

Moles of H₂ = 3.00 mol HCl × (1 mol H₂/2 mol HCl)

Moles of H₂ = 1.50 mol HCl

(b) Calculate the volume of H₂

pV = nRT     Divide both sides by p

 V = (nRT)/p

Data:

n = 1.50 mol

T = 298 K

p = 0.960 atm

Calculation:

V = (1.5 × 0.082 06 × 298)/0.960

V = 38.2 L

3. Volume of gas

pV = nRT

 V = nRT/p

If T and p are constant, (RT)/p = constant = k

V = kn

V ∝ n

The volume is directly proportional to the number of moles of the gas.

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