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Margarita [4]
1 year ago
13

the molecular weight of the nitrous oxide is 44.013 g/mol. assuming standard temperature and pressure, what would be the volume,

in l, of a cylinder containing 3.40 kg of nitrous oxide?
Chemistry
1 answer:
allsm [11]1 year ago
3 0

The volume of the nitrous oxide gas is 1729.3 Liters

<h3>What is the number of moles of gas present in 3.40 kg of nitrous oxide?</h3>

The number of moles of gas present in 3.40 kg of nitrous oxide is determined from the formula below:

Numbers of moles = mass/molar mass

the mass of nitrous oxide = 3.40 kg or 3400 g

the molar mass of nitrous oxide = 44.013 g/mol

Moles of gas = 3400 / 44.013

Moles of gas = 77.25 moles

Using the ideal gas equation to determine the volume of the gas:

PV= nRT

V = nRT/P

where;

  • V is the volume of gas
  • n is the number of moles of gas
  • R is molar Gas constant = 0.082 L.atm/mol/K
  • T is the temperature of the gas

V = 77.25 * 0.082 * 273 / 1

The volume of the gas = 1729.3 Liters

Learn more about ideal gas equation at: brainly.com/question/20212888

#SPJ1

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A it increases is the correct answer.

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Determine the value of Kc for the following reaction if the equilibrium concentrations are as follows: [H2]eq = 0.14 M, [Cl2]eq
pentagon [3]

Question:

A) 12

B) 29

C) 2.1 × 10⁻²

D) 8.7 × 10⁻²

E) 47

Answer:

The correct option is;

E) 47

Explanation:

Kc, which is the equilibrium constant of a chemical reaction is derived by finding the ratio between the product of the equilibrium concentration of the product raised to their respective coefficients to the product of the equilibrium concentration of the reactants also raised to their respective coefficients.

Here we have;

[H₂] = 0.14 M

[Cl₂] = 0.39 M

[HCl] = 1.6

The reaction is given as follows;

H₂ (g) + Cl₂ (g) ⇄ 2HCl (g)

The formula for Kc is given as follows;

Kc = \frac{[HCl]^2}{[H_2][Cl_2]} = \frac{1.6^2}{0.14 \times 0.39}  = 46.886

Therefore, the Kc for the reaction is approximately equal to 47.

3 0
3 years ago
Solid aluminum (Al) and chlorine (C1,) gas react to form solid aluminum chloride (A1CI2). Suppose you have 7.0 mol of Al and 1.0
WITCHER [35]

Answer:

6.3 moles

Explanation:

From the balanced equation of reaction:

2Al + 3Cl_2 --> 2AlCl_3

2 moles of aluminium reacts with 3 moles of chlorine gas to form 2 moles of AlCl3.

Therefore, 1 mole of chlorine will require: 2 x 1/3 = 0.67 mole of aluminium.

Hence, 0.67 mole of aluminium will be needed for 1 mole of chlorine. If 7 moles of aluminium is present, then:

   7 - 0.67 = 6.33 moles of aluminium will be left.

To the nearest 0.1, it means 6.3 moles of aluminium will be left.

4 0
3 years ago
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