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2 years ago

A discarded spray paint can contains only a small volume of the propellant gas at a

Chemistry

1 answer:

Katena32 [7]2 years ago

5 0

Answer:

1.24 × 10³ kPa

Explanation:

Step 1: Given data

Initial pressure of the gas (P₁): 34.5 kPa

Initial volume of the can (V₁): 473 mL

Final pressure of the gas (P₂): ?

Final volume of the can (V₂): 13.16 mL

Step 2: Calculate the final pressure of the gas in the can

If we assume that the gas in the can behaves as an ideal gas and that the temperature remains constant, we can calculate the final pressure of the gas using Boyle's law.

P₁ × V₁ = P₂ × V₂

P₂ = P₁ × V₁ / V₂

P₂ = 34.5 kPa × 473 mL / 13.16 mL = 1.24 × 10³ kPa

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Difference of the enthalpy (of a system) minus the product of the entropy and absolute temperature

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Dihydrogen dioxide decomposes into water and oxygen gas. Calculate the amounts requested if 1.34 moles of dihydrogen dioxide rea

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Answers:

a. Moles of oxygen formed: 0.670 mol

b. Moles of water formed: 1.34 mol

c. Mass of water formed: 24.1 g

d. Mass of oxygen formed: 21.4 g

Explanation:

Dihdyrogen dioxide is the chemical name for a compound made of two hydrogen atoms and two oxide atoms, i.e. H₂O₂, which is also known as hydrogen peroxide or oxygenated water.

The decomposition reaction of dihydrogen dioxide into water and oxygen gas is represented by the balanced chemical equation:

$2H_2O_2(l)\rightarrow 2H_2O(l)+O_2(g)$

The mole ratios derived from that balanced chemical equation are:

2 mol H₂O₂ : 2 mol H₂O : 1 mol O₂

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Set the proportion using the theoretical mole ratio of H₂O₂ to O₂ and the amount of moles of dyhydrogen dioxide that react:

$2\text{ mol }H_2O_2/1\text{ mol }O_2=1.34\text{ mol }H_2O_2/x$

When you solve for x, you get:

x = 1.34 mol H₂O₂ × 1 mol O₂ / 2 mol H₂O₂ = 0.670 mol O₂

b. Moles of water formed

Set the proportion using the theoretical mole ratio of H₂O₂ to H₂O and the amount of moles of dyhydrogen dioxide that react:

$2\text{ mol }H_2O_2/2\text{ mol }H_2O=1.34\text{ mol }H_2O_2/x$

When you solve for x, you get:

x = 1.34 mol H₂O₂ × 2 mol H₂O / 2 mol H₂O₂ = 1.34 mol H₂O

c. Mass of water formed

Using the number of moles of water calculated in the part b., you calculate the mass of water formed, in grams, using the molar mass of water:

Molar mass of water = 18.015 g/mol

Number of moles = mass in grams / molar mass

⇒ mass in grams = number of moles × molar mass

⇒ mass in grams = 1.34 mol × 18.015 g/mol = 24.1 g

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Using the number of moles of oxygen determined in the part a., you calculate the mass in grams using the molar mass of O₂.

Molar mass of O₂ = 32.00 g/mol

mass = molar mass × number of moles

mass = 32.00 g/mol × 0.670 mol = 21.4 g.

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