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3 years ago

A gas sample occupies 4.2 L at a pressure of 101 kPa.

Chemistry

2 answers:

Bond [772]3 years ago

8 0

Answer:

1.8 L

Explanation:

Given data:

Initial volume of gas = 4.2 L

Initial pressure of gas = 101 kpa

Final volume of gas = ?

Final pressure of gas = 235 kpa

Solution:

The given problem will be solved through the Boly's law,

"The volume of given amount of gas is inversely proportional to its pressure by keeping the temperature and number of moles constant"

Mathematical expression:

P₁V₁ = P₂V₂

P₁ = Initial pressure

V₁ = initial volume

P₂ = final pressure

V₂ = final volume

Now we will put the values in formula,

P₁V₁ = P₂V₂

101 kpa × 4.2 L = 235 kpa × V₂

V₂ = 424.2 kpa . L/ 235 kpa

V₂ = 1.8 L

So when pressure is increased the volume of gas becomes 1,8 L.

tekilochka [14]3 years ago

3 0

Answer:

1.8l

Explanation: is the sum

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Answer:- 0.273 kg

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From balanced equation, there is 1:3 mol ratio between aluminum nitrate and sodium chlorate. We will convert grams of aluminum nitrate to moles and then on multiplying it by mol ratio we get the moles of sodium chlorate that could further be converted to grams.

We need molar masses for the calculations, Molar mass of sodium chlorate is 106.44 gram per mole and molar mass of aluminum nitrate is 212.99 gram per mole.

$63.8gAl(NO_3)_3(\frac{1mol}{212.99g})(\frac{3molNaClO_3}{1molAl(NO_3)_3})(\frac{106.44g}{1mol})$

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sodium chlorate solution is 35% m/m. This means 35 g of sodium chlorate are present in 100 g solution. From here, we can calculate the mass of the solution that will contain 95.7 g of sodium chlorate and then the grams are converted to kg.

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Potassium superoxide, KO2, reacts with carbon dioxide to form potassium carbonate and oxygen: This reaction makes potassium supe

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Answer:

There will be produced 0.8448 grams O2

Explanation:

Step 1: Data given

Mass of KO2 = 2.50 grams

Mass of CO2 = 4.50 grams

Molar mass KO2 = 71.1 g/mol

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Step 2: The balanced equation

4 KO2(s) + 2 CO2(g) → 2 K2CO3(s) + 3 O2(g)

Step 3: Calculate moles KO2

Moles KO2 = mass KO2 / Molar mass KO2

Moles KO2 = 2.50 grams / 71.1 g/mol

Moles KO2 = 0.0352 moles

Step 4: Calculate moles of CO2:

Moles CO2 = mass CO2 / Molar mass CO2

Moles CO2 = 4.50 grams / 44.01 g/mol

Moles CO2 = 0.102 moles

Step 5: calculate limiting reactant

For 4 moles of KO2 consumed, we need 2 moles of CO2 to produce 2 moles of K2CO3 and 3 moles of O2

KO2 is the limiting reactant. It will completely be consumed. (0.0352 moles).

CO2 is in excess. There will react 0.0352/2 = 0.0176 moles of CO2

There will remain 0.102 - 0.0176 = 0.0844 moles CO2

Step 6: Calculate moles O2 produced

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For 0.0352 moles KO2 consumed, we have 3/4 * 0.0352 = 0.0264 moles of O2 produced

Step 7: Calculate mass of 02 produced

Mass O2 produced = Moles O2 * Molar mass 02

Mass O2 produced = 0.0264 moles * 32 g/mol

Mass O2 produced = 0.8448 grams O2

There will be produced 0.8448 grams O2

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