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2 years ago

Calculate the volume occupied by 10 moles of hydrogen gas at 145K and 705 Torr.

Chemistry

1 answer:

Norma-Jean [14]2 years ago

4 0

The volume occupied by 10 moles of hydrogen gas at 145K and 705 Torr is 128.01L.

<h3>How to calculate volume?</h3>

The volume of a gas can be calculated using ideal gas law equation as follows:

PV = nRT

Where;

P = pressure
V = volume
n = number of moles
R = gas law constant
T = temperature

P = 705torr = 0.93atm
V = ?
T = 145K
n = 10
R = 0.0821 Latm/molK

0.93 × V = 10 × 0.0821 × 145

0.93V = 119.05

V = 128.01L

Therefore, the volume occupied by 10 moles of hydrogen gas at 145K and 705 Torr is 128.01L.

Learn more about volume at: brainly.com/question/13338592

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3. Given 20g of Barium Hydroxide, how many grams of

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The number of grams of ammonium nitrate (NH₄NO₃) it would take for all the barium hydroxide to react is 18.7g

First, we will write the balanced chemical equation for the reaction

The balanced chemical equation for the reaction is

Ba(OH)₂ + 2NH₄NO₃ → 2NH₄OH + Ba(NO₃)₂

This means, 1 mole of barium hydroxide is required to react with 2 moles of ammonium nitrate

Now, we will calculate the number of moles of barium hydroxide present.

Mass of barium hydroxide (Ba(OH)₂) = 20 g

Using the formula

$Number\ of\ moles = \frac{Mass}{Molar\ mass}$

Molar mass of Ba(OH)₂ = 171.34 g/mol

∴ Number of moles of Ba(OH)₂ present = $\frac{20}{171.34}$

Number of moles of Ba(OH)₂ present = 0.116727 mole

Now,

Since 1 mole of barium hydroxide is required to react with 2 moles of ammonium nitrate

Then,

0.116727 mole of barium hydroxide will react with 2 × 0.116727 mole of ammonium nitrate

2 × 0.116727 = 0.233454 mole

∴ Number of moles of NH₄NO₃ required is 0.233454 mole

Now, for the mass of ammonium nitrate (NH₄NO₃) required

From the formula

Mass = Number of moles × Molar mass

Molar mass of NH₄NO₃ = 80.043 g/mol

∴ Mass of NH₄NO₃ required = 0.233454 × 80.043

Mass of NH₄NO₃ required = 18.68636 g

Mass of NH₄NO₃ required ≅ 18.7g

Hence, the number of grams of ammonium nitrate (NH₄NO₃) it would take for all the barium hydroxide to react is 18.7g

Learn more on determining mass of reactant required here: brainly.com/question/11232389

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