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2 years ago

How many grams of ammonia are present in 5.0 L of a 0.050 M solution

Chemistry

1 answer:

Dennis_Churaev [7]2 years ago

3 0

Answer:

4.25g

Explanation:

Molar concentration or molarity can be calculated thus;

Molarity (M) = number of moles (n) ÷ volume (V)

According to the provided information in this question, Volume = 5.0L, M = 0.050 M

number of moles = molarity × volume

n = 0.050 × 5

n = 0.25moles

Number of moles in a substance = mass (M) ÷ molar mass (MM)

Molar mass of ammonia (NH3) = 14 + 1(3)

= 17g/mol

Mass = moles × molar mass

Mass (g) = 0.25 × 17

Mass = 4.25g

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A sample of an ideal gas has a volume of 2.30 L at 281 K and 1.02 atm. 1.76 atm is the pressure when the volume is 1.41 L and the temperature is 298 K.

<h3>What is Combined Gas Law ?</h3>

This law combined the three gas laws that is (i) Charle's Law (ii) Gay-Lussac's Law and (iii) Boyle's law.

It is expressed as

$\frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2}$

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P₂ = second pressure

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T₁ = first temperature

T₂ = second temperature

Now put the values in above expression we get

$\frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2}$

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$P_{2} = \frac{1.02\ atm \times 2.30\ L \times 298\ K}{281\ K \times 1.41\ L}$

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Thus from the above conclusion we can say that A sample of an ideal gas has a volume of 2.30 L at 281 K and 1.02 atm. 1.76 atm is the pressure when the volume is 1.41 L and the temperature is 298 K.

Learn more about the Combined gas Law here: brainly.com/question/13538773

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