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krok68 [10]
2 years ago
15

A sample of 0.200 moles of nitrogen occupies 0.400 L. Under the same conditions, what number of moles occupies 1.200 L

Chemistry
1 answer:
IRISSAK [1]2 years ago
7 0

The number of mole of nitrogen that occupies 1.2 L under the same condition is 0.6 mole

<h3>Data obtained from the question </h3>
  • Initial mole (n₁) = 0.2 mole
  • Initial volume (V₁) = 0.4 L
  • Final volume (V₂) = 1.2 L
  • Final mole (n₂) =?

<h3>How to determine the final mole </h3>

The final mole can be obtained by using the ideal gas equation as illustrated below:

PV = nRT

Divide both side n

PV / n = RT

Divide both side by P

V / n = RT / P

RT / P = constant

V / n = constant

Thus,

V₁ / n₁ = V₂ / n₂

0.4 / 0.2 = 1.2 / n₂

2 = 1.2 / n₂

Cross multiply

2 × n₂ = 1.2

Divide both side by 2

n₂ = 1.2 / 2

n₂ = 0.6 mole

Learn more about ideal gas equation:

brainly.com/question/4147359

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If you mix cu2+ with? a. nacl, b. nano2, c. h2o , arrange the solutions based on their absorption from highest frequency to lowe
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How many milliliters of 0.150 M NaOH are required to neutralize 85.0 mL of 0.300 M H2SO4 ? The balanced neutralization reaction
nekit [7.7K]

Answer : The volume of NaOH required to neutralize is, 340 mL

Explanation :

To calculate the volume of base (NaOH), we use the equation given by neutralization reaction:

n_1M_1V_1=n_2M_2V_2

where,

n_1,M_1\text{ and }V_1 are the n-factor, molarity and volume of acid which is H_2SO_4

n_2,M_2\text{ and }V_2 are the n-factor, molarity and volume of base which is NaOH.

We are given:

n_1=2\\M_1=0.300M\\V_1=85.0mL\\n_2=1\\M_2=0.150M\\V_2=?

Putting values in above equation, we get:

2\times 0.300M\times 85.0mL=1\times 0.150M\times V_2\\\\V_2=340mL

Hence, the volume of NaOH required to neutralize is, 340 mL

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