4.0 L of He(g), 6.0 L of N2(g), and 10. L of Ar(g), all at 0°C and 1.0 atm, are pumped into an evacuated 8.0 L rigid container, the final pressure in the container at 0°C is

Question

xoxodestinydee7265 xoxodestinydee7265

03-01-2020
Chemistry

contestada

4.0 L of He(g), 6.0 L of N2(g), and 10. L of Ar(g), all at 0°C and 1.0 atm, are pumped into an evacuated 8.0 L rigid container, the final pressure in the container at 0°C is

Respuesta :

Otras preguntas

Answer the questions from the finding God grade 7 book. Please. No links! : Religion

measuring masses of the reactants (Reaction 2) Step 7

What are some of the ways the print revolution transformed Europe?

In what way are the girls in the story and the sister in the poem similar? A. Both the girls and the sister tell a boy that he is making a poor choice in a sit

Aman is adding -17 + 9. He wants to write -17 as the sum of two numbers so that one of the numbers, when added to 9, will equal 0. Write integers in the blanks

Its asking to expand and simplify an expression for the area of this rectangle.

A triangle has an angle that measures 70°. The other two angles are in a ratio of 2:9. What are the measures of those two angles?

Which of these is equal to sin 58°?

(50 points + Brainliest) Create an equivalent expression that includes a set of parentheses that make the value of the expression 13. Remember, you can have 2 o

Determine whether each of the following sequences are arithmetic, geometric or neither. If arithmetic, state the common difference. If geometric, state the comm

anfabba15 anfabba15 · Answer 1 · 2020-01-03T04:07:44+01:00

Answer:

The final pressure in the container at 0°C is 2.49 atm

Explanation:

We apply the Ideal Gases law to know the global pressure.

We need to know, the moles of each:

P He . V He = moles of He . R . 273K

(1atm . 4L) / R . 273K = moles of He → 0.178 moles

P N₂ . V N₂ = moles of N₂ . R . 273K

(1atm . 6L) / R . 273K = moles of N₂ → 0.268 moles

P Ar . V Ar = moles of Ar . R . 273K

(1atm . 10L) / R . 273K = moles of Ar → 0.446 moles

Total moles: 0.892 moles

P . 8L = 0.892 mol . R . 273K

P = ( 0.892 . R . 273K) / 8L = 2.49 atm

R = 0.082 L.atm/mol.K

Eduard22sly Eduard22sly · Answer 2 · 2021-10-15T13:23:37+02:00

The final pressure in the evacuation container at 0 °C is 2.499 atm

To solve this question, we'll begin by calculating the number of mole of each gas. This can be obtained as follow:

For He:

Volume (V) = 4 L

Pressure (P) = 1 atm

Gas constant (R) = 0.0821 atm.L/Kmol

Temperature (T) = 0 °C = 273 K

Number of mole (n) =?

PV = nRT

1 × 4 = n × 0.0821 × 273

4 = n × 22.4133

Divide both side by 22.4133

n = 4 / 22.4133

n = 0.178 mole

For N₂:

Volume (V) = 6 L

Pressure (P) = 1 atm

Gas constant (R) = 0.0821 atm.L/Kmol

Temperature (T) = 0 °C = 273 K

Number of mole (n) =?

PV = nRT

1 × 6 = n × 0.0821 × 273

6 = n × 22.4133

Divide both side by 22.4133

n = 6 / 22.4133

n = 0.268 mole

For Ar:

Volume (V) = 10 L

Pressure (P) = 1 atm

Gas constant (R) = 0.0821 atm.L/Kmol

Temperature (T) = 0 °C = 273 K

Number of mole (n) =?

PV = nRT

1 × 10 = n × 0.0821 × 273

10 = n × 22.4133

Divide both side by 22.4133

n = 10 / 22.4133

n = 0.446 mole

Next, we shall determine the total moles of the gas in the container.

Mole of He = 0.178 mole

Mole of N₂ = 0.268 mole

Mole of Ar = 0.446 mole

Total mole =?

Total mole = 0.178 + 0.268 + 0.446

Total mole = 0.892 mole

Finally, we shall determine the pressure in evacuation container. This can be obtained as follow:

Volume (V) = 8 L

Gas constant (R) = 0.0821 atm.L/Kmol

Temperature (T) = 0 °C = 273 K

Number of mole (n) = 0.892 mole

Pressure (P) =?

PV = nRT

P × 8 = 0.892 × 0.0821 × 273

P × 8 = 19.993

Divide both side by 8

P = 19.993 / 8

P = 2.499 atm

Therefore, the final pressure in the evacuation container at 0 °C is 2.499 atm

Learn more: https://brainly.com/question/15474073