Find the volume in milliliters of 2.00 mol of an ideal gas at 36°C and a pressure of 1120 torr.
Use the ideal gas law and the appropriate Rvalue:
R=0.0821 atmL/mol. K
R= 8.31 kPa • L/mol •K
R= 62.4 torr • L/mol · K
mL

Answers

Answer 1

Answer:

V = 34430 mL

Explanation:

Given data:

Volume in mL = ?

Number of moles of gas = 2.00 mol

Temperature = 36°C (36+273= 309K)

Pressure of gas = 1120 torr

Solution:

Formula:

PV = nRT

V = nRT/P

V = 2.00 mol ×62.4 torr • L/mol · K × 309K / 1120 torr

V = 38563.2 torr • L / 1120 torr

V = 34.43 L

L to mL

34.43 L ×1000 mL / 1 L

34430 mL


Related Questions

If an M&M weighs 0.88 grams, how much would a mole of M&Ms
weigh?

Answers

Answer:

5.30x10²³g

Explanation:

A mole can be understood as an amount of something.

One mole of eggs are 6.022x10²³ eggs

One mole of atoms are 6.022x10²³ atoms

One mole of molecules are 6.022x10²³ molecules

And 1 mole of M&M are 6.022x10²³ M&M. As 1 M&M weighs 0.88g, 1 mole will weigh:

0.88g*6.022x10²³ =

5.30x10²³g

One mole of M&Ms will weigh  5.3 x 10²³ grams.

The given parameters

Weight of an M&M = 0.88 grams

One mole of every substance contains 6.022x10²³ atoms of the substance.

Weight of an M&M, implies an atom of M&M.

1 atom --------- 0.88 grams

 6.022x10²³ atoms = ?

= 0.88 g x  6.022x10²³

= 5.3 x 10²³ grams.

Thus, one mole of M&Ms will weigh  5.3 x 10²³ grams.

Learn more about number of atoms in a substance here: https://brainly.com/question/15193101

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