If 0.214 mol of argon gas occupies a volume of 343.4 mL at a particular temperature and pressure, what volume would 0.375 mol of argon gas occupy under the same conditions?

601.8mL

Explanation:

First, we'll begin by obtaining an equation connecting volume and number of mole together. This is illustrated below:

From the ideal gas equation:

PV = nRT

Divide both side by P

V = nRT/P

Now divide both side by n

V/n = RT/P

Under the same conditions simply means temperature(T), pressure (P) and the gas constant (R) are all constant i.e

RT/P => constant

Therefore, the above equation can be written as:

V1/n1 = V2/n2

Now we can solve for the final volume otherwise known as the new volume as follow:

Data obtained from the question include:

n1 (initial mole) = 0.214 mol

V1 (initial volume) = 343.4 mL

n2 (final mole) = 0.375 mol

V2 (final volume) =?

Applying the the equation V1/n1 = V2/n2, the final volume is obtain as illustrated below:

V1/n1 = V2/n2

343.4/0.214 = V2/0.375

Cross multiply to express in linear form

0.214 x V2 = 343.4 x 0.375

Divide both side by 0.214

V2 = (343.4 x 0.375)/0.214

V2 = 601.8mL

Therefore, 0.375 mol of argon gas will occupy 601.8mL under the same condition.

consider the reaction below.

 no rx

at 500 k, the reaction is at equilibrium with the following concentrations.

[pci5]= 0.0095 m

[pci3] = 0.020

[ci2] = 0.020 m

what is the equilibrium constant for the given reaction?

0.042

0.42

2.4

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