Asample of argon has a volume of 6.0 cm 3 and the pressure is 0.87 atm. if the final temperature is 35 degrees celsius, the final volume is 10 cm 3 and the final pressure is 0.52 atm, what was the initial temperature of the argon

= 36.185 °C

Explanation:

Using the combined gas law;

P1V1/T1 = P2V2/T2

In this case;

P1 = 0.87 atm

V1 = 6.0 cm³

T1 = ?

P2 = 0.52 atm

V2 = 10 cm³

T2 = 35 °C + 273 = 308 K

Therefore;

T1 = P1V1T2/P2V2

   = ( 0.87 × 6.0 × 308)/( 0.52 ×10)

   = 309.185 K

Therefore; Initial temperature = 309.185 K - 273 = 36.185 °C

12.32 L.

Explanation:

The following data were obtained from the question:

Mass of CH4 = 8.80 g

Volume of CH4 =?

Next, we shall determine the number of mole in 8.80 g of CH4. This can be obtained as follow:

Mass of CH4 = 8.80 g

Molar mass of CH4 = 12 + (1×4) = 12 + 4 = 16 g/mol

Mole of CH4 =?

Mole = mass/Molar mass

Mole of CH4 = 8.80 / 16

Mole of CH4 = 0.55 mole.

Finally, we shall determine the volume of the gas at stp as illustrated below:

1 mole of a gas occupies 22.4 L at stp.

Therefore, 0.55 mole of CH4 will occupy = 0.55 × 22.4 = 12.32 L.

Thus, 8.80 g of CH4 occupies 12.32 L at STP.