A 3.74-L cylinder contains 4.17 g of methane, CH4, at a pressure of 3020 mmHg. What is the temperature of the gas

T = 694 K

Explanation:

Given the following data:

V= 3.74 L

mass = m = 4.17 g

P = 3020 mmHg x 1 atm/760 mmHg= 3.97 atm

We use the ideal gases equation to calculate the temperature (T):

PV= nRT ⇒ T= PV/nR

We calculate n (number of moles of methane gas) by dividing the mass into the molar mass of methane (CH₄):

molar mass(CH₄) = MM = (1 x 12 g/mol)+ (4 x 1 g/mol) = 16 g/mol

n = m/MM= 4.17 g/(16 g/mol)= 0.261 mol

Finally, we calculate T:

T= PV/nR = (3.97 atm x 3.74 L)/(0.261 mol x 0.082 L.atm/K.mol) = 694 K

when [oh-] is 10-5  m, and [h3o+] is 10-9  m, the poh is 5 and ph is 9. again in both cases the sum of ph and poh is 14. however, the hydrogen ion concentration is not always going to be equal to exactly 1 x 10 raised to a negative number. for example, we skipped over the value of 2.0 x 10-7.

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