A cylinder of oxygen (O2) gas at STP has a volume of 2.75 L. How many oxygen molecules does it contain?

7.40×10²² molecules.

Explanation:

From the question given above, the following data were obtained:

Volume of oxygen = 2.75 L

Number of molecules of oxygen =?

Next, we shall determine the number of mole of oxygen present. This can be obtained as follow:

22.4 L contains 1 mole of oxygen at STP.

Therefore, 2.75 L will contain = 2.75 / 22.4 = 0.123 mole of oxygen at STP.

Finally, we shall determine the number of molecules of oxygen present. This can be obtained as follow:

From Avogadro's hypothesis:

1 mole of oxygen = 6.02×10²³ molecules

Therefore,

0.123 mole of oxygen = 0.123 × 6.02×10²³

0.123 mole of oxygen = 7.40×10²² molecules.

Thus, the cylinder contains 7.40×10²² molecules of oxygen.

answer: i dont know the answer but this might

explanation:

l

answer : (a) volume of the gas = 8.67 ×   l

                (b) second virial coefficient b = -1.21226 l/mole

solution :   given,     t = 300k

                                p = 20atm

                                z = 0.86

                                n = 8.2 mmoles = 8.2 × moles

                                r =   0.082 l atm/k/mol

        find,     (a) volume of gas   and (b) second virial coefficient b at 300k.

formula used :            

                          and            

(a)            

                    = \times 0.082\times 300}{20}[/tex]

                    = 8.67 ×   l

(b)      

                    = (0.00867-1) × 8.659

                    = - 8.5839 l/mole