What mass of water is produced from the complete combustion of 6.60×10−3 g of methane?

the answer would be c

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

answer is: empirical formula is tl₂o. if we use 100 grams of compound, tha:

m(tl) = 96.2%  · 100 g ÷ 100%.

m(tl) = 96.2 g.

n(tl) = m(tl)  ÷ m(tl). n(tl) = 96.2 g  ÷ 204.4 g/mol.

n(tl) = 0.47 mol. m(o) = 3.77 g. n(o) = m(o)  ÷ m(o). n(o) = 3.77 g  ÷ 16 g/mol. n(ag) = 0.235 mol. n(tl) . n(o) = 0.47 mol : 0.235 mol /0.235. n(tl) . n(o) = 2 : 1.

m(tl₂o) = 204.4 g/mol  · 2 + 16 g/mol = 424.8 g/mol.

m(compound) = 456. 8 g/mol - 424.8 g/mol = 32 g/mol ( 2 oxygens); tl₂o₃.