Achloride of silicon contains 79.1 mass % cl. (a) what is the empirical formula of the chloride?

SiCl₃

Explanation:

To solve this problem, let's assume that we have 100 g of that chloride of silicon. We then would have 79.1 g of Cl and 100 - 79.1=20.9 g of Si.

Then we calculate the amount of moles in the mass of each element, using the elements' atomic weight:

Now we divide the number of moles by the lowest number, in this case, it's the moles of Si:

2.2313 / 0.7443 = 2.9978 => 3

0.7443 / 0.7443 = 1

Thus the empirical formula is SiCl₃

answer; talc and gypsum;

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