Answer these questions fast! im on a time crunch right now

1st image, 17.79 g Mg(NO3)2

2nd image 1.47 mol NH3

3rd image Molar Mass Al(OH)3 = 78 g/mol

Explanation:

1st image

Molar Mass  Mg(NO3)2 = 148. 31 g/mol

0,12 mol Mg(NO3)2 x (148, 31 g Mg(NO3)2 / 1 mol Mg(NO3)2) = 17.79 gMg(NO3)2

2nd image

Molar Mass  NH3 = 17g/mol

25 g NH3 x ( 1 mol NH3/ 17 g NH3) = 1.47 mol NH3

3rd image

Molar Mass Al(OH)3 = 1 Al *(27) + 3 O * (16) + 3 H * (1) = 78 g/mol

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