What amount of heat (in kJ) is required to convert 28.4 g of an unknown solid (MM = 67.44 g/mol) at -15.4 °C to a liquid at 42.7 °C? (specific heat capacity of solid = 1.95 J/g･°C; specific heat capacity of liquid = 1.18 J/g･°C; ∆Hfus = 5.72 kJ/mol; normal freezing point, Tempf = 28.3°C)

mention the spill to the teacher clean the area and throw the broken pieces away

explanation:

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

the answer is a for sure mark brainliest i had got this right on usatestprep