Given δh°f for so2 (g ) (-296.8 kj/mol) and so3 (g ) (-395.7 kj/mol), calculate δh°rxn for the conversion of so2 to so3.

Hey there!:

Given that;

SO2 (g ) = (-296.8 kJ/mol) and SO3 (g )= (-395.7 Kj/mol)

2 SO2(g) + O2(g) > 2 SO3(g)

ΔH reaction = ΔH of product - ΔH of reactant

We know that  ΔH O2 = 0 kj

ΔH reaction = [2 x - 395.7] - [2 x - 296.8 +0.0 KJ]

= - 791.4 - (- 593.6)

= - 791.4 + 593.6

= - 197.8 kJ

Because there are 2 mole of SO2 coverted into 2 mole of SO3 :

=- 197.8 kJ/2 mol

ΔH°rxn  = - 98.9 kJ/ mol

Hope that helps!

the density of gold = 19.3 g/cm3

the relation between density and mass and volume is

density = mass / volume

given :

mass of gold = 13 g

density = 13g / volume = 19.3

volume = 13 /19.3 = 0.67 cm^3

thus volume of gold nugget = 0.67 cm^3

it is b that is the answer.