Question

A) A 9.55−g sample of a piece of diamond at 185 oC was placed in a container containing water at 87.0 oC. If the final temperature of the mixture is 91.5 oC, what is the mass of the mixture?The specific heat capacity of diamond and water are 0.52 J g−1 oC−1 and 4.184 J g−1 oC−1 respectively. Question options: A) 50.1 g B) 125 g C) 24.7 g D) 34.2 g

B)Calculate the enthalpy for the reaction:CH4 + 4Cl2 → CCl4  + 4HCl  △H = ?C +2H2 → CH4              △H = −74.6 kJ?C +2Cl2 → CCl4             △H = −95.7 kJ?H2 + Cl2 → 2HCl            △H = −184.6 ?

Question options:

	A) -354.9 KJ
	B) -205.7 KJ
	C) -390.3 KJ
	D) +354.9 KJ

Answer 1

A) answer : D) 34.2 g

here heat loss by diamond = heat gain by water

(m Cp dT) diamond = ( m Cp dT) water

9.55 x 0.52 x (185 - 91.5 ) = m x 4.184 x (91.5 - 87)

464.321 = 18.828 m

m = 24.66 g

mass of water = 24.66 g

mass of mixture = 24.66 + 9.55 = 34.2 g

mass of mixture = 34.2 g