Question

Help! I didn't understand my lab and can't do the post-lab assignment :(

Answer 1

Heat lost by metal is gained by the water in calorimeter .

m1s1(∆T)1 = m2s2(∆T)2

5.68g× s× (100-27.3) ℃= 17.36g × 4.184J/g.℃ ×(27.3-25.1)℃

s = 0.38697 = 0.387 J/g.℃ (Answer)

(2)

The specific heat capacity of copper is 0.385 J/g.℃ .

Hence the given metal is copper ( Cu) . (Answer)

(3)

Heat supplied by calorimeter = ms∆T = 18.45g × 4.184J/g.℃ ×(25.1 -22.5)℃ = 200.70648 J

∆H_rxn = + 200.7 J. (Answer)

(4)

∆H_rxn = 200.70648J/5.12g = + 39.2 J/g. (Answer)

∆H_rxn = 39.2 J/g × 82.0338g/mol = 3215.76 J/mol

= 3215.8 J/mol. (Answer)

(5)

Since ∆H_rxn is positive , the reaction is endothermic . (Answer)

As temperature of water decreases in calorimeter that means heat is absorbed from the water and utilised in the reaction .