Q.34 The temperature of a 12.58 g sample of calcium carbonate increases from 23.6 °C to 38.2C. If the specific heat of calcium carbonate is 0.82 J/g-K, how many joules of heat are absorbed?

150.6 Joules

Explanation:

Calorimetry problem must be solved by the calorimetry formula.

Q = c . m . ΔT

ΔT = Final T° - Initial T°

c = specific heat (Notice, that in the unit we have K as unit for T°, and the T° is in °C. We don't need to make the conversion, because is a Δ)

For example 100°C - 50°C = 50°C

373K - 323K = 50 K

So let's replace data given: Q = 0.82 J/g.K . 12.58 g . (38.2 - 23.6)

Q = 150.6 Joules

We assume no heat lost by sourrounding

a. the law of proportion states the combination of 2 atoms in a specific ratio to produce multiple compounds such as the given statement of question ‘1st compound contains 88.8 % cu ’ indicates 88.8 g of cu is present along with 11.2 g oxygen in 100 g of compound whereas the statement ‘2st compound contains 79.9 % cu ’ indicates 79.9 g of cu is present along with 20.1 g oxygen in 100 g of compound

1st compound contains 88.8 g cu

for cu

m = 63 g /mol

atomic % = 88.8 /63 = 1.42

for o

m = 16 g /mol

atomic % = 11.2 /16 = 0.7

b. the emperical formula of 1st compound is

cu1.4o0.7

cu2o