10. If 3.5 kJ of energy are added to a 28.2 g sample of iron at 20°C, what is the final temperature of the iron in kelvins? The specific heat of iron
is 0.449 J(g•K).

569K

Explanation:

Q = 3.5kJ = 3500J

mass = 28.2g

∅1 = 20°C = 20 + 273 = 293K

∅2 = x

c = 0.449

Q = mc∆∅

3500 = 28.2×0.449×∆∅

3500 = 12.6618×∆∅

∆∅ = 3500/12.6618

∆∅ = 276.4220

∅2 - ∅1 = 276.4220

∅2 = 276.4220 + ∅1

∅2 = 276.4220 + 293

∅2 = 569.4220K

∅2 = 569K

what are the statements

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