Balance the Following Equations: Instruction: While balancing equation write the physical state of reactants and products as well as any reaction conditions. Write in steps

1) CuSO4 + KI →Cu2I2 + K2SO4 + I2

2) NH3 + O2 →NO + H2O

3)Fe2O3 + CO → Fe + CO2

4) Cu + AgNO3 → Cu(NO3)2 + Ag

5) Pb(NO3)2 + H2SO4 → PbSO4 + HNO3

6) CaCO3 + HCl → CaCl2 + H2O(l) + CO2

7)MnO2 + HCl → MnCl2 + H2O + Cl2​

the answer is b. dilution

atomic mass = 39

number of neutrons = 20

number of protons= 19

electronic configuration = 1s2 2s2 2p6 3s2 3p6 4s1

the number of electron in first shell = 2

number of electrons in n=3 shell = 8

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