1.         avg. mass = 20*0.9092+21*0.0026+22*0.0882;

                              =  20.1790 u

2.         hypothetical avg mass = 0.20*(29+34)+0.80*(29+36);

                               = 64.60 u (numbers given in problem are not real, so it doesn’t match actual atomic mass of Cu

3.         only 2

4.         avg mass = 0.79*24+0.10*(25)+0.11*26;

                        = 24.3 u

5.         0.7735*35 + (1-0.7735)x =3 5.453

                                 x = 37 u

6.         H         1

            O         16

            C         12

7.         There is a typo. It should read: Assume that ¹²C accounts for 99% of carbon. If the rest consists of ¹³C and a little less ¹⁴C, then ¹⁴C makes up what percent of carbon?

(Hint: there seems to be information missing, but if so, you are forgetting to look something up in the periodic table.)

            Let x = percent decimal of ¹³C

0.01  – x = amount of ¹⁴C

So 0.99(12) + 13x + 14(0.01-x) = 12.011

x = 0.009000000000 = 0.9 % ¹³C

amount of ¹⁴C = 0.01 – 0.009 = 0.001 = 0.1%