1. a. -7    b. +2   c.  +4   d.  +3   e. -1

2.  Na₂Ca₂B₁₀O₁₈.16H₂O  <  Na₂O + 2CaO + 5B₂O₃ + 16H₂O   and factoring the left side by 2 =

     NaCaB₅O₉. 8H₂O    next assigning the (OH)_x and yH₂O, accordingly =

     NaCaB₅O₈(OH)₂. 7H₂O

     NaCaB₅O₇(OH)₄. 6H₂O

    NaCaB₅O₆(OH)₆. 5H₂O   answer based on 6:5 (OH)_x:yH₂O as problem defines

3.  Cu₃SO₄(OH)₄ > 3CuO + SO₃ + 2H₂O

      CuO = (239/353) x 100 = 67.3%: SO₃ = (80/353) x 100 = 22.5%:  H₂O = (36/353) x100 = 10.2%

4.  MgO = (31.7/40) = (0.7925/0.266) = 3

     SiO₂ = (63.5/60) = (1.058/0.266) = 4        Mg₃Si₄O₁₁.H₂O <  3MgO + 4SiO₂ + 1H₂O

     H₂O = (4.8/18) = (0.2666/0.266) = 1

5. a. minus1/2a: 4b: 1c      b. (minus8, 1, 4)

6  a.  (minus3, 2, 2)    b.  (2, minus4, 0)  c. (1, 24, minus2)

7. SG = (4 x100 / 6.02 x 10²³  x 22.69 x10^-23) = 2.93

8. a  1.  6       2.  24     3.  12    4.  12

    b. Cu = 12,  S = 4,  H = 16,  O = 32

    c. 1. Potash   2. Soda   3. Lime

    d.  1. Primitive   2. C faced centered    3.  Body Centered (I)

    e.  1. prismatic   2. ditetragonal dipyramidal   3.  trigonal scalenohedral    4. hexoctahedral