Chemistry 3rd Edition by Allan Blackman solution manual

 55.93949 u) + (0.0220 56.9354 u) + (0.0028 57.9333 u) = 55.85 u
 
1.57 As₂O₅. From the problem we find the following ratios:
 
  and 
 
Therefore 3.122 g represents 2/3 as many arsenic atoms as oxygen atoms. To find the relative mass of arsenic to oxygen we can divide this number by 2 to get 1.561 for As and divide 1.000 by 3 to get 0.3333 for oxygen.
 
 =  =
 
       In the unknown compound, there are 1.873 g As for every 1.000 g O. This represents        1.873/4.683 As atoms per O atom, or
 
 =  =
 

1.58 Mass of 1 atom of Mg = 1.66 054 × 10^–24 g × 24.305 = 4.0359 × 10^–23 g
       Mass of 1 atom of Fe = 1.660 54 × 10^–24 g × 55.847 = 9.2736 × 10^–23 g
 
       In 24.305 g of Mg, there will be 6.0222 × 10²³ atoms of Mg
       In 55.847 g of Fe, there will be 6.0221 × 10²³ atoms of Fe
 
The answers are essentially the same. We can therefore assume that in 40.078 g of Ca, there are approximately 6.022 × 10²³ atoms of Ca.
 
1.59      We assume that a typical atom is spherical. The volume of a sphere is given by the equation
Therefore, the volume enclosed by the nucleus is:

The volume enclosed by the atom is:

 
Therefore, there is a factor of 10¹⁵ difference between the volume occupied by the nucleus and the total volume of the atom.
1.60      helium, neon, argon, krypton, xenon and radon exist as discrete atoms X.
 
hydrogen, nitrogen, oxygen, fluorine and chlorine exist as diatomic molecules X₂.
 
1.61      (a)     F^– contains 9 + 1 = 10 electrons
(b)        O₂^– contains (2 ´ 8) + 1 = 17 electrons
(c)        CO₃^2– contains 6 + (3 ´ 8) + 2 = 32 electrons
(d)        Na⁺ contains 11 – 1 = 10 electrons
(e)        PO₄^3– contains 15 + (4 ´ 8) + 3 = 50 electrons
(f)        ClO₄^– contains 17 + (4 ´ 8) + 1 = 50 electrons
 
1.62      (a)     O^2–
(b)        F^–
(c)        Li