Chemistry 3rd Edition by Allan Blackman solution manual

X is therefore aluminium.
 
1.47      2.0158 u. Regardless of the definition, the ratio of the mass of hydrogen to that of oxygen is the same. If ¹²C is assigned a mass of 24 u (twice its accepted value), then hydrogen would also have a mass twice its accepted value.
 
1.48      Taking the mass ratio of ¹⁰⁹Ag to ¹²C and multiplying it by 12, we see that the mass of ¹⁰⁹Ag is 108.90 u.
 
1.49      The average atomic mass is the sum of the mass of each isotope multiplied by its abundance. Therefore, the average atomic mass of antimony = (0.5736 × 120.9038 u) + (0.4264 × 122.9042 u) = 121.8 u
 
 
1.50 (a) 138 neutrons, 88 protons, 88 electrons
       (b) 8 neutrons, 6 protons, 6 electrons
       (c) 124 neutrons, 82 protons, 82 electrons
       (d) 12 neutrons, 11 protons, 11 electrons
 
1.51 (a) 52 neutrons, 38 protons, 38 electrons
       (b) 33 neutrons, 27 protons, 27 electrons
       (c) 15 neutrons, 15 protons, 15 electrons
       (d) 16 neutrons, 14 protons, 14 electrons
 
1.52      Make use of the periodic table to match these correctly:
(a)    Sr: it is the only group 2 element in the list
(b)    In: elements in the same column as Al have similar chemical properties
(c)    Co: transition metals lie in the d block
(d)    Ne: noble gases are in group 18
(e)    Pu: actinoids are in the f block
 
1.53 Consult the periodic table to find examples of various classes of elements.
(a)       Nonmetals lie to the right of the periodic table — they are the elements in green in figure 1.16. Therefore, any three of the following elements will give a correct answer: He, C, N, O, F, Ne, P, S, Cl, Ar, Se, Br, Kr, I, Xe, At, Rn.
(b)       Alkaline earth metals are in group 2. Therefore, any three of the following elements will give a correct answer: Be, Mg, Ca, Sr, Ba, Ra.
(c)       Elements 57–71 are lanthanoids. Therefore, any three of the following elements will give a correct answer: La, Ce, Pr, Nd, Pm, Sm, Eu, Gd, Tb, Dy, Ho, Er, Tm, Yb, Lu.
(d)       Chalcogens are in group 16. Therefore, any three of the following elements will give a correct answer: O, S, Se, Te, Po.
      

Additional exercises
 
1.54 (a) a metal
       (b) ⁵⁵Mn
       (c) 30
       (d) 25
       (e) approximately 4.58 times heavier — ⁵⁵Mn has an atomic mass of 54.94 u
 
1.55      From the data given, we can obtain the mass ratios X : Y, and X : O. Knowing the atomic mass of O allows us to calculate the atomic masses of X and Y.
 
mass ratio X : Y = 1 :  = 1 : 0.171
 
mass ratio X : O = 1 : = 1 : 0.0770
The latter result means that an atom of O is 0.0770 times as heavy as an atom of X. Knowing that the atomic mass of O = 15.999 u, we can calculate the atomic mass of X from the equation:
atomic mass of X = × 15.999 u = 207.8 u
We can now use this value to find the atomic mass of Y. We know that the mass ratio X : Y is 1: 0.171 so an atom of Y weighs  0.171 times the mass of an atom of X. Therefore, the atomic mass of Y = 0.171 × 207.8 u = 35.5 u. Knowing the atomic masses, you can use the periodic table to show that the compound XY₄ is PbCl₄.
 
1.56      55.85 u. The average atomic is the sum of the mass of each isotope multiplied by its abundance. Therefore, the average atomic mass = (0.0580 53.9396 u) + (0.9172