Chemistry 3rd Edition by Allan Blackman solution manual

 
1.24 (a) I    
(b)    W
(c)    Xe
(d)    Sm       
(e)    Pu
 
1.25 ductility
 
1.26 malleability
 
1.27 oxygen (O₂), nitrogen (N₂), fluorine (F₂), chlorine (Cl₂)
 
1.28 mercury and bromine
 
1.29 Lv is more likely, given its position on the periodic table.
 
1.30 Refer to figure 1.16 in the textbook.
 
1.31      Orbital. An orbital is a region in space where there is a non-zero probability of finding an electron.
 
1.32      Both NO and NO₂ must have unpaired spins, because they have a total of 15 and 23 electrons, respectively; both odd numbers. The other molecules all have an even number of electrons.
 
1.33      Atoms in excited states can emit excess energy as light (photons).
 
1.34      Quantisation generally involves energies that are miniscule with respect to the macroscopic world. Therefore, even though quantisation does occur, we cannot relate to it.
 

Review problems
 

1. Compound (c). An authentic sample of methane must have a mass ratio of carbon/hydrogen of 1.000 to 0.336. The only possibility in this list is therefore (c).

 
1.36      Compound (d).  An authentic sample of calcium chloride must have a ratio of 1.00 to 1.77. The only possibility in the list is (d) which has the ratio of the mass of chlorine to calcium of 2.39/1.35 = 1.77.
 
 
1.37      98.5 g chlorine. From the mass ratio given we must multiply the mass of carbon by 11.8 to obtain the mass of chlorine.
 
1.38      The given mass ratio of phosphorus to chlorine is 1.20 : 4.12. Dividing both numbers by 1.20 gives the ‘simpler’ ratio 1.00 : 3.43. Dividing the mass of chlorine by 3.43, we find that for every 6.22 g chlorine there will be 1.81 g phosphorus.
 
1.39      64.0 g of carbon tetrachloride. We know the carbon to chlorine mass ratio is 1.00 : 11.8. Multiplying 5.00 g by 11.8 gives 59.0 g of chlorine. Thus, the mass of carbon tetrachloride will be 5.00 g + 59.0 g = 64.0 g.
 
1.40      55.4 g of phosphorus chloride. From the ratio in problem 1.38, we know that the phosphorus to chlorine mass ratio is 1.00 : 3.43. Multiplying 12.5 g by 3.43 gives 42.9 g chlorine. Thus, the mass of phosphorus chloride formed will be 12.5 g + 42.9 g = 55.4 g.
 
1.41      2.664 g oxygen. If there are twice as many oxygen atoms per carbon atom, there must be twice the mass of oxygen per mass of carbon (2  1.332 g).
 
1.42      The ratio should be 4/2 = 2/1, as required by the formulae of the two compounds. The mass of chlorine in compound 2 would be 2 ´ 0.597 g Cl = 1.19 g Cl.
 
1.43 12 × 1.660 540 2 × 10^–24 g = 1.992 648 2 × 10^–23 g for one ¹²C atom
 
1.44 23 × 1.660 540 2 × 10^–24 g = 3.819 246 6 × 10^–23 g for one ²³Na atom
 
1.45      As there are 2 atoms of nitrogen in a molecule of nitrous oxide, dividing the mass of nitrogen by 2 will give the mass ratio of one atom of nitrogen to one atom of oxygen. This is 0.875 65. Multiplying this by 16.00, the atomic mass of ¹⁶O, gives the atomic mass of nitrogen. Thus 16.00 ´ 0.875 65 = 14.01 u.
 
1.46      We are told that the formula of the compound is X₂O₃ and that in this compound, 1.125 g of X is combined with 1.000 g of O. From these data, we can calculate the ratio of the masses of X and O as being . This means that a single atom of X will be  times as heavy as a single atom of O. We know the atomic mass of O is 15.9994 u, so the atomic mass of X will be  . The unknown element