12 VETERINARY STATE BOARD 



State Avogadro's law. 



Equal volumes of gases, compared under identical conditions of 

 temperature and pressure, contain equal numbers of molecules. 

 Name the elements that enter into the composition of each of the fol- 

 lowing alloys: (a) brass, (b) German silver, (c) soft solder, 

 (d) bell metal. 

 (a) Copper and zinc, (b) copper, zinc and nickel, (c) tin and 

 lead, (d) copper and tin. 

 Is glass a compound or a mixture? To what does green glass owe its 

 color? 

 Glass is a mixture. Green glass owes its color to silicates of iron 

 derived from the impure materials of which it is made. 



What element occurs in all acid compounds? 

 Hydrogen. 



Name two classes of salts and distinguish between the classes named. 



Acid salts are acids in which only a portion of their replaceable 

 hydrogen atoms have been replaced, e.g., KHSO4, potassium hydro- 

 gen sulphate. Acid salts are generally acid in reaction to litmus. 



Basic salts are salts containing a higher proportion of a base 

 than is necessary for the formation of a salt, e.g., Pb(0H)N03, basic 

 lead nitrate. 



What gas is evolved when copper acts on nitric acid? Account for 

 the formation of this gas. 

 Nitric oxide, NO. 



3Cu + 8HNO3 = 3Cu(N03), + 2N0 + 4H,0. 

 Determine how much sulphuric acid and how much copper will be 

 needed to produce 1,000 grammes of copper sulphate by the 

 reaction Cu + zU^SO^ =. CuSO^ + SO^ + 2H,0. [Atomic 

 weight of S = 32, of copper = 63, of O = 16.] 

 See p. 11 regarding determination of percentage composition. 

 Molecular weight of Cu = 63, of sulphuric acid = 196, of copper 

 sulphate = 159; it takes 63 + 196 or 259 parts of copper and 

 sulphuric acid to make 159 parts of copper sulphate. So, 259 • 159 = 

 X : 1000 ; X = 1629. 63/259 of 1629 = 396, and 196/259 of 1629 = 

 1233. Substituting grammes for parts, we have 396 grammes of 

 copper and 1233 grammes of sulphuric acid necessary to make 1000 

 grammes of copper sulphate. 



Find the number of grammes of oxygen that can be prepared from 10 

 grammes of KCIO3. [Atomic weight of K ■= 39, of CI = is 

 of O = 16.] ^^' 



