54 ANSWERS TO PRACTICAL QUESTIONS 



SO 3 = 40 = equivalent of the constituent. 

 FcO.SO,+7HO = 139 = t4 " compound. 



x = weight of the constituent. 

 50 Ibs. = " *' compound. 

 SO S : (FeO.SO 3 -i-7HO) :: x : 50 Ibs. 

 40 : 139 : : x : 50 Ibs. 

 x = 14 /'/gibs. (SO 3.) 



4. The equivalent of the chloride of sodium -(salt) is 58. 5. 

 In 10 Ibs. there are 6 t rr Ibs. of chlorine : what is the equiva* 

 lent of Cl? 



x = equivalent of the given constituent. 

 58.5= " " " compound. 



6 if 7 Ibs. = weight of the given constituent- 

 10 Ibs. = " " " compound. 

 x : 58.5 : : 6 -, f T Ibs. : 10 Ibs. 

 x = 35.5. 



5. In 20 grains of bromide of potassium there are 6^9 

 grains of potassium : the equivalent of potassium being 39, 

 what is the equivalent of the bromide of potassium ? 



89 = equivalent of the given constituent. 

 x= " " " compound. 



6 ~ffe g re = weight of the given constituent. 

 SOgre. = * " " compound. 

 39 : x : : 6 -^ grs. : 20 grs. 

 a? =119. 



6. In 14 Ibs. of iron-rust (Fe^O^ how much O? 



3O = 24 = equivalent of the given constituent. 

 Pe a O 8 = 80 = " " " compound. 



* = weight of the given constituent. 

 14 Ibs. = " " " compound. 

 24 : 80 : : x : 14 Ibs. 

 a; = 4 Vilbs. (O). 



7. In 20 Ibs. of glass (NaO.SiO^ + CaO.SiOJ how many 

 tt>s. of sand (SiOJ ? 



2 SiO 2 = <!0 = equivalent of the given constituent. 

 (NaO.SiO 2 + CaO.SiO 2 ) = 119= " " " coapound. 



x = weight of the given constituent. 

 80 Ibs. = *' ' " compound. 

 60: 119: : x: 20 Ibs. 

 X = 10 ^ Ibs. (SiOj) 



