Identical to the number of substances contained in water, 18 and

given:
m(Н2О) = 18 g
m(H2S) = 17 g
find:
(Н2О) - ?
(H2S) - ?
Solution:
1- way:
М(Н2О) = 18 g/mol
18 g Н2О ----- nbsp;х mol Н2О
18 g Н2О ----- 1 mol Н2О
х=(18 g nbsp; 1 mol)/(18 g) nbsp;nbsp;=1 mol
2- way:
М(Н2О) = 18 g/mol
=m/М nbsp;;(НО) =(m(НО))/(М(НО)) nbsp;= (18 g)/(18 g/mol) nbsp;= 1 mol.
1- way:
М(H2S) = 34 g/mol
17g H2S ----- nbsp;х mol H2S
34 g H2S ----- 1 mol H2S
х= (17 g nbsp; 1 mol)/(34 g) nbsp;nbsp;=0,5 mol
2- way:
М(H2S) = 34 g/mol
=m/М nbsp;;(HS) = m(HS)/М(HS) nbsp;nbsp;= (17 g)/(34 g/mol) nbsp;= 0,5 mol.
Answer:(Н2О) = 1 mol; (H2S) = nbsp;0,5 mol - Misc.