Identical to the number of substances contained in water, 18 and 17 g H2S?
237
380
Ответы на вопрос:
given:
m(Н2О) = 18 g
m(H2S) = 17 g
find:
ν(Н2О) - ?
ν(H2S) - ?
Solution:
1- way:
М(Н2О) = 18 g/mol
18 g Н2О ----- х mol Н2О
18 g Н2О ----- 1 mol Н2О
х=(18 g • 1 mol)/(18 g) =1 mol
2- way:
М(Н2О) = 18 g/mol
ν=m/М ;ν(Н₂О) =(m(Н₂О))/(М(Н₂О)) = (18 g)/(18 g/mol) = 1 mol.
1- way:
М(H2S) = 34 g/mol
17g H2S ----- х mol H2S
34 g H2S ----- 1 mol H2S
х= (17 g • 1 mol)/(34 g) =0,5 mol
2- way:
М(H2S) = 34 g/mol
ν=m/М ;ν(H₂S) = m(H₂S)/М(H₂S) = (17 g)/(34 g/mol) = 0,5 mol.
Answer:ν(Н2О) = 1 mol; ν(H2S) = 0,5 mol - Misc.
m(Н2О) = 18 g
m(H2S) = 17 g
find:
ν(Н2О) - ?
ν(H2S) - ?
Solution:
1- way:
М(Н2О) = 18 g/mol
18 g Н2О ----- х mol Н2О
18 g Н2О ----- 1 mol Н2О
х=(18 g • 1 mol)/(18 g) =1 mol
2- way:
М(Н2О) = 18 g/mol
ν=m/М ;ν(Н₂О) =(m(Н₂О))/(М(Н₂О)) = (18 g)/(18 g/mol) = 1 mol.
1- way:
М(H2S) = 34 g/mol
17g H2S ----- х mol H2S
34 g H2S ----- 1 mol H2S
х= (17 g • 1 mol)/(34 g) =0,5 mol
2- way:
М(H2S) = 34 g/mol
ν=m/М ;ν(H₂S) = m(H₂S)/М(H₂S) = (17 g)/(34 g/mol) = 0,5 mol.
Answer:ν(Н2О) = 1 mol; ν(H2S) = 0,5 mol - Misc.
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