kaziyahf2006
02.10.2019 •
Chemistry
Alaboratory procedure calls for making 590.0 ml of a 1.1 m kno3 solution. how much kno3 in grams is needed?
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Ответ:
C = n/V
n = 1,1M × 0,59 dm³
n = 0,649 mol
M KNO₃ = 39g+14g+16g×3 = 101 g/mol
1 mole 101g
0,649 X
X = 101×0,649
X = 65,549g KNO₃
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(0,1)
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