oblicz, jaką część mola stanowi a) 0,4 g wodorotlenku sodu NaOH
b) 6,3 g kwasu azotowego (V) HNO3
a)
40g NaOH --- 1 mol
0,4g NaOH ---X
X=0,1 mola
b)
63g HNO3 --- 1 mol
6,3g HNO3 ---X
n=x - liczba moli
m=0,4g
n=m/M
M- masa molowa(MNa + MO +MH- z ukladu)
n=0,4/40g/mol
n=0,01mola
b)m=6,3g HNO3
MHNO3= 63g/mol
n=6,3g/63g/mol= 0,1mola
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a)
40g NaOH --- 1 mol
0,4g NaOH ---X
X=0,1 mola
b)
63g HNO3 --- 1 mol
6,3g HNO3 ---X
X=0,1 mola
n=x - liczba moli
m=0,4g
n=m/M
M- masa molowa(MNa + MO +MH- z ukladu)
n=0,4/40g/mol
n=0,01mola
b)m=6,3g HNO3
MHNO3= 63g/mol
n=6,3g/63g/mol= 0,1mola