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Silvania012
@Silvania012
November 2019
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Indique a massa presente nas seguintes amostras :a) 1,5 mol de átomos de ferro (MM- Fe+ 56/gmol
B) 3 mol de átomos de N2. /9/mm-N =14g/mol)
c) 0,2 mol de átomos de C3H6O2.(MM-C=12g/mol/ H=1g/mol/O=16g/mol)
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Rich0031
Nas 3 Letras Basta usar essa fórmula:
n = m/MM em que:
n = quantidade de matéria (mol)
m = massa ( g )
MM = massa molar ( g/mol )
A) n = m/MM
1,5 = m/56
m = 56•1,5
m = 84g
B) n = m/MM ( N2 → 14g/mol • 2 = 28g/mol )
3 = m/28
m = 28•3
m = 84g
C) C3H6O2
C3 = 12g/mol • 3 = 36g/mol
H6 = 1g/mol • 6 = 6g/mol.
O2 = 16g/mol • 2 = 32g/mol
Agora Basta somar:
36g/mol
32g/mol
06g/mol
-------------
74g/mol
n = m/MM
0,2 = m/74
m = 74•0,2
m = 14,8g
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silvania012
Obrigada!
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Lista de comentários
n = m/MM em que:
n = quantidade de matéria (mol)
m = massa ( g )
MM = massa molar ( g/mol )
A) n = m/MM
1,5 = m/56
m = 56•1,5
m = 84g
B) n = m/MM ( N2 → 14g/mol • 2 = 28g/mol )
3 = m/28
m = 28•3
m = 84g
C) C3H6O2
C3 = 12g/mol • 3 = 36g/mol
H6 = 1g/mol • 6 = 6g/mol.
O2 = 16g/mol • 2 = 32g/mol
Agora Basta somar:
36g/mol
32g/mol
06g/mol
-------------
74g/mol
n = m/MM
0,2 = m/74
m = 74•0,2
m = 14,8g