como se escribe la composicion seleniuro de sodio
Respuestas a la pregunta
Contestado por
0
Na2Se
Mm Na2Se = 124,93 g/mol
Na = 2 x 23 = 46 g/mol
Se = 1 x 78,96 = 78,96 g/mol
````````````````````````````````````````
Mm =124.96 g/mol
SODIO:
124.96 g --------- 100 %
46 g --------- x
x = 36.8 %
SELENIO:
124.96 g --------- 100 %
78,96 g --------- x
x = 63.19 %
Mm Na2Se = 124,93 g/mol
Na = 2 x 23 = 46 g/mol
Se = 1 x 78,96 = 78,96 g/mol
````````````````````````````````````````
Mm =124.96 g/mol
SODIO:
124.96 g --------- 100 %
46 g --------- x
x = 36.8 %
SELENIO:
124.96 g --------- 100 %
78,96 g --------- x
x = 63.19 %
Otras preguntas