jaka masę wyrażoną w gramach ma 0,25 mola tlenku siarki(VI) ?
SO3 - 32g/mol + 16g/mol x 3 = 80g/mol
1mol - 80g/mol
0.25 mol - x
x = 80g/mol : 0.25mol = 20 g/mol
x= 20g/mol
n=0,25 mola
m=?
M SO3=32+48=80
n=m/M
m=Mn
m=80*0,25=20 gram
" Life is not a problem to be solved but a reality to be experienced! "
© Copyright 2013 - 2024 KUDO.TIPS - All rights reserved.
SO3 - 32g/mol + 16g/mol x 3 = 80g/mol
1mol - 80g/mol
0.25 mol - x
x = 80g/mol : 0.25mol = 20 g/mol
x= 20g/mol
n=0,25 mola
m=?
M SO3=32+48=80
n=m/M
m=Mn
m=80*0,25=20 gram