Question

Calculati in conditii normale:
a) masa cantarita de 10 moli HCL
b) volumul ocupat de 56 g CO
c) numarul de atomi din 34 g NH3

Answer 1

Răspuns:

Explicație:

10moli HCl

MHCl= 1+ 35,5=36,5------> 36,5g/moli

m=n .M

m=10moli. 36,5g/moli=365g HCl

56g CO

VCO

MCO=12+16=28------->28g/moli

n=m:M

n= 56g: 28g/moli=2moli CO

VCO=2moli . 22,4L/moli=44,8 L

34g NH3

nr. atomi

-in molecula  NH3  se afla 4atomi [ 1N+3H]

MNH3=14 +3=17------->17g/moli

n=34g : 17g/moli=2moli NH3

1mol NH3---------4 . 6,023.10²³atomi

2moli------------------x

x=2 .4 .6,023.10²³atomi=48,184.10²³atomi