Question

Ajutor la chimie! Dau coroana​

Answer 1

Răspuns:

Explicație:

4.

nr.moli

300g Al2S3

n= m : M

M=2.27 +3.32=150------->150g/moli

n= 300g  : 150g/moli=  2 moli Al2S3

10g CaCO3

40+12+3.16=100--------->100g/moli

n= 10g : 100g/moli =0 ,1 moli CaCO3

80g NaOH

M=23+16+1=40--------->40g/moli

n=80g : 40g/moli=2moli NaOH

5.

g subst.

0,2 moli Na2SO4

m = n . M

M=2.23+ 32 +4.16=142-------->142g/moli

m= 0,2moli . 142g/moli=28,4 g Na2SO4

5moli HNO3

M=1+14+3.16=63-------->63g/moli

m=5moli .63g/moli= 315g HNO3

10moli Ca(OH)2

M=40+ 2.16 +2=74------->74g/moli

m= 10moli . 74g/moli=740g Ca(OH)2

6.

g azot

340g NH3

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

17g HNO3---------14gN

340g NH3-----------X

X=340. 14 :17=280 g N

7.

Comp.proc . a K2CO3

M=2.39+ 12 +3.16=138-------->138g/moli

138 g K2CO3--------78g K---------12g C---------48g O

100g---------------------x-------------------y----------------z

x=100.78:138=57,52% K

y=100.12:138 =8,7% C

Z=100.48 : 138=34,78% O

8.

nr. molecule

36g H2O

MH2O=2+16=18------->18g/moli

n= 36g : 18g/moli=2moli H2O

1moli H2O-----------6,023.10²³molec. H2O

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

x=2 . 6,023.10²³=12,046  .10²³ molec. H2O

9.

nr. neutroni

71 g Cl2

MCl2 = 2 .35,5 =71--------> 71g/moli

Z=17          17p ,17e  ,18n

A=35

1mol Cl2 -----------18 . 6,023.10²³  n