Question

1. Calculează masa molară pentru următoarele substanțe chimice:

Answer 1

Răspuns:

Explicație:

1.

MAgNO3= 108 + 14 + 3.16=170-------> 170g/moli

MNaCl= 32 + 35,5= 58,5 -------> 58,5g/moli

MK3PO4=3.39 + 31 + 4.16=212-------> 212g/moli

MH3PO4=3 + 31 + 4.16=98-----> 98g/moli

MMgBr2 = 24 + 2.80=184-----> 184g/moli

MCa(OH)2= 40 + 2.16 + 2=74------> 74g/moli

MCu(NO3)2=64 + 2.14 + 6.16=188------> 188g/moli

MCaCO3= 40 + 12 + 3.16=100-------> 100g/moli

MFe2(SO4)3= 2 . 56 + 3 . 32 + 12 . 16=

MNH4NO3= 2 . 14 + 4 + 3 .16=

2.

1260 g HNO3

n= m : M

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

n= 1260 g : 63g/mol i=20 moli HNO3

1mol HNO3---------6,022.10²³molecule HNO3

20 moli-------------------x

x= 20 . 6,022.10²³molecule =

7,3g HCl

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

n= 7,3g : 36,5 g/moli=0,2 moli HCl

1mol HCl--------6,022.10²³molecule HCl

0,2moli---------------------y

y= 0,2  . 6,022.10²³molecule =