Question

Se dau substanțele:
-dioxid de carbon
-carbonat de magneziu
-hidroxid de zinc
SE CERE:
a)formula chimica
b)tipul de substanța
c)calculează masa molara
d)formula procentuală
V A R O G AJUTAȚI-MĂ!!!!!!


D A U C O R O A N A !!!!

Answer 1

Răspuns:

Explicație:

CO2

M=12 + 2.16=44----> 44g/moli

44gCO2----------12g C------------32g O

100g CO2--------------x-------------y

x= 100.12 :44= 27,27% C

y= 100.32 : 44=72,73% O

MgCO3

M= 24 +12 + 3.16=84----> 84g/moli

84g MgCO3-------------24gMg----------12 g C--------48g O

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

x= 100 .24 : 84= 28,57% Mg

y= 100 .12: 84=14,29%C

z= 100 .48 . 84= 57,14%

Zn(OH)2

M= 65 + 2.16 + 2= 99-----> 99g/moli

99g Zn(OH)2---------------65g Zn-------------32gO------------2gH

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

x= 100. 65 : 99=65,66%Zn

y= 100.32 : 99= 32,32% O

z= 100. 2 : 99=2,02% H