Ex. 3 va rog mult! Dau coroana!

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Lauraandreea2219ask Lauraandreea2219ask · Answer 1

a) M Ca(NO3)2= A Ca+ 2A N+ 6A O

=40+2×14+6×16

=40+28+96

=68+96

=164 u.a.m => miu= 164 g/mol

164 g Ca(NO3)2....... 1 ion Ca^2+

X g Ca(NO3)2....... 8 ioni Ca^2+

164/x= 1/8 => x=164×8= 1312 g Ca(NO3)2

b) 164 g Ca(NO3)2...... 2 moli N

Y g Ca(NO3)2...... 0,04 moli N

164/y= 2/0,04 => y= 164×0,04/2= 6,56/2= 3,28 g

Ca(NO3)2

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Misi23ask Misi23ask · Answer 2

Răspuns:

Explicație:

Hristos a inviat!

Problema 3

Propun o alta metoda de rezolvare

a) 1 mol Ca(NO₃)₂........................1 mol Ca²⁺

a.................................................8 moli Ca²⁺

a=8 moli Ca(NO₃)₂

1 mol..................164g Ca(NO₃)₂

8 moli...................b

b=8×164=1312 g Ca(NO₃)₂

b) 1 mol Ca(NO₃)₂.......................2 moli atomi N

c................................................0,04 moli atomi N

c= 0,04/2=0,02 moli Ca(NO₃)₂

1mol..........................164 g Ca(NO₃)₂

0,02 moli........................d

d=0,02×164=3,28g Ca(NO₃)₂

unde M Ca(NO₃)₂ = 40+(14+16×3)×2=40+62×2=40+124=164 g/mol