Question

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dau coroana​

Answer 1

1.ACa=40

MCO2=12+2*16=44

MHNO3=1+14+3*16=63

2.mCa=80g

n=m/A=80g/40 g/mol=2 moli

m=3,2 g

n=m/A=3,2 g/32g/mol=0,1 moil

Answer 2

1.

A(Ca) = 40 g/mol,

M(CO2) = 1x12+2x16 = 44 g/mol,

M(HNO3) = 1x1+1x14+3x16 = 63 g/mol

2.

A(Ca) = 40 g/mol

niu = masa/M = 80/40 = 2 moli Ca

40 g Ca ................... 1x6,022x10la23 atomi Ca

80 g Ca ................... n = 80x1x6,022x10la23/40

= 12,044x10la23 atomi Ca sau 1,2044x10la24 atomi Ca

A(S) = 32 g/mol

niu = masa/M= 3,2/32 = 0,1 moli S

32 g S ................... 1x6,022x10la23 atomi S

3,2 g S ................... n = 3,2x1x6,022x10la23/32

= 0,6022x10la23 atomi S sau 6,022x10la22 atomi S