Question

Help please!!!
1. how many atoms of carbon are present in 7.02 grams of carbon tetrafluoride?
2. how many grams of fluorine are present in 7.56x10^22 molecules of carbon tetrafluoride?

Answer 1

Răspuns:

1molC..........12g

X................7,02g

x= niu= 0,585molC

sau       niu=m/A,C

1mol C...........6,023x10²³atomi

0,585mol...............x

x=N=3,52X10²³atomi

sau   N=niuxN,A (numarul lui Avogadro)

niu= N/N,A=7,56X10²²/6,023x10²³ mol=0,1255molCF4

m=niuxM,    unde  M,CF4=88g/mol        m=0,1255x88g=11,04gCF4

88gCF4...........76gF

11,04g..................X

X=9,53gF

Explicație: