Question

Sa se determine formula chimica a substantei :
a) un raport masic Li:N:O=7:14:48
c) cu urmatoarea compozitie procentuală masica:Ca-24,69%,H-1.23%,C-14.81%,O-59,25%
f) care este baza unui metal trivalent care contine 34,61% metal ​
Ajutati-ma repede!!

Answer 1

Răspuns:

a. Li:N:O=7:14:48

69 g sol..... 7 g Li...... 14 g N...... 48 g O

100g sol..... x g Li...... y g N....... z g O

x= 100*7/69=10,14 g Li :7=1,44:1,44= 1 atg

y= 100*14/69=20,28 g N:14=1,44:1,44= 1 atg

z= 100*48/69= 69,56 g O:16=4,34:1,44=3 atg

F- LiNO₃ - azotat de litiu

b. %Ca= 24,69:40=0,61:0,61= 1 atg

%H= 1,23:1= 1,23:0,61=2 atg

%C= 14,81:12=1,23: 0,61= 2 atg

%O=59,25:16= 3,7:0,61= 6 atg

=> CaH₂C₂O₆- Ca(HCO₃)₂ - sare potasica

c. Me (OH)₃ - formula chimica

μ= A Me+ 3*A O+ 3*A H= A Me+ 3*16+3= 51+x, x= A Me

100 %..... 34,61% metal

(51+x)g ...... x g metal

100x= (51+x)*34,61

<=> 100x= 1765,11+ 34,61x

<=> 65,39x= 1765,11

<=> x= 27 g=> A Me= 27 g/mol => Me= Al - aluminiu