Question

va rog mult
vreau sa imi scrieti:
1)raportul numarului de atomi
2)raportul de masă
3)masa moleculara,molul,masa molara
4)compoziția procentuala prin ambele metode
pentru:
K2SO4
Mg3(PO4)
Fe(OH)3
BaCO3
VA ROG MULT DE TOT
E urgent
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urgent
ofer corooaaanaaa
promit
Va roggg

Answer 1

K2SO4

raport atomic=r.a.=2K:S:40=2:1:4
raport de masa=r.m.=2AK:AS:4AO=2.39:32:4.16=78:32:64=39:16:32
                                                                                 ( se simplif. cu 2)
M=78+32+64=174                 174 g ----------1 molK2SO4
Compozitie proc. prin masa molec.

M=174
174 g K2SO4----------78 g K---------32 g S--------64g O
100g------------------------x--------------y---------------z
x=100.78:174=44,82 la suta K
Y=100.32:174=18,39 la suta S
Z=100.64:174=36,78 la suta O


Compoz. proc. prin raport de masa
se aduna partile raportului
39+16+32=87
87g K2SO4--------39g K------16g S--------32gO
100 g -----------------X------------Y---------------Z
X=100.39:87=44,82 la suta K
Y=18,39 la suta S
Z=36,78 la suta O


Mg3(PO4)2                                                                                  A=nr.de masa
r.a.=3Mg:2P:8O=3:2:8
r.m=3AMg:2AP:8AO=3.24:2.31:8.16=72:62:128=36:31:64

M=72+62+128=262               262 g ---------1mol Mg3(PO4)2

COMP .PROC. prin masa .molec
262 g Mg3(PO4)2-------72 g Mg--------62gP------128 g O
100 g-------------------------x--------------------y--------z
x=100.72:262=
y=100.62:262=
z=100.128:262=

Comp.proc .prin raport de masa
se aduna partile raportului
36+31+64=131
131g Mg3(PO4)2--------36 g Mg--------31g P-------64g O
100g----------------------------X--------------Y---------------Z
X=100.36:131=
Y=100.31:131=
Z=100.64:131=

Pentru celelalte calculeaza singur , dupa model.Asa vei sti daca ai inteles.