Question

Va rog sa ma ajutati la chimie dau 40 de puncte... Vreau rezolvare completă daca se poate !

Answer 1

a) MSO₂ = AS + 2AO = 32+2×16 = 64g/mol

raport de masa ⇒ 32:32 = 1:1

2 parti masice.............1 parte S..........1 parte O
100g...........................x.......................y

x = 100×1/2 =  50% S
y = 50% O

b) MH₃PO₄ = 98g/mol

raport de masa ⇒ 3:31:64

98 parti masice...........3 parti H........32 parti P..........64 parti O
100g...........................x....................y........................z

x = 3×100/98 = Calculeaza !
y = 100×32/98 = Calculeaza !
z = 64×100/98 = Calculeaza !

c) MMg(OH)₂ = AMg + 2(AO+AH) = 24+2(16+1) = 58g/mol

raport de masa ⇒ 24:32:2 = 12:16:1

12+16+1 = 29 parti masice

29 parti masice........12 parti Mg........16 parti O.......1 parte H
100g.........................x..........................y...................z

x = 12×100/29 = Calculeaza !
y= 1600/29  = Calculeaza !
z = 100/29 = Calculeaza !

Se deseneaza axa numerica (cele doua axe x si y).. Pe axa y se reprezinta procentul si pe axa x gramele substantelor (in functie de masa molara sau raportul de masa se pune valoarea maxima ).. Sau in functie de preferinta.. Tot in stilul asta se face reprezentarea grafica..

Answer 2

$M_{SO_2}=32+(2*16)=32+32=64\ g/mol \\ S:O=32:32=1:1\\ \\ S=\frac{32*100}{64}=50\% \\ O=\frac{(2*16)*100}{64}=\frac{3200}{64}=50\%$
$M_{H_3PO_4}=(3*1)+31+(4*16)=3+31+64=98\ g/mol \\ H:P:O=3:31:64 \\ \\ H=\frac{(3*1)*100}{98}=\frac{300}{98}=3,06\% \\ \\ P=\frac{31*100}{98}=31,63\% \\ \\ O=\frac{(4*16)*100}{98}=\frac{6400}{98}=65,30\%$
$M_{Mg(OH)_2}=24+(2*17)=24+34=58\ g/mol \\ Mg:O:H=24:32:2=12:16:1 \\ \\ Mg=\frac{24*100}{58}=41,37\% \\ \\ O=\frac{(2*16)*100}{58}=\frac{3200}{58}=55,17\% \\ \\ H=\frac{200}{58}=3,44\%$