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andrey1shpirko
@andrey1shpirko
April 2021
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Пожалуйста, решите! Срочно надо! Даю 77 балов! Желательно чтобы решение было написано на бумаге и сфотографировано!
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Alexei78
3.1
дано
m(ppaNaNO3) = 600 g
W(NaNO3)=10%
m(H2O)=40 g
-----------------------
W1(NaNO3)-?
m(NaNO3)=m(ppaNaNo3)*W(NaNO3) / 100%= 600*10%/100% = 60 g
m(H2O)=600 - 60 = 540 g
m(общ H2O )= 540+40=540 g
W1(NaNO3) = 60/540*100% = 11.11 %
ответ 11.11%
3.2
дано
m(ppa) = 120 g
W(соли)=3%
+m(соли)=80 g
------------------------
W1(соли)-?
m(соли) = 120*3%/100% = 3.6 g
m(h2O)= 120-3.6= 116.4 g
m(общ соли) = 3.6+80 = 83.6g
W1(соли)=83.6 / 120 * 100% = 69.66%
ответ 69.66%
3.3
CaCL2+2AgNO3-->2AgCL↓+Ca(NO3)2
1 + 2 + 2 + 1 = 6
Ca⁺²+2CL⁻+2Ag⁺+2NO⁻₃-->2AgCL↓+Ca⁺²+2NO⁻₃
2CL⁻+2Ag⁺-->2AgCL↓
2 votes
Thanks 1
andrey1shpirko
Большое спасибо!
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Answers & Comments
дано
m(ppaNaNO3) = 600 g
W(NaNO3)=10%
m(H2O)=40 g
-----------------------
W1(NaNO3)-?
m(NaNO3)=m(ppaNaNo3)*W(NaNO3) / 100%= 600*10%/100% = 60 g
m(H2O)=600 - 60 = 540 g
m(общ H2O )= 540+40=540 g
W1(NaNO3) = 60/540*100% = 11.11 %
ответ 11.11%
3.2
дано
m(ppa) = 120 g
W(соли)=3%
+m(соли)=80 g
------------------------
W1(соли)-?
m(соли) = 120*3%/100% = 3.6 g
m(h2O)= 120-3.6= 116.4 g
m(общ соли) = 3.6+80 = 83.6g
W1(соли)=83.6 / 120 * 100% = 69.66%
ответ 69.66%
3.3
CaCL2+2AgNO3-->2AgCL↓+Ca(NO3)2
1 + 2 + 2 + 1 = 6
Ca⁺²+2CL⁻+2Ag⁺+2NO⁻₃-->2AgCL↓+Ca⁺²+2NO⁻₃
2CL⁻+2Ag⁺-->2AgCL↓