Oblicz zawartość procentową tlenu w P2O5 i K2CO3.
mP2O5=2*31u+5*16u=62u+80u=142u
%O=80u/142u * 100%
%O=56,3%
mK2CO3=2*39u+12u+3*16u=78u+12u+48u=138u
%O=48u/138u * 100%
%O=34,8%
mP2O5 = 30x2 + 16x5 = 60u + 80u = 140u
100% - 140 u
x% - 80 u = 100 x 80 :140 (skróć 0 z 140 i 80 ) =800:14 =(znak równań napisz falowany) =ok. 57 %
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mK2CO3 = 39x2 + 12 +16x3 = 78u + 12u + 48u = 138u
100%-138u
x%-48u = 100 x 48 :138 = 4800 : 138 = (znowu napisz falowany znak = ) = ok.34%
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mP2O5=2*31u+5*16u=62u+80u=142u
%O=80u/142u * 100%
%O=56,3%
mK2CO3=2*39u+12u+3*16u=78u+12u+48u=138u
%O=48u/138u * 100%
%O=34,8%
mP2O5 = 30x2 + 16x5 = 60u + 80u = 140u
100% - 140 u
x% - 80 u = 100 x 80 :140 (skróć 0 z 140 i 80 ) =800:14 =(znak równań napisz falowany) =ok. 57 %
-----------------------------------------------------------------------------------------
mK2CO3 = 39x2 + 12 +16x3 = 78u + 12u + 48u = 138u
100%-138u
x%-48u = 100 x 48 :138 = 4800 : 138 = (znowu napisz falowany znak = ) = ok.34%