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nastyaqwert
@nastyaqwert
July 2022
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помогите пожалуйста решить задачупосле прокаливания 10,08 г соединения состава (NH4)2X2O7 получено 6.08 г оксида X2O3/установите элемент X
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Anzhelika009
(NH4)2X2O7=N2+X2O3+4H2O
1) n((NH4)2X2O7)=m((NH4)2X2O7)/M((NH4)2X2O7)=10.08/(148+2x)
2) n(X2O3)=m(X2O3)/M(X2O3)=6.08/(2x+48)
3) n(X2O3)=n((NH4)2X2O7) ------------- 10.08/(148+2x)= 6.08/(2x+48)
10.08*(2x+48)=6.08*(148+2x)
20.16x + 483.84= 899.84+12.16x
8x= 416 x=52 (Cr)
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Answers & Comments
1) n((NH4)2X2O7)=m((NH4)2X2O7)/M((NH4)2X2O7)=10.08/(148+2x)
2) n(X2O3)=m(X2O3)/M(X2O3)=6.08/(2x+48)
3) n(X2O3)=n((NH4)2X2O7) ------------- 10.08/(148+2x)= 6.08/(2x+48)
10.08*(2x+48)=6.08*(148+2x)
20.16x + 483.84= 899.84+12.16x
8x= 416 x=52 (Cr)