Cân bằng phương trình sau:

\(xKCl{O_3}\;\mathop \to \limits^{{t^o}} \;yKCl + z{O_2}\)

Giá trị biểu thức: \(A = x - 2y + z\) là:

A. \(I = \dfrac{{40}}{3}\left( A \right),\,{I_1} = \dfrac{{80}}{3}\left( A \right),\,{I_2} = \dfrac{{100}}{3}\left( A \right)\).

B. \(I = \dfrac{{80}}{7}\left( A \right),\,{I_1} = \dfrac{{40}}{7}\left( A \right),\,{I_2} = \dfrac{{100}}{7}\left( A \right)\).

C. \(I = \dfrac{{40}}{11}\left( A \right),\,{I_1} = \dfrac{{80}}{11}\left( A \right),\,{I_2} = \dfrac{{120}}{11}\left( A \right)\).

D. \(I = \dfrac{{40}}{7}\left( A \right),\,{I_1} = \dfrac{{80}}{7}\left( A \right),\,{I_2} = \dfrac{{120}}{7}\left( A \right)\).

A. \(I = \dfrac{{40}}{3}\left( A \right),\,{I_1} = \dfrac{{80}}{3}\left( A \right),\,{I_2} = \dfrac{{100}}{3}\left( A \right)\).

B. \(I = \dfrac{{80}}{7}\left( A \right),\,{I_1} = \dfrac{{40}}{7}\left( A \right),\,{I_2} = \dfrac{{100}}{7}\left( A \right)\).

C. \(I = \dfrac{{40}}{11}\left( A \right),\,{I_1} = \dfrac{{80}}{11}\left( A \right),\,{I_2} = \dfrac{{120}}{11}\left( A \right)\).

D. \(I = \dfrac{{40}}{7}\left( A \right),\,{I_1} = \dfrac{{80}}{7}\left( A \right),\,{I_2} = \dfrac{{120}}{7}\left( A \right)\).

24Ω và 12Ω

12Ω và 24Ω

2Ω và 1Ω

3Ω và 2Ω

\(I = \dfrac{{16}}{{11}}\left( A \right),\,{I_1} = \dfrac{8}{{11}}\left( A \right),\,{I_2} = \dfrac{8}{{11}}\left( A \right).\)

\(I = \dfrac{8}{{11}}\left( A \right),\,{I_1} = \dfrac{{16}}{{11}}\left( A \right),\,{I_2} = \dfrac{{16}}{{11}}\left( A \right).\)

\(I = \dfrac{{24}}{7}\left( A \right),\,{I_1} = \dfrac{8}{7}\left( A \right),\,{I_2} = \dfrac{{16}}{7}\left( A \right).\)

\(I = \dfrac{{24}}{{11}}\left( A \right),\,{I_1} = \dfrac{8}{{11}}\left( A \right),\,{I_2} = \dfrac{{16}}{{11}}\left( A \right).\)

A. \({x^3}-{x^2} + x + 1.\)

B. \({x^3} + 2{x^2} - x - 2.\)

C. \(2{x^3}-{x^2} + 2x - 1.\)

D. \({x^3}-{x^2} - x - 2.\)

A. 32, 30, 34

B. 32, 33, 31

C. 32, 34, 30 

D. 31, 32, 30 

A.250 g, 150 g, 625 g.

B.125 g, 250 g, 625 g.

C.125 g, 350 g, 525 g.

D.250 g, 125 g, 625 g.