For 2HCl(g) ==> H2(g) + Cl2(g) the Keq = [H2][Cl2]/[HCl]^2
Its 0
All chemical reactions obey the law of conservation of matter.
67.2 mL.
The oxidation half-reaction is: Fe => Fe+3 + 3e-, and the reduction half-reaction is: F2 + 2e- => 2 F-1. For a complete equation, the oxidation half-reaction as written must be multiplied by 2 and added to the reduction half-reaction as written multiplied by 3 to result in an overall reaction of 2 Fe + 3 F2 = 2 FeF3.
Cl2 because its mass is approx. 35 so multiply by 2 and then its 70 & thats the highest mass of all of them
Keq = [H2O][CO] [H2][CO2]
Its 0
Keq=[H2][Cl2]/[HCl]2
HI will be consumed. The reaction will proceed to the left. More I2 will form.
In:(H2)g oxidation state: 0 In:(O2)g oxidation state: 0 In:(H2O)l oxidation state: H: +1 and O: -2
All chemical reactions obey the law of conservation of matter.
67.2 mL.
Please provide more information/context/clarification to help us answer this question. You can post your response in this answer text by clicking "Edit."
Please provide more information/context/clarification to help us answer this question. You can post your response in this answer text by clicking "Edit."
The oxidation half-reaction is: Fe => Fe+3 + 3e-, and the reduction half-reaction is: F2 + 2e- => 2 F-1. For a complete equation, the oxidation half-reaction as written must be multiplied by 2 and added to the reduction half-reaction as written multiplied by 3 to result in an overall reaction of 2 Fe + 3 F2 = 2 FeF3.
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Cl2 because its mass is approx. 35 so multiply by 2 and then its 70 & thats the highest mass of all of them