Answer:
1) The value of Kc:
C. remains the same.
2) The value of Qc:
A. is greater than Kc.
3) The reaction must:
B. run in the reverse direction to restablish equilibrium.
4) The concentration of N2 will:
B. decrease.
Explanation:
Hello,
In this case, by means of the Le Chatelier's principle which is based on the shift a chemical reaction could have under some modifications, we have:
1) The value of Kc:
C. remains the same, since it just depend the reaction's thermodynamics as it is computed via:
2) The value of Qc:
A. is greater than Kc, since the reaction quotient is:
Thus, the lower the concentration of ammonia, the higher Qc, making Qc>Kc.
3) The reaction must:
B. run in the reverse direction to restablish equilibrium, since ammonia was withdrawn and should be regenerated to reach the equilibrium.
4) The concentration of N2 will:
B. decrease, since less reactant is forming the products.
Best regards.
Answer:
I think its might be 1 because the ionic numbers for CA is +2 and for P its +3
Answer:
I am looking for this as well. Can someone please help!
Explanation:
Answer:
65
Explanation:
estion
Al comenzar la reacción: N2(g) + 2O2(g) ------> 2NO2(g) existe 1 mol de N2 y 2 moles de O2 y al
finalizarla está presente una mezcla formada por 2,2 moles en total, ¿cuál es el rendimiento para la
reacción?
Answer:
= 72.73%
Explanation:
The percentage by mass of an element is given by;
% element = total mass of element in compounds/molar mass of compound × 100
The mass of oxygen in carbon dioxide = 32 g
Molar mass of CO2 = 44 g
Therefore;
% of O2 = 32/44 × 100%
<u>= 72.73%</u>