Answer:
C + O2 → CO2
Explanation:
C + O2 → CO ----------------- (1)
from equ (1) on reactant side, C has 1 mole, O has 2 moles
from equ (1) on product side, C has 1 mole, O has 1 mole
Thus, to balance the equation, O should have 2 moles
C + O2 → CO2
Answer: pH = 2,897 , basic![[H+][OH-] = 10^{-14} ==> [H+] = \frac{10^{-14}}{7,89*10^{-12} } =\frac{1}{789} \\pH= -lg([H+]) = 2,897 \\pH basic](https://tex.z-dn.net/?f=%5BH%2B%5D%5BOH-%5D%20%3D%2010%5E%7B-14%7D%20%3D%3D%3E%20%5BH%2B%5D%20%3D%20%5Cfrac%7B10%5E%7B-14%7D%7D%7B7%2C89%2A10%5E%7B-12%7D%20%7D%20%3D%5Cfrac%7B1%7D%7B789%7D%20%5C%5CpH%3D%20-lg%28%5BH%2B%5D%29%20%3D%202%2C897%20%5C%5CpH%3C7%20%3D%3D%3E%20basic)
Explanation:
I think the answer is number (4)
Answer:
According to Le Chatelier's principle, increasing the reaction temperature of an exothermic reaction causes a shift to the left and decreasing the reaction temperature causes a shift to the right.
Explanation:
C6H12O6(s) + 6O2(g) ⇌6CO2(g) + 6H2O(g)
We are told that the forward reaction is exothermic, meaning heat is removed from the reacting substance to the surroundings.
According to Le Chatelier's principle,
1. for an exothermic reaction, on increasing the temperature, there is a shift in equilibrium to the left and formation of the product is favoured.
2. if the temperature of the system is decreased, the equilibrium shifts to right and the formation of the reactants is favoured.
3. if the reaction temperature is kept constant, the system is at equilibrium and there is no shift to the right nor to the left.
