The true statement is that after reaching equilibrium, the rate of forming products and reactants is the same.
<h3>What is true about the given reaction?</h3>
The given reaction shows a reaction between A and B to form CD
The reaction is a reversible reaction.
A reversible reaction is a reaction which can proceed in either of two ways where the reactants can react to form the product and also the products an break down to form the reactants.
In the reaction given, as the concentration of A and b decreases, the concentration of CD increases and vice versa.
At equilibrium, the rate of formation of CD is equal to the the rate of decomposition of CD.
Therefore, the true statement is that after reaching equilibrium, the rate of forming products and reactants is the same.
In conclusion, a reaction at equilibrium has the forward and backward reactions occurring at the sane rate.
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Answer:
the correctanswer is a a and b
Answer:
pH = 2 and pOH = 12
Explanation:
Given [OH⁻] = 1 x 10⁻¹²M
pOH = -log[OH⁻] = -log(1 x 10⁻¹²) = - ( -12 ) = 12
pH + pOH = 14 => pH = 14 - pOH = 14 - 12 = 2
I can see two answers, I’d go with D, but all neutral atoms, of the same element would have the same number of outer electrons. However, if you consider that some of the atoms might be ions, that would eliminate B.
