The greatest amount of AB would be produced if the equilibrium constant of the reaction is equal to 
. Hence, option D is correct.
<h3>What is an equilibrium constant?</h3>
A number that expresses the relationship between the amounts of products and reactants present at equilibrium in a reversible chemical reaction at a given temperature.
The equilibrium constant expression is a mathematical relationship that shows how the concentrations of the products vary with the concentration of the reactants.
If the value of K is greater than 1, the products in the reaction are favoured. If the value of K is less than 1, the reactants in the reaction are favoured.
Hence, option D is correct.
Learn more about the equilibrium constant here:
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Answer:
The molarity of the solution increases.
Explanation:
Molarity is the measure of the concentration of the solute in the solution. In this case, the solvent is the sugar solution and the solute is the sugar.
If sugar is ADDED to the already sugary solution, then there would be more sugar. Therefore, the sugar (solute) would increase in number.
This means that the answer is the third choice: The molarity of the solution increases.
The answer would not be the first or second choice because there isn't anything in the question that implies water. It just says sugar solution.
The answer is not the last choice because the sugar concentration does not decrease after you have added more sugar to it. It increases.
Lead fluoride hope this helps
Answer:
The bombarding particle is a Proton
Explanation:
A Nuclear transmutation reaction occurs when radioactive element decay, usually converting them from one element/isotope into another element. Transmutation is the process which causes decay, generally, alpha or beta.
¹⁶₈O(P,alpha) ¹³₇N, can be written as
¹⁶₈O + x goes to ¹³₇N + ⁴₂He
Where x can be anything, balancing the equation in order to give us the correct amount of proton number and nucleus number
16 + x = 13 + 4
x = 17 – 16 = 1, Hence we can say that x = ¹₁P
<u>¹⁶₈O + ¹₁P goes to ¹³₇N + ⁴₂He</u>
Here we can clearly see the bombarding particle is ¹₁P (proton). The ejected particle being ⁴₂He which is also known as an alpha particle
