The amount of the solute present in the given solution is called the concentration. The best way to represent the concentration of the solution is ![\rm [K_{2}CrO_{4}].](https://tex.z-dn.net/?f=%5Crm%20%5BK_%7B2%7DCrO_%7B4%7D%5D.)
<h3>What is molar concentration?</h3>
Molar concentration is the molarity of the solution that is the measure of the concentration of the solute dissolved in the solution.
The formula for calculating molar concentration is given as,
The concentration of any substance is represented in the square bracket like ![\rm [X]](https://tex.z-dn.net/?f=%5Crm%20%5BX%5D)
or
Therefore, option B. ![\rm [K_{2}CrO_{4}]](https://tex.z-dn.net/?f=%5Crm%20%5BK_%7B2%7DCrO_%7B4%7D%5D)
is the representation of the concentration.
Learn more about the molarity here:
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Is there a picture of the isotope or?- becaue I can’t help if I don’t have a visual.
Answer:
Pottasium reacts with water vigorously and the reation is exothermic. The heat released causes the hydrogen released to ignite
Explanation:
The response would become spontaneous if the value of ΔG° was negative.
According to the estimated value of ΔG°, it is shown that ΔG° value decreases as temperature value increases. The value shifts from being more favorable to being less favorable. It would appear that the value of ΔG° would be negative at a specific temperature, causing the reaction to occur spontaneously.
The reaction is in an equilibrium state if ΔG = 0. If ΔG < 0, the reaction is spontaneous in the direction written. The relationship between terms from the equilibrium is paralleled by the relevance of the sign of a change in the Gibbs free energy.
Learn more about ΔG° here:
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The rate constant of the reaction K we can get it from this formula:
K=㏑2/ t1/2 and when we have this given (missing in question):
that we have one jar is labeled t = 0 S and has 16 yellow spheres inside and the jar beside it labeled t= 10 and has 8 yellow spheres and 8 blue spheres and the yellow spheres represent the reactants A and the blue represent the products B
So when after 10 s and we were having 16 yellow spheres as reactants and becomes 8 yellow and 8 blue spheres as products so it decays to the half amount so we can consider T1/2 = 10 s
a) by substitution in K formula:
∴ K = ㏑2 / 10 = 0.069
The amount of A (the reactants) after N half lives = Ao / 2^n
b) so no.of yellow spheres after 20 s (2 half-lives) = 16/2^2 = 4
and the blue spheres = Ao - no.of yellow spheres left = 16 - 4 = 12
c) The no.of yellow spheres after 30 s (3 half-lives) = 16/2^3 = 2
and the blue spheres = 16 - 2 = 14
