Answer: It's a redox reaction
Explanation:
I'm going on a limb here, but Carbon is a definite. <span />
This is a straightforward dilution calculation that can be done using the equation
where <em>M</em>₁ and <em>M</em>₂ are the initial and final (or undiluted and diluted) molar concentrations of the solution, respectively, and <em>V</em>₁ and <em>V</em>₂ are the initial and final (or undiluted and diluted) volumes of the solution, respectively.
Here, we have the initial concentration (<em>M</em>₁) and the initial (<em>V</em>₁) and final (<em>V</em>₂) volumes, and we want to find the final concentration (<em>M</em>₂), or the concentration of the solution after dilution. So, we can rearrange our equation to solve for <em>M</em>₂:

Substituting in our values, we get
![\[M_2=\frac{\left ( 50 \text{ mL} \right )\left ( 0.235 \text{ M} \right )}{\left ( 200.0 \text{ mL} \right )}= 0.05875 \text{ M}\].](https://tex.z-dn.net/?f=%5C%5BM_2%3D%5Cfrac%7B%5Cleft%20%28%2050%20%5Ctext%7B%20mL%7D%20%5Cright%20%29%5Cleft%20%28%200.235%20%5Ctext%7B%20M%7D%20%5Cright%20%29%7D%7B%5Cleft%20%28%20200.0%20%5Ctext%7B%20mL%7D%20%5Cright%20%29%7D%3D%200.05875%20%5Ctext%7B%20M%7D%5C%5D.)
So the concentration of the diluted solution is 0.05875 M. You can round that value if necessary according to the appropriate number of sig figs. Note that we don't have to convert our volumes from mL to L since their conversion factors would cancel out anyway; what's important is the ratio of the volumes, which would be the same whether they're presented in milliliters or liters.
Decay constant of the process 1×10^(-12) day^(-1).
<h3>What is decay constant?</h3>
A radioactive nuclide's probability of decay per unit time is known as its decay constant, which is expressed in units of s1 or a1. As a result, as shown by the equation dP/P dt =, the number of parent nuclides P declines with time t. Nuclear forces are about 1,000,000 times more powerful than electrical and molecular forces in their ability to bind protons and neutrons. The strength of the bonds holding the radioactive element are likewise indifferent to the decay probabilities and's, in addition to being unaffected by temperature and pressure. The decay constant is related to the nuclide's T 1/2 half-life by T 1/2 = ln 2/.
To know more about decay constant:
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All of them are compounds.