Answer: Valence electrons
Chemists use reactions to generate a desired product. For the most part, a reaction is only useful if it occurs at a reasonable rate. For example, a reaction that took 8,000 years to complete would not be a desirable way to produce brake fluid. However, a reaction that proceeded so quickly that it caused an explosion would also not be useful (unless the explosion was the desired result). For these reasons, chemists wish to be able to control reaction rates. In order to gain this control, we must first know what factors affect the rate of a reaction. We will discuss some of these factors in this section.
To solve this equation you can use a basic molarity formula and substitute the correct values.
The formula for molarity is M = mol/L.
Start by changing your mL to L. To do this, divide 25 by 1000.
This results in:
.025 L
You can now rearrange your molarity formula to isolate what you are solving for, which is moles.
This leaves your formula as:
ML = mol
Now, plug in your converted volume and given molarity into the formula.
This leaves you with:
(0.100)(0.025) = mol
Multiply to find your final value for moles.
This results in:
0.0025 mol of HNO3
I hope this is what you’re looking for! :)
Answer: The half-life of a first-order reaction is, 
Explanation:
All the radioactive reactions follows first order kinetics.
Rate law expression for first order kinetics is given by the equation:
![k=\frac{2.303}{t}\log\frac{[A_o]}{[A]}](https://tex.z-dn.net/?f=k%3D%5Cfrac%7B2.303%7D%7Bt%7D%5Clog%5Cfrac%7B%5BA_o%5D%7D%7B%5BA%5D%7D)
where,
k = rate constant = ?
t = time taken = 440 s
= initial amount of the reactant = 0.50 M
[A] = left amount = 0.20 M
Putting values in above equation, we get:


The equation used to calculate half life for first order kinetics:

Putting values in this equation, we get:

Therefore, the half-life of a first-order reaction is, 
<span>At room temperature and atmospheric pressure, nothing happens when the two gasses are mixed. However, at high temperature and pressure (450C, 200atm), in the presence of an iron oxide catalyst, the production of ammonia is thermodynamically advantageous.</span>