Te is the correct ok is this
Answer: It's equal to 10^(-2.3), or 0.00501 M, or 5.01 * 10^-3 moles/Liter
Explanation:
Well, pH = - log[H+]
Or, in words, pH is equal to -1 multiplied by the logarithm (base 10) of the hydrogen ion concentration.
So you have 2.3 = -log[H+]. We want to isolate the H+, so let's start simplifying the right hand side of the equation. First, we multiply both sides by -1.
-2.3=log[H+]
Now, the definition of a logarithm says that if the log (base 10) of [H+] is -2.3, then 10 raised to the -2.3 power is [H+]
So on each side of the equation, we raise 10 to the power of that side of the equation.
10^(-2.3) = 10^(log[H+])
and because 10^log cancels out...
10^(-2.3) = [H+]
Now we've solved for [H+], the hydrogen ion concentration!
<span>The half-life of 9 months is 0.75 years.
2.0 years is 2.0/0.75 = 2.67 half-lives.
Each half-life represents a reduction in the amount remaining by a factor of two, so:
A(t)/A(0) = 2^(-t/h)
where A(t) = amount at time t
h = half-life in some unit
t = elapsed time in the same unit
A(t)/A(0) = 2^(-2.67) = 0.157
15.7% of the original amount will remain after 2.0 years.
This is pretty easy one to solve. I was happy doing it.</span>
Answer:
Explanation:
Hello there!
Unfortunately, the question is not given in the question; however, it is possible for us to compute the equilibrium constant as the problem is providing the concentrations at equilibrium. Thus, we first set up the equilibrium expression as products/reactants:
![K=\frac{[NO_2]^2}{[NO]^2[O_2]}](https://tex.z-dn.net/?f=K%3D%5Cfrac%7B%5BNO_2%5D%5E2%7D%7B%5BNO%5D%5E2%5BO_2%5D%7D)
Then, we plug in the concentrations at equilibrium to obtain the equilibrium constant as follows:
In addition, we can infer this is a reaction that predominantly tends to the product (NO2) as K>>>>1.
Best regards!
Answer:
Only the H is unbalanced.
Explanation:
There are 4 H's and 2 of everything else
