Since the pH of the substance is 5, the substance contains more of hydrogen ions.
<h3>What is pH of a solution?</h3>
The pH of a a solution is the negative logarithm to base ten of the hydrogen ions concentration of the solution.
![pH = -log[H^{+}]](https://tex.z-dn.net/?f=pH%20%3D%20-log%5BH%5E%7B%2B%7D%5D)
A neural solution has pH 7.
A solution with pH below 7 has more hydrogen ions while a solution with pH above 7 has more hydroxide ions.
The pH of a substance contains a hydrogen ion concentration of 1.0 x 10-5 M is 5.
Therefore, the substance contains more of hydrogen ions.
Learn more about pH at: brainly.com/question/940314
Dipole-dipole interactions, and London dispersion interactions
Answer:
D
Explanation:
the chart would show that the student got the best overall results from type X, as compared to the other ones
Answer:
![[H^+]=0.000285](https://tex.z-dn.net/?f=%5BH%5E%2B%5D%3D0.000285)

Explanation:
In this, we can with the <u>ionization equation</u> for the hydrazoic acid (
). So:

Now, due to the Ka constant value, we have to use the whole equilibrium because this <u>is not a strong acid</u>. So, we have to write the <u>Ka expression</u>:
![Ka=\frac{[H^+][N_3^-]}{[HN_3]}](https://tex.z-dn.net/?f=Ka%3D%5Cfrac%7B%5BH%5E%2B%5D%5BN_3%5E-%5D%7D%7B%5BHN_3%5D%7D)
For each mol of
produced we will have 1 mol of
. So, we can use <u>"X" for the unknown</u> values and replace in the Ka equation:
![Ka=\frac{X*X}{[HN_3]}](https://tex.z-dn.net/?f=Ka%3D%5Cfrac%7BX%2AX%7D%7B%5BHN_3%5D%7D)
Additionally, we have to keep in mind that
is a reagent, this means that we will be <u>consumed</u>. We dont know how much acid would be consumed but we can express a<u> subtraction from the initial value</u>, so:

Finally, we can put the ka value and <u>solve for "X"</u>:



So, we have a concentration of 0.000285 for
. With this in mind, we can calculate the <u>pH value</u>:
![pH=-Log[H^+]=-Log[0.000285]=3.55](https://tex.z-dn.net/?f=pH%3D-Log%5BH%5E%2B%5D%3D-Log%5B0.000285%5D%3D3.55)
I hope it helps!
For every 2 moles of CO and 1 mole of O2, 2 moles of CO2 will be produced.
The ratio is 2:1:2.
Hope this helps! :)