Answer:
VC = 18
Explanation:
Since L is the midpoint and you have LV, you know that LC is also 9.
Answer:
an increasng atomic number/ increasing proton count
Explanation:
It's made up of two different metals bonded together, which expand by different amounts as they heat up. As the temperature changes, the bimetallic strip curves more or less tightly (contracts or expands) and the pointer, attached to it, moves up or down the scale.
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!