It’s the second won the air layer one
Which atoms? ~~~~~~~~~~~~~~~~~~~~~~~~~`
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!
Answer:
51 J
Explanation:
The air inside a bicycle tire pump has 27 joules of heat conducted away. By convention, when heat is released, it takes the negative sign, so Q = -27 J.
77.9 joules of work done are being done on the air inside a bicycle tire pump. By convention, when work is being done on the system, it takes the positive sign, so W = 77.9 J
We can calculate the change in the internal energy (ΔU) using the following expression.
ΔU = Q + W
ΔU = (-27 J) + 77.9 J
ΔU = 51 J
The answer should be A, the number of electrons increases by 1, from left to right.
Across the periods (horizontal direction) in periodic table, each element has one more electron then the one before it. The number of period determines the number of occupied electron shells of the elements in that period. Which means, period 1 elements have 1 occupied electron shell, period 2 has 2, period 3 has 3 etc.
Therefore, the answer is A, we always go from left to right, and plus since period 2 elements only have 2 occupied electrons so their 2nd electron shell electrons must increase throughout the period.