Answer:

Explanation:
pH is derived from the concentration of hydronium ions in a solution. Hydrocyanic acid is HCN.
First, we shall figure out the moles of HCN:

If HCN was a strong acid:
HCN has a 1:1 ratio of H+ ions, the moles of H+ is also the same.
To find the molarity, we now divide by Liters. This gets us:

Finally, we plug it into the definition of pH:
![pH = -log[H^{+} ]](https://tex.z-dn.net/?f=pH%20%3D%20-log%5BH%5E%7B%2B%7D%20%5D)


However, since HCN is a weak acid, it only partially dissociates. The
of HCN is
.
![K_a = \frac{[H^+][A^-]}{[HA]}](https://tex.z-dn.net/?f=K_a%20%3D%20%5Cfrac%7B%5BH%5E%2B%5D%5BA%5E-%5D%7D%7B%5BHA%5D%7D)
We can use an ice table to determine that when x = H+,

![[H^+] = 8.83*10^{-6}](https://tex.z-dn.net/?f=%5BH%5E%2B%5D%20%3D%208.83%2A10%5E%7B-6%7D)
![pH = -log[H^{+} ]](https://tex.z-dn.net/?f=pH%20%3D%20-log%5BH%5E%7B%2B%7D%20%5D)


Answer:
Honey
Explanation:
It’s honey because the item that was dropped in it is almost at the top and that means there is something sticky or force there holding it back!
1 CH4 (g) + 2 O2 (g) -----> CO2 (g) + 2H2O(l) ΔH= - 890 kJ
1 mol 2 mol
1) If ΔH has minus, it means "release". We need only "release" choices.<span>
2) From reaction
1 mol </span>CH4 (g) "releases" ΔH= - 890 kJ - We do not have this choice.
2 mol O2 (g) "release" ΔH= - 890 kJ, so
1 mol O2 (g) "release" ΔH= - 445 kJ
Correct answer is B.
Since
21.2 g H2O was produced, the amount of oxygen that reacted can be obtained
using stoichiometry. The balanced equation was given: 2H₂ + O₂ → 2H₂O and
the molar masses of the relevant species are also listed below. Thus, the
following equation is used to determine the amount of oxygen consumed.
Molar mass of H2O = 18
g/mol
Molar mass of O2 = 32
g/mol
21.2 g H20 x 1 mol
H2O/ 18 g H2O x 1 mol O2/ 2 mol H2O x 32 g O2/ 1 mol O2 = 18.8444 g O2
<span>We then determine that
18.84 g of O2 reacted to form 21.2 g H2O based on stoichiometry. It is
important to note that we do not need to consider the amount of H2 since we can
derive the amount of O2 from the product. Additionally, the amount of H2 is in
excess in the reaction.</span>
Hello!
The basic equations to solve this is
pH = -log[H+]
pOH = -log[OH-]
pH + pOH = 14
------------------------------------------------------------------------------------------------------
Find pHpH = -log(1 * 10^-1)
pH = 1
------------------------------------------------------------------------------------------------------
Find pOH1 + pOH = 14
pOH = 13
------------------------------------------------------------------------------------------------------
Find OH-[OH-] = 10^(-pOH)
[OH-] = 1 * 10^-13mo/L
The answer is
![[OH-] = 1 * 10^{-13} mol/L](https://tex.z-dn.net/?f=%5BOH-%5D%20%3D%201%20%2A%2010%5E%7B-13%7D%20mol%2FL)
Hope this helps!