Answer:
a) [H₃O⁺] = 1.8x10⁻⁵ M
b) pH = 4.75
c) % rxn = 3.5x10⁻³ %
Explanation:
a) The dissociation reaction of HCN is:
HCN(aq) + H₂O(l) ⇄ H₃O⁺(aq) + CN⁻(aq)
0.5 M - x x x
The dissociation constant from the above reactions is given by:


By solving the above quadratic equation we have:
x = 1.75x10⁻⁵ M = 1.8x10⁻⁵ M = [H₃O⁺] = [CN⁻]
Hence, the [H₃O⁺] is 1.8x10⁻⁵ M.
b) The pH is equal to:
Then, the pH of the HCN solution is 4.75.
c) The % reaction is the % ionization:

Therefore, the % reaction or % ionization is 3.5x10⁻³ %.
I hope it helps you!
Answer:
Option B:Publishing scientific journals
Explanation:
We are told that Lindsey is trying to gain credibility for her studies.
Since she completed her experiment and discussed her finding with colleagues, the most logical next step would be to publish scientific journals. This is because the other options given are not steps that should be taken because she has completed the research and therefore has no need to speak at a conference next nor even create new charts which they must have done during the research. No need for her to make sure the topic is popular.
Option B is correct
Answer:The electron has a negative charge and the proton has a positive charge, and these charges work against each other to make the electromagnetic force that holds the entire atom together.
Explanation:
This is the answer
C)
<em>O</em><em>n</em><em>l</em><em>y</em><em> </em><em>v</em><em>a</em><em>lence</em><em> </em><em>electrons</em><em> </em><em>are involved in forming chemical bonds betwen two atoms</em><em>.</em>
Hope this helped you- have a good day bro cya)
Answer:
A mixture of 100. mL of 0.1 M HC3H5O3 and 50. mL of NaOH
Explanation:
The pH of a buffer solution is calculated using following relation

Thus the pH of buffer solution will be near to the pKa of the acid used in making the buffer solution.
The pKa value of HC₃H₅O₃ acid is more closer to required pH = 4 than CH₃NH₃⁺ acid.
pKa = -log [Ka]
For HC₃H₅O₃
pKa = 3.1
For CH₃NH₃⁺
pKa = 10.64
pKb = 14-10.64 = 3.36 [Thus the pKb of this acid is also near to required pH value)
A mixture of 100. mL of 0.1 M HC3H5O3 and 50. mL of NaOH
Half of the acid will get neutralized by the given base and thus will result in equal concentration of both the weak acid and the salt making the pH just equal to the pKa value.