Solution:
Benzoic acid is C6H5COOH
In finding pH
C6H5COOH(aq) <=> C6H5COO^- + H^+ pKa = 4.19, pKa = -logKa so Ka = 10^(-4.19)
Ka = 6.45 x 10^-6
[C6H5COO^-] = x = [H^+]; [C6H5COOH] = 0.5 - x (we are able to make an estimate of [C6H5COOH] = 0.5.
Ka = [H^+][C6H5COO^-]/[C6H5COOH] = x^2/(0.5 - x) = 6.45 x 10^-6
Now,
According to the quadratic equation. x^2 = 3.23 x 10^-5 - 6.45 x 10^-6x
x^2 + (6.45 x 10^-6)x - 3.23 x 10^-5 = 0
enter a = 1, b = 0.00000645, c = 0.0000323
x = 5.68 x 10^-3 = 0.00568 M expression is [C6H5COOH] = 0.5 M is the correct answer.
[H^+] = 0.00568 M, so pH = -log(0.00568 M ) = 2.25
This is the required solution.
Answer
36.45 u; X = Cl
Explanation:
The atomic mass of X is the <em>weighted averag</em>e of the atomic masses of its isotopes.
We multiply the atomic mass of each isotope by a number representing its relative importance (i.e., its percent of the total).
Set up a table for easy calculation
.
0.7577 × 34.969 u = 26.496 u
0.2423 × 36.965 u = <u> 8.9566 u
</u>
TOTAL = 36.45 u
X = Cl
Step 1
<em>The reaction involved:</em>
CO + 2 H2 → CH3OH (completed and balanced)
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Step 2
<em>Data provided:</em>
19.7 g H2 (the limiting reactant)
Excess reactant = CO
144.5 g CH3OH = actual yield
----
<em>Data needed:</em>
The molar masses of:
H2) 2.00 g/mol
CH3OH) 32.0 g/mol
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Step 3
The theoretical yield:
By stoichiometry,
CO + 2 H2 → CH3OH (The molar rate between H2 and CH3OH = 2:1)
2 x 2.00 g H2 --------- 32.0 g CH3OH
19.7 g H2 --------- X
X = 19.7 g H2 x 32.0 g CH3OH/2 x 2.00 g H2
X = 157.6 g CH3OH (The theoretical yield)
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Step 4
The % yield is defined as follows:
Answer: d. 93% (it is the nearest value in comparison to my result)
Oxidation refers to a change in electrons (In this case, loss of electrons). It has no impact on the number of neutrons, or for that matter on any sub-atomic particles in the nucleus. So the number of neutrons (3) remains the same
