Answer:
52.9 KJmol-1
Explanation:
From;
log(k2/k1) = Ea/2.303 * R (1/T1 - 1/T2)
The temperatures must be converted to Kelvin;
T1 = 25° C + 273 = 298 K
T2= 35°C + 273 = 308 K
R= gas constant = 8.314 JK-1mol-1
Substituting values;
log 2 = Ea/2.303 * 8.314 (1/298 - 1/308)
Ea = 52.9 KJmol-1
The solution would be like this for this specific problem:
Given:
pH of a 0.55 M hypobromous
acid (HBrO) at 25.0 °C = 4.48
[H+] = 10^-4.48 = 3.31 x
10^-5 M = [BrO-] <span>
Ka = (3.31 x 10^-5)^2 / 0.55 = 2 x 10^-9</span>
To add, Hypobromous Acid does not require acid
adjustment, which is necessary for chlorine-based product and is stable and
effective in pH ranges of 5-9.<span>
</span>Hypobromous Acid combines with organic
compounds to form a bromamine. Chlorine also combines with the same organic
compounds to form a chloramine. <span>It is also
one of the least expensive intervention antimicrobial compounds available.</span>
The concentration of the hydroxide ion given that the concentration of the hydronium ion 1.0 times 10^-14 M. The reverse mathematical method used to determine the pOH can be used to get the hydroxide ion concentration from the pOH. How many hydroxide ions are there in a solution with a pOH of 5.70, for instance.
Calculate 10-5.70, or "inverse" log, on a calculator (- 5.70). It indicates that one hydroxide ion is produced by one part of the NaOH solution. Because of this, the molar concentration of hydroxide ions in the solution is the same as the molar concentration of the NaOH.
To learn more about hydroxide, click here.
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