Answer:
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HEY mate here your answer
CoBr2 . 6 h2o = Cobalt(II) Bromide Hexahydrate
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>
Answer:
<span>In the addition of hbr to 1-butyne the electrophile in the first step of the mechanism is <u>Hydrogen atom of HBr</u>.
Explanation:
In this reaction first of all HBr approaches the triple bond. A Pi Complex (weak inter-molecular interactions) is formed between the two molecules. And the triple bond attacks the partial positive hydrogen atom creating a negative charge on Bromine along with positive charge on itself (Sigma Complex). In second step the negative Bromide attacks the positive carbon of Butyne.</span>
