It is converted into a CARBOXYLIC ACID
Haloform reaction mechanism.
First step
The base (hydroxide ion) takes out the alpha hydrogen producing enolate. Then, the reaction between the enolate and the halogen occurs, leading to the formation of the halogenated ketone along with the halogens corresponding anion.
Second step
Step 1 is repeated twice to yield a tri-halogenated ketone. The net reaction till the formation of the tri-halogenated ketone can be written as
The hydroxide ion acts as a nucleophile and attacks the electrophilic carbon which is doubly bonded to oxygen. This carbon-oxygen double bond becomes a single bond making the oxygen atom anionic. This makes the reformation of the carbon oxygen double bond favourable and the carbon attached to three halogens is displaced, leaving us with the carboxylic acid. An acid base reaction ensues, the carboxylic acid donates a proton to the tri-halomethyl anion giving the required haloform product.
<span>Nitrogen gas is converted to nitrate compounds by nitrogen-fixing bacteria in soil turns nitrogen gas into root nodules. Nitrogen is the most commonly limiting nutrient in plants. Legumes use nitrogen fixing bacteria, specifically symbiotic rhizobia bacteria, within their root nodules to counter the limitation.</span>
Answer:
2.61 L
Explanation:
p1v1 = p2v2
p1v1 / p2 = v2
98 * 2 / 75 = v2 = 2.61 L
If the concentration of the hydrogen ions, increase then the solution would become more and more acidic and thus have a lower pH value than compared to the pH value before at the 1.0 X 10^-2 M.
For Ca(OH)2, Ksp = [Ca2+][OH-]^2
You have your Ksp as 6.5 x 10^-6. Your [OH-] comes almost entirely from the 0.10 mol of NaOH, since Ca(OH)2 barely dissolves. Your [OH-] is therefore 0.10 M (since you have 1 L of solution).
6.5 x 10^-6 = [Ca2+](0.10)^2
Solve for [Ca2+]:
6.5 x 10^-6 / (0.10)^2 = [Ca2+]
[Ca2+] = 0.00065 M
The maximum concentration of [Ca2+] is 0.00065 M, and you have 0.0010 M Ca(OH)2, so you’ll end up with 0.00065 M Ca2+ in solution.
