it allows plants to survive and continue to reproduce
Answer:
9.36
Explanation:
Sodium formate is the conjugate base of formic acid.
Also,
for sodium formate is 
Given that:
of formic acid = 
And, 
So,
Concentration = 0.35 M
HCOONa ⇒ Na⁺ + HCOO⁻
Consider the ICE take for the formate ion as:
HCOO⁻ + H₂O ⇄ HCOOH + OH⁻
At t=0 0.35 - -
At t =equilibrium (0.35-x) x x
The expression for dissociation constant of sodium formate is:
![K_{b}=\frac {[OH^-][HCOOH]}{[HCOO^-]}](https://tex.z-dn.net/?f=K_%7Bb%7D%3D%5Cfrac%20%7B%5BOH%5E-%5D%5BHCOOH%5D%7D%7B%5BHCOO%5E-%5D%7D)
Solving for x, we get:
x = 0.44×10⁻⁵ M
pOH = -log[OH⁻] = -log(0.44×10⁻⁵) = 4.64
pH + pOH = 14
So,
<u>pH = 14 - 4.64 = 9.36</u>
Answer:
gde
Explanation:
We are attempting to synthesize 1-butyne from 1-chlorobutane. Since 1-chlorobutane is a primary alkyl halide, 1-butene is formed when 1-chlorobutane is reacted with a bulky base such as t -BuOK or t -BuOH in presence of strong heat. This is an E2 reaction.
Secondly, the 1-butene is reacted with bromine in carbon tetrachloride. The vicinal dihalide (1,2-dibromobutane) is formed. This can now undergo further elimination reactions in the presence of sodamide and strong heat to yield 1-butyne which is the desired product. These reactions involve the elimination of the first HBr molecule to give an alkenyl bromide. A second elimination now gives the terminal alkyne.
Answer: For example, if electricity is passed through molten lead bromide, the lead bromide is broken down to form lead and bromine. This is what happens during electrolysis: Positively charged ions move to the negative electrode during electrolysis. ... Negatively charged ions move to the positive electrode during electrolysis.
Explanation:
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