LiAlH₄ is used as strong reducing agent. The hydride produced acts as a nucleophile and attacks the electrophillic carbon of carbonyl group. In given example two reactions are observed.
Reaction 1: In first step LiAlH₄ when treated with
Ester produces corresponding
Aldehyde and alcohol.
Reaction 2: This aldehyde produced further reacts with LiAlH₄ to give a reduced product (
Alcohols). In this case Butanol is produced.
The reaction schemes are shown as felow,
Hey there!:
Given % of Mn=59.1% means 59.1 g of Mn present in 100 g of manganese fluoride.
Molar mass of Mn= 54.938 g/mol
Moles of Mn = mass / molar mass
59.1 /54.938 => 1.07 ≈ 1 mol.
and % of F=40.9% means 40.9 g of of F present in 100 g of manganese fluoride.
Molar mass of F=18.998 g/mol
Moles of F :
40.9 / 18.999 => 2.15 mol ≈ 2 mol.
The mole ratio between Mn:F= 1 : 2
Therefore the empirical formula of manganese fluoride:
=> MnF2=Mn1F2
Hope that helps!
Answer:
An exothermic process releases heat, causing the temperature of the immediate surroundings to rise.
Explanation:
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When we have Kw = 1.2x10^-15
and according to Kw formula:
Kw = [OH-] [H+]
by substitution:
1.2x10^-15 = [OH-][H+]
when the solution is neutral that means:
[OH-] = [H+]
∴1.2X10^-15 = X^2
∴X = 3.46 x 10^-8
∴[OH-] = [H+] = 3.46x10^-8