Aldehydes and ketones having
α-hydrogen atoms, undergoes aldol condensation, in present of base (NaOH).
The initial product formed during this reaction is
β-hydroxy alcohol, which then undergoes dehydration to form
α,β-unsaturated aldehyde or ketone.
In present case, 3,3-dimethyl-2-butanone has 2α-hyrogen atom, while methylcyclopentane-1-carbaldehyde has 1α-hydrogen atom. So the major product formed during cross aldol condensation reaction of these reactants is:
5-hydroxy-4,4-dimethly-1-(2-methylcyclopentyl)pent-1-en-3-one. The complete reaction product formed is shown below.
Answer: 58.44g
Explanation: The molar mass of NaCl is 58.44g.
Answer:
pH of resulting solution = 7.98
Explanation:
The balanced equation
HA + NaOH - Na+ + A- + H2O
Number of moles of A = Number of moles of HA = Number of moles of NaOH
= 35.8/1000 * 0.020 = 0.000716 mol
Initial concentration of A = 0.000716/0.0608 = 0.01178 M
pKb = 14 – pKa = 14 -3.9 = 10.1
Kb = 10^{-Kb} = 10^{-10.1} = 7.943 * 10^-11
Kb = [HA][OH-]/[A-]
Kb = a^2/(0.01178 -a) = 7.943 * 10^-11
a^2 + 7.943 * 10^-11 a – 9.357 * 10^-13 = 0
a = 9.673 * 10^-7
OH- = a = 9.673 * 10^-7 M
pOH = -log [OH-] = -log (9.673 * 10^-7) = 6.02
pH = 14-6.02 = 7.98
Answer:
(5.00M)(X) = (0.120L)(0.470M)
X = (0.120)(0.470)/(5.00)
0.01128
Explanation:
Hey there!:
Volume in mL :
1.71 L * 1000 => 1710 mL
Density = 0.921 g/mL
Therefore:
Mass = Density * Volume
Mass = 0.921 * 1710
Mass = 1574.91 g