The scheme is shown below, the steps involved are as follow,
Step one: Reduction: The carbonyl group of given compound on reduction using
Wolf Kishner reagent converts the carbonyl group into -CH₂- group.
Step two: Epoxidation: The double bond present in starting compound when treated with
m-CPBA (<span>meta-Chloroperoxybenzoic acid) gives corrsponding epoxide.
Step three: Reduction: The epoxide is reduced to alcohol on treatment with
Lithium Aluminium Hydride (LiAlH</span>₄)<span> followed by hydrolysis.
Step four: Oxidation: The hydroxyl group (alcohol) is
oxidized to carbonyl (ketonic group) using oxidizing agent
Chromic acid (H</span>₂CrO₄).
if you have 1mol of NO. how many molecules of NO are there
Answer:
6.02 x 10²³ molecules
Explanation:
Given parameters:
Number of moles of NO = 1 mole
Unknown:
Number of molecules in NO;
Solution:
A mole of compound contains the Avogadro's number of particles.
1 mole of a substance contains 6.02 x 10²³ molecules
So, 1 mole of NO contained 6.02 x 10²³ molecules
Answer:
b. The shorter the half-life, the more dangerous the radioisotope.
Answer:
False.
Explanation:
A machine does not decrease the work needed. It simply makes the work easier or changes how the work is done.
Answer:
pH = 2.69
Explanation:
The complete question is:<em> An analytical chemist is titrating 182.2 mL of a 1.200 M solution of nitrous acid (HNO2) with a solution of 0.8400 M KOH. The pKa of nitrous acid is 3.35. Calculate the pH of the acid solution after the chemist has added 46.44 mL of the KOH solution to it.</em>
<em />
The reaction of HNO₂ with KOH is:
HNO₂ + KOH → NO₂⁻ + H₂O + K⁺
Moles of HNO₂ and KOH that react are:
HNO₂ = 0.1822L × (1.200mol / L) = <em>0.21864 moles HNO₂</em>
KOH = 0.04644L × (0.8400mol / L) = <em>0.0390 moles KOH</em>
That means after the reaction, moles of HNO₂ and NO₂⁻ after the reaction are:
NO₂⁻ = 0.03900 moles KOH = moles NO₂⁻
HNO₂ = 0.21864 moles HNO₂ - 0.03900 moles = 0.17964 moles HNO₂
It is possible to find the pH of this buffer (<em>Mixture of a weak acid, HNO₂ with the conjugate base, NO₂⁻), </em>using H-H equation for this system:
pH = pKa + log₁₀ [NO₂⁻] / [HNO₂]
pH = 3.35 + log₁₀ [0.03900mol] / [0.17964mol]
<h3>pH = 2.69</h3>