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Nezavi [6.7K]
1 year ago
6

Arrange the procedural steps, from start to finish, that are required to prepare indigo from nitrobenzaldehyde and acetone in ba

se, and then test its ability as a dye.
Chemistry
1 answer:
lions [1.4K]1 year ago
5 0

Acetone has α-hydrogens (on both sides) and thus can be deprotonated to give a nucleophilic enolate anion. The aldehyde carbonyl is much more electrophilic than that of a ketone, and therefore reacts rapidly with the enolate.

<h3>What is nitrobenzaldehyde?</h3>
  • Synthesis. The synthesis of 3-nitrobenzaldehyde is achieved via nitration of benzaldehyde, which yields especially the meta-isomer. Creation allocation is about 19% for the ortho-, 72% for the meta- and 9% for the para isomers.
  • Acetone, propanone, or dimethyl ketone, is an organic combination with the formula (CH3)2CO. It is the easiest and smallest ketone. It is a colorless, highly volatile, and combustible liquid with a characteristic aromatic odor.
  • Nitration of benzene with conc nitric acid and conc sulphuric acid gives nitrobenzene. Chlorination with chlorine in the existence of anhydrous aluminum chloride gives meta nitro chlorobenzene.

To learn more about sulphuric acid, refer to:

brainly.com/question/4018599

#SPJ4

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Anika [276]
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</em></h2><h2><em> </em></h2><h2><em>Electrons: 3 </em></h2><h2><em> </em></h2><h2><em>Beryllium </em></h2><h2><em> </em></h2><h2><em>Atomic Number: 4 </em></h2><h2><em> </em></h2><h2><em>Symbol: Be </em></h2><h2><em> </em></h2><h2><em>Atomic Mass: 9 </em></h2><h2><em> </em></h2><h2><em>Protons: 4 </em></h2><h2><em> </em></h2><h2><em>Neutrons: 5 </em></h2><h2><em> </em></h2><h2><em>Electrons: 4 </em></h2><h2><em> </em></h2><h2><em>Boron </em></h2><h2><em> </em></h2><h2><em>Atomic Number: 5 </em></h2><h2><em> </em></h2><h2><em>Symbol: B </em></h2><h2><em> </em></h2><h2><em>Atomic Mass: 11 </em></h2><h2><em> </em></h2><h2><em>Protons: 5 </em></h2><h2><em> </em></h2><h2><em>Neutrons: 6 </em></h2><h2><em> </em></h2><h2><em>Electrons: 5 </em></h2><h2><em> </em></h2><h2><em>Carbon </em></h2><h2><em> </em></h2><h2><em>Atomic Number: 6 </em></h2><h2><em> </em></h2><h2><em>Symbol: C </em></h2><h2><em> </em></h2><h2><em>Atomic Mass: 12 </em></h2><h2><em> </em></h2><h2><em>Protons: 6 </em></h2><h2><em> </em></h2><h2><em>Neutrons: 6 </em></h2><h2><em> </em></h2><h2><em>Electrons: 6 </em></h2><h2><em> </em></h2><h2><em>Nitrogen </em></h2><h2><em> </em></h2><h2><em>Atomic Number: 7 </em></h2><h2><em> </em></h2><h2><em>Symbol: N </em></h2><h2><em> </em></h2><h2><em>Atomic Mass: 14 </em></h2><h2><em> </em></h2><h2><em>Protons: 7 </em></h2><h2><em> </em></h2><h2><em>Neutrons:7 </em></h2><h2><em> </em></h2><h2><em>Electrons: 7 </em></h2><h2><em> </em></h2><h2><em>Oxygen </em></h2><h2><em> </em></h2><h2><em>Atomic Number: 8 </em></h2><h2><em> </em></h2><h2><em>Symbol: O </em></h2><h2><em> </em></h2><h2><em>Atomic Mass: 16 </em></h2><h2><em> </em></h2><h2><em>Protons: 8 </em></h2><h2><em> </em></h2><h2><em>Neutrons: 8 </em></h2><h2><em> </em></h2><h2><em>Electrons: 8 </em></h2><h2><em> </em></h2><h2><em>Fluorine </em></h2><h2><em> </em></h2><h2><em>Atomic Number: 9 </em></h2><h2><em> </em></h2><h2><em>Symbol: F </em></h2><h2><em> </em></h2><h2><em>Atomic Mass: 19 </em></h2><h2><em> </em></h2><h2><em>Protons: 9 </em></h2><h2><em> </em></h2><h2><em>Neutrons: 10 </em></h2><h2><em> </em></h2><h2><em>Electrons: 9 </em></h2><h2><em> </em></h2><h2><em>Neon </em></h2><h2><em> </em></h2><h2><em>Atomic Number: 10 </em></h2><h2><em> </em></h2><h2><em>Symbol: Ne </em></h2><h2><em> </em></h2><h2><em>Atomic Mass: 20 </em></h2><h2><em> </em></h2><h2><em>Protons: 10 </em></h2><h2><em> </em></h2><h2><em>Neutrons: 10 </em></h2><h2><em> </em></h2><h2><em>Electrons: 10 </em></h2><h2><em> </em></h2><h2><em>Sodium </em></h2><h2><em> </em></h2><h2><em>Atomic Number: 11 </em></h2><h2><em> </em></h2><h2><em>Symbol: Na </em></h2><h2><em> </em></h2><h2><em>Atomic Mass: 23 </em></h2><h2><em> </em></h2><h2><em>Protons: 11 </em></h2><h2><em> </em></h2><h2><em>Neutrons: 12 </em></h2><h2><em> </em></h2><h2><em>Electrons: 11 </em></h2><h2><em> </em></h2><h2><em>Magnesium </em></h2><h2><em> </em></h2><h2><em>Atomic Number: 12 </em></h2><h2><em> </em></h2><h2><em>Symbol: Mg </em></h2><h2><em> </em></h2><h2><em>Atomic Mass: 24 </em></h2><h2><em> </em></h2><h2><em>Protons: 12 </em></h2><h2><em> </em></h2><h2><em>Neutrons: 12 </em></h2><h2><em> </em></h2><h2><em>Electrons: 12 </em></h2><h2><em> </em></h2><h2><em>Aluminum </em></h2><h2><em> </em></h2><h2><em>Atomic Number: 13 </em></h2><h2><em> </em></h2><h2><em>Symbol: Al </em></h2><h2><em> </em></h2><h2><em>Atomic Mass: 27 </em></h2><h2><em> </em></h2><h2><em>Protons: 13 </em></h2><h2><em> </em></h2><h2><em>Neutrons: 14 </em></h2><h2><em> </em></h2><h2><em>Electrons: 13 </em></h2><h2><em> </em></h2><h2><em>Silicon </em></h2><h2><em> </em></h2><h2><em>Atomic Number: 14 </em></h2><h2><em> </em></h2><h2><em>Symbol: Si </em></h2><h2><em> </em></h2><h2><em>Atomic Mass: 28 </em></h2><h2><em> </em></h2><h2><em>Protons: 14 </em></h2><h2><em> </em></h2><h2><em>Neutrons: 14 </em></h2><h2><em> </em></h2><h2><em>Electrons: 14 </em></h2><h2><em> </em></h2><h2><em>Phosphorus </em></h2><h2><em> </em></h2><h2><em>Atomic Number: 15 </em></h2><h2><em> </em></h2><h2><em>Symbol: P </em></h2><h2><em> </em></h2><h2><em>Atomic Mass: 31 </em></h2><h2><em> </em></h2><h2><em>Protons: 15 </em></h2><h2><em> </em></h2><h2><em>Neutrons: 16 </em></h2><h2><em> </em></h2><h2><em>Electrons: 15 </em></h2><h2><em> </em></h2><h2><em>Sulfur </em></h2><h2><em> </em></h2><h2><em>Atomic Number: 16 </em></h2><h2><em> </em></h2><h2><em>Symbol: S </em></h2><h2><em> </em></h2><h2><em>Atomic Mass: 32 </em></h2><h2><em> </em></h2><h2><em>Protons: 16 </em></h2><h2><em> </em></h2><h2><em>Neutrons: 16 </em></h2><h2><em> </em></h2><h2><em>Electrons: 16 </em></h2><h2><em> </em></h2><h2><em>Chlorine </em></h2><h2><em> </em></h2><h2><em>Atomic Number: 17 </em></h2><h2><em> </em></h2><h2><em>Symbol: Cl </em></h2><h2><em> </em></h2><h2><em>Atomic Mass: 35 </em></h2><h2><em> </em></h2><h2><em>Protons: 17 </em></h2><h2><em> </em></h2><h2><em>Neutrons: 18 </em></h2><h2><em> </em></h2><h2><em>Electrons: 17 </em></h2><h2><em> </em></h2><h2><em>Argon </em></h2><h2><em> </em></h2><h2><em>Atomic Number: 18 </em></h2><h2><em> </em></h2><h2><em>Symbol: Ar </em></h2><h2><em> </em></h2><h2><em>Atomic Mass: 40 </em></h2><h2><em> </em></h2><h2><em>Protons: 18 </em></h2><h2><em> </em></h2><h2><em>Neutrons: 22 </em></h2><h2><em> </em></h2><h2><em>Electrons: 18 </em></h2><h2><em> </em></h2><h2><em>Potassium </em></h2><h2><em> </em></h2><h2><em>Atomic Number: 19 </em></h2><h2><em> </em></h2><h2><em>Symbol: K </em></h2><h2><em> </em></h2><h2><em>Atomic Mass: 39 </em></h2><h2><em> </em></h2><h2><em>Protons: 19 </em></h2><h2><em> </em></h2><h2><em>Neutrons: 20 </em></h2><h2><em> </em></h2><h2><em>Electrons: 19 </em></h2><h2><em> </em></h2><h2><em>Calcium </em></h2><h2><em> </em></h2><h2><em>Atomic Number: 20 </em></h2><h2><em> </em></h2><h2><em>Symbol: Ca </em></h2><h2><em> </em></h2><h2><em>Atomic Mass: 40 </em></h2><h2><em> </em></h2><h2><em>Protons: 20 </em></h2><h2><em> </em></h2><h2><em>Neutrons: 20 </em></h2><h2><em> </em></h2><h2><em>Electrons: 20</em> </h2>

4 0
3 years ago
When 1.2383 g of an organic iron compound containing Fe, C, H, and O was burned in O2, 2.3162 g of CO2 and 0.66285 g of H2O were
uysha [10]

Answer:

\boxed{\text{C$_{15}$H$_{21}$FeO$_{6}$}}

Explanation:

Let's call the unknown compound X.

1. Calculate the mass of each element in 1.23383 g of X.

(a) Mass of C

\text{Mass of C} = \text{2.3162 g } \text{CO}_{2}\times \dfrac{\text{12.01 g C}}{\text{44.01 g }\text{CO}_{2}}= \text{0.632 07 g C}

(b) Mass of H

\text{Mass of H} = \text{0.66285 g }\text{H$_{2}$O}\times \dfrac{\text{2.016 g H}}{\text{18.02 g } \text{{H$_{2}$O}}} = \text{0.074 157 g H}

(c)Mass of Fe

(i)In 0.4131g of X

\text{Mass of Fe} = \text{0.093 33 g Fe$_{2}$O$_{3}$}\times \dfrac{\text{111.69 g Fe}}{\text{159.69 g g Fe$_{2}$O$_{3}$}} = \text{0.065 277 g Fe}

(ii) In 1.2383 g of X

\text{Mass of Fe} = \text{0.065277 g Fe}\times \dfrac{1.2383}{0.4131} = \text{0.195 67 g Fe}

(d)Mass of O

Mass of O = 1.2383 - 0.632 07 - 0.074 157 - 0.195 67 = 0.336 40 g

2. Calculate the moles of each element

\text{Moles of C = 0.63207 g C}\times\dfrac{\text{1 mol C}}{\text{12.01 g C }} = \text{0.052 629 mol C}\\\\\text{Moles of H = 0.074157 g H} \times \dfrac{\text{1 mol H}}{\text{1.008 g H}} = \text{0.073 568 mol H}\\\\\text{Moles of Fe = 0.195 67 g Fe} \times \dfrac{\text{1 mol Fe}}{\text{55.845 g Fe}} = \text{0.003 5038 mol Fe}\\\\\text{Moles of O = 0.336 40} \times \dfrac{\text{1 mol O}}{\text{16.00 g O}} = \text{0.021 025 mol O}

3. Calculate the molar ratios

Divide all moles by the smallest number of moles.

\text{C: } \dfrac{0.052629}{0.003 5038}= 15.021\\\\\text{H: } \dfrac{0.073568}{0.003 5038} = 20.997\\\\\text{Fe: } \dfrac{0.003 5038}{0.003 5038} = 1\\\\\text{O: } \dfrac{0.021025}{0.003 5038} = 6.0006

4. Round the ratios to the nearest integer

C:H:O:Fe = 15:21:1:6

5. Write the empirical formula

\text{The empirical formula is } \boxed{\textbf{C$_{15}$H$_{21}$FeO$_{6}$}}

5 0
3 years ago
The closeness of a measurement to its true value is a measure of its _____.
svlad2 [7]
The closeness of a measurement to its true value is a measure of its accuracy. This term is the degree of which a certain measurement conforms to the correct value or the standard value. It is not the same with the term precision. Precision, on the other hand, is a measure used to characterize the closeness of the data measured.
7 0
3 years ago
Read 2 more answers
How many grams of h2o are needed to produce 45g of NO
schepotkina [342]

We need (i) the stoichiometric equation, and (ii) the equivalent mass of dihydrogen.
Explanation:
1
2
N
2
(
g
)
+
3
2
H
2
(
g
)
→
N
H
3
(
g
)
11.27

g
of ammonia represents
11.27
⋅
g
17.03
⋅
g
⋅
m
o
l
−
1

=

?
?

m
o
l
.
Whatever this molar quantity is, it is clear from the stoichiometry of the reaction that 3/2 equiv of dihydrogen gas were required. How much dinitrogen gas was required?
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2 years ago
What are two factors that affect an objects kinetic energy
Bad White [126]
Potential energy and height; best guess;)
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3 years ago
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