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a year ago

Chemistry Amines


In the above reaction, 3.9 g of benzene on nitration gives 4.92 g of nitrobenzene. The percentage yield of nitrobenzene in the above reaction is _______ % (Round off to the nearest integer).
(Given atomic mass : C: 12.0u, H:1.0u, O:16.0u, N:14.0u)
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V
Vishal Baghel

Contributor-Level 10

Moles of benzene = 3.9 g / 78 g/mol . This would produce (3.9 / 78) moles of nitrobenzene in 100% conversion.
Produced moles of nitrobenzene = 4.92 g / 123 g/mol .
% yield = [ (4.92 / 123) / (3.9 / 78) ] * 100 = [ (4.92 * 100 * 78) / (123 * 3.9) ] = 80.0%
Ans = 80

New answer posted

a year ago

Chemistry Solutions

The mole fraction of a solute in a 100 molal aqueous solution is _______ *10-2. (Round off to the nearest integer).
[Given : Atomic masses : H : 1.0u, O : 16.0u]
0 Follower 9 Views

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Vishal Baghel

Contributor-Level 10

Molality = (mole of solute * 1000) / wt of solvent (gm)
100 = (n_solute * 1000) / [ (1 - n_solute) * 18]
(1 - n_solute) / n_solute = 1000 / (100 * 18) = 10/18
18 (1 - n_solute) = 10 n_solute
18 - 18 n_solute = 10 n_solute
18 = 28 n_solute
n_solute = 18 / 28? 0.6428 = 64.28 * 10? ²
Ans = 64 (Rounded off)

New answer posted

a year ago

Chemistry Structure of Atom

A certain orbital has n= 4 and m? = -3. The number of radial nodes in this orbital is _______ (Round off to the nearest integer).
0 Follower 5 Views

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Vishal Baghel

Contributor-Level 10

For n = 4, the possible values of l are 0, 1, 2, 3.
For l = 3, and m = -3.
Radial nodes = (n - l - 1) = (4 - 3 - 1) = 0
Ans = 0

New answer posted

a year ago

Chemistry Redox Reactions

15 mL of aqueous solution of Fe2+ in acidic medium completely reacted with 20 ml of 0.03 M aqueous Cr2O72-. The molarity of the Fe2+ solution is _______ *10-2 M (Round off to the nearest integer).
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Vishal Baghel

Contributor-Level 10

Cr? O? ²? + Fe²? - (H? )-> Cr³? + Fe³?
(n=6) (n=1)
Meq Cr? O? ²? = Meq Fe²?
20 * 0.03 * 6 = 15 * M * 1
M = (20 * 0.03 * 6) / 15 = 0.24 = 24 * 10? ²
Ans = 24

New answer posted

a year ago

Chemistry Class 12th

The reaction of white phosphorous on boiling with alkali in inert atmosphere resulted in the formation of product 'A'. The reaction of 1 mol of 'A' with excess of AgNO₃ in aqueous medium gives _______ mol(s) of Ag. (Round off to the nearest integer).
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Vishal Baghel

Contributor-Level 10

P? + 3NaOH + 3H? O → PH? + 3NaH? PO? (A)
Here, NaH? PO? is a reducing agent.
NaH? PO? + AgNO? + H? O → Ag + HNO? + NaH? PO?
Equivalents of NaH? PO? = equivalents of Ag (n-factor of NaH? PO? = 4)
1 mole * 4 = n moles of Ag * 1
So, moles of Ag = 4.

New answer posted

a year ago

Chemistry Class 11th

0.01 moles of a weak acid HA (Ka = 2.0 *10-6) is dissolved in 1.0 L of 0.1 M HCl solution. The degree of dissociation of HA is _______ *10-5. (Round off to the nearest integer).
[Neglect volume change on adding HA. Assume degree of dissociation <<1]
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Vishal Baghel

Contributor-Level 10

For the weak acid HA in the presence of strong acid HCl:
Ka = [ (Cα + 0.1) * Cα] / [C (1-α)] ≈ (0.1 * 10? ²α) / 10? ² = 0.1α
Given Ka = 2 * 10?
2 * 10? = 10? ¹ * α
α = 2 * 10?
Ans = 2

New answer posted

a year ago

Chemistry Class 11th

The pressure exerted by a non- reactive gaseous mixture of 6.4 g of methane and 8.8 g of carbon dioxide in a 10 L vessel at 27°C is _______ kPa. (Round off to the nearest integer).
[ Assume gases are ideal, R = 8.314 J mol-1 K-1 Atomic masses: C: 12.0 u, H: 1.0 u, O: 16.0 u]
0 Follower 2 Views

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Vishal Baghel

Contributor-Level 10

Mole of CH? = 6.4 / 16 = 0.4 and mole of CO? = 8.8 / 44 = 0.2
Total mole = (0.4 + 0.2) = 0.6 mole of a non-reacting mixture of gas
Using Ideal Gas Law; P = nRT / V
P = (0.6 * 8.314 * 300) / 10 = 149.65 kPa
Ans = 150 (Rounded off)

New answer posted

a year ago

Chemistry Chemistry Chemical Kinetics

For a certain first order reaction 32% of the reactant is left after 570s. The rate constant of this reaction is _______ *10⁻³ s⁻¹. (Round off to the nearest integer).
[Given: log₁₀2 = 0.301, ln10 = 2.303]
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Vishal Baghel

Contributor-Level 10

Kt = 2.303 log? (A? /A? )
K * 570 = 2.303 log? (100/32)
K = [2.303 / 570] [log?10² - log?2? ]
K = [2.303 / 570] [2 - 5 * 0.301] = [2.303 / 570] * 0.495 = 0.002 = 2.0 * 10? ³ s? ¹
Ans = 2

New answer posted

a year ago

Chemistry Class 11th

The oxygen dissolved in water exerts a partial pressure of 20 kPa in the vapour above water. The molar solubility of oxygen in water is _______ *10-5 mol dm-3. (Round off to the nearest integer).
[ Given : Henry's law constant = KH = 8.0 *104 kPa for O2. Density of water with dissolved oxygen = 1.0kg dm-3]
0 Follower 20 Views

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Vishal Baghel

Contributor-Level 10

Partial Pressure of O? = K? * solubility (K? = Henry's constant)
Solubility = PO? / K? = 20 / (8.0 * 10? ) = 2.5 * 10? = 25 * 10? M
Ans = 25

New answer posted

a year ago

Chemistry Class 11th

Which of the following is an aromatic compound?
0 Follower 27 Views

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Vishal Baghel

Contributor-Level 10

In an aromatic compound, overall delocalization is processed by the electrons according to Hückel's rule. For the compound in question, the total number of delocalized electrons is 6.

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