Class 11th
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New answer posted
7 months ago
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
7 months agoContributor-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
7 months agoContributor-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
7 months agoContributor-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
7 months agoContributor-Level 9
According to principle of equi-partition of Energy, the average energy per molecules associated with each degree of freedom is (1/2)kT.
New answer posted
7 months agoContributor-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
7 months agoContributor-Level 9
m (dv/dt) = P ⇒ ∫v dv = ∫ (P/m) dt ⇒ v = (2Pt/m)¹/²
⇒ ∫dx = ∫ (2P/m)¹/² t¹/² dt ⇒ x = (2P/m)¹/² (2/3)t³/² ⇒ x ∝ t³/²
New answer posted
7 months agoContributor-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
7 months agoContributor-Level 10
For a bouncing object with initial height h = 5m and coefficient of restitution e = 0.9 (so e² =0.81):
Total distance traveled, d = h + 2e²h + 2e? h + . = h * (1 + e²) / (1 - e²)
Total time taken, t = √* (2h/g)* + 2√* (2e²h/g)* + 2√* (2e? h/g)* + . = √* (2h/g)* * (1 + e) / (1 - e)
Average speed = d/t = √* (gh/2)* * (1 + e²) / (1 + e)² = 5 * (1.81) / (1.9)² = 2.50 m/s
New answer posted
7 months agoContributor-Level 9
Using conservation of Angular momentum along axis of rotation, we can write
Mr²ω = (Mr² + 2mr²)ω? ⇒ ω? = Mω / (M + 2m)
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