Chemistry

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)

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

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

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.

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

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)

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

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

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.

At the point of intersection in an Ellingham diagram? G for two processes becomes equal, so? G for the reduction becomes zero. A sudden increase in the slope indicates a change in the state of the metal oxide, i.e., from solid to liquid or liquid to vapor.