Answer:
2.1 × 10⁻¹ M
2.0 × 10⁻¹ m
Explanation:
Molarity
The molar mass of aniline (solute) is 93.13 g/mol. The moles corresponding to 3.9 g are:
3.9 g × (1 mol/93.13 g) = 0.042 mol
The volume of the solution is 200 mL (0.200 L). The molarity of aniline is:
M = 0.042 mol/0.200 L = 0.21 M = 2.1 × 10⁻¹ M
Molality
The moles of solute are 0.042 mol.
The density of the solvent is 1.05 g/mL. The mass corresponding to 200 mL is:
200 mL × 1.05 g/mL = 210 g = 0.210 kg
The molality of aniline is:
m = 0.042 mol/0.210 kg = 0.20 m = 2.0 × 10⁻¹ m
In the given above, we have two densities which are 0.89 g/mL and 0.72 g/mL. We are also given that the liquids are immiscible. After the settlement of the liquids, they will form two layers.
The heavier substance, the one which has a higher density will be at the bottom and the lighter substance, the one which has a lower density will be at the top layer.
Answer:
0.297 °C
Step-by-step explanation:
The formula for the <em>freezing point depression </em>ΔT_f is
ΔT_f = iK_f·b
i is the van’t Hoff factor: the number of moles of particles you get from a solute.
For glucose,
glucose(s) ⟶ glucose(aq)
1 mole glucose ⟶ 1 mol particles i = 1
Data:
Mass of glucose = 10.20 g
Mass of water = 355 g
ΔT_f = 1.86 °C·kg·mol⁻¹
Calculations:
(a) <em>Moles of glucose
</em>
n = 10.20 g × (1 mol/180.16 g)
= 0.056 62 mol
(b) <em>Kilograms of water
</em>
m = 355 g × (1 kg/1000 g)
= 0.355 kg
(c) <em>Molal concentration
</em>
b = moles of solute/kilograms of solvent
= 0.056 62 mol/0.355 kg
= 0.1595 mol·kg⁻¹
(d) <em>Freezing point depression
</em>
ΔT_f = 1 × 1.86 × 0.1595
= 0.297 °C
Answer:
B. Cu + 4HNO3 → Cu(NO3)2 + 2H2O + 2NO2
Explanation:
Hello,
In this case, we should understand oxidizing agents as those substances able to increase the oxidation state of another substance, therefore, in B. reaction we notice that copper oxidation state at the beginning is zero (no bonds are formed) and once it reacts with nitric acid, its oxidation states raises to +2 in copper (II) nitrate, thus, in B. Cu + 4HNO3 → Cu(NO3)2 + 2H2O + 2NO2 nitritc acid is acting as the oxidizing agent.
Moreover, in the other reactions, copper (A.), sodium (C. and D.) remain with the same initial oxidation state, +2 and +1 respectively.
Regards.
