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:
The answer to your question is A.
Pure substances can not be broken down into others, so they cannot be molecules
Explanation:
Correct answer is <span>Fuels do not have to be purchased to generate power.</span>
Answer:
The standard potential, E cell, for this galvanic cell is 0.5670V
Explanation:
Ni²⁺(aq) + 2e⁻ → Ni(s) E red = - 0.23V ANODE
Cu²⁺(aq) + 2e- → Cu(s) E red = + 0.337V CATHODE
ΔE° = E cathode - E anode
ΔE° = 0.337V - (0.23V) = 0.5670 V
