Answer:
The answer is partial charge 8+.
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: 1.) His action is not justified because he could have done something else like talking with someone.
2.) Some of the negative consequences could be Mr. King could lose his job or worse not be able to teach again.
3.) Some other options were having him retake the test or talking to the head of the school, or many other things.