Answer:
1045
Answer: 1045 J of energy was released on cooling the down the water from 20 °C to 10 °C.
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
Accurate data means the data experimentally obtained are close to the true value. Precise data means the data obtained are close to one another. In this case, the data are close to the true value which is 1.2 and the data are relatively close to one another. Hence the set is both accurate and precise.
Hello,
<span>3. The parents’ phenotypes are expressed equally in the offspring’s phenotype.</span>
Magnesium(?)
<span>2 HCl + Mg ? MgCl2 + H2</span>
