Answer:
The freezing point of the solution is - 4.39 °C.
Explanation:
We can solve this problem using the relation:
<em>ΔTf = (Kf)(m),</em>
where, ΔTf is the depression in the freezing point.
Kf is the molal freezing point depression constant of water = -1.86 °C/m,
density of water = 1 g/mL.
<em>So, the mass of 575 mL is 575 g = 0.575 kg.</em>
m is the molality of the solution (m = moles of solute / kg of solvent = (465 g / 342.3 g/mol)/(0.575 kg) = 2.36 m.
<em>∴ ΔTf = (Kf)(m</em>) = (-1.86 °C/m)(2.36 m) = <em>- 4.39 °C.</em>
<em>∵ The freezing point if water is 0.0 °C and it is depressed by - 4.39 °C.</em>
<em>∴ The freezing point of the solution is - 4.39 °C.</em>
When trying to clean muddy, dirty river water, FILTRATION would work best
Answer:
The normal amount of disaccharide would be produced, but fewer monosaccharides would be produced.
Explanation:
The first reaction, the conversion of starch into disaccharides, is catalyzed by the enzyme amylase. <u>Since amylase is present in a normal amount, a normal amount of disaccharides will be produced.</u>
In the second reaction, these disaccharides will be transformed into monosaccharides by a disaccharidase. However, since t<u>here is less disaccharidase, there will be fewer monosaccharides produced than if it was a normal amount of amylase.</u>
The answer is (1). The electrons transfer form one reactant to another means that there is the change of valence. For other choices, the valence of reactant does not change.
