Answer
a
Explanation:
the giraffe doesn't need carbon dioxide
Answer:
3.74g of ethylene glycol must be added to decrease the freezing point by 0.400°C
Explanation:
One colligative property is the freezing point depression due the addition of a solute. The equation is:
ΔT=Kf*m*i
<em>Where ΔT is change in temperature = 0.400°C</em>
<em>Kf is freezing point constant of the solvent = 1.86°C/m</em>
<em>m is molality of the solution (Moles of solute / kg of solvent)</em>
<em>And i is Van't Hoff constant (1 for a nonelectrolyte)</em>
Replacing:
0.400°C =1.86°C/m*m*1
0.400°C / 1.86°C/m*1 = 0.215m
As mass of solvent is 280.0g = 0.2800kg, the moles of the solute are:
0.2800kg * (0.215moles / 1kg) = 0.0602 moles of solute must be added.
The mass of ethylene glycol must be added is:
0.0602 moles * (62.10g / mol) =
3.74g of ethylene glycol must be added to decrease the freezing point by 0.400°C
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Answer:
333.3mL
Explanation:
Using the formula as follows:
C1V1 = C2V2
Where;
C1 = initial concentration (M)
C2 = final concentration (M)
V1 = initial volume (mL)
V2 = final volume (mL)
According to the information provided in this question,
C1 = 4.00M
C2 = 1.50M
V1 = 125mL
V2 = ?
Using C1V1 = C2V2
4 × 125 = 1.5 × V2
500 = 1.5V2
V2 = 500/1.5
V2 = 333.3mL
Therefore, the CuSO4 solution needs to be diluted to 333.3mL to make 1.50 M solution.
