The Answer to your question would be A
Answer:
the fuel efficiency in kilometers per liter is 16.561 kilometer per liter
Explanation:
The computation of the full efficiency in kilometers per liter is shown below:
39.0 miles ÷ gallon = (39.0 miles ÷ gallon) × (1.6094 km ÷ 1 miles) × (1 gallon ÷ 3.79 L)
Now cut the opposite miles and gallons
So, the fuel efficiency would be
= 16.561 kilometers per liter
Hence, the fuel efficiency in kilometers per liter is 16.561 kilometer per liter
Oxidation happens at the anode and reduction happens at the cathode.<span />
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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