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
<em />
Answer:
The answer is: 
Explanation:
Scientific notation is of the kind: 
The given calculation:
Therefore, the answer in scientific notation is:
Answer: 2.7kg/L
Explanation:
1 gram per cubic centimeter g/cm3= 1.00 kilograms per liter kg/l
Plants combine water and Carbon dioxide from the air to make glucose for themselves, and giving off oxygen in the process. So CO2 and H2O would be on ther reactant side (left) with O2 and C6H12O6 being on the product side (right).
What is the answer to this question
