Answer:
The correct option is;
X, W, Y, Z
Explanation:
The parameters given are;
Spring (S), Spring Constant (N/m)
W, 24
X, 35
Y, 22
Z, 15
The equation for elastic potential energy, 
, is 
The above equation can also be written as 
Where:
k = The spring constant in (N/m)
x = The spring extension
Therefore, since the elastic potential energy, , of the spring is directly proportional to the spring constant, k, we have the springs with higher spring constant will have higher elastic potential energy, , therefore the correct order is as follows;
X > W > Y > Z
Mass of X₂O₇ = 54,9g
2x + 33,6g = 54,9g
2x = 54,9g - 33,6g
2x = 21,3g | :2
x = 10,65g/mol
Answer:
The molality of the solution is 0.3716 mol/kg
The number of moles of solute is 0.0157 mol
The molecular weight of the solute is 129.30 g/mol
The molar mass of the solute is 129.32 g/mol
Explanation:
m (molality of the solution) = ∆T/Kf = (43.17 - 40.32)/7.67 = 0.3716 mol/kg
Number of moles of solute = molality × mass of solvent in kilogram = 0.3716 × 0.04219 = 0.0157 mol
Molecular weight of solute = mass/number of moles = 2.03/0.0157 = 129.3 g/mol
When Kf = 7.66 °C.kg/mol
Molar mass = 2.03 ÷ (2.85/7.66 × 0.04219) = 129.32 g/mol
Answer:
333.7g of antifreeze
Explanation:
Freezing point depression in a solvent (In this case, water) occurs by the addition of a solute. The law is:
ΔT = Kf × m × i
Where:
ΔT is change in temperature (0°C - -20°C = 20°C)
Kf is freezing point depression constant (1.86°C / m)
m is molality of solution (moles solute / 0.5 kg solvent -500g water-)
i is Van't Hoff factor (1, assuming antifreeze is ethylene glycol -C₂H₄(OH)₂)
Replacing:
20°C = 1.86°C / m × moles solute / 0.5 kg solvent × 1
5.376 = moles solute
As molar mass of ethylene glycol is 62.07g/mol:
5.376 moles × (62.07g / 1mol) = <em>333.7g of antifreeze</em>.
