An aqueous solution of potassium sulfate exhibits colligative properties. Colligative properties are properties that depends on the concentration of a substance in a solution. These properties are freezing point depression, vapor pressure lowering, osmotic pressure and boiling point elevation. For this problem we use the concept of freezing point depression since we are given the freezing point of the solution. Freezing point depression is as:
ΔT = -k(f) x m x i
-2.24 - 0 = -1.86 x m x 3
<span>m = 0.4014
Thus, the molality of the solution is 0.4014.</span>
Answer:
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Explanation:
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The answer is 18.02 g/mol
Explanation:
We have to find the number of moles of N₂ that are present in a sample that has a volume of 40.0 L at STP.
STP means Standard Conditions of Temperature and Pressure. These conditions are 273.15 K and 1 atm. We know that 1 mol of N₂ will occupy 22.4 L. We can use that ratio to find the answer to our problem.
1 mol of N₂ = 22.4 L
moles of N₂ = 40.0 L * 1 mol/(22.4 L)
moles of N₂ = 1.79 mol
Answer: 1.79 moles of nitrogen are present.
