Answer:
Option a.
0.01 mol of CaCl₂ will have the greatest effect on the colligative properties, because it has the biggest i
Explanation:
To determine which of the solute is going to have a greatest effect on colligative properties we have to consider the Van't Hoff factor (i)
These are the colligative properties:
ΔP = P° . Xm . i → Lowering vapor pressure
ΔT = Kb . m . i → Boiling point elevation
ΔT = Kf . m . i → Freezing point depression
π = M . R . T → Osmotic pressure
Van't Hoff factor are the numbers of ions dissolved in the solution. For nonelectrolytes, the i values 1.
CaCl₂ and KNO₃ are two ionic solutes. They dissociate as this:
CaCl₂ → Ca²⁺ + 2Cl⁻
We have 1 mol of Ca²⁺ and 2 chlorides, so 3 moles of ions → i = 3
KNO₃ → K⁺ + NO₃⁻
We have 1 mol of K⁺ and 1 mol of nitrate, so 2 moles of ions → i = 2
Option a, is the best.
D
Is the answer because of the ratio of you fart and the smell in ur nose
Sound waves are waves of growing larger and smaller, making it seem elastic through substances, such as air.
Answer:
Vf = 1.22 mL
Explanation:
If we assume that the pressure is constant and the number of moles does not change, we can say that the volume of the gas is modified in a directly ratio, to the Absolute Temperature.
Let's convert the values:
91°C + 273 = 364K
0.9°C + 273 = 273.9K
Volume decreases if the temperature is decreases
Volume increases if the T° increases
V₁ / T₁ = V₂ / T₂ → 1.63mL /364K = V₂ / 273.9K
V₂ = (1.63mL /364K) . 273.9K → 1.22 mL
I believe that it would be Al1N1.