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.
Answer:
Yellow seeds
Explanation:
Dominant traits are those that are inherited unchanged in a hybridization like the yellow seeds.
Is it inches millimeters meters ?
Answer:
34.02 g.
Explanation:
Hello!
In this case, since the gas behaves ideally, we can use the following equation to compute the moles at the specified conditions:
Now, since the molar mass of a compound is computed by dividing the mass over mass, we obtain the following molar mass:
So probably, the gas may be H₂S.
Best regards!
Answer:
0.0611M of HNO3
Explanation:
<em>The concentration of the NaOH solution must be 0.1198M</em>
<em />
The reaction of NaOH with HNO3 is:
NaOH + HNO3 → NaNO3 + H2O
<em>1 mole of NaOH reacts per mole of HNO3.</em>
That means the moles of NaOH used in the titration are equal to moles of HNO3.
<em>Moles HNO3:</em>
12.75mL = 0.01275L * (0.1198mol / L) = 0.0015274 moles NaOH = Moles HNO3.
In 25.00mL = 0.025L -The volume of the aliquot-:
0.00153 moles HNO3 / 0.025L =
<h3> 0.0611M of HNO3</h3>
