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.
The equation for energy of a photon is E=hv where v equals frequency and h equals the Planck constant (6.626X10^-34). So since you've been given frequency you can just plug in frequency to find the total energy in joules.
E=(3.55X10^17)(6.626X10^-34)
E=2.35223X10^-16
Not sure how many significant figures you needed. Hope this helped.
Buffers - mixtures of conjugate acid and conjugate base at ±1 pH unit from pH = pKa. Resistant to changes in pH in response to small additions of H+ or OH-. ... Polyprotic acids - dissociation of each H+ can be treated separately if the pKa values are different
The partial pressure of hydrogen is 0.31 atm
calculation
find the number of hydrogen moles the container, that is
25/100 x 6.4 =1.6 moles of hydrogen
find the partial pressure for hydrogen in 1.6 moles
that is 6.4 moles= 1.24 atm
1.6 moles= ?
by cross multiplication
1.6moles x1.24 atm/ 6.4 moles= 0.31 atm
