solution:
the change in the boiling point is given as,
dTbp =2.30°c
elevation constant for the solvent is given by,
kb=0.512°c/m

= 4.49m
Answer:
#Molecules XeF₆ = 2.75 x 10²³ molecules XeF₆.
Explanation:
Given … Excess Xe + 12.9L F₂ @298K & 2.6Atm => ? molecules XeF₆
1. Convert 12.9L 298K & 2.6Atm to STP conditions so 22.4L/mole can be used to determine moles of F₂ used.
=> V(F₂ @ STP) = 12.6L(273K/298K)(2.6Atm/1.0Atm) = 30.7L F₂ @ STP
2. Calculate moles of F₂ used
=> moles F₂ = 30.7L/22.4L/mole = 1.372 mole F₂ used
3. Calculate moles of XeF₆ produced from reaction ratios …
Xe + 3F₂ => XeF₆ => moles of XeF₆ = ⅓(moles F₂) = ⅓(1.372) moles XeF₆ = 0.4572 mole XeF₆
4. Calculate number molecules XeF₆ by multiplying by Avogadro’s Number (6.02 x 10²³ molecules/mole)
=> #Molecules XeF₆ = 0.4572mole(6.02 x 10²³ molecules/mole)
= 2.75 x 10²³ molecules XeF₆.
Stored water is gravitational
For Ca(OH)2, Ksp = [Ca2+][OH-]^2
You have your Ksp as 6.5 x 10^-6. Your [OH-] comes almost entirely from the 0.10 mol of NaOH, since Ca(OH)2 barely dissolves. Your [OH-] is therefore 0.10 M (since you have 1 L of solution).
6.5 x 10^-6 = [Ca2+](0.10)^2
Solve for [Ca2+]:
6.5 x 10^-6 / (0.10)^2 = [Ca2+]
[Ca2+] = 0.00065 M
The maximum concentration of [Ca2+] is 0.00065 M, and you have 0.0010 M Ca(OH)2, so you’ll end up with 0.00065 M Ca2+ in solution.
Number one is the correct answer