Explanation:
Vapor pressure is defined as the pressure exerted by vapors or gas on the surface of a liquid.
Vapor pressure is inversely proportional to the number of solute particles. Hence, more will be the solute particles lower will be the vapor pressure and vice-versa.
(a) 
It dissociates to give two particles.
(b) 
Total number of particles it give upon dissociation are 1 + 2 = 3. Hence, it gives 3 particles.
(c) 
Total number of particles it give upon dissociation are 1 + 3 = 4. Hence, it gives 4 particles.
(d) Surcose being a cobvalent compound doe not dissociate into ions. Therefore, there will be only 1 particle is present.
(e) 
Total number of particles it give upon dissociation are 1 + 1 = 2. Hence, it gives 2 particles.
First, let us find the corresponding amount of moles H₂ assuming ideal gas behavior.
PV = nRT
Solving for n,
n = PV/RT
n = (6.46 atm)(0.579 L)/(0.0821 L-atm/mol-K)(45 + 273 K)
n = 0.143 mol H₂
The stoichiometric calculations is as follows (MW for XeF₆ = 245.28 g/mol)
Mass XeF₆ = (0.143 mol H₂)(1 mol XeF₆/3 mol H₂)(245.28 g/mol) = <em>11.69 g</em>
The formula to be used for this problem is as follows:
E = hc/λ, where h is the Planck's constant, c is the speed of light and λ is the wavelength. Also 1 aJ = 10⁻¹⁸ J
0.696×10⁻¹⁸ = (6.62607004×10⁻³⁴ m²·kg/s)(3×10⁸ m/s)/λ
Solving for λ,
λ = 2.656×10⁻⁷ m or <em>0.022656 nm</em>
Answer:

Explanation:
Hello!
In this case, since the dissolution of copper (I) chloride is:

And its equilibrium expression is:
![Ksp=[Cu^+][Cl^-]](https://tex.z-dn.net/?f=Ksp%3D%5BCu%5E%2B%5D%5BCl%5E-%5D)
We can represent the molar solubility via the reaction extent as
, however, since there is 0.050 M KCl we immediately add such amount to the chloride ion concentration since KCl is readily ionized; therefore we write:

Thus, solving for
, we obtain:

By using the quadratic equation, we obtain:

Clearly, the solution is
because no negative results are
allowed. Therefore, the molar solubility is:

Best regards!
C5H12 + 8 O2 → 5 CO2 + 6 H2O
8 molecules of O2 are required.