Answer is: 48,25 torr.
Raoult's Law: p = x(solv) · p(solv)
p - <span>vapour pressure of a solution.
</span>x(solv) - <span>mole fraction of the solvent.
</span>p(solv) - <span>vapour pressure of the pure solvent.
</span>n(ethanol) = 950g ÷ 46,07g/mol = 20,62 mol.
x(solv) = moles of solvent ÷ total number of moles
x(solv) = 20,62 ÷ 21,77 = 0,965.
p = 0,965 ·50,0 torr = 48,25 torr.
Profundal
profundal zone is the coldest and deepest zone
We are going to use this equation:
ΔT = - i m Kf
when m is the molality of a solution
i = 2
and ΔT is the change in melting point = T2- 0 °C
and Kf is cryoscopic constant = 1.86C/m
now we need to calculate the molality so we have to get the moles of NaCl first:
moles of NaCl = mass / molar mass
= 3.5 g / 58.44
= 0.0599 moles
when the density of water = 1 g / mL and the volume =230 L
∴ the mass of water = 1 g * 230 mL = 230 g = 0.23Kg
now we can get the molality = moles NaCl / Kg water
=0.0599moles/0.23Kg
= 0.26 m
∴T2-0 = - 2 * 0.26 *1.86
∴T2 = -0.967 °C
Weather can be predicted only as probable, not definite because the daily weather conditions depends on winds and storms.
It is common observation that weather forecasts are often given as a probability and never in definite terms.
This is because, the weather condition at anytime depends on the temperature of the earth's atmosphere which causes air masses to move leading to wind.
The temperature of the earth's atmosphere changes frequently hence weather conditions also change frequently.
Learn more: brainly.com/question/21209813
Answer:
The concentration of a saturated solution of CuF₂ in aqueous 0.20 M NaF is 4.0×10⁻⁵ M.
Explanation:
Consider the ICE take for the solubility of the solid, CuF₂ as:
CuF₂ ⇄ Cu²⁺ + 2F⁻
At t=0 x - -
At t =equilibrium (x-s) s 2s
The expression for Solubility product for CuF₂ is:
![K_{sp}=\left [ Cu^{2+} \right ]\left [ F^- \right ]^2](https://tex.z-dn.net/?f=K_%7Bsp%7D%3D%5Cleft%20%5B%20Cu%5E%7B2%2B%7D%20%5Cright%20%5D%5Cleft%20%5B%20F%5E-%20%5Cright%20%5D%5E2)
Given s = 7.4×10⁻³ M
So, Ksp is:
Ksp = 1.6209×10⁻⁶
Now, we have to calculate the solubility of CuF₂ in NaF.
Thus, NaF already contain 0.20 M F⁻ ions
Consider the ICE take for the solubility of the solid, CuF₂ in NaFas:
CuF₂ ⇄ Cu²⁺ + 2F⁻
At t=0 x - 0.20
At t =equilibrium (x-s') s' 0.20+2s'
The expression for Solubility product for CuF₂ is:
Solving for s', we get
<u>s' = 4.0×10⁻⁵ M</u>
<u>The concentration of a saturated solution of CuF₂ in aqueous 0.20 M NaF is 4.0×10⁻⁵ M.</u>
