If + + + 8 8 x 2 = 6 O + 6 = 48 + = 6 + 6 +

What amount of sucrose (C12H22O11) should be added to 5.83 mol water to lower the vapor pressure of water at 50 °C to 72.0 torr?

Triss [41]

Answer:

The amount of sucrose that must be added is 1.66 moles

Explanation:

Colligative property of lowering vapor pressure has this formula:

Vapor pressure of pure solvent (P°) - Vapor pressure of solution = P° . Xm

We have both vapor pressure (pure solvent and solution9, so let's determine the ΔP

ΔP = 92.6 Torr - 72 Torr = 20.6 Torr

Let's add the data in the formula

20.6 Torr = 92.6 Torr . Xm

Xm = Mole fraction of solute → (mol of solute/ mol of solute + mol of solvent)

Mol of solvent = 5.83 mol (data from the problem)

Therefore Xm = 20.6 Torr / 92.6 Torr → 0.222

Let's find out the moles of solute (our unknown value)

0.22 = moles of solute / moles of solute + 5.83 moles of solvent

0.222 (moles of solute + 5.83 moles of solvent) = moles of solute

0.222 moles of solute + 1.29 moles of solvent = moles of solute

1.29 moles of solvent = moles of solute - 0.222 moles of solute

1.29 moles = 0.778 moles of solute

1.29 / 0.778 = moles of solute → 1.66 moles

3 0

3 years ago

calculate the wavelength of light associated with the transition from n=1 to n=3 in the hydrogen atom?

inessss [21]

Answer:

$\Delta E=E_{final}-E_{initial}$

$\Delta E=-1312[\frac{1}{(n_f^2)}-\frac {1}{(n_i^2 )}]KJ mol^{-1}$

$\Delta E=-1312[\frac{1}{3^2)}-\frac {1}{(1^2 )}]KJ mol^{-1}$

$\Delta E=-1312[\frac{1}{(9)}-\frac {1}{(1 )}]KJ mol^{-1}$

$\Delta E=-1312[0.111-1]KJ mol^{-1}$

$\Delta E=1166 KJ mol^{-1}$

$\frac{=1166,000 \mathrm{J}}{6.022 \times 10^{23} \text { photons }}$

$=193623 \times 10^{-23} \frac {J}{photon}$

$\Delta E=1.93623 \times 10^{-18} \frac {J}{photon}$

$\Delta E=\frac {h\times c}{\lambda} \\\\=\frac {(6.626\times 10^{-34} J s \times 3 \times 10^8 ms^{-1})}{\lambda}$

h is planck's constant

c is the speed of light

λ is the wavelength of light

$\lambda =\frac {h\times c}{\Delta E}\\\\=\frac {(6.626\times10^{-34} J s\times3 \times 10^8 ms^{-1})}{(1.93623\times10^{-18} J/photon)}$

Wavelength

$\lambda =10.3 \times 10^{-8} m \times \frac {(10^9 nm)}{1m} =103 nm (Answer)$

Thus, the wavelength of light associated with the transition from n=1 to n=3 in the hydrogen atom is 103 nm.

7 0

3 years ago

The PH solution of NH3 of 0.950molar solution is 11.612. find the Kb

Vaselesa [24]

Answer:

1.8 x 10⁻⁵

Explanation:

NH3(aq) + H2O(l) ⇄ NH4⁺(aq) + OH⁻(aq)

I 0.95 0 0

C -x +x +x

E 0.95-x x x

Kb= [NH₄⁺] [OH⁻] / ( NH₃) = x²/ (0.95-x )

P(OH) = 14-PH = 14-11.612 = 2.388

(OH)⁻¹ = 10⁻²°³⁸⁸ = 4.09 x 10⁻³ = x

Kb = (4.09 x 10⁻³)²/ (0.95-4.09 x 10⁻³)

= 1.8 x 10⁻⁵

8 0

3 years ago

You can put out a chip pan fire by putting a damp cloth over the pan. explain why this works.

Free_Kalibri [48]

Because it cuts off the oxygen

4 0

3 years ago

Fe2o3(s)+3h2(g)→2fe(s)+3h2o(l) Calculate the mass of H2O

777dan777 [17]

Answer:

mass of H₂O = 54.06 g

Explanation:

balanced chemical equation

Fe₂O₃(s) + 3H₂(g) → 2Fe(s) + 3H₂O(l)

given data

moles of H₂O = 3 mol

mas of H₂O = ?

Molar mass of H₂O = 18.02 g/mol

we find the mass of H₂O by:

mass of H₂O = moles × molar mass

mass of H₂O = 3 mol × 18.02 g/mol

mass of H₂O = 54.06 g

7 0

3 years ago