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3 years ago

Calculate the ph of a 0.17 m solution of c6h5nh3no3 (kb for c6h5nh2 = 3.8 x 10-10). record your ph value to 2 decimal places.

1 answer:

6 0

Mass of the solution = 0.17m
Kb for C6H5NH2 = 3.8 x 10^-10
We know Ka for C6H5NH2 = 1.78x10^-11
We have Kw = Ka x Kb => Ka = Kw / Kb
=> (C2H5NH2)(H3O^+)/(C2H5NH3^+) => 1.78x10^-11 = K^2 / 0.17
K^2 = 3 x 10^-12 => K = 1.73 x 10^-6.
pH = -log(Kw(H3O^+)) = -log(1.73 x 10^-6) = 5.76

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The energy E of the electron in a hydrogen atom can be calculated from the Bohr formula: E = R_y/n^2 In this equation R_y stands

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The wavelength of the line in the emission line spectrum of hydrogen caused by the transition of the electron for the given energy levels is $5.23\times 10^{-5} m$

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$E=\frac{R_y}{n^2}$

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n = principal quantum number of the orbital

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$E_{10}=\frac{2.18\times 10^{-18} J}{10^2}=2.18\times 10^{-20} J$

Energy difference between both the levels will corresponds to the energy of the wavelength of the line which can be calculated by using Planck's equation.

$E'=E_{10}-E_{11}=2.18\times 10^{-20} J-1.80\times 10^{-20} J$

$=E'=0.38\times 10^{-20} J$

$\lambda =\frac{hc}{E'}$ (Planck's' equation)

$\lambda = \frac{6.626\times 10^{-34} Js\times 3\times 10^8 m/s}{0.38\times 10^{-20} J}$

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2 years ago

The vapor pressure of pure water at 25 °c is 23.8 torr. What is the vapor pressure (torr) of water above a solution prepared by

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The vapour pressure of the solution is 23.4 torr.

Use Raoult’s Law to calculate the vapour pressure:

p₁ = χ₁p₁°

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χ₁ = the mole fraction of the solvent

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The formula for vapour pressure lowering Δp is

Δp = p₁° - p₁

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Step 1. Calculate the mole fraction of glucose

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χ₂ = n₂/(n₁ + n₂) = 0.1000/(0.1000 + 5.272) = 0.1000/5.372 = 0.018 62

Step 2. Calculate the vapour pressure lowering

Δp = χ₂p₁° = 0.018 62 × 23.8 torr = 0.4430 torr

Step 3. Calculate the vapour pressure

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The structure of monobromo derivatives of pentane that is 1-bromopentane, 2-bromopentane, and 3-bromopentane and having molecular formula, $C_5H_1_1Br$ is shown in the image.

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