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4 years ago

Calculate the hydrogen-ion concentration (H+] for the aqueous solution in which [OH-1 is 1 x 10-12 mol/L.

Chemistry

2 answers:

Artemon [7]4 years ago

7 0

Answer:If we have [H+][OH-]= Kw = 1.0 x 10^-14

Then [H+]= Kw/ [OH-]= 1.0x 10^-14/ 1 x 10^-11 =1 x 10^-3 mol/L

And here is the solution - as you can see it is an acidic one :

pH = - log [H+]= - log 1 x 10^-3 = 3 < 7

Explanation:

Vlad1618 [11]4 years ago

5 0

Answer:

[H+] = 0.01mol/L

Explanation:

p[OH] = -log[OH-]

[OH-] = 1*10^-12

pOH = -Log[1*10^-12]

pOH = 12.

But pH + pOH = 14

pH = 14 - pOH

pH = 14 - 12

pH = 2

pH = -Log[H+]

2 = -Log[H+]

Take the anti log of both sides

10⁻² = [H+]

[H+] = 0.01

The hydrogen ion concentration is 0.01mol/L

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The partial pressure of O2 in air at sea level is 0.21atm. The solubility of O2 in water at 20∘C, with 1 atm O2 pressure is 1.38

adell [148]

Answer:

$1.21x10^{-3}$ M

Explanation:

Henry's law relational the partial pressure and the concentration of a gas, which is its solubility. So, at the sea level, the total pressure of the air is 1 atm, and the partial pressure of O2 is 0.21 atm. So 21% of the air is O2.

Partial pressure = Henry's constant x molar concentration

0.21 = Hx1.38x $10^{-3}$

H = $\frac{0.21}{1.38x10^{-3} }$

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For a pressure of 665 torr, knowing that 1 atm = 760 torr, so 665 tor = 0.875 atm, the ar concentration is the same, so 21% is O2, and the partial pressure of O2 must be:

P = 0.21*0.875 = 0.1837 atm

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4 years ago

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

Katarina [22]

Answer:

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 $5.23\times 10^{-5} m$

Explanation:

Given :

The energy E of the electron in a hydrogen atom can be calculated from the Bohr formula:

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

$R_y=2.18\times 10^{-18} J$ = Rydberg energy

n = principal quantum number of the orbital

Energy of 11th orbit = $E_{11}$

$E_{11}=\frac{2.18\times 10^{-18} J}{11^2}=1.80\times 10^{-20} J$

Energy of 10th orbit = $E_{10}$

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

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)

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

$\lambda = 5.2310\times 10^{-5} m\approx 5.23\times 10^{-5} m$

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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