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Advocard [28]
2 years ago
6

What is the pH of a solution of KOH with a hydroxide concentration of [OH⁻] = 1.10 x 10⁻⁴

Chemistry
1 answer:
olga2289 [7]2 years ago
3 0

Answer:

p[H+] = 10.042

Explanation:

As we know that

pKw = pH + pOH......eq (1)

we will calculate the pH of OH- and then we will calculate the pH of H+

So p[OH-] = - log [1.10 * 10^{-4}]

Solving the right side of the equation, we get

p[OH-]

= - [-3.958]\\= 3.958

Now we know that

pKw = 14.0

Substituting the value of pOH in the above equation, we get -

14.0 = p[H+] + 3.958\\p[H+] = 14 - 3.958\\p[H+] = 10.042

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Answer:

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2 years ago
What is the wavenumber of the radiation emitted when a hydrogen
LUCKY_DIMON [66]

Answer: Wavenumber of the radiation emitted  is 0.08\times 10^{8}m^{-1}

Explanation:

The relationship between wavelength and energy of the wave follows the equation:

E=\frac{hc}{\lambda}

where,

E = energy of the radiation = 1.634\times 10^{-18}J

h = Planck's constant  = 6.626\times 10^{-34}Js

c = speed of light = 3\times 10^8m/s

\lambda = wavelength of radiation = ?

Putting values in above equation, we get:

1.634\times 10^{-18}J=\frac{(6.626\times 10^{-34}Js)\times (3\times 10^8m/s)}{\lambda}\\\\\lambda=12.16\times 10^{-8}m

\bar {\nu}=\frac{1}{\lambda}=\frac{1}{12.16\times 10^{-8}}=0.08\times 10^{8}m^{-1}

Thus wavenumber of the radiation emitted  is 0.08\times 10^{8}m^{-1}

8 0
3 years ago
Calcium bicarbonate (Ca(HCO3)2) Ca:H:C:O = 1:2:2 ___ Lithium sulfide (Li:2s) Li:S = 2:__
Anastaziya [24]

Answer: The ratio of atoms in calcium bicarbonate ; Ca : H : C : O = 1:2:2:6.

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Explanation:

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In a molecular formula of calcium carbonate there are:

Number of Calcium atoms = 1

Number of Hydrogen atom = 1 × 2 = 2

Number of Carbon atoms = 1 × 2 = 2

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So, Ca : H : C : O = 1 : 2 : 2 : 6

In lithium sulfide :Li_2S

In a molecular formula of lithium sulfide there are:

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8 0
3 years ago
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Explanation:

4 0
2 years ago
Read 2 more answers
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4.32x10^2/20
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3 years ago
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