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Lady bird [3.3K]
4 years ago
10

The values used in the scale of pH and pOH are derived from a system designed by ______. Gordonsen Sorenson Curie Dalton

Chemistry
1 answer:
yaroslaw [1]4 years ago
3 0

Sorenson

Explanation:

The values used in the scale of pH and pOH are derived from a system designed by Sorenson. Søren Peter Lauritz Sørensen, a Danish chemist introduced the system of pH and pOH for describing the alkalinity and acidity of substances.

  • The pH and pOH scale is logarithmic scale that ranks the acidity and bascity of compounds.
  • pH is the negative logarithm of the concentration of  hydrogen/hydroxonium ions in solution i.e

              pH = -log₁₀{H⁺]

  • pOH is the negative log of the concentration of the hydroxyl ions in a solution i.e

             pOH = -log₁₀{OH⁻]

Learn more:

calculating pH: brainly.com/question/12985875

pH scale: brainly.com/question/11063271

#learnwithBrainly

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The number of electron domains in a molecule or ion is the number of bonds (double and triple bonds count as one domain) plus the number of nonbonding (lone) electron pairs.

Explanation:

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What is the minimum (non-zero) thickness of a benzene (n = 1.501) thin film that will result in constructive interference when v
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Explanation:

At each reflecting surface (benzene and glass) there will be 180 degree phase change.

Now, for constructive interference the optical path in benzene is \lambda.

Formula to calculate thickness of a benzene thin film is as follows.

     Optical path length through benzene (\lambda) = 2 \times n \times d

Hence, substituting the given values into the above formula as follows.

    Optical path length through benzene = 2 \times n \times d

                   d = \frac{\text{Optical path length through benzene}}{2 \times n}

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3 years ago
What is the molality of 49 grams of NaCl in 469 grams of H2O (water)
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Answer : The molality of sodium chloride is, 1.79 m

Explanation :

To calculate the molality of solution, we use the equation:

\text{Molality}=\frac{m_{solute}\times 1000}{M_{solute}\times W_{solvent}\text{ (in grams)}}

Where,

m_{solute} = Given mass of solute (sodium chloride) = 49 g

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Now put all the given values in above equation, we get:

\text{Molality of }NaCl=\frac{49g\times 1000}{58.5g/mol\times 469kg}\\\\\text{Molality of }NaCl=1.79mol/kg=1.79m

Hence, the molality of sodium chloride is, 1.79 m

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4 years ago
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