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Mariulka [41]
3 years ago
12

What is the [OH-] in a solution with a POH of 7.86? [OH-] = [?] x 10!?)

Chemistry
1 answer:
gogolik [260]3 years ago
5 0
<h3>Answer:</h3>

1.380 × 10^-8 M

<h3>Explanation:</h3>

The pH is the measure of alkalinity or acidity of a substance.

It is the negative logarithm of H+ ions concentration [H+].

pH = -log[H+]

pOH on the other hand is negative logarith of OH- concentration [OH-].

pOH = -log[OH-]

In this case;

pOH = 7.86

We can calculate the [OH-]

pOH = -log[OH-]

But, pOH = 7.86

pOH = -log[OH-]

-log[OH-] = 7.86

[OH-] = Antilog -7.86

       = 1.380 × 10^-8 M

Therefore, [OH-] is 1.380 × 10^-8 M

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What is the pH value of Potassium hydroxide
RSB [31]

Answer:

from 10-14

Explanation:

the pH of potassium hydroxide is extremely high and is a strong base although the exact value depends on the concentration of the base in water.

3 0
2 years ago
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What happens in a chemical change that does not occur in a physical change?
Andrej [43]

Answer:

In a chemical reaction, there is a change in the composition of the substances in question; in a physical change there is a difference in the appearance, smell, or simple display of a sample of matter without a change in composition. Although we call them physical "reactions," no reaction is actually occurring.

6 0
3 years ago
In an experiment to study the photoelectric effect, a scientist measures the kinetic energy of ejected electrons as afunction of
crimeas [40]

Answer:

a) v₀ = 4.41 × 10¹⁴ s⁻¹

b) W₀ = 176 KJ/mol of ejected electrons

c) From the graph, light of frequency less than v₀ will not cause electrons to break free from the surface of the metal. Electron kinetic energy remains at zero as long as the frequency of incident light is less than v₀.

d) When frequency of the light exceeds v₀, there is an increase of electron kinetic energy from zero steadily upwards with a constant slope. This is because, once light frequency exceeds, v₀, its energy too exceeds the work function of the metal and the electrons instantaneously gain the energy of incident light and convert this energy to kinetic energy by breaking free and going into motion. The energy keeps increasing as the energy and frequency of incident light increases and electrons gain more speed.

e) The slope of the line segment gives the Planck's constant. Explanation is in the section below.

Explanation:

The plot for this question which is attached to this solution has Electron kinetic energy on the y-axis and frequency of incident light on the x-axis.

a) Wavelength, λ = 680 nm = 680 × 10⁻⁹ m

Speed of light = 3 × 10⁸ m/s

The frequency of the light, v₀ = ?

Frequency = speed of light/wavelength

v₀ = (3 × 10⁸)/(680 × 10⁻⁹) = 4.41 × 10¹⁴ s⁻¹

b) Work function, W₀ = energy of the light photons with the wavelength of v₀ = E = hv₀

h = Planck's constant = 6.63 × 10⁻³⁴ J.s

E = 6.63 × 10⁻³⁴ × 4.41 × 10¹⁴ = 2.92 × 10⁻¹⁹J

E in J/mol of ejected electrons

Ecalculated × Avogadros constant

= 2.92 × 10⁻¹⁹ × 6.023 × 10²³

= 1.76 × 10⁵ J/mol of ejected electrons = 176 KJ/mol of ejected electrons

c) Light of frequency less than v₀ does not possess enough energy to cause electrons to break free from the metal surface. The energy of light with frequency less than v₀ is less than the work function of the metal (which is the minimum amount of energy of light required to excite electrons on metal surface enough to break free).

As evident from the graph, electron kinetic energy remains at zero as long as the frequency of incident light is less than v₀.

d) When frequency of the light exceeds v₀, there is an increase of electron kinetic energy from zero steadily upwards with a constant slope. This is because, once light frequency exceeds, v₀, its energy too exceeds the work function of the metal and the electrons instantaneously gain the energy of incident light and convert this energy to kinetic energy by breaking free and going into motion. The energy keeps increasing as the energy and frequency of incident light increases and electrons gain more speed.

e) The slope of the line segment gives the Planck's constant. From the mathematical relationship, E = hv₀,

And the slope of the line segment is Energy of ejected electrons/frequency of incident light, E/v₀, which adequately matches the Planck's constant, h = 6.63 × 10⁻³⁴ J.s

Hope this Helps!!!

5 0
4 years ago
A pharmacist has calculated that a patient requires 30 mmol of phosphate and 80 meq of potassium to be added to the pn. how many
aalyn [17]

Since potassium and phosphate is what we are to find for and they are both found in the potassium phosphate solution, therefore we solve for this one first on the basis of the phosphate.

The formula for finding the volume given the concentration and number of moles is:

Volume = number of moles / concentration in Molarity

Volume potassium phosphate required = 30 mmol phosphate / (3 mmol / mL)

<u>Volume potassium phosphate required = 10 mL</u>

This would also contain potassium in amounts of:

Amount of potassium in potassium phosphate = 10 mL (4.4 meg / mL)

Amount of potassium in potassium phosphate = 44 meg

 

Therefore the potassium chloride required is:

Volume of potassium chloride = (80 meg – 44 meg) / (2 meg / mL)

<span><u>Volume of potassium chloride = 72 mL</u></span>

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