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polet [3.4K]
3 years ago
10

Calculate the hydroxide ion concentration [OH-] for a solution with a pH of 6.10

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

There are different formula you need to keep in mind when solving for [OH-]

Given that pH = 6.10

pH + pOH = 14

6.10 + pOH = 14

pOH = 7.9

[OH-] = 10^(-pOH)

[OH-] = 10^(-7.9)

[OH-] = 0.000000013

[OH-] = 1.3 x 10^-8


<h2><u>Answer: [OH-] = 1.3 x 10^-8</u></h2>
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If something has more mass does it mean it has more atoms?
MrMuchimi

Answer:

I think so

Explanation:

There are atone in every particle so the more particles, the more atoms

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How many moles of ethane gas, C2H6(g), will occupy 15.0 L at 30 degrees Celsius and 5.0 atm?
Alenkinab [10]

Answer:

3 mol

Explanation:

Given data:

Volume of ethane = 15.0 L

Temperature = 30°C

Pressure = 5.0 atm

Number of moles of ethane = ?

Solution:

PV = nRT

P= Pressure

V = volume

n = number of moles

R = general gas constant = 0.0821 atm.L/ mol.K  

T = temperature in kelvin

Now we will convert the temperature.

30+273 = 303 K

5.0 atm × 15.0 L = n×0.0821 atm.L/ mol.K   × 303 K

75 atm.L =  n× 24.87 atm.L/ mol

n = 75 atm.L / 24.87 atm.L/ mol

n = 3 mol

8 0
3 years ago
an electron in the 3rd shell of an Aluminium atom moves to the first shell in a bombardment process. Calculate the frequency of
Liula [17]

Following the quantic theory, the energy of a photon equals the radiation frequency multiplied by the universal constant. ν = 2.923x10¹⁵ Hz. E = 3.09x10¹⁵Hz.

<h3>What is quantum mechanic?</h3>

It is the branch of physics that studies objects and forces at a very low scale, at atoms, subatoms, and particles levels.

Quantum mechanics states that the elemental particles that constitute matter -electrons, neutrons, protons- have the properties of a wave and a particle.

It emerges from the quantic theory exposed by Max Planck (1922), in which he affirmed that light propagates in energy packages or photons.

He discovered the Universal Planck constant, h, used to calculate the energy of a photon.

He stated that the energy of a photon (E) equals the radiation frequency (ν) multiplied by the universal constant (h).

E = νh

In the exposed example, we need to calculate the energy required to change from the 3rd shell to the first shell.

To do it, we should know that the energy in a level (Eₙ) equals the energy associated to an electron in the most inferior energy level (E₁) divided by the square of the shell number (n²).

Eₙ = E₁ / n²

E₁ is a constant. We can express it in <em>Joules </em>or <em>electroVolts </em>

  • E₁ = -2.18x10⁻¹⁸ J
  • E₁ = -13.6 eV

So, let us calculate the energy at level 1 and 3

Eₙ = E₁ / n²

  • E₁ =  -2.18x10⁻¹⁸ J / 1² =<u>  -2.18x10⁻¹⁸</u><u> J</u>

        E₁ =  -13.6 eV / 1² =<u>  -13.6 </u><u>eV</u>

  • E₃ =  -2.18x10⁻¹⁸ J / 3² =  -2.18x10⁻¹⁸ J / 9 =<u> - 2.42x10⁻¹⁹ </u><u>J</u>

        E₃ =  -13.6 eV / 3² =  -13.6 eV / 9 = <u>- 1.51 </u><u>eV</u>

The change of energy can be calculated in two ways,

<u>Option 1</u>

ΔE = E₁ - E₃ = 2.18x10⁻¹⁸ - 2.42x10⁻¹⁹ =<u> 1.93x10⁻¹⁸</u><u>J</u>

ΔE = E₁ - E₃ = 13.6 - 1.51 = <u>12.09 </u><u>eV</u>

<u>Option 2</u>

ΔE = -2.18x10⁻¹⁸ J (1/nf² - 1/ni²)

ΔE =-13.6 eV (1/nf² - 1/ni²)

Where nf is the final level and ni is the initial level. When the electron passes from its initial level to its final level it is called electronic transition.

  • ni = 3
  • nf = 1

ΔE = -2.18x10⁻¹⁸ J (1/nf² - 1/ni²)

ΔE = -2.18x10⁻¹⁸ J (1/1² - 1/3²)

ΔE = -2.18x10⁻¹⁸ J (1 - 0.111)

ΔE = -2.18x10⁻¹⁸ J (0.888)

<u>ΔE</u><u> = - 1.937x10⁻¹⁸ </u><u>J</u>

or

ΔE = -13.6 eV  (1/nf² - 1/ni²)

ΔE = -13.6 eV  (1/1² - 1/3²)

ΔE = -13.6 eV  (1 - 0.111)

ΔE = -13.6 eV  (0.888)

<u>ΔE</u><u> = -12.08</u><u> eV</u>

This is the energy required for the electron to go from n= 3 to n = 1. The negative sign (-) means energy (as light or photons) released or emitted.

<u />

If we want to express the result in Hz, we just need to make a conversion.

1Hz ⇔ 6.626x10⁻³⁴J ⇔ 4.136x10¹⁵ eV.

The energy required for the electron to go from n= 3 to n = 1 is <u>3.09x10¹⁵ </u><u>Hz</u><u>.</u>

Now, we need to calculate the frequency, ν. This is, how many times the wave oscillates back and foward per second.

To do it, we will use the universal Planck constant, h, and the absolute value of the energy, E.

ν = E/h = 1.937x10⁻¹⁸ J / 6.626x10⁻³⁴ Js = 2.923x10¹⁵ 1/s =  <u>2.923x10¹⁵ Hz</u>.

<u>Answer</u>:

  • Frequency, ν = E/h = <u>2.923x10¹⁵ </u><u>Hz</u>.
  • Energy, E = <u>3.09x10¹⁵ </u><u>Hz</u><u>.</u>

You can learn more about quantum mechanic at

brainly.com/question/11855107

brainly.com/question/23780112

brainly.com/question/11852353

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soldier1979 [14.2K]

Answer:

Should be either mendele or fermium

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Explanation: <u><em>really trying to help! If it’s wrong super sorry!!!</em></u>

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