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3 years ago

How are electrons added to orbitals in an electron configuration?

Chemistry

1 answer:

Elenna [48]3 years ago

4 0

Answer:

Electrons are added to atomic orbitals in order from low energy bottom of the graph to high (top of the graph according to the Aufbau principle. Principle energy levels are color coded, while sublevels are grouped together and each circle represents an orbital capable of holding two electrons

Explanation:

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A chemistry student is given 650. mL of a clear aqueous solution at 33.°C. He is told an unknown amount of a certain compound X

Debora [2.8K]

Answer:

No, you can not calculate the solubility of X in water at 17 0C.

Explanation:

Solubility refers to the amount of a substance that dissolves in 1000 L of water.

To calculate the solubility of a solute in water, all the water is evaporated and the solid is carefully collected, washed, dried and weighed. The mass of solid obtained can now be used to calculate the solubility of the solute in water as long as there was no loss in mass of solid during the experiment.

In this case, the student threw away part of the solid that precipitated. As a result of this, the mass of solid obtained at the end of the experiment is not exactly the total mass of solute that dissolved in the solvent. Hence, the solubility of X in water at 17 0C can not be accurately calculated.

7 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 Joules or electroVolts

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 = -2.18x10⁻¹⁸ J

E₁ = -13.6 eV / 1² = -13.6 eV

E₃ = -2.18x10⁻¹⁸ J / 3² = -2.18x10⁻¹⁸ J / 9 = - 2.42x10⁻¹⁹ J

E₃ = -13.6 eV / 3² = -13.6 eV / 9 = - 1.51 eV

The change of energy can be calculated in two ways,

Option 1

ΔE = E₁ - E₃ = 2.18x10⁻¹⁸ - 2.42x10⁻¹⁹ = 1.93x10⁻¹⁸J

ΔE = E₁ - E₃ = 13.6 - 1.51 = 12.09 eV

Option 2

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)

ΔE = - 1.937x10⁻¹⁸ J

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)

ΔE = -12.08 eV

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.

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 3.09x10¹⁵ Hz.

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 = 2.923x10¹⁵ Hz.

Answer:

Frequency, ν = E/h = 2.923x10¹⁵ Hz.
Energy, E = 3.09x10¹⁵ Hz.

You can learn more about quantum mechanic at

brainly.com/question/11855107

brainly.com/question/23780112

brainly.com/question/11852353

5 0

2 years ago

I need help with number two

Komok [63]

The temperature is held constant at (b) and (d). At these points, the substance is changing states. B is changing from solid to liquid and D is changing from liquid to gas

3 0

3 years ago

10. A 20.00 mL sample of 0.150 mol/L ammonia (NH3(aq)) is titrated to the equivalence point by 20.0 mL of a solution of 0.150 mo

Natalija [7]

Answer:

$\large \boxed{\rm a)\, NH_{3}(aq) + \text{HI}(aq) \, \longrightarrow \, \,$ NH_{4}^{+}(aq) +\text{I}^{-}(aq);\,b)\,11.22;\, c)\, 5.19}$

Explanation:

a) Balanced equation

The balanced chemical equation for the titration is

$\large \boxed{\rm NH_{3}(aq) + \text{HI}(aq) \, \longrightarrow \, \,$ NH_{4}^{+}(aq) +\text{I}^{-}(aq)}$

b) pH at start

For simplicity, let's use B as the symbol for NH₃.

The equation for the equilibrium is

$\rm B + H_{2}O \, \rightleftharpoons\,BH^{+} + OH^{-}$

(i) Calculate [OH]⁻

We can use an ICE table to do the calculation.

B + H₂O ⇌ BH⁺ + OH⁻

I/mol·L⁻¹: 0.150 0 0

C/mol·L⁻¹: -x +x +x

E/mol·L⁻¹: 0.150 - x x x

$K_{\text{b}} = \dfrac{\text{[BH}^{+}]\text{[OH}^{-}]}{\text{[B]}} = 1.8 \times 10^{-5}\\\\\dfrac{x^{2}}{0.150 - x} = 1.8 \times 10^{-5}$

Check for negligibility:

$\dfrac{0.150 }{1.8 \times 10^{-5}} = 8300 > 400\\\\x \ll 0.150$

(ii) Solve for x

$\dfrac{x^{2}}{0.150} = 1.8 \times 10^{-5}\\\\x^{2} = 0.150 \times 1.8 \times 10^{-5}\\x^{2} = 2.7 \times 10^{-6}\\x = \sqrt{2.7 \times 10^{-6}}\\x = \text{[OH]}^{-} = 1.64 \times 10^{-3} \text{ mol/L}$

(iii) Calculate the pH

$\text{pOH} = -\log \text{[OH}^{-}] = -\log(1.64 \times 10^{-3}) = 2.78\\\\\text{pH} = 14.00 - \text{pOH} = 14.00 - 2.78 = \mathbf{11.22}\\\\\text{The pH of the solution at equilibrium is } \large \boxed{\mathbf{11.22}}$

(c) pH at equivalence point

(i) Calculate the moles of each species

$\text{Moles of B} = \text{Moles of HI} = \text{20.00 mL} \times \dfrac{\text{0.0150 mmol}}{\text{1 mL}} = \text{3.00 mmol}$

B + HI ⇌ BH⁺ + I⁻

I/mol: 3.00 3.00 0

C/mol: -3.00 -3.00 +3.00

E/mol/: 0 0 3.00

(ii) Calculate the concentration of BH⁺

At the equivalence point we have a solution containing 3.00 mmol of NH₄I

Volume = 20.00 mL + 20.00 mL = 40.00 mL

$\rm [BH^{+}] = \dfrac{\text{3.00 mmol}}{\text{40.00 mL}} = \text{0.0750 mol/L}$

(iii) Calculate the concentration of hydronium ion

We can use an ICE table to organize the calculations.

BH⁺+ H₂O ⇌ H₃O⁺ + B

I/mol·L⁻¹: 0.0750 0 0

C/mol·L⁻¹: -x +x +x

E/mol·L⁻¹: 0.0750 - x x x

$K_{\text{a}} = \dfrac{K_{\text{w}}} {K_{\text{b}}} = \dfrac{1.00 \times 10^{-14}}{1.8 \times 10^{-5}} = 5.56 \times 10^{-10}\\\\\dfrac{x^{2}}{0.0750 - x} = 5.56 \times 10^{10}\\\\\text{Check for negligibility of }x\\\dfrac{0.0750}{5.56 \times 10^{-10}} = 1.3 \times 10^{6} > 400\\\\\therefore x \text{ $\ll$ 0.0750}$

$\dfrac{x^{2}}{0.0750} = 5.56 \times 10^{-10}\\\\x^{2} = 0.0750 \times 5.56 \times 10^{-10}\\x^{2} = 4.17 \times 10^{-11}\\x = \sqrt{4.17 \times 10^{-11}}\\\rm [H_{3}O^{+}] =x = 6.46 \times 10^{-6}\, mol \cdot L^{-1}$

(iv) Calculate the pH

$\text{pH} = -\log{\rm[H_{3}O^{+}]} = -\log{6.46 \times 10^{-6}} = \large \boxed{\mathbf{5.19}}$

The titration curve below shows the pH at the beginning and at the equivalence point of the titration.

8 0

3 years ago

Calculate the volume of 0.0500 m edta needed to titrate ca caco3

ehidna [41]

EDTA is a complex used to determine water hardness. The reaction between EDTA and CaCO₃ is in 1:1 ratio. To solve this problem, we must know the moles of CaCO₃ present (otherwise we can't solve the problem). Suppose you have 0.5 moles of CaCO₃. Since the ratio is 1:1, it would also need 0.5 moles EDTA. Then,

Volume = Moles/Molarity = 0.5 moles/0.05 M = 10 L of EDTA

7 0

4 years ago