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yanalaym [24]
3 years ago
4

Select the set of quantum numbers that represents each electron in a ground‑state be atom.

Chemistry
1 answer:
lukranit [14]3 years ago
3 0

Answer:

There are four electronsin a beryllium atom. The four set of quantum numbers for the ground-state beryllium atom are:

  • n = 1, ℓ = 0, mℓ = 0, ms = +1/2
  • n = 1, ℓ = 0, and mℓ = , ms = -1/2
  • n = 2, ℓ = 0, mℓ = 0, ms = +1/2
  • n = 2, ℓ = 0, and mℓ = , ms = -1/2

Explanation:

<em>Be</em> is the chemical symbol for beryllium.

The atomic number of beryllium is 4.

Hence, beryllium atoms have 4 protons and 4 electrons.

The electron configuration for the ground-state Be atom, is:

  • 1s² 2s²

So, you have to find the sets of quantum numbers for the two 1s and the two 2 s electrons.

<u>For the two 1s electrons</u>:

  • <u>Principal quantun number, n</u>.

The principal quantum number, n,  is the main energy level and is indicated by the number in front of the letter (kind of orbital).

So, for the two 1s electrons, n = 1

  • <u>Orbital Angular Momentum Quantum Number, ℓ</u>

The angular moment quantum number, ℓ, is the shape of the orbital.

The possible numbers for ℓ are from 0 to n- 1. Orbital s means ℓ = 0.

  • <u>Magnetic Quantum Number, mℓ</u>

The magnetic quantum number, mℓ, represents the space orientation of the orbital.

The possible numbers for mℓ are from - ℓ to + ℓ. Hence, for the two `s electrons mℓ is 0.

So far you have: n = 1, ℓ = 0, and mℓ = 0.

  • <u>Spin quantum number, s or ms</u>

The spin quantum number, related to the rotation of the electron, may be only +1/2 or -1/2.

So, now you have the two sets for these two electrons:

  • n = 1, ℓ = 0, mℓ = 0, ms = +1/2
  • n = 1, ℓ = 0, and mℓ = , ms = -1/2

<u>For the two 2s electrons:</u>

  • n = 2

  • ℓ = 0 (because the orbital is 2)

  • mℓ = 0 (because ℓ = 0)

  • ms = +1/2 or -1/2.

So, the two sets for the two s electrons are:

  • n = 2, ℓ = 0, mℓ = 0, ms = +1/2
  • n = 2, ℓ = 0, and mℓ = , ms = -1/2
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Answer:

I. dissolving more solute in an unsaturated solution

Explanation:

A saturated solution is the solution which contains maximum concentration of solute dissolved in solvent. ​

However, if the solution is being provided with energy (heating) , it will dissolve more heat and the solution that cannot even dissolve more solute after heating is super saturated solution.

<u>Both saturated and supersaturated solutions have maximum of their capacity to dissolve the solute and hence further energy requires energy and these processes are non-spontaneous.</u>

<u>An unsaturated solution can dissolve more solute easily and thus is a spontaneous process.</u>

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Consider the following balanced equation for the following reaction:
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<u>Answer:</u> The amount of carbon dioxide formed in the reaction is 47.48 grams

<u>Explanation:</u>

For the given chemical equation:

15O_2(g)+2C_6H_5COOH(aq.)\rightarrow 14CO_2(g)+6H_2O(l)

To calculate the actual yield of carbon dioxide, we use the equation:

\%\text{ yield}=\frac{\text{Actual yield}}{\text{Theoretical yield}}\times 100

Percentage yield of carbon dioxide = 83 %

Theoretical yield of carbon dioxide = 1.30 moles

Putting values in above equation, we get:

83=\frac{\text{Actual yield of carbon dioxide}}{1.30moles}\times 100\\\\\text{Actual yield of carbon dioxide}=\frac{1.30\times 83}{100}=1.079moles

To calculate the number of moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}

Molar mass of carbon dioxide = 44 g/mol

Moles of carbon dioxide = 1.079 moles

Putting values in above equation, we get:

1.079mol=\frac{\text{Mass of carbon dioxide}}{44g/mol}\\\\\text{Mass of carbon dioxide}=(1.079mol\times 44g/mol)=47.48g

Hence, the amount of carbon dioxide formed in the reaction is 47.48 grams

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When lithium nitride, Li 3 N ( s ) , is treated with water, H 2 O ( l ) , ammonia, NH 3 ( g ) , is produced. Predict the formula
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Answer : The formula of the gas produced is, PH_3 (phosphine gas)

Explanation :

According to the question, when sodium phosphide is treated with water then it react to give phosphine gas and sodium hydroxide.

The balanced chemical reaction will be:

Na_3P(aq)+3H_2O(l)\rightarrow PH_3(g)+3NaOH(aq)

By Stoichiometry of the reaction we can say that:

1 mole of sodium phosphide reacts with 3 moles of water to give 1 mole of phosphine gas and 3 moles of sodium hydroxide.

Thus, the formula of the gas produced is, PH_3 (phosphine gas)

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When 0.300 g of a diprotic acid was titrated with 0.100 M LiOH, 40.0 mL of the LiOH solution was needed to reach the second equi
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Answer:

  • <em><u>H₂C₄H₄O₆</u></em>

Explanation:

The clue of this question is to find the molar mass of the <em>diprotic acid</em> and compare witht the molars masses of the choices' acid to identify the formula of the diprotic acid.

The procedure is:

  1. Find the number of moles of the base: LiOH
  2. Use stoichionetry to infere the number of moles of the acid.
  3. Use the formula molar mass = mass in grams / number of moles, to find the molar mass of the diprotic acid.
  4. Compare and conclude.

<u>Solution:</u>

<u>1. Number of moles of the base, LiOH:</u>

  • M = n / V in liter ⇒ n = M × V = 0.100 M × 40.0 ml × 1 liter / 1,000 ml = 0.004 mol LiOH.

<u>2. Stoichiometry:</u>

Since this a neutralization reaction of a diprotic acid with a mono hydroxide base (LiOH), the mole ratio at the second equivalence point is: 2 mol of base / 1 mole of acid; because each mole of LiOH releases 1 mol of OH⁻, while each mole of diprotic acid releases 2 mol of H⁺.

Hence, 0.004 mol LiOH × 1 mol acid / 2 mol LiOH = 0.002 mol acid.

<u>3. Molar mass of the acid:</u>

  • molar mass = mass in grams / number of moles = 0.300 g / 0.002 mol = 150. g/mol

<u>4. Molar mass of the possible diprotic acids:</u>

a. H₂Se: 2×1.008 g/mol + 78.96  g/mol = 80.976 g/mol

b. H₂Te: 2×1.008 g/mol + 127.6  g/mol = 129.616 g/mol

c. H₂C₂O₄ ≈ 2×1.008 g/mol + 2×12.011 g/mol + 4×15.999 g/mol ≈ 90.034 g/mol

d. H₂C₄H₄O₆ = 6×1.008 g/mol +  4×12.011 g/mol + 6×15.999 g/mol = 150.086 g/mol ≈ 150 g/mol.

<u>Conclusion:</u> since the molar mass of H₂C₄H₄O₆ acid is 150 g/mol, you conclude that is the diprotic acid whose 0.300 g were titrated with 40.0 ml of 0.100 M LiOH solution.

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