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Mandarinka [93]
2 years ago
9

Write the electron configurations of P and Cl using both spdf and orbital box diagrams.

Chemistry
1 answer:
Alenkinab [10]2 years ago
5 0

Answer:

Explanation:

Here, we want to write the electronic configurations for the given atoms

Spdf Notation:

Here, we start by knowing the atomic number of the atoms. The atomic number is the same as the number of electrons for an electrically neutral atom

For Phosphorus:

The number of electrons is 15

We have the spdf notation as follows:

1s^2\text{ 2s}^2\text{ 2p}^6\text{ 3s}^2\text{ 3p}^3

For Chlorine, the number of electrons is 17

We have the spdf notation as follows:

1s^2\text{ 2s}^2\text{ 2p}^6\text{ 3s}^2\text{ 3p}^5

Orbital Diagram:

For the orbital diagram, we simply fill in the orbitals as shown in the spdf notation

Each subshell contains 2 maximum number of electrons of opposite spin

For Phosphorus, we have it that:

For Chlorine, we have it that:

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PLEASE SOMEONE! The density of the acetic acid solution is 1.05 g/mL. Calculate the %(m/m) of the acetic acid solution (convert
tigry1 [53]

Answer:

The Percentage concentration of acetic acid = 1.39 %

Explanation:

Density of acetic acid solution = 1.05 g/mL

Volume of acetic acid solution = 0.1 L = 100 mL

From the formula, Density = mass / volume; mass = density × volume

Mass of acetic acid solution = 1.05 g/mL × 100 mL = 105 g

Molar concentration of acetic acid solution = 0.243 mol/L

Molar mass of acetic acid, CH₃COOH = (12 × 2 + 1 ×4 + 16 ×2) = 60 g/mol

From the formula, mass concentration = molar concentration × molar mass

Mass concentration of acetic acid, CH₃COOH = 0.243 mol/L × 60 g/mol = 14.58 g/L

In one liter of acetic acid solution, there are 14.58 g of acetic acid. Therefore, in 0.1 L, there will be 14.58 × 0.1 = 1.458 g of acetic acid.

Percentage concentration of acetic acid = mass of acetic acid / mass of acetic acid solution × 100%

Percentage concentration of acetic acid = (1.458 / 105) × 100% = 1.39 %

The Percentage concentration of acetic acid = 1.39 %

8 0
3 years ago
The atomic mass and abundance of Cr-50 is 49.946 amu and 4.3%. The atomic mass and abundance of Cr-52 is 51.941 amu and 83.8%. T
Sati [7]
Multiply the mass by the abundance and add each. Make sure to convert the percentage into a decimal. (49.946 * .043) + (51.941 * .838) + (52.941 * .095) + (53.939 * .024) = 51.99 round up using sig figs and the answer is c. 52.00 amu
6 0
3 years ago
Calculate the pH of a solution prepared by mixing: (Show your work for these calculations) pk of acetic acid is 4.75 a. two mole
kupik [55]

Answer:

Explanation:

To calculate pH you need to use Henderson-Hasselbalch formula:

pH = pka + log₁₀ \frac{[A^-]}{[HA]}

Where HA is the acid concentration and A⁻ is the conjugate base concentration.

The equilibrium of acetic acid is:

CH₃COOH ⇄ CH₃COO⁻ + H⁺ pka: 4,75

Where <em>CH₃COOH </em>is the acid and <em>CH₃COO⁻ </em>is the conjugate base.

Thus, Henderson-Hasselbalch formula for acetic acid equilibrium is:

pH = 4,75 + log₁₀ \frac{[CH_{3}COO^-]}{[CH_{3}COOH]}

a) The pH is:

pH = 4,75 + log₁₀ \frac{[2 mol]}{[2 mol]}

<em>pH = 4,75</em>

<em></em>

b) The pH is:

pH = 4,75 + log₁₀ \frac{[2 mol]}{[1mol]}

<em>pH = 5,05</em>

<em></em>

I hope it helps!

7 0
4 years ago
Assume the solubility of benzoic acid in ice-cold water is 1.70 g/L and the solubility of benzoic acid in hot water is 68.0 g/L.
lana [24]

Answer:

The volume of water needed to recrystallize 0.700 g of benzoic acid is 10.29 mL.

Explanation:

For the complete recrystalization,

the amount of hot water should be such that, the benzoic acid is completely soluble in it.

As we are given that the solubility of benzoic acid in hot water is 68.0 g/L. Now we have to determine the volume of water is needed to recrystallize 0.700 g of benzoic acid.

we conclude that,

As, 68.0 grams of benzoic acid soluble in 1 L of water.

So, 0.700 grams of benzoic acid soluble in  of water.

The volume of water needed = 0.01029 L = 10.29 mL

conversion used : (1 L = 1000 mL)

Therefore, the volume of water needed to recrystallize 0.700 g of benzoic acid is 10.29 mL.

4 0
4 years ago
How many mL of 2M stock solution would I use to prepare 0.500 L of 0.1 M NaCl?
nordsb [41]

Answer:

25 mL

Explanation:

Step 1: Given data

  • Concentration of the concentrated solution (C₁): 2 M
  • Volume of the concentrated solution (V₁): ?
  • Concentration of the diluted solution (C₂): 0.1 M
  • Volume of the diluted solution (V₂): 0.500 L

Step 2: Calculate the volume of the concentrated NaCl solution

We will use the dilution rule.

C₁ × V₁ = C₂ × V₂

V₁ = C₂ × V₂ / C₁

V₁ = 0.1 M × 0.500 L / 2 M

V₁ = 0.025 L = 25 mL

4 0
3 years ago
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