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1 year ago

The element with a completely filled p-subshell is:

Chemistry

1 answer:

tigry1 [53]1 year ago

4 0

The element with the completely filled p-subshell is Ar (Option C)

<h3>What is electronic configuration?</h3>

This is the arrangement of elecartons in the atomic orbitals of an element. The modern electronic configuration makes use of the s, p, d, f orbital notation where

Sharp (s) = Maximum of 2 electrons
Principal (p) = Maximum of 6 electrons
Diffuse (d) = Maximum of 10 electrons
Fundamental (f) = Maximum of 14 electrons

<h3>How to determine the element with the completely filled p-orbital</h3>

To obtain the correct answer, we shall write the electronic configuration of each element given in the question. This is illustrated below

Na (11) => 1s² 2s²2p⁶ 3s¹
P (15) => 1s² 2s²2p⁶ 3s²3p³
Ar (18) => 1s² 2s²2p⁶ 3s²3p⁶
Al (12) => 1s² 2s²2p⁶ 3s²

From the above illustrations, we can see that only Ar has completely filled p-subshell.

Thus, Ar (Option C ) is the correct answer to the question

Complete question

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Learn more about electronic configuration:

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What type of solution would have a pOH of 7

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Answer: None of these, because it is a neutral solution

Explanation:

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4 0

3 years ago

Read 2 more answers

Triangles in an octagon

elena-s [515]

Answer:

what is the specific question

Explanation:

3 0

3 years ago

MM H2O2 = 34.02 g/mol MM H2O = 18.02 g/mol MM O2 = 32 g/mol

Kaylis [27]

The grams of oxygen that are produced is 228.8 grams

<em>calculation</em>

2H₂O₂ → 2H₂O +O₂

Step 1: use the mole ratio to determine the moles of O₂

from equation above H₂O₂:O₂ is 2:1

therefore the moles of O₂ = 14.3 moles ×1/2 = 7.15 moles

Step 2: find mass of O₂

mass = moles × molar mass

= 7.15 moles × 32 g/mol =228.8 g

3 0

3 years ago

A hypothetical covalent molecule, x–y, has a dipole moment of 1.44 d and a bond length of 163 pm. calculate the partial charge o

Aneli [31]

As we know,
1 D = 3.34 × 10⁻³⁰ C.m
So,
1.44 D = ?
Solving for 1.44 D,
= (3.34 × 10⁻³⁰ C.m × 1.44 D) ÷ 1 D

1.44 D = 4.80 × 10⁻³⁰ C.m

Dipole Moment is given as,

Dipole Moment = q  × r
Solving for q,
q = Dipole Moment / r ------ (1)
Where,
Dipole Moment =  4.80 × 10⁻³⁰ C.m

r = 163 pm = 1.63 × 10⁻¹⁰ m

Putting values in eq. 1,

q = 4.80 × 10⁻³⁰ C.m / 1.63 × 10⁻¹⁰ m

q = 2.94 × 10⁻²⁰ C

As,
1.602 × 10⁻¹⁹ C = 1 e⁻
So,
2.94 × 10⁻²⁰ C = X e⁻

Solving for X,

X = (2.94 × 10⁻²⁰ C × 1 e⁻) ÷ 1.602 × 10⁻¹⁹ C

= 0.183 e⁻

Result:
So one element is containing + 0.183 e⁻ while the other element is containing - 0.183 e⁻.

4 0

3 years ago

Please help me with this chemistry problem

Margarita [4]

Answer:

50 g of K₂CO₃ are needed

Explanation:

How many grams of K₂CO₃ are needed to make 500 g of a 10% m/m solution?

We analyse data:

500 g is the mass of the solution we want

10% m/m is a sort of concentration, in this case means that 10 g of solute (K₂CO₃) are contained in 100 g of solution

Therefore we can solve this, by a rule of three:

In 100 g of solution we have 10 g of K₂CO₃

In 500 g of solution we may have, (500 . 10) / 100 = 50 g of K₂CO₃

6 0

3 years ago