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FromTheMoon [43]
3 years ago
7

How many electrons can be held in a sublevel l = 3?

Chemistry
2 answers:
Stells [14]3 years ago
8 0

Answer : The number of electrons held in sub-level l = 3 can be, 14 electrons

Explanation :

Azimuthal Quantum Number : It describes the shape of the orbital. It is represented as 'l'. The value of l ranges from 0 to (n-1). For l = 0,1,2,3... the orbitals are s, p, d, f...

Magnetic Quantum Number : It describes the orientation of the orbitals. It is represented as m_l. The value of this quantum number ranges from (-l\text{ to }+l). When l = 2, the value of m_l will be -2, -1, 0, +1, +2.

Spin Quantum number : It describes the direction of electron spin. This is represented as m_s The value of this is +\frac{1}{2} for upward spin and -\frac{1}{2} for downward spin.

As we are given that,

l=3

m_l=-3,-2,-1,0,1,2,3

m_s=+\frac{1}{2}\text{ and }-\frac{1}{2}  (For each sub-shell)

From this we conclude that, there are 7 orbitals and each orbital contains 2 electrons. So, the number of electrons held in sub-level l = 3 are, 7\times 2=14 electrons.

Hence, the number of electrons held in sub-level l = 3 can be, 14 electrons

muminat3 years ago
4 0
There can be two electrons in one orbital maximum. The s sublevel has just one orbital, so can contain 2 electrons max. The p sublevel has 3 orbitals, so can contain 6 electrons max. The d sublevel has 5 orbitals, so can contain 10 electrons max.

Hope this helped
Have a nice day :)
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A compound is found to contain 11.21 % hydrogen and 88.79 % oxygen by mass. What is the empirical formula for this compound?
Artemon [7]

Answer:

H₂O

Explanation:

The empirical formular of the compound is obtained using the following steps;

Step 1: Divide the percentage composition by the atomic mass

Hydrogen = 11.21 / 1 = 11.21

Oxygen = 88.79  / 16 = 5.55

Step 2: Divide by the lowest number

Hydrogen  = 11.21 / 5.55 = 2.02 ≈ 2

Oxygen = 5.55 / 5.55 = 1

This means the ratio of the elements is 2 : 1

The empirical formular (simplest formular of a compound) of the compound is;

H₂O

7 0
3 years ago
Read 2 more answers
Which statement accurately describes binary star systems?
Yakvenalex [24]

Answer:

The correct answer is D.They have stars that may appear to wobble.

Explanation:

Binary star - Wikipediaen.wikipedia.org › wiki › Binary_star

''Astronomers have discovered some stars that seemingly orbit around an empty space. Astrometric binaries are relatively nearby stars which can be seen to wobble around a point in space, with no visible companion. The same mathematics used for ordinary binaries can be applied to infer the mass of the missing companion''

<em>-Wikipediaen.wikipedia.org › wiki › Binary_star</em>

<em>-Edge 2021</em>

<em>might be a lil late but YW

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4 0
3 years ago
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A 3.452 g sample containing an unknown amount of a Ce(IV) salt is dissolved in 250.0-mL of 1 M H2SO4. A 25.00 mL aliquot is anal
SOVA2 [1]

Answer:

1,812 wt%

Explanation:

The reactions for this titration are:

2Ce⁴⁺ + 3I⁻ → 2Ce³⁺ + I₃⁻

I₃⁻ + 2S₂O₃⁻ → 3I⁻ + S₄O₆²⁻

The moles in the end point of S₂O₃⁻ are:

0,01302L×0,03428M Na₂S₂O₃ = 4,463x10⁻⁴ moles of S₂O₃⁻. As 2 moles of S₂O₃⁻ react with 1 mole of I₃⁻, the moles of I₃⁻ are:

4,463x10⁻⁴ moles of S₂O₃⁻×\frac{1molI_{3}^-}{2molS_{2}O_{3}^-} = 2,2315x10⁻⁴ moles of I₃⁻

As 2 moles of Ce⁴⁺ produce 1 mole of I₃⁻, the moles of Ce⁴⁺ are:

2,2315x10⁻⁴ moles of I₃⁻×\frac{2molCe^{4+}}{1molI_{3}^-} = 4,463x10⁻⁴ moles of Ce(IV). These moles are:

4,463x10⁻⁴ moles of Ce(IV)×\frac{140,116g}{1mol} = <em>0,0625 g of Ce(IV)</em>

As the sample has a 3,452g, the weight percent is:

0,0625g of Ce(IV) / 3,452g × 100 = <em>1,812 wt%</em>

I hope it helps!

5 0
3 years ago
What do we assume about the volume of the actual molecules themselves in a sample of gas, compared to the bulk volume of the gas
Firlakuza [10]

We assume that the volume of the molecules themselves in a gas sample is negligible compared to the bulk volume of the gas sample: this helps us to explain why gases are so compressible.

4 0
3 years ago
If nitrogen-13 has a half life of 2.5 years, how much remains from a 100g sample after 7.5 years
Sonbull [250]

Answer:

12.50g

Explanation:

T½ = 2.5years

No = 100g

N = ?

Time (T) = 7.5 years

To solve this question, we'll have to find the disintegration constant λ first

T½ = In2 / λ

T½ = 0.693 / λ

λ = 0.693 / 2.5

λ = 0.2772

In(N/No) = -λt

N = No* e^-λt

N = 100 * e^-(0.2772*7.5)

N = 100*e^-2.079

N = 100 * 0.125

N = 12.50g

The sample remaining after 7.5 years is 12.50g

5 0
3 years ago
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