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alexgriva [62]
1 year ago
9

Write a full set of quantum numbers for the following:

Chemistry
1 answer:
bogdanovich [222]1 year ago
8 0

It is known that the maximum value of ml is equal to the value of l. But the minimum value of n is as follows.

                   n = l + 1

where n = principal quantum number  

               l = azimuthal quantum number

Values of n can be 1, 2, 3, 4, and so on. Whereas the values of l are 0 for s, 1 for p, 2 for d, 3 for f, and so on.

Also, "m" is known as a magnetic quantum number whose values can be equal to -l and +l.

The electronic configuration of Li is. So here, n = 2, l = 0, m = 0 and s = ±.

Electronic configuration of is. So here, n = 4, l = 1, m = -1, 0, +1, and s = ± .

Electronic configuration of is. So here, n = 5, l = 1, m =  -1, 0, +1, and s = ± The electronic configuration of B is. So here, n = 2, l = 1, m = -1, 0, +1, and s = ±.

<h3>What is quantum number?</h3>
  • Quantum numbers in quantum physics and chemistry explain the values of conserved quantities in a quantum system's dynamics.
  • Quantum numbers are quantities that can be precisely known at the same time as the system's energy and are related to the eigenvalues of operators that commute with the Hamiltonian and their corresponding eigenspaces.
  • A base state of a quantum system is fully described by the specification of all of its quantum numbers, which can theoretically be measured collectively.
<h3>What purpose do quantum numbers serve?</h3>
  • Because they can be used to determine an atom's electron configuration and the likely placement of its electrons, quantum numbers are significant.
  • The atomic radius and other properties of atoms, such as ionization energy, are also understood using quantum numbers.

Learn more about quantum numbers here:

brainly.com/question/16977590

#SPJ4

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A valid lewis structure of __________ cannot be drawn without violating the octet rule. a valid lewis structure of __________ ca
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A valid Lewis structure of IF3 cannot be drawn without violating the octet rule.

Answer: IF3 (Iodine Trifluoride)

This is because, I (Iodine) and F (Fluorine) both have odd number of valence electrons (7) which also means that there are too many valence electrons in the valence shell.
5 0
3 years ago
What element is being oxidized in the following redox reaction?
DIA [1.3K]

Answer:

The element that has been oxidized is the N

Explanation:

Zn²⁺(aq) + NH₄⁺(aq) → Zn(s) + NO₃⁻(aq)

See all the oxidation states:

Zn²⁺  → acts with +2

In ammonia, H acts with +1 and N with -3

Zn(s), acts with 0. In all the elements in ground state, the oxidation state is 0.

Zn changed from 2+ to 0. The oxidation number, has decreased.

This element has been reduced.

NO₃⁻ (aq) it's a ion, from nitric acid.

N acts with +5

O acts with -2

The global charge is -1

The N, has increased the oxidation state, so this element is the one oxidized.

8 0
3 years ago
A 825 g iron block is heated to 352 degrees C and is placed in an insulated container (of negligible heat capacity) containing 4
Stella [2.4K]

Answer : The final equilibrium temperature of the water and iron is, 537.12 K

Explanation :

In this problem we assumed that heat given by the hot body is equal to the heat taken by the cold body.

q_1=-q_2

m_1\times c_1\times (T_f-T_1)=-m_2\times c_2\times (T_f-T_2)

where,

c_1 = specific heat of iron =  560 J/(kg.K)

c_1 = specific heat of water = 4186 J/(kg.K)

m_1 = mass of iron = 825 g

m_2 = mass of water = 40 g

T_f = final temperature of water and iron = ?

T_1 = initial temperature of iron = 352^oC=273+352=625K

T_2 = initial temperature of water = 20^oC=273+20=293K

Now put all the given values in the above formula, we get:

(825\times 10^{-3}kg)\times 560J/(kg.K)\times (T_f-625K)=-(40\times 10^{-3}kg)\times 4186J/(kg.K)\times (T_f-293K)

T_f=537.12K

Therefore, the final equilibrium temperature of the water and iron is, 537.12 K

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3 years ago
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Explanation:

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D, It is a flow of protons, is the best answer. Electricity is the flow of electrons, not protons.
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