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stepladder [879]
4 years ago
5

Give all possible ml values for orbitals that have each of the following: (a) l = 3; (b) n = 2; (c) n = 6, l = 1.

Chemistry
1 answer:
larisa86 [58]4 years ago
6 0

Answer : All possible values of 'ml' for the following orbitals are:

(a) At l = 3,  m_l=+3,+2,+1,0,-1,-2,-3

(b) l = 0, 1

At l = 0,  m_l=0

At l = 1,  m_l=+1,0,-1

(c) At l = 1,  m_l=+1,0,-1

Explanation:

There are 4 quantum numbers :

Principle Quantum Numbers : It describes the size of the orbital. It is represented by n. n = 1,2,3,4....

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_sThe value of this is +\frac{1}{2} for upward spin and -\frac{1}{2} for downward spin.

(a) l = 3 then the value of 'ml' is,

At l = 3,  m_l=+3,+2,+1,0,-1,-2,-3

(b) n = 2 then the value of 'ml' is,

l = 0, 1

At l = 0,  m_l=0

At l = 1,  m_l=+1,0,-1

(c) n = 6 and l = 1 then the value of 'ml' is,

n = 6

l = 0, 1, 2, 3, 4, 5

At l = 1,  m_l=+1,0,-1

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<em>...is titrating 212.7 mL of a 0.6800 M solution of hydrazoic acid (HN3) with a 0.2900 M solution of KOH. The p Ka of hydrazoic acid is 4.72. Calculate the pH of the acid solution after the chemist has added 571.6 mL of the KOH solution to it</em>.

To solve this question we need to know that hidrazoic acid reacts with KOH as follows:

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0.1658 moles - 0.1446 moles =

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In 212.7mL + 571.6mL = 784.3mL = 0.7843L

The molarity of KOH = [OH-] is:

0.0212 mol KOH / 0.7843L = 0.027M = [OH-]

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pOH = -log 0.027M

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