Which of the following is not a valid set of quantum numbers
2 answers:
Answer: C. n = 3, l = 3, m = 3
Explanation:
1. Principle Quantum Numbers: This quantum number describes the size of the orbital. It is represented by n. n = 1,2,3,4....
2. Azimuthal Quantum Number: This quantum number 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...
3. Magnetic Quantum Number: This quantum number describes the orientation of the orbitals. It is represented as . The value of this quantum number ranges from
. When l = 2, the value of
will be -2, -1, 0, +1, +2.
A. n = 2, l = 1 , m = 0 is possible.
B. n = 1, l = 0, m = 0 is possible.
C. n = 3, l = 3, m = 3 is not possible as for n = 3, l = 0, 1 and 2 and m = 0,+1, +2, -1 and -2.
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Answer:
We can not swim in 1.00 × 10²⁷ molecules of water
Explanation:
The given number of molecules of water = 1.00 × 10²⁷ molecules
The Avogadro's number, , gives the number of molecules in one mole of a substance
≈ 6.0221409 × 10²³ molecules/mol
Therefore
Therefore, we have;
The number of moles of water present in 1.00 × 10²⁷ molecules, n = (The number of molecules of water) ÷
∴ n = (1.00 × 10²⁷ molecules)/(6.0221409 × 10²³ molecules/mol) = 1,660.53902857 moles
The mass of one mole of water = The molar mass of water = 18.01528 g/mol
The mass, 'm', of water in 1,660.53902857 moles of water is given as follows;
Mass = (The number of moles of the substance) × (The molar mass of the substance)
∴ The mass of the water in the given quantity of water, m = 1,660.53902857 moles × 18.01528 g/mol ≈ 29.9150756 kg.
The density pf water, ρ = 997 kg/m³
Volume = Mass/Density
∴ The volume of the water present in the given quantity of water, v = 29.9150756 kg/(997 kg/m³) ≈ 30.0050909 liters
The volume of the water present in 1.00 × 10²⁷ molecules of water ≈ 30.0 liters
The average volume of a human body = 62 liters
Therefore, we can not swim in the given quantity of 1.00 × 10²⁷ molecules = 30.0 liters water