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Tju [1.3M]
3 years ago
7

3. According to Hund's rule, what's the expected magnetic behavior of vanadium (V)?

Physics
1 answer:
ivanzaharov [21]3 years ago
6 0

Answer:

Diamagnetic

Explanation:

Hunds rule states that electrons occupy each orbital singly first before pairing takes place in degenerate orbitals. This implies that the most stable arrangement of electrons in an orbital is one in which there is the greatest number of parallel spins(unpaired electrons).

For vanadium V ion, there are 18 electrons which will be arranged as follows;

1s2 2s2 2p6 3s2 3p6.

All the electrons present are spin paired hence the ion is expected to be diamagnetic.

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krok68 [10]

Answer:

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Explanation:

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3 years ago
A jet airliner moving initially at 889 mph
Eduardwww [97]

Answer:

1500 mph

Explanation:

Take east to be +x and north to be +y.

The x component of the velocity is:

vₓ = 889 cos 0° + 830 cos 59°

vₓ = 1316.5 mph

The y component of the velocity is:

vᵧ = 889 sin 0° + 830 sin 59°

vᵧ = 711.4 mph

The speed is found with Pythagorean theorem:

v² = vₓ² + vᵧ²

v² = (1316.5 mph)² + (711.4 mph)²

v = 1496 mph

Rounded to two significant figures, the jet's speed relative to the ground is 1500 mph.

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8 0
3 years ago
Explain why machine can"t be <br>100% efficient practically?​
Zigmanuir [339]

Answer:

Heat

Explanation:

Heat is the main component in why everything is not 100% efficient. Heat is a byproduct of everything and cannot be avoided.

5 0
3 years ago
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A metallic wire has a diameter of 4.12mm. When the current in the wire is 8.00A, the drift velocity is 5.40×10−5m/s.What is the
podryga [215]

Answer:

6.9\times 10^{28}m^{-3}

Explanation:

We are given that

Diameter of wire=d=4.12 mm

Radius of wire=rr=\frac{d}{2}=\frac{4.12}{2}=2.06mm=2.06\times 10^{-3} m

1mm=10^{-3} m

Current=I=8 A

Drift velocity=v_d=5.4\times 10^{-5} m/s

We have to find the density of free electrons in the metal

We know that

Density of electron=n=\frac{I}{v_deA}

Using the formula

Density of free electrons=\frac{8}{5.4\times 10^{-5}\times 1.6\times 10^{-19}\times 3.14\times (2.06\times 10^{-3})^2}

By using Area of wire=\pi r^2

\pi=3.14\\e=1.6\times 10^{-19} C

Density of free electrons=6.9\times 10^{28}m^{-3}

3 0
2 years ago
Read 2 more answers
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