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crimeas [40]
3 years ago
15

Which hybridization of orbitals is present in beryllium fluoride, BeF2?

Chemistry
1 answer:
alisha [4.7K]3 years ago
4 0

Answer is: beryllium fluoride has sp hybridization.

Beryllium (Be) is metal from group 2 and fluorine (F) is nonmetal from group 17.

Electron configuration of beryllium atom: ₄Be: 1s² 2s².

Electron configuration of fluorine atom: ₉F 1s² 2s² 2p⁵.

In beryllium, one electron from 2s orbital gets excitted to the level of 2p orbital; s and p orbital are hybridized and form two sp-hybridised  orbitals arranged in a linear shape.

Two fluorine atoms come closer to beryllium atom and sp hybridised orbitals of Be and p orbital of F get overlapped with each other and make bonds.

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A thermally isolated system is made up of a hot piece of aluminum and a cold piece of copper; the aluminum and the copper are in
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Answer:

  • <u>Copper</u><em> is the object that experiences the greater temperature change.</em>

Explanation:

<em>Thermally isolated system </em>means that the system does not exchange thermal energy with the surroundings.

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In consequence, the law of conservation of energy states that the heat lost by the hot substance will be gained by the cold matter.

In equations, that is:

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Now, the gain or loss or heat of a substance, Q, is related with the mass (m), the specific heat (Cs), and the cahnge of temperature (ΔT), per the equation:

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  • [m × Cs × ΔT ] aluminum = [m × Cs × ΔT ] copper.

Under the reasoning assumption that the masses of aluminum and copper are equal, the equations is simplified to:

  • [Cs × ΔT ] aluminum = [Cs × ΔT ] copper.

  • Cs aluminum / Cs copper = ΔT copper / ΔT aluminum

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  • Cs aluminum / Cs copper > 2 ΔT copper / ΔT aluminum

  • ΔT copper / ΔT aluminum > 2

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Hence, the <u>conclusion</u> is that the object that experiences the greater temperature change is copper (the one with the lower specific heat), under the assumption that both objects have the same amount of matter (mass).

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