Explanation:
To delineate the the nature of the bonds that would be formed between the two elements, let us first write the electronic configuration of the two species;
Be = 2, 2
F = 2, 7
Beryllium is a metal with two valence electrons whereas fluorine is a halogen with seven valence electrons.
When Be loses two electrons it becomes isoelectronic with He;
Be → Be²⁺ + 2e⁻
Also, when fluorine gains an electron, it becomes isoelectronic with Ne;
F + e⁻ → F⁻
This loss and gain of electrons between the two elements creates an electrostatic attraction them and they enter into an electrovalent bond.
Hence;
Be²⁺ + 2F⁻ → BeF₂
Ideal gas law:
PV=nRT ⇒ V=nRT / P
P=pressure=1 atm
V=volume
n=number moles=2.10 moles
R=0,082 Atm l/ºK mol
T=temperature=273 K
V=(2.10 moles*0.082 (atm l)/º(K mol)*237ºK) / 1 atm=47.01 litres
47.1 L
Answer:
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