Which type of bonding is usually exhibited when the electronegativity difference between two atoms is 1.1?
1 answer:
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Answer:
The answers are;
1. 8.2 liters
2. 1214.84 ml
3. 318.027 K
4. 4.00 l.
Explanation:
1. Boyle's law states that the volume of given mass of gas is inversely proportional to its pressure at constant temperature
that is
P₁·V₁ = P₂·V₂
Where:
P₁ = Initial pressure = 40.0 mm Hg
V₁ = Initial volume = 12.3 liters
P₂ = Final pressure = 60.0 mm Hg
V₂ = Final volume = Required
From P₁·V₁ = P₂·V₂, V₂ is given by
= 8.2 l
The volume reduces to V₂ = 8.2 liters
2. Here Charles law states that
T₁ = Initial temperature = 27.0 °C = 300.15 K
V₁ = Initial volume = 900.0 mL
T₂ = Final temperature = 132.0 °C = 405.15 K
V₂ = Final volume = Required
Therefore =
= 1214.84 ml
V₂ = 1214.84 ml
3. Gay-Lussac's Law states that
Where:
P₁ = Initial pressure = 15.0 atmospheres
T₁ = Initial temperature = 25.0 °C = 298.15 K
P₂ = Final pressure = 16.0 atmospheres
T₂ = Final temperature = Required
∴
= = 318.027 K
T₂ = 318.027 K
4. Avogadro's law states that,
Equal volume of all gases at the same temperature and pressure contain equal number of molecules.
Therefore if 5.00 moles of gas occupies 2.00 l volume, then
1 moles will occupy 2.00/5 l volume and
10 moles will occupy 2.00/5 × 10 or 4.00 l volume.