Calculate the number of moles in 369 grams of CaoCl2? a) 1 b) 2 c) 3 d) 4
1 answer:
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Answer:
4.99*10²³ molecules of N₂O₄ are in 76.3 g of N₂O₄
Explanation:
Avogadro's Number is the number of particles that make up a substance (usually atoms or molecules) and that can be found in the amount of one mole of said substance. Its value is 6.023*10²³ particles per mole. Avogadro's number represents a quantity without an associated physical dimension. Avogadro's number applies to any substance.
You know that the molar mass of N₂O₄ is 92.02 g/mol, and you have 76.3 g. Then you can apply the following rule of three: 92.02 grams are present in 1 mole of the compound, 76.3 grams in how many moles are they?
amount of moles= 0.83 moles
Then, you can apply another rule of three: if by definition of Avogadro's number 1 mole of the compound has 6.023*10²³ molecules, 0.83 moles of the compound, how many molecules will it have?
amount of molecules= 4.99*10²³
<u><em>4.99*10²³ molecules of N₂O₄ are in 76.3 g of N₂O₄</em></u>
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.