The equation of state for a hypothetical ideal gas is known as the ideal gas law, sometimes known as the general gas equation. i.e. PV = nRT or P1V1 = P2V2.
- According to the ideal gas law, the sum of the absolute temperature of the gas and the universal gas constant is equal to the product of the pressure and volume of one gram of an ideal gas.
- Robert Boyle, Gay-Lussac, and Amedeo Avogadro's observational work served as the basis for the ideal gas law. The Ideal gas equation, which simultaneously describes every relationship, is obtained by combining all of their observations into a single statement.
- When applying the gas constant R = 0.082 L.atm/K.mol, pressure, volume, and temperature should all be expressed in units of atmospheres (atm), litres (L), and kelvin (K).
- At high pressure and low temperature, the ideal gas law basically fails because molecule size and intermolecular forces are no longer negligible but rather become significant considerations.
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Using the mass/volume percentage method for percentages of the solution, you simply divide the grams of solute by the volume of the solution and multiply by 100 to get your percentage.
(75.0g/250mL)•100 = 30.0% solute
Photosynethesis, respiration, and combustion.
Answer:
0.558mole of SO₃
Explanation:
Given parameters:
Molar mass of SO₃ = 80.0632g/mol
Mass of S = 17.9g
Molar mass of S = 32.065g/mol
Number of moles of O₂ = 0.157mole
Molar mass of O₂ = 31.9988g/mol
Unknown:
Maximum amount of SO₃
Solution
We need to write the proper reaction equation.
2S + 3O₂ → 2SO₃
We should bear in mind that the extent of this reaction relies on the reactant that is in short supply i.e limiting reagent. Here the limiting reagent is the Sulfur, S. The oxygen gas would be in excess since it is readily availbale.
So we simply compare the molar relationship between sulfur and product formed to solve the problem:
First, find the number of moles of Sulfur, S:
Number of moles of S = 
Number of moles of S = 
= 0.558mole
Now to find the maximum amount of SO₃ formed, compare the moles of reactant to the product:
2 mole of Sulfur produced 2 mole of SO₃
Therefore; 0.558mole of sulfur will produce 0.558mole of SO₃
1 mols of Aluminium ion forms 1 mol aluminium phosphate
Molar mass of AlPO_4
Moles of AlPO_4
- 61µg/106
- 0.000061/106
- 5.75×10^{-7}
- 57.5µmol
Moles of Al3+=57.5µmol
