Unburned hydrocarbon on reacting with oxygen undergoes combustion reaction. However, the activation energy of this reaction is significantly high. When a catalyst like Pd is added to the reaction system, it provides active sites for the reaction to occur. It acts are a heterogeneous catalyst. It is pertinent of note that catalyst is refereed as heterogeneous, when it exist in different phase as compared to reactant and products. In present case, reactants and products are in gas phase, while catalyst is in solid phase. Due to availability of larger surface area at active site of Pd, activation energy of reaction decreases and decrease in activation energy favors higher reaction rates.
The given compound is Aluminum sulfate, Al2(SO4)3:
Molar masses:
Aluminum = 27 g/mol
Sulfur = 32 g/mol
Oxygen = 16 g/mol
The total molar mass is 342 g/mol
The ratio by mass of the elements:
Aluminum = 27*2/342
= 0.16
Sulfur = (32*3)/342
= 0.28
Oxygen = (16*12)/342
= 0.56
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Answer:
distillation
Explanation:
Actually, distillation is used to separate liquids from nonvolatile solids, as in the separation of alcoholic liquors from fermented materials, or in the separation of two or more liquids having different boiling points, as in the separation of gasoline, kerosene, and lubricating oil from crude oil.
Answer:
Por ejemplo, el petróleo o el carbón son ejemplos de recursos no renovables porque, aunque se forman mediante un proceso natural, este necesita demasiado tiempo. ... Esos combustibles fósiles provienen de materia orgánica, pero tardan cientos de miles de años en producirse.
The answer is 4.9 moles.
Solution:
Using the equation for boiling point elevation Δt,
Δt = i Kb m
we can rearrange the expression to solve for the molality m of the solution:
m = Δt / i Kb
Since we know that pure water boils at 100 °C, and the Ebullioscopic constant Kb for water is 0.512 °C·kg/mol,
m = (105°C - 100°C) / (2 * 0.512 °C·kg/mol)
= 4.883 mol/kg
From the molality m of the solution of salt added in a kilogram of water, we can now find the number of moles of salt:
m = number of moles / 1.0kg
number of moles = m*1.0kg
= (4.883 mol/kg) * (1.0kg)
= 4.9 moles
