Answer:
0.145 moles de AlBr3.
Explanation:
¡Hola!
En este caso, al considerar la reacción química dada:
Al(s)+Br2(l)⟶AlBr3(s)
Es claro que primero debemos balancearla como se muestra a continuación:
2Al(s)+3Br2(l)⟶2AlBr3(s)
Así, calculamos las moles del producto AlBr3 por medio de las masas de ambos reactivos, con el fin de decidir el resultado correcto:
Así, inferimos que el valor correcto es 0.145 moles de AlBr3, dado que viene del reactivo límite que es el aluminio.
¡Saludos!
Answer:
75 mg
Explanation:
We can write the extraction formula as
x = m/[1 + (1/K)(Vaq/Vo)], where
x = mass extracted
m = total mass of solute
K = distribution coefficient
Vo = volume of organic layer
Vaq = volume of aqueous layer
Data:
m = 75 mg
K = 1.8
Vo = 0.90 mL
Vaq = 1.00 mL
Calculations:
For each extraction,
1 + (1/K)(Vaq/Vo) = 1 + (1/1.8)(1.00/0.90) = 1 + 0.62 = 1.62
x = m/1.62 = 0.618m
So, 61.8 % of the solute is extracted in each step.
In other words, 38.2 % of the solute remains.
Let r = the amount remaining after n extractions. Then
r = m(0.382)^n.
If n = 7,
r = 75(0.382)^7 = 75 × 0.001 18 = 0.088 mg
m = 75 - 0.088 = 75 mg
After seven extractions, 75 mg (99.999 %) of the solute will be extracted.
A joule times a second :)
Answer:The answer is C) The total mass of C and D only.
Answer:
The mole fraction of codeine is 5.4%
Explanation:
Mole fraction = Mole of solute / Total moles (Mole solute + Mole solvent)
Solute (Codeine)
Molar mass 299.36 g/m
Mass / Molar mass = Mole → 46.85 g /299.36 g/m = 0.156 moles
Solvent (Ethanol)
Molar mass 46.07 g/m
125.5 g / 46.07 g/m = 2.724 moles
Mole fraction (Codeine) 0.156 / (0.156 + 2.724) → 0.054
