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.
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Answer:
eg=linear, mg=linear
Explanation:
First of all, it must be stated that most triatomic molecules are either linear or bent. This depends on the electron geometry of the molecule and the number of bonding groups, multiple bonds and lone pairs present.
CO2 contains four regions of electron density and two bonding groups. For a specie containing two bonding groups, a linear molecular geometry is expected with an angle of 180°.
For a specie having two bonding groups and no lone pairs with multiple bonds, the expected electron geometry is also linear.
Due to the other pesticides have outgrown their strength on today’s insects. All must be accompanied by mass paperwork for safety.
When dealing with making diluted solutions from concentrated solutions, we can use the following formula
c1v1 = c2v2
where c1 and v1 are the concentration and volume of the concentrated solution respectively.
c2 and v2 are the concentration and volume of the diluted solution respectively
substituting these values in the above formula,
20 mL x 0.200 M = C x 250.0 mL
C = 0.0160 M
