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:
20 mL OF 6 M HYDROCHLORIC ACID WILL BE NEEDED
Explanation:
M1 V1 = M2 V2
M1 = Molarity of sodium hydroxide = 3 M
V1 = volume of sodium hydroxide = 40 mL
M2 = Molarity of hydrochloric acid = 6 M
V2 = Volume of hydrochloric acid = unknown
Rearranging the equation, we have:
V2 = M1 V1 / M2
V2 = 3 * 40 mL / 6
V2 = 120 / 6
V2 = 20 mL
To precipitate the benzoic acid by 6 M of hydrochloric acid, 20 mL volume will be needed.
are dry ice (solid carbon dioxide), iodine, arsenic, and naphthalene (the stuff mothballs are made of).
1- First we will get the value of POH from:
PH + POH = 14
and we have PH = 12.01
∴ POH = 14 - 12.01 = 1.99
2- Then we need to get the concentration of OH:
when POH = -㏒[OH]
1.99 = -㏒[OH]
∴[OH] = 0.01 M
now we have the concentration at Equ by subtracting from the initial concentration of OH = 0.24 M
∴ [OH] = 0.24 - 0.01 = 0.23 M
∴ Kb = (0.01)^2 / 0.22977 M
∴ Kb value = 4.4 x 10^-4
