A single atom of an element has 11 protons, 11 electrons, and 12 neutrons. Which element is it? a V b Na c Mg d Se
1 answer:
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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:
Option C
Explanation:
The chemical reactions which are involved while solving this problem is there in the file attached and each chemical reaction is represented by a certain equation number
Lattice energy for rubidium chloride ( RbCl) is represented by the equation 6
Equation 1 represents the change in enthalpy for formation of RbCl
Equation 2 represents the sublimation reaction of rubidium
Equation 3 represents the ionization enthalpy of rubidium
Equation 4 represents the enthalpy of atomization of chlorine which means it describes the bond enthalpy of Cl2 molecule
Equation 5 represents the electron affinity of chlorine
To find the lattice energy for RbCl we have to use all the equations from 1 to 5 so that at last we get the equation 6
We have to perform operations such as
Equation 1 - equation 2 - equation 3 - equation 4 - equation 5
By performing these operations the intermediate compounds gets cancelled and at last we get equation 6
So Equation 1 ≡ ΔH = -431 kJ/mol
Equation 2 ≡ Rb(s) ---> Rb(g) = 85.8 kJ/mol
Equation 3 ≡ IE1(Rb) = 397.5 kJ/mol
Equation 4 ≡ BE(Cl2) = 226 kJ/mol
Equation 5 ≡ Electron Affinity Cl = -332 kJ/mol
Value corresponding to the equation 6 will be the value of lattice energy of RbCl and the value is -695·3 kJ/mol
∴ Lattice energy for rubidium chloride is approximately -695 kJ/mol
