How are elements and compounds are different
1 answer:
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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
Answer:
1) 1,1,1-trichloropropane
2) 1,1,2-trichloropropane
3) 1,2,2-trichloropropane
4) 1,2,3-trichloropropane
Explanation:
For this question, we must remember that isomers are molecules that have the <em>same formula but different structure</em>s. For the formula we can draw a <u>linear chain of three carbons</u> and change the position of the chlorine atoms in the carbon chain.
With this in mind, if we put all the chlorine atoms on the same carbon we will get <u>1,1,1-trichloropropane</u>. If we change an atom from chlorine to carbon 2 we will obtain <u>1,1,2-trichloropropane</u>. If we move another chlorine atom to carbon two we will get <u>1,2,2-trichloropropane</u>. Finally, if we put a chlorine atom in each carbon we will obtain <u>1,2,3-trichloropropane</u>.
See figure one for further explanations
I hope it helps!
