When bromine water is shaken with a liquid organic compound it is rapidly
1 answer:
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Molar mass of :
O2 = 16 * 2 = 32 g/mol
CO2 = 12 + 16 * 2 = 44 g/mol
<span>Balanced chemical equation :
</span>
1 CH4 + 2 O2 = 1 CO2 + 2 H2O
↓ ↓
2 moles 1 mole
2* 32 g O2 ----------> 1* 44 g CO2
x g O2 ------------> 10.0 g CO2
44 x = 2 * 32*10.0
44 x = 640
of O2
O2 = 16 * 2 = 32 g/mol
CO2 = 12 + 16 * 2 = 44 g/mol
<span>Balanced chemical equation :
</span>
1 CH4 + 2 O2 = 1 CO2 + 2 H2O
↓ ↓
2 moles 1 mole
2* 32 g O2 ----------> 1* 44 g CO2
x g O2 ------------> 10.0 g CO2
44 x = 2 * 32*10.0
44 x = 640
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
