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
The lattice energy for rubidium chloride (RbCl) is calculated by substituting the given values into the equation derived from Hess's Law. The calculated lattice energy is found to be -695 kJ/mol.
In this question, you are asked to select the lattice energy for rubidium chloride (RbCl). The lattice energy can be calculated using various given energies including enthalpy of formation (ΔHf), electron affinity (Cl), enthalpy of sublimation, ionization energy, and bond dissociation energy. Using Hess's Law, this can be summed up as:
ΔHf(RbCl) = [Sublimation Energy (Rb) + Ionization Energy (Rb) + 0.5 × Bond Energy (Cl₂) + Electron Affinity (Cl)] - Lattice Energy (RbCl)
By rearranging this formula, we find that the Lattice Energy (RbCl) = [Sublimation Energy (Rb) + Ionization Energy (Rb) + 0.5 × Bond Energy (Cl₂) + Electron Affinity (Cl)] - ΔHf(RbCl). Substituting in the given values, we find the lattice energy to be -695 kJ/mol. Therefore, the correct option is C. -695.
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Answer:
3 : 1
Explanation:
Let the rate of He be R1
Molar Mass of He (M1) = 4g/mol
Let the rate of O2 be R2
Molar Mass of O2 (M2) = 32g/mol
Recall:
R1/R2 = √(M2/M1)
R1/R2 = √(32/4)
R1/R2 = √8
R1/R2 = 3
The ratio of rate of effusion of Helium to oxygen is 3 : 1
b) cellular respiration
c) anaerobic respiration
d) cellular division
0.23 mol
Ob
0.38 mol
0.28 mol
Od
0.47 mol
Answer:
0.23 mol
Explanation:
Answer:
Protons: 50, Neutrons: 75, Electrons: 50
Explanation:
The atomic number, which in this case is 50, determines the number of protons and electrons an atom has. Take the atomic number and subtract the number of protons to get the number of neutrons.
B. Surface tension
C. Adhesion
D. Capillary action
Answer: The star's outer layers drifts away.
Explanation: When a star starts running out of fuel the pressure in the core becomes very high which pushes the outer layers of the star.
A sun like star becomes a red giant and after throwing away the mass, the core contracts to becomes a white dwarf.
A massive star becomes super-giant and then the core becomes a neutron star or a black-hole.
Answer:
The answer is d
Explanation: