What is the process of separating liquids by boiling points?
2 answers:
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Answer:
a) Binding Energy = 1084.266 MeV
b) Binding energy per nucleon = 8.28 MeV
Explanation:
Binding Energy,
Where ΔM = Mass defect of Iodine-131
The mass of iodine-131 = 130.906124 u.
Number of protons in Iodine-131 = 53
Number of neutrons in Iodine-131 = 78
Mass of a proton = 1.007276 u
Mass of neutron = 1.008664 u
Total mass of the 53 protons in Iodine-131 = (53*1.007276)
Total mass of the 53 protons in Iodine-131 = 53.39 u
Total mass of the 78 protons in Iodine-131 = (78*1.008664)
Total mass of the 78 protons in Iodine-131 = 78.68 u
Binding Energy = 1.164 * 931.5
Binding Energy = 1084.266 MeV
b) Binding Energy per nucleon
The mass of the nucleon of iodine 131 = 130.906124
Binding Energy per nucleon = 1084.266/130.906124
Binding energy per nucleon = 8.28 MeV
Answer:
Explanation:
Given the amount of heat absorbed and the amount of substance in moles, we may calculate the heat of vaporization. Heat of vaporization is defined as the amount of heat per 1 mole of substance required to evaporate that specific substance.
Based on the value of heat of vaporization, we will identify the substance. Firstly, let's calculate the heat of vaporization:
Secondly, let's use any table for heat of vaporization values for substances. We identify that the heat of vaporization of is 18.7 kJ/mol
Answer : The electron configurations consistent with this fact is, (b) [Kr] 4d¹⁰
Explanation :
Electronic configuration : It is defined as the representation of electrons around the nucleus of an atom.
Number of electrons in an atom are determined by the electronic configuration.
Paramagnetic compounds : They have unpaired electrons.
Diamagnetic compounds : They have no unpaired electrons that means all are paired.
The given electron configurations of Palladium are:
(a) [Kr] 5s²4d⁸
In this, there are 2 electrons in 's' orbital and 8 electrons in 'd' orbital. From the partial orbital diagrams we conclude that 's' orbital are paired but 'd' orbital are not paired. So, this configuration shows paramagnetic.
(b) [Kr] 4d¹⁰
In this, there are 10 electrons in 'd' orbital. From the partial orbital diagrams we conclude that electrons in 'd' orbital are paired. So, this configuration shows diamagnetic.
(c) [Kr] 5s¹4d⁹
In this, there are 1 electron in 's' orbital and 9 electrons in 'd' orbital. From the partial orbital diagrams we conclude that 's' orbital and 'd' orbital are not paired. So, this configuration shows paramagnetic.
