Answer:8
Everything after the decimal place is a significant figure here
Answer:
When electron jumps from high energy level to lower energy level.
Explanation:
The elctronic transition from one energy level to another energy level within the atom, always involve energy transitions.
The energy released or absorbed by electronic transition is always discrete and is called as " Photon". It means when electron jumps from when energy level to another energy level the energy released or absorbed is treated as photon emitted or absorbed.
When an electron jumps from higher energy level to a lower energy level, a photon of specific wavelength and specific energy is emitted in other words we can say that energy is released or emitted.
The energy of photon emitted or absorbed is easily calculated using Rydberg Formula which is simply the energy difference between the two energy levels and is given as under;
Ephoton = Eo ( 1 / n1^2 - 1 / n2 ^ 2)
In the above formula n1 is the initial energy level of electron and n2 is the final energy level of electron.
Eo = 13.6 eV ( Here "o" in Eo is in subscripts)
In n1 and n2 1 and 2 are in the subscripts.
^ represents that the disgits after them are exponents.
So by just putting the values of energy levels n1 and n2 we can easily calculate the value of energy of photon ( energy due to electronic transition) and compare the results that which transition will give high energy photon and which will give low energy photon.
Answer:
+523 kJ.
Explanation:
The following data will be used to calculate the average C-S bond energy in CS2(l).
S(s) ---> S(g)
ΔH = 223 kJ/mol
C(s) ---> C(g)
ΔH = 715 kJ/mol
Enthalpy of formation of CS2(l)
ΔH = 88 kJ/mol
CS2(l) ---> CS2(g)
ΔH = 27 kJ/mol
CS2(g) --> C(g) + 2S(g)
So we must construct it stepwise.
1: C(s) ---> C(g) ΔH = 715 kJ
2: 2S(s) ---> 2S(g) ΔH = 446 kJ
adding 1 + 2 = 3
ΔH = 715 + 446
= 1161 kJ
3: C(s) + 2S(s) --> C(g) + 2S(g) ΔH = 1161 kJ
4: C(s) + 2S(s) --> CS2(l) ΔH = 88 kJ
adding (reversed 3) from 4 = 5
ΔH = -1161 + 88
= -1073 kJ
5: C(g) + 2S(g) --> CS2(l) ΔH = -1073 kJ
6: CS2(l) ---> CS2(g) ΔH = 27 kJ
adding 5 + 6 = 7
ΔH = -1073 + 27
= -1046 kJ
7. C(g) + 2S(g) --> CS2(g) ΔH = -1046 kJ
Reverse and divide by 2 for C-S bond enthalpy
= -(-1046)/2
= +523 kJ.
Except for the Middle East :)
Answer:
Potassium chloride, KCl , is an ionic compound formed by the electrostatic force of attraction that holds the potassium cations and the chlorine anions together. Potassium, K , is located in group 1 of the Periodic Table.
Explanation:
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