Absorbed photon energy
Ea = hc/λ.. (Planck's equation)
Ea = hc / 92.05^-9m
<span>Energy emitted
Ee = hc/ 1736^-9m </span>
Energy retained ..
∆E = Ea - Ee = hc(1/92.05<span>^-9 - 1/1736^-9) </span>
<span>∆E = (6.625^-34)(3.0^8) (1.028^7)
∆E = 2.04^-18 J </span>
<span>Converting J to eV (1.60^-19 J/eV)
∆E = 2.04^-18 / 1.60^-19
∆E = 12.70 eV </span>
<span>Ground state (n=1) energy for Hydrogen = - 13.60eV </span>
<span>New energy state = (-13.60 + 12.70)eV = -0.85 eV </span>
<span>Energy states for Hydrogen
En = - (13.60 / n²) </span>
n² = -13.60 / -0.85 = 16
n = 4
From the calculation, the standard free energy of the system is -359kJ.
<h3>What is the standard free-energy?</h3>
The standard free-energy is the energy present in the system. We have to first obtain the cell potential using the formula;
Ereduction - E oxidation = 0.96 V - 0.34 V = 0.62 V
Using the formula;
ΔG = -nFEcell
ΔG =-(6 * 96500 * 0.62)
ΔG =-359kJ
Learn more about free energy:brainly.com/question/15319033
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This is an acid – base reaction and this always result a salt and water
in a neutralization reaction. <span>
The salt that is formed will be calcium bromide (calcium
is located in group 2 so calcium bromide has a formula of CaBr2)
so essentially we got:
HBr + Ca(OH)2 ------> CaBr2 + H2O </span>
balancing the elements: <span>
<span>2HBr(aq) + Ca(OH)2(aq) --------> CaBr2(aq) +
2H2O(l)</span></span>
Answer:
sp²
Explanation:
You need to look at how many electron orbitals around the atom. Looking at the structure below, you can see that there are three electron orbitals. This gives you an sp² hybridization.
The answer is c. hg (mercury)
