Answer:
0.2193 μm
Explanation:
The reaction showing the Photodissociation of ozone (O3) is given below as:
O₃ + hv --------------------------> O₂ + O⁺
H° (142.9) (0) (438kJ/mol).
The first thing to do here is to determine the change in the enthalpy of the total reaction, this can be done by subtracting the change in the enthalpy of the reactant from the change in enthalpy in the product. Hence, we have:
ΔH° = [438 kJ/mol + 247.5 kJ/mol] - (142.9) = 542.6 KJ/mol.
This value, that is 542.6 KJ/mol will then be used in the determination of the value for the maximum wavelength that could cause this photodissociation.
Therefore, the maximum wavelength could cause this photodissociation ≤ h × c/ E = [ 1.199 × 10⁻⁴]/ 542.6 = 2.193 × 10⁻⁷ = 0.2193 μm
Answer:
A wave with small amplitude.
Explanation:
Amplitude us inversely proportional to energy so the smaller the amplitude, the higher the energy.
Hope it helps.
1. The resulting concentration will be 0.00044 mol/L
2. The minimum mass of sodium sulfite to add will be 0.4032 grams.
<h3>Stoichiometric problems</h3>
1. Using m1v1=m2v2
m1 = 0.01 mol/L, v1 = 20 mL, v2 = 450 mL
m2 = m1v1/v2 = 0.01 x 20/450 = 0.00044 mol/L
2. 
Mole ratio of the reactants = 1:1
Mole of 80 mL, 0.0400 mol/L Ca(NO3)2 = 80/1000 x 0.0400 = 0.0032 mol
Equivalent mole of Na2SO3 = 0.0032 moles
Mass of 0.0032 moles Na2SO3 = 0.0032 x 126 = 0.4032 grams
Thus, the minimum mass of sodium sulfite to be added must be 0.4032 grams.
More on stoichiometric calculations can be found here: brainly.com/question/27287858
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Answer:
Explanation:
mol solute / L soln. M = 5.50 mol / .600 L. M = 9.16 molar
There would be 0.8315 moles
