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8_murik_8 [283]
3 years ago
8

The 1995 Nobel Prize in Chemistry was shared by Paul Crutzen, F. Sherwood Rowland, and Mario Molina for their work concerning th

e formation and decomposition of ozone in the stratosphere. Rowland and Molina hypothesized that chlorofluorocarbons (CFCs) in the stratosphere break down upon exposure to UV radiation, producing chlorine atoms. Chlorine was previously identified as a catalyst in the breakdown of ozone into oxygen gas. Using the enthalpy of reaction for two reactions with ozone, determine the enthalpy of reaction for the reaction of chlorine with ozone. (1)ClO(g)+O3(g)⟶Cl(g)+2O2(g)Δ????∘rxn=−122.8 kJ(2)2O3(g)⟶3O2(g)Δ????∘rxn=−285.3 kJ(3)O3(g)+Cl(g)⟶ClO(g)+O2(g) Δ????∘rxn= ?
Chemistry
1 answer:
Klio2033 [76]3 years ago
6 0

Answer:

The enthalpy of the reaction of the chlorine with ozone is -162.5 kJ/mol.

Explanation:

ClO(g)+O_3(g)\rightarrow Cl(g)+2O_2(g),\Delta H_{rxn}=-122.8 kJ/mol...(1)

2O_3(g)\rightarrow 3O_2(g),\Delta H_{rxn}=-285.3 kJ/mol...(2)

O_3(g)+Cl(g)\rightarrow ClO(g)+O_2(g),\Delta H_{rxn}=x...(3)

The enthalpy of reaction for the reaction of chlorine with ozone can be calculated by using Hess's law:

(2 )- (1) = (3)

x=-285.3 kJ/mol - (-122.8 kJ/mol)=-162.5 kJ/mol

The enthalpy of the reaction of the chlorine with ozone is -162.5 kJ/mol.

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5 0
3 years ago
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3 0
3 years ago
A particular laser consumes 130.0 Watts of electrical power and produces a stream of 2.67×1019 1017 nm photons per second.
solniwko [45]

The missing question is:

<em>What is the percent efficiency of the laser in converting electrical power to light?</em>

The percent efficiency of the laser that consumes 130.0 Watt of electrical power and produces a stream of 2.67 × 10¹⁹ 1017 nm photons per second, is 1.34%.

A particular laser consumes 130.0 Watt (P) of electrical power. The energy input (Ei) in 1 second (t) is:

Ei = P \times t = 130.0 J/s \times 1 s = 130.0 J

The laser produced photons with a wavelength (λ) of 1017 nm. We can calculate the energy (E) of each photon using the Planck-Einstein's relation.

E = \frac{h \times c }{\lambda }

where,

  • h: Planck's constant
  • c: speed of light

E = \frac{h \times c }{\lambda } = \frac{6.63 \times 10^{-34}J.s  \times 3.00 \times 10^{8} m/s }{1017 \times 10^{-9} m }= 6.52 \times 10^{-20} J

The energy of 1 photon is 6.52 × 10⁻²⁰ J. The energy of 2.67 × 10¹⁹ photons (Energy output = Eo) is:

\frac{6.52 \times 10^{-20} J}{photon} \times 2.67 \times 10^{19} photon = 1.74 J

The percent efficiency of the laser is the ratio of the energy output to the energy input, times 100.

Ef = \frac{Eo}{Ei} \times 100\% = \frac{1.74J}{130.0J} \times 100\% = 1.34\%

The percent efficiency of the laser that consumes 130.0 Watt of electrical power and produces a stream of 2.67 × 10¹⁹ 1017 nm photons per second, is 1.34%.

You can learn more about lasers here: brainly.com/question/4869798

8 0
3 years ago
what volume of a 0.138 m potassium hydroxide solution is required to neutralize 26.0 ml of a 0.205 m nitric acid solution?
Law Incorporation [45]

A neutralization titration is a chemical response this is used to decide the composition of an answer and what kind of acid or base is in it. This is a way of volumetric analysis and the formula is (M1V1 = M2V2).

Utilize the titration method of M1V1 = M2V2 in view that we're given the concentrations of every compound and the quantity of KOH. Let: M1 = 0.138M, V1 = x, M2 = 0.205M, V2 = 26.0 ML.

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  • V1= initial volume
  • M2 = final mass
  • V2= final volume
  • (M1V1 = M2V2)
  • (0.138)(V1) = (0.205)x(26.0)
  • V2=(0.205)x(26.0)\ 0.138
  • V2 = 47.10 M/L
  • The final value of Volume needed for neutralization of nitric acid solution is  V2 = 47.10 M/L

Read more about the neutralization:

brainly.com/question/23008798

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4 0
1 year ago
What is the mass of an atom that has 9 protons, 11 neutrons, and 10 electrons
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Explanation:

6 0
3 years ago
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