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LUCKY_DIMON [66]
3 years ago
11

1. Chlorofluorocarbons (CFCs) A carbon-chlorine bond in the CFC molecule can be broken by sunlight, leaving a highly reactive fr

ee radical which then goes on to destroy the surrounding ozone molecules. The energy of a C-Cl bond is 328 kJ/mole. Calculate the wavelength of light needed to break a bond in a single molecule. In which region of the spectrum (infrared, visible, UV) does this wavelength fall
Chemistry
1 answer:
Reptile [31]3 years ago
8 0

Answer: The wavelength for this photon is 365 nm. The wavelength corresponds to UV region.

Explanation:

The relationship between wavelength and energy of the wave follows the equation:

E=\frac{Nhc}{\lambda}

E= energy

N = avogadros number

\lambda = wavelength of the wave

h = Planck's constant  = 6.626\times 10^{-34}Js

c = speed of light = 3\times 10^8m/s

328\times 10^3J=\frac{6.023\times 10^{23}\times 6.626\times 10^{-34}\times 3\times 10^8m/s}{\lambda}

\lambda=3.65\times 10^{-7}m=365nm   1m=10^9nm

Thus wavelength for this photon is 365 nm. The wavelength of 365 nm corresponds to UV region.

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Which of these happens when the warm water gets to the North and South Poles
Airida [17]

Answer:It creates a heat source on the cold water and the molecules act against each other and then it becomes steam.

Explanation:I had to take this on a test.

6 0
3 years ago
Marian claims that Food A has slightly more than three times the Calories as Food B. Johan claims that Food B has more Calories.
Liula [17]

Explanation:

In order to justify Marian's statement we have to look at the '' Amount per serving calories'' ⇒

In food label A we can see that this value is 160 calories

In food label B we can see that this value is 50 calories

⇒ 160 calories is slightly more than three times 50 calories

Otherwise If we want to justify Johan statement we need to look at the '' serving size '' ⇒

In food label A we can see that the serving size is 1 cup (237 mL)

In food label B we can see that the serving size is \frac{1}{4} cup (56g)

Working with the information of the food label A we can write the following expression :

\frac{1Cup}{160calories}=\frac{\frac{1}{4}Cup}{x}  ⇒ Looking at the value of ''x'' ⇒

x=160calories(\frac{1}{4})=40calories

x=40calories

If we look at the same amount of portion volume :

In \frac{1}{4} cup of food A we have 40 calories

In \frac{1}{4} cup of food B we have 50 calories

We could conclude that Food B has more calories.

That's how both claims could both be justified.

8 0
4 years ago
What term is best associated with protons or neutrons escaping to become more stable?
Gre4nikov [31]
The correct answer is C
3 0
3 years ago
The equilibrium constant, K, for the following reaction is 2.44×10-2 at 518 K: PCl5(g) PCl3(g) + Cl2(g) An equilibrium mixture o
Volgvan

Answer:

[PCl₅] = 0.5646M

[PCl₃] = 0.1174M

[Cl₂] = 0.1174M

Explanation:

In the reaction:

PCl₅(g) ⇄ PCl₃(g) + Cl₂(g)

K equilibrium is defined as:

<em>K = 2.44x10⁻² = [PCl₃] [Cl₂] / [PCl₅]</em>

The initial moles of each compound when volume is 15.3L are:

PCl₅ = 0.300mol/L×15.3L = 4.59mol

Cl₂ = 8.55x10⁻²mol/L×15.3L = 1.308mol

PCl₃ = 8.55x10⁻²mol/L×15.3L = 1.308mol

At 8.64L, the new concentrations are:

[PCl₅] = 4.59mol / 8.64L = 0.531M

[PCl₃] = 1.308mol / 8.64L = 0.151M

[Cl₂] = 1.308mol / 8.64L = 0.151M

At these conditions, reaction quotient, Q, is:

Q = [0.151M] [0.151M] / [0.531M]

Q = 4.29x10⁻²

As Q > K, <em>the reaction will shift to the left producing more reactant, </em>that means equilibrium concentrations are:

[PCl₅] = 0.531M + X

[PCl₃] = 0.151M - X

[Cl₂] = 0.151M - X

<em>Where X is reaction coordinate.</em>

Replacing in K expression:

2.44x10⁻² = [0.151M - X] [0.151M - X] / [0.531M + X]

1.296x10⁻² + 2.44x10⁻²X = 0.0228 - 0.302X + X²

<em>0 = 9.84x10⁻³ - 0.3264X + X²</em>

Solving for X:

X = 0.293 → False solution. Produce negative concentrations

<em>X = 0.0336M → Right solution.</em>

Replacing:

[PCl₅] = 0.531M + 0.0336

[PCl₃] = 0.151M - 0.0336

[Cl₂] = 0.151M - 0.0336

<h3>[PCl₅] = 0.5646M</h3><h3>[PCl₃] = 0.1174M</h3><h3>[Cl₂] = 0.1174M</h3>
4 0
4 years ago
A pure substance is found to contain 53.7% fluorine and 46.3% xenon by mass. What is the empirical formula of this substance?
evablogger [386]

Answer: XF8

Explanation:

Empirical Formular shows the simplest ratio of elements in a compound.

 Xe = 46.3%             F  = 53.7%

Divide the percentage composition of each element by the atomic  mass.

Xe = 46.3/ 131.3                      F= 53.7/ 19

      = 0.353( approx)               =  2.826 (approx)

Divide through with the smallest of the answers gotten in previous step.

 Xe = 0.353 / 0.353                F = 2.826/ 0.353

       =  1                                       = 8.0

Empirical formular = XF8

3 0
3 years ago
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