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alexira [117]
2 years ago
10

The substance chlorine monoxide, ClO(g), is important in atmospheric processes that lead to the depletion of the ozone layer. Th

e ClO molecule has an experimental dipole moment of 1.24 D and the Cl−O bond length is 1.60 Å.
Part A: Determine the magnitude of the charges on the Cl and O atoms in units of the electronic charge, e.

Part B: Based on the electronegativities of the elements, which atom would you expect to have a partial negative charge in the ClO molecule?
Atom O or Atom Cl

Part C: Using formal charges as a guide, propose the dominant Lewis structure for the molecule. (Draw a diagram)

Part D: The anion ClO− exists. What is the formal charge on the Cl for the best Lewis structure for ClO−?
Chemistry
1 answer:
adoni [48]2 years ago
7 0

Answer:

B = Based on the electronegativities of the elements, which atom would you expect to have a partial negative charge in the ClO molecule?

Atom O or Atom Cl

Explanation:

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5pts. Chemistry multiple choice​
aleksandrvk [35]

Answer:

C. Xenon

Explanation:

The question only requires you to use your visual skills to identify the unknown gas. Look at the spectra of unknown gas and xenon, are they not identical? SInce they are same, the unknown sample must be xenon.

4 0
2 years ago
Read 2 more answers
determine the frequency and wavelength (in nm) of the light emitted when the e- fell from n=4 and n=2
Lostsunrise [7]

Answer:

Frequency = 6.16 ×10¹⁴ Hz

λ = 4.87×10² nm

Explanation:

In case of hydrogen atom energy associated with nth state is,

En =  -13.6/n²

For n = 2

E₂ = -13.6 / 2²

E₂ = -13.6/4

E₂ = -3.4 ev

Kinetic energy of electron = -E₂ = 3.4 ev

For n = 4

E₄ = -13.6 / 4²

E₄ = -13.6/16

E₄ = -0.85 ev

Kinetic energy of electron = -E₄ = 0.85 ev

Wavelength of radiation emitted:

E = hc/λ = E₄ - E₂

hc/λ = E₄ - E₂

by putting values,

6.63×10⁻³⁴Js × 3×10⁸m/s / λ = -0.85ev   - (-3.4ev )

6.63×10⁻³⁴ Js× 3×10⁸m/s / λ = 2.55 ev

λ = 6.63×10⁻³⁴ Js× 3×10⁸m/s  /2.55ev

λ = 6.63×10⁻³⁴ Js× 3×10⁸m/s  /2.55× 1.6×10⁻¹⁹ J

λ = 19.89 ×10⁻²⁶ Jm / 2.55× 1.6×10⁻¹⁹ J

λ = 19.89 ×10⁻²⁶ Jm / 4.08×10⁻¹⁹ J

λ = 4.87×10⁻⁷ m

m to nm:

4.87×10⁻⁷ m ×10⁹nm/1 m

4.87×10² nm

Frequency:

Frequency = speed of electron / wavelength

by putting values,

Frequency = 3×10⁸m/s /4.87×10⁻⁷ m

Frequency = 6.16 ×10¹⁴ s⁻¹

s⁻¹ = Hz

Frequency = 6.16 ×10¹⁴ Hz

3 0
3 years ago
True or false The atomic number of an element is always more than the mass number of that element.
Veseljchak [2.6K]
The sentence is true
8 0
3 years ago
State what would be observed when the following pairs of reagents are mixed in a test tube.
Mandarinka [93]

Answer:

(i). C6H2COOH and Na2CO3(aq)

observation: <u>Bubbles</u><u> </u><u>of</u><u> </u><u>a</u><u> </u><u>colourless</u><u> </u><u>gas</u><u> </u><u>(</u><u>carbon</u><u> </u><u>dioxide</u><u> </u><u>gas</u><u>)</u>

(ii) CH3CH2CH2OH and KMnO4 /H

observation: <u>The</u><u> </u><u>orange</u><u> </u><u>solution</u><u> </u><u>turns</u><u> </u><u>green</u><u>.</u>

[<em>This</em><em> </em><em>is</em><em> </em><em>because</em><em> </em><em>oxidation</em><em> </em><em>of</em><em> </em><em>propanol</em><em> </em><em>to</em><em> </em><em>propanoic</em><em> </em><em>acid</em><em> </em><em>occurs</em>]

(iii) CH3CH2OH and CH3COOH + conc. H2SO4

observation: <u>A</u><u> </u><u>sweet</u><u> </u><u>fruity</u><u> </u><u>smell</u><u> </u><u>is</u><u> </u><u>formed</u><u>.</u>

[<em>This</em><em> </em><em>is</em><em> </em><em>because</em><em> </em><em>an</em><em> </em><em>ester</em><em>,</em><em> </em><em>diethylether</em><em> </em><em>is</em><em> </em><em>formed</em><em>]</em>

(iv) CH3CH = CHCH3 and Br2 /H2O

observation: <u>a</u><u> </u><u>brown</u><u> </u><u>solution</u><u> </u><u>is</u><u> </u><u>formed</u><u>.</u>

4 0
3 years ago
An unknown substance is obtained by police. The forensics laboratory has
timurjin [86]

Answer:

70.0 %

Explanation:

Step 1: Given data

  • Mass of nitrogen (mN): 74.66 g
  • Mass of the compound (mNxOy): 250 g

Step 2: Calculate the mass of oxygen (mO) in the compound

The mass of the compound is equal to the sum of the masses of the elements that form it.

mNxOy = mN + mO

mO = mNxOy - mN

mO = 250 g - 74.66 g = 175 g

Step 3: Determine the percent composition of oxygen in the sample

We will use the following expression.

%O = mO / mNxOy × 100%

%O = 175 g / 250 g × 100% = 70.0 %

5 0
3 years ago
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