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3 years ago

When a molecule absorbs IR electromagnetic energy, what are the vibrational modes affected? Select one or more correct answers.

Chemistry

1 answer:

tester [92]3 years ago

4 0

Answer:

Stretching and bending vibrational modes are affected

Explanation:

Frequency region for IR (infra red) radiation is 4000-400 $cm^{-1}$

Stretching and bending frequencies of covalent bonds in a molecule lies within range of IR radiation.

When a molecule is excited with IR radiation, frequencies related to stretching and bending mode of covalent bond resonates with incoming frequency of radiation by obeying some certain quantum mechanical selection rule.Therefore we can identify those stretching/bending mode and thereby identify a particular covalent bond.

So, option (C) and (D) are correct

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. The electron affinity for an oxygen atom is negative. What statement is probably true regarding

kramer

The best and most correct answer among the choices provided by your question is the second choice or letter B.

<span>Regarding the second electron affinity for an oxygen, i. e., the electron affinity for O-, it is much larger and negative.</span>

I hope my answer has come to your help. Thank you for posting your question here in Brainly. We hope to answer more of your questions and inquiries soon. Have a nice day ahead!

7 0

3 years ago

Redox 1/2 reaction Cu(s) + 2 AgC2H3O2(aq) = Cu(C2H3O2)2(aq) + 2 Ag(s)

Anna11 [10]

Explanation:

Reaction:

Cu + 2AgC₂H₃O₂ → Cu(C₂H₃O₂)₂ + 2Ag

The problem is to split the reaction into oxidation and reduction halves:

The oxidation half is the sub-reaction that undergoes oxidation

The reduction half is the one that undergoes reduction:

The ionic equation:

Cu + 2Ag⁺ + 2C₂H₃O₂⁻ → Cu²⁺ + 2C₂H₃O₂⁻ + 2Ag

Oxidation half:

Cu → Cu²⁺ + 2e⁻

Reduction half:

2Ag⁺ + 2e⁻ → 2Ag

C₂H₃O₂⁻ is neither oxidized nor reduced in the reaction.

learn more:

Oxidation state brainly.com/question/10017129

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4 years ago

1.) The process for converting ammonia to nitric acid involves the conversion of NH3 to

Firdavs [7]

Answer:

a) 1.39 g ; b) O₂ is limiting reactant, NH₃ is excess reactant; c) 0.7 g

Explanation:

We have the masses of two reactants, so this is a limiting reactant problem.

We will need a balanced equation with masses, moles, and molar masses of the compounds involved.

1. Gather all the information in one place with molar masses above the formulas and masses below them.

MM: 17.03 32.00 30.01

4NH₃ + 5O₂ ⟶ 4NO + 6H₂O

Mass/g: 1.5 1.85

2. Calculate the moles of each reactant

$\text{moles of NH}_{3} = \text{1.5 g NH}_{3} \times \dfrac{\text{1 mol NH}_{3}}{\text{17.03 g NH}_{3}} = \text{0.0881 mol NH}_{3}\\\\\text{moles of O}_{2} = \text{1.85 g O}_{2} \times \dfrac{\text{1 mol O}_{2}}{\text{32.00 g O}_{2}} = \text{0.057 81 mol O}_{2}$

3. Calculate the moles of NO we can obtain from each reactant

From NH₃:

The molar ratio is 4 mol NO:4 mol NH₃

$\text{Moles of NO} = \text{0.0881 mol NH}_{3} \times \dfrac{\text{4 mol NO}}{\text{4 mol NH}_{3}} = \text{0.0881 mol NO}$

From O₂:

The molar ratio is 4 mol NO:5 mol O₂

$\text{Moles of NO} = \text{0.057 81 mol O}_{2}\times \dfrac{\text{4 mol NO}}{\text{5 mol O}_{2}} = \text{0.046 25 mol NO}$

4. Identify the limiting and excess reactants

The limiting reactant is O₂ because it gives the smaller amount of NO.

The excess reactant is NH₃.

5. Calculate the mass of NO formed

$\text{Mass of NO} = \text{0.046 25 mol NO}\times \dfrac{\text{30.01 g NO}}{\text{1 mol NO}} = \textbf{1.39 g NO}$

6. Calculate the moles of NH₃ reacted

The molar ratio is 4 mol NH₃:5 mol O₂

$\text{Moles reacted} = \text{0.057 81 mol O}_{2} \times \dfrac{\text{4 mol NH}_{3}}{\text{5 mol O}_{2}} = \text{0.046 25 mol NH}_{3}$

7. Calculate the mass of NH₃ reacted

$\text{Mass reacted} = \text{0.046 25 mol NH}_{3} \times \dfrac{\text{17.03 g NH}_{3}}{\text{1 mol NH}_{3}} = \text{0.7876 g NH}_{3}$

8. Calculate the mass of NH₃ remaining

Mass remaining = original mass – mass reacted = (1.5 - 0.7876) g = 0.7 g NH₃

8 0

3 years ago

Give the nuclear symbol for the isotope of silicon for which a=29? Enter the nuclear symbol for the isotope (e.G., 42he).

quester [9]

The mass number is the summation of number of proton and neutron present in a nucleus of an atom. For the neutral atom the number of positive charge (number of proton) must be equal to the number of electrons. The number of electrons present in an atom is the atomic number of the atom. The standard way to express the mass number (a) and atomic number (m) of a atom (say X) is $x^{a}_{m}$ . Now for silicon number of electron or atomic number is 14. And the mass number (a) given 29. Thus the expression nucleus of silicon will be $Si^{29}_{14}$

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3 years ago

What determines the size and shape of craters

Mumz [18]

Size, speed and angle of the falling object determine the size shape and complexity of the resulting crater

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4 years ago

Read 2 more answers