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taurus [48]
3 years ago
14

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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Consider the halogenation of ethene, where x is a generic halogen: h2c=ch2(g)+x2(g)→h2xc−ch2x(g) you may want to reference (page
KengaRu [80]
Consider the halogenation of ethene is as follows:
CH₂=CH₂(g) + X₂(g) → H₂CX-CH₂X(g)
We can expect that this reaction occurring by breaking of a C=C bond and forming of two C-X bonds.
When bond break it is endothermic and when bond is formed it is exothermic.
So we can calculate the overall enthalpy change as a sum of the required bonds in the products:
Part a) 
C=C break = +611 kJ
2 C-F formed = (2 * - 552) = -1104 kJ
Δ H = + 611 - 1104 = - 493 kJ

2C-Cl formed = (2 * -339) = - 678 kJ
ΔH = + 611 - 678 = -67 kJ

2 C-Br formed = (2 * -280) = -560 kJ
ΔH = + 611 - 560 = + 51 kJ

2 C-I Formed = (2 * -209) = -418 kJ
ΔH = + 611 - 418 = + 193 kJ

Part b)
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8 0
3 years ago
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A sample of calcium phosphate was found to have a mass of 125.3 g. How many molecules were contained in the sample?
Viktor [21]

The answer for the following problem is mentioned below.

  • <u><em>Therefore number of molecules(N) present in the calcium phosphate sample are  19.3 × 10^23 molecules.</em></u>

Explanation:

Given:

mass of calcium phosphate (Ca_{3}(PO_{4} )_{2} ) = 125.3 grams

We know;

molar mass of calcium phosphate  (Ca_{3}(PO_{4} )_{2} ) = (40×3) + 3 (31 +(4×16))

molar mass of calcium phosphate  (Ca_{3}(PO_{4} )_{2} ) = 120 + 3(95)

molar mass of calcium phosphate  (Ca_{3}(PO_{4} )_{2} )  = 120 +285 = 405 grams

<em>We also know;</em>

No of molecules at STP conditions(N_{A}) = 6.023 × 10^23 molecules

To solve:

no of molecules present in the sample(N)

We know;

\frac{m}{M} =\frac{N} }{}N÷N_{A}

\frac{405}{125.3} =\frac{N}{6.023*10^23}

N =(405×6.023 × 10^23) ÷ 125.3

N = 19.3 × 10^23 molecules

<u><em>Therefore number of molecules(N) present in the calcium phosphate sample are  19.3 × 10^23 molecules</em></u>

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3 years ago
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