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Vanyuwa [196]
3 years ago
8

Calculate the percentage difference in the fundamental vibrational wavenumbers of 23Na35Cl and 23Na37Cl on the assumption that t

heir force constants are the same. The mass of 23Na is 22.9898mu.
Chemistry
1 answer:
stealth61 [152]3 years ago
4 0

Answer:

1.089%

Explanation:

From;

ν =1/2πc(k/meff)^1/2

Where;

ν = wave number

meff = reduced mass or effective mass

k = force constant

c= speed of light

Let

ν =1/2πc (k/meff)^1/2  vibrational wave number for 23Na35 Cl

ν' =1/2πc(k'/m'eff)^1/2 vibrational wave number for 23Na37 Cl

The between the two is obtained from;

ν' - ν /ν  = (k'/m'eff)^1/2 - (k/meff)^1/2 / (k/meff)^1/2

Therefore;

ν' - ν /ν = [meff/m'eff]^1/2 - 1

Substituting values, we have;

ν' - ν /ν = [(22.9898 * 34.9688/22.9898 + 34.9688) * (22.9898 + 36.9651/22.9898 * 36.9651)]^1/2  -1

ν' - ν /ν = -0.01089

percentage difference in the fundamental vibrational wavenumbers of 23Na35Cl and 23Na37Cl;

ν' - ν /ν * 100

|(-0.01089)|  × 100 = 1.089%

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A solid with a density of 3.57g/mL and mass of 19.5g is added to a graduated cylinder that contain 23.2mL of water. What is the
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Answer:

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3 0
3 years ago
What amount of carbon dioxide (in moles) is produced from the reaction of 2.24 moles of ethanol with excess oxygen?
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Answer : The amount of carbon dioxide produced is, 197.12 grams.

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First we have to calculate the moles carbon dioxide.

From the balanced chemical reaction, we conclude that

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So, 2.24 mole of ethanol react to give 2\times 2.24=4.48 moles of carbon dioxide

Now we have to calculate the mass of carbon dioxide.

\text{Mass of }CO_2=\text{Moles of }CO_2\times \text{Molar mass of }CO_2

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Therefore, the amount of carbon dioxide produced is 197.12 grams.

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2 years ago
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