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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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<u>Explanation:</u>

The given chemical equation follows:

3Cu^{2+}(aq.)+2Al\rightarrow 2Al^{3+}(aq.)+2Au(s)

<u>Oxidation half reaction:</u> Al(aq.)\rightarrow Al^{3+}(aq.)+3e^-;E^o_{Al^{3+}/Al}=1.66V       ( × 2)

<u>Reduction half reaction:</u> Cu^{2+}(aq.)+2e^-\rightarrow Cu(s);E^o_{Cu^{2+}/Cu}=0.16V       ( × 3)

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To calculate the E^o_{cell} of the reaction, we use the equation:

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To calculate the EMF of the cell, we use the Nernst equation, which is:

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E_{cell} = electrode potential of the cell = ? V

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T = temperature = 42^oC=[42+273]K=315K

F = Faraday's constant = 96500

[Al^{3+}]=1.63M

[Cu^{2+}]=3.43M

Putting values in above equation, we get:

E_{cell}=1.82-\frac{2.303\times 8.314\times 315}{2\times 96500}\times \log(\frac{(1.63)^2}{(3.43)^3})\\\\E_{cell}=1.86V

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