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goldenfox [79]
3 years ago
12

4. A new compound X has been made, and a 5 x 10-4 Molar solution of it gives an absorbance of 0.405 at the maximum absorbance of

265 nm. The path length of the UV cell is 0.5cm.
Use Beer’s Law to calculate the Molar Absorptivity coefficient of Compound X at this wavelength, quoting the correct units.
Chemistry
1 answer:
vesna_86 [32]3 years ago
4 0

Answer:

1620 cm⁻¹·M⁻¹

Explanation:

<em>Beer's Law </em>states that:

A = ε * b * C

Where A is the absorbance, ε is the Molar Absorptivity coefficient, b is the path length, and C is the molar concentration.

<u>Use the given data</u>:

0.405 = ε * 0.5 cm * 5x10⁻⁴ M

And <u>solve for ε</u>:

ε = 1620 cm⁻¹·M⁻¹

So the Molar Absorptivity coefficient of Compound X at this 265 nm is 1620 cm⁻¹·M⁻¹.

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Answer:

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A gas mixture with 4 mol of Ar, x moles of Ne, and y moles
maks197457 [2]

Answer:

a) \Delta G_{mixing}=\frac{R*T}{12}*[4*ln (1/3) +x*ln (x/12) +(8-x)*ln ((8-x)/12)]

b) x=4

c) \Delta G_{max}=-2721.9 J/mol

Explanation:

Gas mixture:

n_{Ar}= 4 mol

n_{Ne}= x mol

n_{Xe}= y mol

n_{tot}= n_{Ar} + n_{Ne} + n_{Xe}=3*n_{Ar}

n_{Ne} + n_{Xe}=2*n_{Ar}

x + y=8 mol

y=8 mol- x

Mol fractions:

x_{Ar}=\frac{4 mol}{12 mol}=1/3

x_{Ne}=\frac{x mol}{12 mol}=x/12

x_{Xe}=\frac{8 - x mol}{12 mol}=(8-x)/12

Expression of \Delta G_{mixing}

\Delta G_{mixing}=R*T*\sum_{i]*x_i*ln (x_i)

\Delta G_{mixing}=R*T*[1/3*ln (1/3) +x/12*ln (x/12) +(8-x)/12*ln ((8-x)/12)]

\Delta G_{mixing}=\frac{R*T}{12}*[4*ln (1/3) +x*ln (x/12) +(8-x)*ln ((8-x)/12)]

Expression of \Delta G_{max}

\frac{d \Delta G_{mixing}}{dx}=0

\frac{d \Delta G_{mixing}}{dx}=\frac{R*T}{12}*[ln (x/12)+12-ln ((8-x)/12)-12]

0=\frac{R*T}{12}*[ln (x/12)-ln ((8-x)/12)

0=[ln (x/12)-ln ((8-x)/12)

ln (x/12)=ln ((8-x)/12)

x=(8-x)

x=4

\Delta G_{max}=\frac{8.314*298}{12}*[4*ln (1/3) +4*ln (4/12) +(8-4)*ln ((8-4)/12)]

\Delta G_{max}=\frac{8.314*298}{12}*[4*ln (1/3) +4*ln (1/3) +(4)*ln (1/3)]

\Delta G_{max}=\frac{8.314*298}{12}*[12*ln (1/3)]

\Delta G_{max}=-2721.9 J/mol

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3 years ago
Benzoic acid, c6h5cooh, has a ka = 6.4 x 10-5. what is the concentration of h3o+ in a 0.5 m solution of benzoic acid? 3.2 x 10-5
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The answer for this issue is: 
The chemical equation is: HBz + H2O <- - > H3O+ + Bz- 
Ka = 6.4X10^-5 = [H3O+][Bz-]/[HBz] 
Let x = [H3O+] = [Bz-], and [HBz] = 0.5 - x. 
Accept that x is little contrasted with 0.5 M. At that point, 
Ka = 6.4X10^-5 = x^2/0.5 
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The molecular formula of glucose is C₆H₁₂O₆ and the mass is 180.156 g/mol. It is an aldohexose that contains an aldehydic functional group. In its structure, there are six oxygen atoms, six carbon atoms, and twelve hydrogen atoms.

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