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Butoxors [25]
3 years ago
15

Th e molar absorption coeffi cient of a substance dissolved in water is known to be 855 dm3 mol−1 cm−1 at 270 nm. To determine t

he rate of decomposition of this substance, a solution with a concentration of 3.25 mmol dm−3 was prepared. Calculate the percentage reduction in intensity when light of that wavelength passes through 2.5 mm of this solution.
Chemistry
1 answer:
Olegator [25]3 years ago
3 0

Answer : The percentage reduction in intensity is 79.80 %

Explanation :

Using Beer-Lambert's law :

A=\epsilon \times C\times l

A=\log \frac{I_o}{I}

\log \frac{I_o}{I}=\epsilon \times C\times l

where,

A = absorbance of solution

C = concentration of solution = 3.25mmol.dm^{3-}=3.25\times 10^{-3}mol.dm^{-3}

l = path length = 2.5 mm = 0.25 cm

I_o = incident light

I = transmitted light

\epsilon = molar absorptivity coefficient = 855dm^3mol^{-1}cm^{-1}

Now put all the given values in the above formula, we get:

\log \frac{I_o}{I}=(855dm^3mol^{-1}cm^{-1})\times (3.25\times 10^{-3}mol.dm^{-3})\times (0.25cm)

\log \frac{I_o}{I}=0.6947

\frac{I_o}{I}=10^{0.6947}=4.951

If we consider I_o = 100

then, I=\frac{100}{4.951}=20.198

Here 'I' intensity of transmitted light = 20.198

Thus, the intensity of absorbed light I_A = 100 - 20.198 = 79.80

Now we have to calculate the percentage reduction in intensity.

\% \text{reduction in intensity}=\frac{I_A}{I_o}\times 100

\% \text{reduction in intensity}=\frac{79.80}{100}\times 100=79.80\%

Therefore, the percentage reduction in intensity is 79.80 %

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You notice that the water in your friend's swimming pool is cloudy and that the pool walls are discolored at the water line. A q
kicyunya [14]

Answer:

3,78 mL of 12,0wt% H₂SO₄

Explanation:

The equilibrium in water is:

H₂O (l) ⇄ H⁺ (aq) + OH⁻ (aq)

The initial concentration of [H⁺] is 10⁻⁸ M and final desired concentration is [H⁺] = 10^{-7,2}

Thus, you need to add:

[H⁺] = 10^{-7,2} - 10^{-8} = <em>5,31x10⁻⁸ M</em>

The total volume of the pool is:

7,00 m × 18,0 m ×1,50 m = 189 m³ ≡ 189000 L

Thus, moles of H⁺ you need to add are:

5,31x10⁻⁸ M × 189000 L = 1,00<em>x10⁻² moles of H⁺</em>

These moles comes from

H₂SO₄ → 2H⁺ +SO₄²⁻

Thus:

1,00x10⁻² moles of H⁺ × \frac{1 H_{2}SO_{4} moles}{2H^+ mole} = <em>5,00x10⁻³ moles of H₂SO₄</em>

These moles comes from:

5,00x10⁻³ moles of H₂SO₄ × \frac{98,1g}{1mol} × \frac{100 gsolution}{12 g H_{2}SO_{4} } × \frac{1mL}{1,080 g} = <em>3,78 mL of 12,0wt% H₂SO₄</em>

<em></em>

I hope it helps!

5 0
3 years ago
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