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Amiraneli [1.4K]
3 years ago
6

The molar absorptivity of a tyrosine residue at 280 nm is 2000 M-1cm-1, while for tryptophan it is 5500 M-1cm-1. A protein has b

een isolated that is known to contain one tyrosine residue and an unknown number of tryptophans. A 1.0 micromolar solution of this protein is placed in a 1.0 cm cuvette and the absorbance at 280 nm is measured as 0.024. How many tryptophans are in the protein
Chemistry
1 answer:
adelina 88 [10]3 years ago
7 0

Answer:

There are 4 tryptophans in the protein.

Explanation:

According to question,  protein contains one tyrosine residue and say x number of tryptophans.

Concentration of protein solution = 1.0 micromolar = 1.0\times 10^{-6} Molar

Molar absorptivity of a protein solution : \epsilon

\epsilon = \epsilon _{tyro}+\epsilon _{tryp}

=1\times 2000 M^{-1}cm^{-1}+x\times 5500 M^{-1}cm^{-1}

Length of the cuvette = l = 1.0 cm

Absorbance of protein solution at 280 nm = A = 0.024

A=\epsilon \times l\times c ( Beer-Lambert's law)

0.024=(1\times 2000 M^{-1}cm^{-1}+x\times 5500 M^{-1}cm^{-1})\times 1 cm\times 1.0\times 10^{-6} M

Solving for x :

x = 4

There are 4 tryptophans in the protein.

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Step 1 : Write balanced chemical equation

The balanced chemical equation for the reaction between iodine gas and chlorine gas is given below.

I_{2}  (g) + Cl_{2}  (g) ----->  2 ICl (g)

Step 2 : Set up ICE table

We will set up an ICE table for the above reaction

Following points are considered while drawing ICE table

- The initial concentration of product is assumed as 0

- The change in concentration (C) is assumed as x. Change (x) is negative for reactants and positive for products

- The coefficients in balanced equation are considered while writing C values

Check attached file for ICE table

Step 3 : Set up equilibrium constant equation

The equation for equilibrium constant can be written as

K_{eq} = \frac{[ICl]^{2}}{[I_{2}][Cl_{2}]}

Step 4 : Solving for x

Keq at 298.15 K is given as 81.9

Let us plug in the equilibrium values (E) for I₂, Cl₂ and ICl from ICE table

81.9 = \frac{(2x)^{2}}{(0.437-x) (0.269-x)}

81.9 = \frac{(2x)^{2}}{x^{2} -0.706x + 0.118}

(2x)^{2} = 81.9 [ x^{2} -0.706x + 0.118]

4x^{2} = 81.9x^{2} -57.8x +9.66

77.9x^{2} -57.8x+9.66 = 0

Solving the above equation using quadratic formula we get

x = 0.488 or x = 0.254

The value 0.488 cannot be used because the change (C) cannot be greater that initial concentration of the reactants.

Therefore the change in concentration of the gases during the reaction is 0.254 M

Hence, x = 0.254 M

From the ICE table, we know that the equilibrium concentration of ICl is 2x

[ICl]_{eq} = 2 ( 0.254) = 0.508 M

The concentration of ICl when the reaction reaches equilibrium is 0.508 M

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Answer:

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Explanation:

<u>Step 1: Define</u>

6.7 × 10²⁴ molecules NO₂ (Nitrogen dioxide)

<u>Step 2: Define conversions</u>

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Molar Mass of NO₂ - 14.01 + 2(16.00) = 46.01 g/mol

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<u>Step 4: Check</u>

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