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Rasek [7]
4 years ago
5

(b) Data has been collected to show that at a given wavelength in a 1 cm pathlength cell, Beer's Law for the absorbance of Co2+

is linear. If a 0.135 M solution of Co2+ has an absorbance of 0.350, what is the concentration of a solution with an absorbance of 0.460?
Chemistry
1 answer:
OlgaM077 [116]4 years ago
3 0

Answer : The concentration of a solution with an absorbance of 0.460 is, 0.177 M

Explanation :

Using Beer-Lambert's law :

A=\epsilon \times C\times l

where,

A = absorbance of solution

C = concentration of solution

l = path length

\epsilon = molar absorptivity coefficient

From this we conclude that absorbance of solution is directly proportional to the concentration of solution at constant path length.

Thus, the relation between absorbance and concentration of solution will be:

\frac{A_1}{A_2}=\frac{C_1}{C_2}

Given:

A_1 = 0.350

A_2 = 0.460

C_1 = 0.135 M

C_2 = ?

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

\frac{0.350}{0.460}=\frac{0.135}{C_2}

C_1=0.177M

Therefore, the concentration of a solution with an absorbance of 0.460 is, 0.177 M

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4 0
4 years ago
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Iteru [2.4K]

Answer:

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3 years ago
Read 2 more answers
Partner Namets) For each of the following reactions carried out: Write the balanced chemical equation, the full ionic equation,
torisob [31]

Answer : The full balanced ionic equation will be,

2KI(aq)+Pb(NO_3)_2(aq)\rightarrow 2KNO_3(aq)+PbI_2(s)

Reactants are lead nitrate and potassium iodide.

Products are lead iodide and potassium nitrate.

The spectator ions are, K^+,NO_3^-

Explanation :

Complete ionic equation : In complete ionic equation, all the substance that are strong electrolyte and present in an aqueous are represented in the form of ions.

Net ionic equation : In the net ionic equations, we are not include the spectator ions in the equations.

Spectator ions : The ions present on reactant and product side which do not participate in a reactions. The same ions present on both the sides.

When potassium iodide react with lead nitrate then it gives potassium nitrate and lead iodide as a product.

The full balanced ionic equation will be,

2KI(aq)+Pb(NO_3)_2(aq)\rightarrow 2KNO_3(aq)+PbI_2(s)

The ionic equation in separated aqueous solution will be,

2K^+(aq)+2I^{-}(aq)+Pb^{2+}(aq)+2NO_3^{-}(aq)\rightarrow PbI_2(s)+2K^+(aq)+2NO_3^{-}(aq)

In this equation, K^+\text{ and }NO_3^- are the spectator ions.

By removing the spectator ions from the balanced ionic equation, we get the net ionic equation.

The net ionic equation will be,

Pb^{2+}(aq)+2I^{-}(aq)\rightarrow PbI_2(s)

4 0
4 years ago
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