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kozerog [31]
2 years ago
7

A concentrated solution of compound A in the same solvent was diluted from an initial volume of 2.00 mL to a fi nal volume of 25

.00 mL and then had an absorbance of 0.733. What is the concentration of A in the concentrated solution?
Chemistry
1 answer:
Andre45 [30]2 years ago
4 0

Answer:

The concentration of A in the concentrated solution is 0.05864

Explanation:

Given

Volume 1 = V1 = 2.00mL

Volume 2 = V2 = 25.00mL

Absorbance 1 = A1 = ?

Absorbance 2 = A2 = 0.733

Beer's law states that the concentration of a chemical solution is directly proportional to its absorbance.

i.e.

V = kA where V = Volume (Concentration)

A = Absorbance

k = Constant

Make k the subject of formula

k = V/A

V1/A1 = V2/A2

Make A1 the subject of the formula

A1 = V1 * A2/V2

Substitute the values of V1, A2 and V2

A1 = 2 * 0.733/25

A1 = 0.05864

So, the concentration of A in the concentrated solution is 0.05864

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What would happen if I removed a resistor from a parallel circuit? What would
barxatty [35]

Answer:

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natka813 [3]

Answer:

Take a look at the attachment below

Explanation:

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<em>Hope that helps!</em>

5 0
3 years ago
How would you prepare 3.5 L of a 0.9M solution of KCl?
PolarNik [594]
V=3,5L\\&#10;Cm=0,9M\\&#10;M_{KCl}=74\frac{g}{mol}\\\\&#10;C_{m}=\frac{n}{V}\\\\&#10;n=\frac{m}{M}\\\\&#10;C_{m}=\frac{m}{MV} \ \ \ \Rightarrow  \ \ \ m=C_{m}MV\\\\&#10;m=0,9\grac{mol}{L}*74\frac{g}{mol}*3,5L=233,1g

B. Add 233 g of KCl to a 3.5 L container; then add enough water to dissolve the KCl and fill the container to the 3.5 L mark. 
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A flexible container at an initial volume of 5.120 L contains 8.500 mol of gas.
SOVA2 [1]

Answer:

30.05 mol

Explanation:

solving the proportion

V1 / n1 = V2 / n2

5.120 L            18.10 L

–––––––– = ––––––

8.500 mol         x

x = 30.05 mol

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