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Anarel [89]
2 years ago
10

If the absorbance of a KMnO4 solution of unknown concentration is 0.633, calculate the concentration of KMnO4 in the solution.

Chemistry
1 answer:
solong [7]2 years ago
3 0

The molar Concentration of KMnO₄ is 0.000219 M

Concentration is the abundance of a constituent divided by means of the overall extent of an aggregate. numerous styles of mathematical description may be outstanding: mass awareness, molar awareness, variety concentration, and quantity awareness.

y is absorbance

x is the molar concentration of KMnO_4

y = 4.84E + 03x - 2.26E - 01

0.833 = 4.84 * 10⁺⁰³ x - 2.26 * 10⁻¹

1.059 = 4.84 * 10⁺⁰³ x

X = 0.000219 M

Hence, The molar Concentration of KMnO₄ is 0.000219 M

Learn more about concentration here:-brainly.com/question/14469428

#SPJ9

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The answer is Solid.

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4 0
3 years ago
A 51.24-g sample of Ba(OH)2 is dissolved in enough water to make 1.20 liters of solution. How many mL of this solution must be d
g100num [7]

Answer:

0.40 L

Explanation:

Calculation of the moles of Ba(OH)_2 as:-

Mass = 51.24 g

Molar mass of Ba(OH)_2 = 171.34 g/mol

The formula for the calculation of moles is shown below:

moles = \frac{Mass\ taken}{Molar\ mass}

Thus,

Moles= \frac{51.24\ g}{171.34\ g/mol}

Moles= 0.2991\ mol

Volume = 1.20 L

The expression for the molarity is:

Molarity=\frac{Moles\ of\ solute}{Volume\ of\ the\ solution}

Molarity=\frac{0.2991\ mol}{1.20\ L}=0.24925\ M

Thus,

Considering

Molarity_{working\ solution}\times Volume_{working\ solution}=Molarity_{stock\ solution}\times Volume_{stock\ solution}

Given  that:

Molarity_{working\ solution}=0.100\ M

Volume_{working\ solution}=1\ L

Volume_{stock\ solution}=?

Molarity_{stock\ solution}=0.24925\ M

So,  

0.100\ M\times 1\ L=0.24925\ M\times Volume_{stock\ solution}

Volume_{stock\ solution}=\frac{0.100\times 1}{0.24925}\ L=0.40\ L

<u>The volume of 0.24925M stock solution added = 0.40 L </u>

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3 years ago
A car engine produces 604 kJ of mechanical energy from fuel capable of producing 2416 kJ of energy. The efficiency of the engine
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I think it’s 40% but I’m not sure
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Please help me out !!
raketka [301]
27Hq)+3816(7/ 23q09)
8 0
2 years ago
Consider the reaction below.
saw5 [17]

This answer to this question is a rule that is applied to any reaction taken at dynamic equilibrium, with respect to 500 K. In other words, you can say that this reaction is of no use to us -

In a chemical equilibrium, it is known that the forward and reverse reactions occur at equal rates. At this point the concentrations of products and reactants remain constant, or in other words do not change

<u><em>Solution = Option C</em></u>

6 0
3 years ago
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