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Natali [406]
3 years ago
12

How many grams of kclo3 can be dissolved in 100 grams of water at 30 degrees Celsius

Chemistry
1 answer:
kondaur [170]3 years ago
8 0
<span>The solubility of KClO</span>₃ : ( 10.1 / 100 g water ) at 30ºC

10.1 g ------------ 100 g ( H₂O )
    ? g ------------- 100 g ( H₂O )

Mass of KClO₃ :

100 * 10.1 / 100

1010 / 100 = 10.1 g of KClO₃

hope this helps!
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Read 2 more answers
Calculate the Zn conc. of Zn/Zn++ // Cl/Cl- 0.1M Emf=2.21v
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Answer:

Option a. = 0.01 M

Explanation:

To do this, we need to gather the data:

E = 2.21 V

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Zn(s) + Cl₂(g) <-------> Zn²⁺(aq) + 2Cl⁻(aq)       Q = [Zn] [Cl]²

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E = E° - (0.059/n) logQ

E = E° - (0.059/n) ln([Cl⁻]² * [Zn²⁺])   (1)

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First, we need to write the semi equation of oxidation and reduction, and get the standard potential of the cell:

Zn(s) --------> Zn²⁺(aq) + 2e⁻       E₁° = 0.76 V

Cl₂(g) + 2e⁻ -----------> 2Cl⁻(aq)   E₂° = 1.36 V

---------------------------------------------------------------

Zn(s) + Cl₂(g) -------> Zn²⁺(aq) + 2Cl⁻(aq)    E° = 0.76 + 1.36 = 2.12 V

Now, let's replace in (1) and then, solve for [Zn]:

2.21 = 2.12 - (0.059/2) log ([0.1]² * [Zn])

2.21 - 2.12 = -0.0295 log (0.01[Zn])

- 0.09 / 0.0295 = log (0.01[Zn])

-3.0508 = log (0.01[Zn])

10^(-3.0508) = 0.01[Zn]

8.8961x10⁻⁴ = 0.01[Zn]

[Zn²⁺] = 0.08896 M

This value can be rounded to 0.1 M. so the correct option will be option A.

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