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yuradex [85]
3 years ago
7

Using the volumes of sodium thiosulfate solution you just entered, the mass of bleach sample, and the average molarity of the so

dium thiosulfate solution entered earlier, calculate the mass percent of NaClO for each bleach sample. Enter the calculated mass percent of NaClO in each of the 3 acceptable trials (For the first trial, the volume of sodium thiosulfate =10.1, mass of bleach is 0.496g, and average molarity of sodium thiosulfate=0.042M)
Chemistry
1 answer:
dedylja [7]3 years ago
4 0

Answer:

3.18 (w/w) %

Explanation:

In the problem, you can find mass of NaClO knowing the reaction of NaClO with Na₂S₂O₃ is:

NaClO + 2Na₂S₂O₃ + H₂O → NaCl + Na₂S₄O₆ +2NaOH + NaCl

<em>Where 1 mole of NaClO reacts with 2 moles of Na₂S₂O₃</em>

<em> </em>Moles of thiosulfate in the titration are:

0.0101L ₓ (0.042mol / L) = 4.242x10⁻⁴ moles of Na₂S₂O₃

Thus, moles of NaClO in the initial solution are:

4.242x10⁻⁴ moles of Na₂S₂O₃ ₓ (1mol NaClO / 2 mol Na₂S₂O₃) = 2.121x10⁻⁴ moles NaClO

As molar mass of NaClO is 74.44g/mol, mass of 2.121x10⁻⁴ moles are:

2.121x10⁻⁴ moles ₓ (74.44g / mol) = <em>0.0158g of NaClO</em>

As mass of bleach is 0.496g, mass percent is:

0.0158g NaClO / 0.496g bleach ₓ 100 =

<h3>3.18 (w/w) % </h3>
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zalisa [80]

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Half-life of the first order reaction is 20 min. Rate constant can be calculated as follows:

K=\frac{0.6932}{t_{1/2}}=\frac{0.6932}{20 min}=0.03466 min^{-1}

The rate expression for first order reaction is as follows:

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7 0
3 years ago
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The reaction between butanoic acid and ethanol is:

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Answer:

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Explanation:

8.647 + 45.969

or rewrite for easier look:

45.969 +

 8.647 =

54.616

Hope this helped :3

 

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