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yuradex [85]
4 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]4 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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Here's what I get  

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Thus, the removal of two electrons from Cu gives you Cu²⁺.

I found the ionization energies of Cu and used them to create those of Cu²⁺ (see table and graph below).

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2.6(b) A sample of 2.00 mol CH3OH(g) is condensed isothermally and reversibly to liquid at 64°C. The standard enthalpy of vapori
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Answer:

The value of W is 5.602 kJ, Q is -70.6 kJ, change in U is -65 kJ, and change in H is -70.3 kJ.

Explanation:

Based on the given information, the mass of CH3OH given is 64 grams, which is condensed isothermally and reversibly to liquid at 64 degrees C. The given standard enthalpy of vaporization of methanol at 64 degrees C is 35.3 kJ per mole.

The moles of CH3OH can be determined by using the formula,  

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= 64.0 grams / 32.0 grams per mole

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In condensation heat is given off, thus, it is an exothermic process, hence, q will be -70.6 kJ

The work or W can be calculated by using the formula,  

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= 55.3 L

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W = 5.602 kJ

Now U can be calculated using the formula,  

U = q + W

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Thus, q = -70.6 kJ, W = 5.602 kJ, U = -65 kJ, and ΔH = -70.3 kJ.  

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