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kirill [66]
3 years ago
15

5.00 ml of commercial bleach was diluted to 100.0 ml. 25.0 ml of the diluted sample was titrated with 4.56 ml of 0.100 m s2o3 2-

. what is the concentration of hypochlorite in the original bleach solution? assume the density of the commercial bleach is 1.08 g/ml. calculate the average percent by mass of naclo in the commercial bleach.
Chemistry
1 answer:
aev [14]3 years ago
3 0

The solution would be like this for this specific problem:

 

4 NaOCl + S2O3{2-} + 2 OH{-} → 2 SO4{2-} + H2O + 4 NaCl

<span>(0.00456 L) x (0.100 mol/L S2O3{2-}) x (4 mol NaOCl / 1 mol S2O3{2-}) x (100.0 mL / 25 mL) x </span><span>
<span>(74.4422 g NaClO/mol) = 0.54313 g </span></span>

<span>(5.00 mL) x (1.08 g/mL) = 5.40 g solution </span>

(0.54313 g) / (5.40 g) = 0.101 = 10.1%

 

So, the average percent by mass of NaClO in the commercial bleach is 10.1%.

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Answer:

1.4 g/cm3

Explanation:

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Mass = 21g

Volume = 15cm3

Density = 21/15 = 1.4

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Plants contain xylem and phloem tissues for carrying water and food. What organ system in animals performs a similar function as
Nuetrik [128]

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3 0
3 years ago
What is the product if an atom of Po-209<br> undergoes alpha decay?
g100num [7]
<h3>Answer:</h3>

Lead-205 (Pb-205)

<h3>Explanation:</h3>

<u>We are given;</u>

  • An atom of Po-209

We are supposed to identify its product after an alpha decay;

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  • When an element undergoes an alpha decay, the mass number decreases by 4 while the atomic number decreases by 2.
  • Therefore, when Po-209 undergoes alpha decay it results to the formation of a product with a mass number of 205 and atomic number of 82.
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7 0
4 years ago
He rate constant of a reaction is 4.55 × 10−5 l/mol·s at 195°c and 8.75 × 10−3 l/mol·s at 258°c. what is the activation energy o
Xelga [282]

Answer : The activation energy of the reaction is, 17.285\times 10^4kJ/mole

Solution :  

The relation between the rate constant the activation energy is,  

\log \frac{K_2}{K_1}=\frac{Ea}{2.303\times R}\times [\frac{1}{T_1}-\frac{1}{T_2}]

where,

K_1 = initial rate constant = 4.55\times 10^{-5}L/mole\text{ s}

K_2 = final rate constant = 8.75\times 10^{-3}L/mole\text{ s}

T_1 = initial temperature = 195^oC=273+195=468K

T_2 = final temperature = 258^oC=273+258=531K

R = gas constant = 8.314 kJ/moleK

Ea = activation energy

Now put all the given values in the above formula, we get the activation energy.

\log \frac{8.75\times 10^{-3}L/mole\text{ s}}{4.55\times 10^{-5}L/mole\text{ s}}=\frac{Ea}{2.303\times (8.314kJ/moleK)}\times [\frac{1}{468K}-\frac{1}{531K}]

Ea=17.285\times 10^4kJ/mole

Therefore, the activation energy of the reaction is, 17.285\times 10^4kJ/mole

8 0
3 years ago
Read 2 more answers
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sladkih [1.3K]
Heat of vaporization of water will be required as water is already at it's boiling point thus heat required will be 540*10=5400 cal
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3 years ago
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