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san4es73 [151]
3 years ago
15

Suppose that a student doing this experiment ran into some procedural and calculation difficulties. State the effect each of the

following would have on the calculated value of R determined from this experiment. Briefly explain. (1) After the reaction was completed, the student failed to withdraw the rod to its original position before taking the second buret reading. (2) The student neglected to take into account the pressure of water vapor in the system when doing the calculations. (3) The hydrogen peroxide solution had decomposed slightly after standing for a while, so its concentration
Chemistry
2 answers:
Bingel [31]3 years ago
5 0
The only one i can make sense out of would be: <span>
(1) "</span><span>After the reaction was completed, the student failed to withdraw the rod to its original position before taking the second buret reading." </span><em>this would affect the volume, since the rod is submerged in the solution, it would affect the reading (it would appear as a greater volume).</em>
VLD [36.1K]3 years ago
4 0

Answer:

1.- The volume will be incorrect.

2.- The volume will be incorrect.

3.-  will decrease.

Explanation:

1.- If the rod is not in the correct position, the volume inside the buret will continue to leak, so the reading of the volume that is suppose to be show the equilibrium will be incorrect.

2.- The pressure of water vapor will be different depending the environmental pressure, in this way, if the student do not check either the temperature or the pressure, the calculations will be different, because the liquids behave different at different temperatures.

3.- If a substance get decomposed after a while, the concentration starts to decrease, this is important for some processes like doing hair dyes, in which the amount of hydrogen peroxide is to important.

Hope this info is useful

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Explanation:

The given data is as follows.

         Temperature of dry bulb of air = 25^{o}C

          Dew point = 10^{o}C = (10 + 273) K = 283 K

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Using Antoine's equation we get the following.

            ln (P^{s}_{a}) = 16.26205 - \frac{3799.887}{T(K) - 46.854}

            ln (P^{s}_{a}) = 16.26205 - \frac{3799.887}{283 - 46.854}

                               = 0.17079

                   P^{s}_{a} = 1.18624 kPa

As total pressure (P_{t}) = atmospheric pressure = 760 mm Hg

                                   = 101..325 kPa

The absolute humidity of inlet air = \frac{P^{s}_{a}}{P_{t} - P^{s}_{a}} \times \frac{18 kg H_{2}O}{29 \text{kg dry air}}

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Hence, air leaving the humidifier has a has an absolute humidity (%) of 0.07 kg H_{2}O/ kg dry air.

Therefore, amount of water evaporated for every 1 kg dry air entering the humidifier is as follows.

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Hence, calculate the amount of water evaporated for every 100 kg of dry air as follows.

                0.06265 kg \times 100

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Thus, we can conclude that kg of water the must be evaporated into the air for every 100 kg of dry air entering the unit is 6.265 kg.

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