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shtirl [24]
3 years ago
6

Complete the sentence. When an acid is added to water, hydroxide ions __________.

Chemistry
2 answers:
kenny6666 [7]3 years ago
8 0

Answer:

The hydroxide ions decrease.

Explanation:

I got it right on the quiz. This is what I saw. Read this, "Adding water to an acid or base will change its pH. Water is mostly water molecules so adding water to an acid or base reduces the concentration of ions in the solution. When an acidic solution is diluted with water the concentration of H + ions decreases and the pH of the solution increases towards 7."

Hope this helps! Tell me if this is wrong just incase.

yarga [219]3 years ago
4 0

Answer:

decrease or stay the same idk but i'm pretty sure its decrease

Explanation:

When an acid dissolves in a solution like water the hydrogen ions increase

When a base dissolves in a solution the hydroxide ions increase

it is not the first one from the evidence shown above

it is not the last one because hydroxide could not just disappear well maybe if you waited for 1,000 years or if you added so much acid there was no water left but they simply say added to water

<u>the reason i think its </u><u>decrease instead of stay the same</u><u> is because as the hydrogen ions increase shouldn't the hydroxide ions decrease since an acid is acidic and that can dissolve metals.</u>

have a nice day!

<u />

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Air at 25°C with a dew point of 10 °C enters a humidifier. The air leaving the unit has an absolute humidity of 0.07 kg of water
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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

At the dew point temperature, the first drop of water condenses out of air and then,

        Partial pressure of water vapor (P_{a}) = vapor pressure of water at a given temperature (P^{s}_{a})

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

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

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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}}

                  \frac{1.18624}{101.325 - 1.18624} \times \frac{18 kg H_{2}O}{29 \text{kg dry air}}

                 = 0.00735 kg H_{2}O/ kg dry air

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.

                 0.07 kg - 0.00735 kg

              = 0.06265 kg H_{2}O for every 1 kg dry air

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

                0.06265 kg \times 100

                  = 6.265 kg

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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