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Blizzard [7]
3 years ago
13

5. _______ with the largest atoms and the smallest number of valence electrons and _______ with the smallest atoms and the great

est number of valence electrons are the most reactive.
Chemistry
2 answers:
kogti [31]3 years ago
7 0

We have to fill the gap

The correct answer is: Metals with the largest atoms and the smallest number of valence electrons and non-metals with the smallest atoms and the greatest number of valence electrons are the most reactive.

Reactivity of any species depends on ease of release or gain of electron. The species which can release electron easily from valence shell, the species is more reactive whereas the species which can accept electron to its valence shell easily is also more reactive.


Olegator [25]3 years ago
6 0

Explanation:

Metals are the substances that contain less number of valence electrons. Whereas as non-metals are the substances that contain more number of valence electrons.

Also, when we move down a group then there occurs increase in size of atoms due to addition of electrons.

So, outer shell of atoms move away from the nucleus. Hence, force of attraction between nucleus and valence electrons of metals decreases. So, metals are able to lose an electron easily with increase in size.

On the other hand, non-metals find it difficult to attract an electron when increase in size of atom occurs.

Thus, we can conclude that metals with the largest atoms and the smallest number of valence electrons and non-metals with the smallest atoms and the greatest number of valence electrons are the most reactive.

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Answer:

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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
madreJ [45]

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})

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}}

                  \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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