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Leviafan [203]
3 years ago
15

The ease with which a raw material can be molded , flattened , or bent is known as its ?

Chemistry
2 answers:
Leya [2.2K]3 years ago
6 0
A. Malleability

Ductility is being able to be drawn into wires
Elasticity is being able to resist stress
Resilience is being able to spring back into shape (kind of like elasticity)
Angelina_Jolie [31]3 years ago
4 0

Answer:

Malleability

Explanation:

Malleability is the capability of a substance, especially a metal, to be reshaped or molded into a different shape. For people in the field of chemistry, the malleability of a metal provides an important means of describing the specific features of a metal and relating it to the arrangement of the atoms within the metal.

Malleability in metals happens due to the metallic bonds that keep the atoms in place. Metallic bonds, is made up of  a 'sea' of electrons that easily move from atom to another, allow the metal atoms to slide past each other if a force is applied. The force can come from a blow from a hammer, the impact from a fall, high pressure from being squeezed, or from a collision.

The degree of malleability differs widely among metals as well as mixtures of different metals, also known as alloys. Multiple factors can affect the malleability of a metal or alloy, but two fundamentally important factors are the strength of the metallic bond and the temperature.

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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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        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

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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}}

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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.

                 0.07 kg - 0.00735 kg

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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

                  = 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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