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4.5 moles of carbon monoxide is hoe many grams
1 answer:
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Explanation:
The given data is as follows.
Temperature of dry bulb of air =
Dew point = = (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 = vapor pressure of water at a given temperature
Using Antoine's equation we get the following.
= 0.17079
= 1.18624 kPa
As total pressure = atmospheric pressure = 760 mm Hg
= 101..325 kPa
The absolute humidity of inlet air =
= 0.00735 kg / kg dry air
Hence, air leaving the humidifier has a has an absolute humidity (%) of 0.07 kg / 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 for every 1 kg dry air
Hence, calculate the amount of water evaporated for every 100 kg of dry air as follows.
= 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.
Answer:
Kinetic energy is directly proportional to mass
Explanation:
Kinetic energy is directly proportional to the mass of an object and also directly proportional to the square of the velocity of that object:
Notice that if we keep velocity constant and only increase the mass of a object, the kinetic energy of that object would increase, as we've already emphasized the direct relationship between the kinetic energy term and the mass term.
Let's take a simple example: assume that object 1 and object 2 are both moving at the same velocity but object 1 has a much lower mass than object 2. According to the equation, object 1 has lower kinetic energy. This object can then transform all of its kinetic energy into some other form, say, heat the ground. The heat transferred will be significantly lower than by the object 2 moving at the same velocity but having a much greater mass.