The picture shows a model of a cell.
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:- C. Hafnium.
Solution:- Mass of the sample is 46.0 g and it's volume is .
From mass and volume, we can calculate it's density using the formula:
On the basis of the density, this substance could either be mercury or hafnium. Since the substance is a solid at room temperature where as mercury is liquid. So, it can't be mercury.
The right choice is C) Hafnium.