Answer:
(a) The proportion of dry air bypassing the unit is 14.3%.
(b) The mass of water removed is 1.2 kg per 100 kg of dry air.
Explanation:
We can express the proportion of air that goes trough the air conditioning unit as 
and the proportion of air that is by-passed as 
, being 
.
The amount of water that goes into the drier inlet has to be 0.004 kg/kg, and can be expressed as:
Replacing the first equation in the second one we have
(b) Of every kg of dry air feed, 85.7% goes in to the air conditioning unit.
It takes (0.016-0.002)=0.014 kg water per kg dry air feeded.
The water removed of every 100 kg of dry air is
It can also be calculated as the difference in humiditiy between the inlet and the outlet: (0.016-0.004=0.012 kgW/kDA) and multypling by the total amount of feed (100 kgDA).
100 * 0.012 = 1.2 kgW
<h3>
<u>Answer;</u></h3>
= 930.23 mL
<h3><u>Explanation</u>;</h3>
Using the combined gas law;
P1V1/T1 = P2V2/T2
Where; P1 = 600 kPa, V1 = 800 mL, and T1 = -25 +273 = 258 K, and
V2= ?, P2 = 1000 kPa, and T2 = 227 +273 = 500 K
Thus;
V2 = P1V1T2/T1P2
= (600 ×800 ×500) / (258 × 1000)
= 930.23 mL
Answer:
The answer is
<h2>19.31 g/cm³</h2>
Explanation:
The density of a substance can be found by using the formula
From the question
mass of gold = 475.09g
volume = 24.6 cm³
The density of the gold is
We have the final answer as
<h3>19.31 g/cm³</h3>
Hope this helps you
68.7g
molar mass of Ba(137.327)x (.500)= 68.6635
.500 has 3 sig figs.
Answer:
Si.
Explicación:
Sí, podemos predecir si una solución acuosa será un buen conductor de corriente eléctrica si conocemos el soluto que se disuelve en ese líquido y su concentración o cantidad. Hay algunas sustancias que se ionizan cuando se agregan en un líquido como el agua. Debido a esta ionización, será un buen conductor de electricidad. Por ejemplo, la adición de una gran cantidad de cloruro de sodio en agua ioniza y hace que el agua sea un buen medio conductor de electricidad.
