It would be sadness because people have been going thru alot of things over the years
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
Answer:
4.9 × 10²³ molecules
Explanation:
Given data:
Number of molecules = ?
Number of moles of oxygen = 0.815 mol
Solution:
The given problem will solve by using Avogadro number.
It is the number of atoms , ions and molecules in one gram atom of element, one gram molecules of compound and one gram ions of a substance.
The number 6.022 × 10²³ is called Avogadro number.
1 mole = 6.022 × 10²³ molecules
0.815 mol × 6.022 × 10²³ molecules / 1 mol
4.9 × 10²³ molecules
oxygen -17 and oxygen-18 are isotopes of oxygen
<em><u>Explanation</u></em>
Isotope is two or more forms of the same element that has equal amount of proton but different number of neutrons in the nuclei.
oxygen -17 and oxygen-18 are isotope of oxygen since they have the equal amount of protons but different number of neutrons in their nuclei.
that is oxygen 17 has 8 protons and 9 neutrons while oxygen 18 has 8 protons and 10 neutrons.
Answer:
The correct answer is 1.33 x 10⁻⁵ M
Explanation:
The concentration of the stock solution is: C= 1.33 M
In the first dilution, the student added 1 ml of stock solution to 9 ml of water. The total volume of the solution is 1 ml + 9 ml = 10 ml. So, the first diluted concentration is:
C₁= 1.33 M x 1 ml/10 ml = 1.33 M x 1/10 = 0.133 M
The second dilution is performed on C₁. The student added 1 ml of 0.133 M solution to 9 ml of water. Again, the total volume is 1 ml + 9 ml = 10 ml. The second diluted concentration is:
C₂= 0.133 M x 1 ml/10 ml = 0.133 M x 1/10= 0.0133 M
Since the student repeated the same dilution process 3 times more (for a total of 5 times), we have to multiply 5 times the initial concentration by the factor 1/10:
Final concentration = initial concentration x 1/10 x 1/10 x 1/10 x 1/10 x 1/10
= initial concentration x (1/10)⁵
= 1.33 M x 1 x 10⁻⁵
= 1.33 x 10⁻⁵ M
