Answer:
68,2%
Explanation:
Supposing the initial salt concentration of lake Parsons is the same of non-isolated lakes, 6,67L, and the change of salt concentration in isolated lake is just for water evaporation it is possible to write:
6,67gL⁻¹×X = 21gL⁻¹×Y
<em>-Where X is the initial water and Y is the water that remains in the isolated lake-</em>
Thus:
6,67X = 21Y
0,318 = Y/X
0,318 is the ratio of water that remains between total water. To obtain the ratio of evaporated water:
1-0,318 = 0,682
In percentage: <em>68,2%</em>
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I hope it helps!
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Answer:
select all of them except they are the biggest
Answer:
1223.38 mmHg
Explanation:
Using ideal gas equation as:
![PV=nRT](https://tex.z-dn.net/?f=PV%3DnRT)
where,
P is the pressure
V is the volume
n is the number of moles
T is the temperature
R is Gas constant having value = ![62.3637\text{ L.mmHg }mol^{-1}K^{-1}](https://tex.z-dn.net/?f=62.3637%5Ctext%7B%20L.mmHg%20%7Dmol%5E%7B-1%7DK%5E%7B-1%7D)
Also,
Moles = mass (m) / Molar mass (M)
Density (d) = Mass (m) / Volume (V)
So, the ideal gas equation can be written as:
![PM=dRT](https://tex.z-dn.net/?f=PM%3DdRT)
Given that:-
d = 1.80 g/L
Temperature = 32 °C
The conversion of T( °C) to T(K) is shown below:
T(K) = T( °C) + 273.15
So,
T = (32 + 273.15) K = 305.15 K
Molar mass of nitrogen gas = 28 g/mol
Applying the equation as:
P × 28 g/mol = 1.80 g/L × 62.3637 L.mmHg/K.mol × 305.15 K
⇒P = 1223.38 mmHg
<u>1223.38 mmHg must be the pressure of the nitrogen gas.</u>