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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The earths moon is most like going to differ in size or even color
Answer:

In which [Ag+] in negligibly small and the concentration of each reactant is 1.0 M
The answer is A) PO43- < NO3- < Na+
Explanation:
Ag+ is removed from the solution just like PO43-, so there are just 2 possible answers at this point: a or b. Then we can notice that Na3PO4 releases 3 moles of Na+ and just 1 mole of NO3-
We have 100mL of each reactant with the same concentration for both (1.0 M) so:
(0.1)(1)(3)= 0.3 mol Na+
(0.1)(1)= 0.1 mol NO3-
so PO43- < NO3- < Na+
Answer:
The answer to your question is: 6.55 x 10 ²³ atoms of Br
Explanation:
CH2Br2 = 37.9 g
MW CH2Br2 = (12 x 1) + (2 x 1) + (80 x 2) = 174 g
174 g of CH2Br2 ------------------ 160 g of Br2
37.9 g of CH2Br2 --------------- x
x = 37.9 x 160/174 = 34.85 g of Br
1 mol of Br ----------------- 160 g Br2
x ---------------- 174 g Be2
x = 174 x 1 /160 = 1.088 mol of Br2
1 mol of Br ----------------- 6.023 x 10 ²³ atoms
1.088 mol of Br ------------- x
x = 1.088 x 6.023 x 10 ²³ / 1 = 6.55 x 10 ²³ atoms
Answer : The amount of oxygen gas collected are, 0.217 mol
Explanation :
Using ideal gas equation :

where,
P = pressure of gas =
(1 atm = 760 torr)
V = volume of gas = 5 L
T = temperature of gas = 
n = number of moles of gas = ?
R = gas constant = 0.0821 L.atm/mol.K
Now put all the given values in the ideal gas equation, we get:


Thus, the amount of oxygen gas collected are, 0.217 mol