Answer:
Explanation:
It will be better to use solvents that are lighter than water, because their density has an influence on the miscibility . This will give you a better separation during extraction.
Answer:
c) 22
Explanation:
Let's consider the following balanced equation.
N₂(g) + 3 H₂(g) ----> 2 NH₃(l)
According to the balanced equation, 34.0 g of NH₃ are produced by 1 mol of N₂. For 170 g of NH₃:
According to the balanced equation, 34.0 g of NH₃ are produced by 3 moles of H₂. For 170 g of NH₃:
The total gaseous moles before the reaction were 5.00 mol + 15.0 mol = 20.0 mol.
We can calculate the pressure (P) using the ideal gas equation.
P.V = n.R.T
where
V is the volume (50.0 L)
n is the number of moles (20.0 mol)
R is the ideal gas constant (0.08206atm.L/mol.K)
T is the absolute temperature (400.0 + 273.15 = 673.2K)
The epicenter was located somewhere on a circle centered at Recording station X, with a radius of 250 km.<span>
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Answer:
11482 ppt of Li
Explanation:
The lithium is extracted by precipitation with B(C₆H₄)₄. That means moles of Lithium = Moles B(C₆H₄)₄. Now, 1 mole of B(C₆H₄)₄ produce the liberation of 4 moles of EDTA. The reaction of EDTA with Mg²⁺ is 1:1. Thus, mass of lithium ion is:
<em>Moles Mg²⁺:</em>
0.02964L * (0.05581mol / L) = 0.00165 moles Mg²⁺ = moles EDTA
<em>Moles B(C₆H₄)₄ = Moles Lithium:</em>
0.00165 moles EDTA * (1mol B(C₆H₄)₄ / 4mol EDTA) = 4.1355x10⁻⁴ mol B(C₆H₄)₄ = Moles Lithium
That means mass of lithium is (Molar mass Li=6.941g/mol):
4.1355x10⁻⁴ moles Lithium * (6.941g/mol) = 0.00287g. In μg:
0.00287g * (1000000μg / g) = 2870μg of Li
As ppt is μg of solute / Liter of solution, ppt of the solution is:
2870μg of Li / 0.250L =
<h3>11482 ppt of Li</h3>
Answer:
SO2(g) and H2S(g)
Explanation:
Sulphuric acid (H2SO4) is very important in industries and is often referred to as the best chemical due to its multiple uses. It is corrosive and used in making fertilizers and explosives. It can also be used to make paints, detergents etc.
Sulphuric acid reacts with Sodium iodide to give Hydrogen iodide and Sodium hydrogen Sulphate.
NaI (s) + H2SO4 (l) =HI (g) + NaHSO4 (s)
The Hydrogen iodide then reacts with the Sulphuric acid to give Iodine,Hydrogen sulphide and water
HI (g) + H2SO4 (l) = 4I2(s) + H2S (g) + 4 H2O (l)
The sulphuric acid can also undergo reduction reaction with hydrogen iodide to give Iodine, Sulphur dioxide and Water.
2 HI (g) + H2SO4 (l) = I2(s) + SO2 (g) + H2O (l)
