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:
25.99mL is the volume internal volume of the flask
Explanation:
<em>To complete the question:</em>
<em>The temperature of the water was measured to be 21ºC. Use this data to find the internal volume of the stoppered flask</em>
<em />
The flask was filled with water, that means the internal volume of the flask is equal to the volume that the water occupies.
To find the volume of the water you need to find the mass and by the use of density of water at 21ºC (0.997992g/mL), you can find the volume of the flask, thus:
Mass water = Mass filled flask - Mass of clean flask
Mass water = 60.167g - 34.232g
Mass water = 25.935g of water.
To convert this mass to volume:
25.935g × (1mL / 0.997992g) =
<h3>25.99mL is the volume internal volume of the flask</h3>
Answer:
Explanation:
The following equation relates to Born-Haber cycle
Where
is enthalpy of formation
S is enthalpy of sublimation
B is bond enthalpy
is ionisation enthalpy of metal
is electron affinity of non metal atom
is lattice energy
Substituting the given values we have
-435.7 = 79.2 + 1/2 x 242.8 + 418.7 - 348 +U_L
= - 707 KJ / mol
Answer:
C₂H₂O₃
Explanation:
The empirical formula of a compound is derived bu finding the whole ratios of the constituent elements.
In succinic acid, the ratios of carbon to hydrogen to oxygen is calculated as follows:
<u>% mass</u>
Carbon- 40.60
Hydrogen - 5.18
Oxygen - 54.22
<u>RAM</u>
Carbon -12
Oxygen - 15.994
Hydrogen -1.008
<u>No of moles elements in the compound</u>
Carbon = 40.60/12=3.3833
Oxygen = 54.22/15.994= 3.39
Hydrogen= 5.18/1.008 = 5.1389
Mole ratios of the individual elements we divide by the smallest value of the number of moles.
Carbon: Hydrogen : Oxygen
3.3833/3.3833:3.39/3.3833:5.1389/3.3833
=1:1:1.5
We can multiply the value by 2 to get the whole number ratio.
=2:2:3
The empirical formula will be:
C₂H₂O₃
The answer is 200 g.
If the molar mass of CaCl2 is 110.98 g/mol, this means there are 110.98 g in 1 L of 1 M solution.
Let's find how many g of CaCl2 are present in 0.720 M.
110.98 g : 1 M = x : 0.720 M
x = 110.98 g * 0.720 M : 1 M
x = 79.90 g
So there are 79.90 g in 0.720 M. In other words, in 1 L of 0.720 M solution there will be 79.90 g.
Now, we need to prepare ten beakers with 250 mL of solutions:
10 * 250 mL = 2500 mL = 2.5 L
79.90 g : 1 L = x : 2.5 L
x = 79.90 g * 2.5 L : 1 L
x = 199.75 g ≈ 200 g
