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>
The ionic equation is as below
Ca^2+(aq) + SO4^2-(aq) ---> CaSO4(s)
EXPLANATION
K2SO4(aq) +Cai2(aq) ---> CaSO4(s) + Ki (aq)
ionic equation
= 2K^+(aq) + SO4^2-(aq) + Ca^2+(aq) + 2i^-(aq) --->CaSO4(s) + 2K^+(aq) +2 i^-(aq)
cancel the spectator ions that is 2k^+ and 2i^-
The net ionic equation is therefore
= Ca^2+(aq) + SO4^2-(aq) ----> CaSO4(s)
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Answer:
The answer to your question is T1 = 384.7 °K
Explanation:
Data
Volume 1 = V1 = 45.7 l
Temperature 1 = T1 = ?
Volume 2 = V2 = 33.9 l
Temperature 2 = T2 = 12.4°C
To solve this problem use Charles' law
V1/T1 = V2/T2
T1 = V1T2/V2
-Convert temperature to °K
T2 = 12.4 + 273 = 285.4°K
-Substitution
T1 = (45.7 x 285.4) / 33.9
-Simplification
T1 = 13042.8 / 33.9
-Result
T1 = 384.7 °K
Answer:
The amount of NaF produced is doubled.
(d) is correct option.
Explanation:
Given that,
A 2 mole sample of F₂ reacts with excess NaOH according to the equation.
The balance equation is

If the reaction is repeated with excess NaOH but with 1 mole of F₂
The balance equation is

Hence, The amount of NaF produced is doubled.
(d) is correct option.