To counter the removal of A the equilibrium change by <u>s</u><em>hifting toward the left</em>
<em> </em><u><em>explanation</em></u>
<u><em> </em></u>If the reaction is at equilibrium and we alter the condition a new equilibrium state is created
<u><em> </em></u>The removal of A led to the shift of equilibrium toward the left since it led to less molecules in reactant side which favor the backward reaction.( equilibrium shift to the left)
Answer:
Explanation:
Molarity of acid(volume of acid)(# of H ions)= molarity of base(volume of base)(# of OH ions)
M(v)(#)=M(v)(#)
sulfuric acid sodium hydroxide
H2SO4 NaOH
(3)(11.6)(2)=M(25)(1)
M=2.784
First we need to find the number of moles of both K and O reacted
K - 0.779 g / 39 g/mol
= 0.02 mol
the mass of O₂ reacted = 1.417 g - 0.779 g = 0.638 g
O₂ moles = 0.638 g / 32 g/mol
= 0.02 mol
the number of both K and O₂ moles reacted are equal
therefore stoichiometry of K to O₂ reacted are 1:1
then the formula of potassium superoxide is KO₂
It's a because if you add them together you till get 1.40
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>
