Molar mass of LiBr ( lithium bromide ) = <span>86.845 g/mol
</span><span>Amount of calculation to moles of lithium bromide (LiBr)
86.845 g ------------- 1 mole
100 g ----------------- ?
n = m / M
n = 100 / 86.845
n = 1.1514 moles of LiBr
V ---------------------1.1514 moles
1 L -------------------- 4 M
1.1514 * 1 / 4
V = 1151.4 / 4
V = 0,28785 L
hope this helps!
hope this helps !</span>
D to E
Is the answer that you're looking for
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:
Explanation:
2Al(s) + 3 2 O2(g) → Al2O3(s) And given the stoichiometry ...and EXCESS dioxygen gas...we would get 6.25⋅ mol of alumina. the which represents a mass... ...6.25 ⋅ mol ×101.96 ⋅ g ⋅ mol−1 molar mass of alumina ≡ 637.25 ⋅ g.
