I think the answer is C for this question
The balanced chemical reaction is written as:
4Al + 3O2 = 2Al2O3
To determine the mass of oxygen gas that would react with the given amount of aluminum metal, we use the initial amount and relate this amount to the ratio of the substances from the chemical reaction. We do as follows:
moles Al = 16.4 g ( 1 mol / 26.98 g ) = 0.61 mol Al
moles O2 = 0.61 mol Al ( 3 mol O2 / 4 mol Al ) = 0.46 mol O2
mass O2 = 0.46 mol O2 ( 32.0 g / mol ) = 14.59 g O2
Therefore, to completely react 16.4 grams of aluminum metal we need a minimum of 14.59 grams of oxygen gas.
I'm not 100% sure on this, but I would go with C) NaCl.
NaCl is a salt, and that is used to melt the ice on the roads. Hope this helps!
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>
