Explanation:
It is known that 1 gram contains 1000 milligrams. And, mathematically we can represent it as follows.
or 
So, when we have to convert grams into milligrams then we simply multiply the digit with 1000. And, if we have to convert a digit from milligrams to grams then we simply divide it by 1000.
Answer:
1.505×10^23 atoms of lead
Explanation:
Volume of lead in the lungs = total volume of lungs = 5.60L
1 mole = 22.4L
5.6L of lead = 5.6/22.4 = 0.25 mole
From Avogadro's law
1 mole of lead contains 6.02×10^23 atoms of lead
0.25 mole of lead = 0.25×6.02×10^23 = 1.505×10^23 atoms of lead
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>
3.84 - 1.43 = 2.41
2.41g of table sugar
% mass = ( (mass of element) / (total mass) ) * 100
% mass = (2.41 / 3.84) * 100
% mass = (0.6276) * 100
% mass = 62.76
62.76%
The product of a reaction between these two elements is 
.
Explanation:
The oxidation state of an ion in a compound is equal to its charge.
The aluminum having a charge of +3 because oxidation state is +3
The oxide is having charge of -2
The product of these reactants will produce a chemical compound.
The compound formed is i.e Aluminium oxide. The compound while getting formed will share the charge and cation A+ will have the charge of anion and anion will have the charge of cation. This will result in a compound as there should be a neutral charge on the compound formed.
The <em>+</em><em>3 charge of the cation Al+ will go to anion oxide O2- and the charge of anion -2 will go with cation Al+. </em>
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