0.370 mol metal oxide = 55.45 g
<span>1 mol = 55.45/0.370 = 149.86 g </span>
<span>in 1 mol there are 3 mol O = 16 * 3 = 48 g of O </span>
<span>there is 48/149.86 * 100% O in the sample </span>
<span>the sample has 48/149.86 * 0.370 = 0.119 g O</span>
Answer:
D
Explanation:
We must study the reaction pictured in the question closely before we begin to attempt to answer the question.
Now, the reaction is a free radical reaction. This implies that only one electron is transferred. The transfer of one electron is shown using a half arrow rather than a full arrow. The both species are radicals (odd electron species) and contribute one electron each.
Hence we must show electron movements in both species using a half arrow.
Lead (II) acetate trihydrate is the systematic name for the formula Pb(C₂H₃O₂)₂ . 3H₂O.
<h3>What is Molecular Formula ?</h3>
The chemical formula that gives total number of atoms of each element in one molecule of a compound is called Molecular Formula.
<h3>What is Oxidation State ?</h3>
Oxidation state is also known as oxidation number. It is defined as the atom is equal to the total number of electrons which have been removed from the element in order to form chemical bond with other atom.
Now find the oxidation state of Pb in Pb(C₂H₃O₂)₂ .3H₂O
Assume the oxidation state of Pb in Pb(C₂H₃O₂)₂ .3H₂O be x
x + 2 × (-1) + 3 × 0 = 0
x - 2 + 0 = 0
x = 2
Oxidation state of Pb is +2 or (II)
Thus from the above conclusion we can say that The systematic name for the formula Pb(C₂H₃O₂)₂ .3H₂O is Lead (II) acetate trihydrate.
Learn more about the Molecular Formula here: brainly.com/question/15960587
#SPJ4
Answer:
1 in. =2.54 cm
3*(10^-4)*2.54=7.62*10^-4
Explanation:
When this happens, light changes speed and the light ray bends, either toward or away from what we call the normal line, an imaginary straight line that runs perpendicular to the surface of the object.
