Al₂O has a higher percentage of aluminium.
Explanation:
To solve this problem, we have to compare the molar mass of the aluminium in each compound to one another as percentage of the whole compound:
Molar mass of Al₂O = 2(27) + 16 = 70g/mol
Molar mass of Al(NO₃)₃ = 27 + 3[14 + 3(16)] = 213g/mol
Percentage by mass of Al in Al₂O =
x 100 = 77%
Percentage by mass of Al in Al(NO₃)₃ =
x 100 = 12.7%
Al₂O has a higher percentage of aluminium.
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Answer:
Chemical equations need to be balanced so that they follow the law of conservation of mass.

If every oxygen ion is combined with an aluminium ion has a charge of -2,the charge of each aluminum ion would be -3.
Uncharged Aluminum atom must need to lose it's electrons,in order to form the bond with oxygen which has vacant orbitals
ion
atom that has a positive or negative charge because it lost or gained one or more electrons
chemical bond
the attractive force that holds atoms or ions together
ionic bond
a chemical bond in which one atom loses an electron and the other atom gains electrons to form ions
chemical formula
a combination of chemical symbols and numbers to represent a substance
covalent bond
bond formed by the sharing of electrons between atoms
Answer:
110 degree
Explanation:
This is because Hybridization of an s orbital with all three p orbitals (px , py, and pz) results in four sp3 hybrid orbitals. sp3 hybrid orbitals are oriented at bond angle of 109.5 degrees from each other. This 109.5 degrees gives an arrangement of tetrahedral geometry
The mass defect for the isotope thorium-234 if given mass is 234.04360 amu is 1.85864 amu.
<h3>How do we calculate atomic mass?</h3>
Atomic mass (A) of any atom will be calculated as:
A = mass of protons + mass of neutrons
In the Thorium-234:
Number of protons = 90
Number of neutrons = 144
Mass of one proton = 1.00728 amu
Mass of one neutron = 1.00866 amu
Mass of thorium-234 = 90(1.00728) + 144(1.00866)
Mass of thorium-234 = 90.6552 + 145.24704 = 235.90224 amu
Given mass of thorium-234 = 234.04360 amu
Mass defect = 235.90224 - 234.04360 = 1.85864 amu
Hence required value is 1.85864 amu.
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