Answer:
The answer to your question is 75%
Explanation:
Data
Theoretical production = 4 moles
Experimental production = 3 moles
Percent yield = ?
Formula
Substitution
Result
Percent yield = 75 %
Answer:
They have electrons in their 3d- and 4s-orbital for bond formation.
Explanation:
d- metals or transition metal are metal which form ion with partially filled d-orbital. Examples are iron and manganese.
The metals have 2 electrons in their 4s orbital. If only this is used for bonding, they will form compounds where they have oxidation State of +2 as seen in MnO.
If two 4s and one of 3d electrons are used, oxidation state of +3 is formed as seen in FeCl3.
If two 2s electron I used with two 3d electrons, compound with oxidation state of +4 is formed as seen in MnO2
Answer:
Rb: [Kr] 5s
Step-by-step explanation:
Rb is element 37, the first element in Period 5.
It has one valence electron, so its valence electron configuration is 5s.
The noble gas configuration uses the symbol of the previous noble gas as a shortcut for the electron configurations of the inner electrons.
The preceding noble gas is Kr, so the electron configuration is Rb: [Kr] 5s.
Answer:
Empirical formula is CH₄
Molecular formula = C₂H₈
Explanation:
Mass of carbon = 37.5 g
Mass of hydrogen = 12.5 g
Molecular weight = 32 g/mol
Molecular formula = ?
Empirical formula = ?
Solution:
Number of gram atoms of C = 37.5 g /12g/mol = 3.125
Number of gram atoms of H = 12.5 g / 1.008 g/mol= 12.4
Atomic ratio:
C : H
3.125/3.125 : 12.4 /3.125
1 : 4
C : H : = 1 : 4
Empirical formula is CH₄
Molecular formula:
Molecular formula = n (empirical formula)
n = molar mass of compound / empirical formula mass
n = 32 / 16
n = 2
Molecular formula = n (empirical formula)
Molecular formula = 2 ( CH₄)
Molecular formula = C₂H₈
The two forms of oxygen, O2 and O3 is "<span>They have different molecular structures and different properties."</span>
