Answer:
ewewr2qq2we4e2eq2we4e2eq2we4e2eq2we4e2eq2we4e2eq2we4e2eq2we4e2eq2we4e2eq2
Explanation:
asw342253463421q412qq24q2asw342253463421q412qq24q2asw342253463421q412qq24q2asw342253463421q412qq24q2asw342253463421q412qq24q2asw342253463421q412qq24q2asw342253463421q412qq24q2asw342253463421q412qq24q2asw342253463421q412qq24q2asw342253463421q412qq24q2asw342253463421q412qq24q2asw342253463421q412qq24q2asw342253463421q412qq24q2asw342253463421q412qq24q2asw342253463421q412qq24q2asw342253463421q412qq24q2
Answer:
Higher molar mass compounds will be less soluble than lower molar mass molecules of the same type.
Explanation:
Bigger Mass = slower/less soluble
Small Mass = faster/more soluble
Increases mass by 1 amu. Neutrons have mass.
Answer:
It obeys rule
Explanation:
Benzene is an aromatic hydrocarbon because it obeys Hückel's rule.
Types of Bonds can be predicted by calculating the
difference in electronegativity.
If, Electronegativity difference is,
Less
than 0.4 then it is Non Polar Pure Covalent
Between 0.4 and 1.7 then it is Polar Covalent
Greater than 1.7 then it is Ionic
For Br and Br,
E.N of Bromine = 2.96
E.N of Bromine = 2.96
________
E.N Difference
0.00 (Non Polar/Pure Covalent)
For N and O,
E.N of Oxygen = 3.44
E.N of Nitrogen = 3.04
________
E.N Difference
0.40 (Non Polar/Pure Covalent)
For P and H,
E.N of Hydrogen = 2.20
E.N of Phosphorous = 2.19
________
E.N Difference 0.01 (Non Polar/Pure Covalent)
For K and O,
E.N of Oxygen = 3.44
E.N of Potassium = 0.82
________
E.N Difference 2.62 (Ionic)
