Heat has been released or lost. Odor may be present. Absence of a substance may be converted to something else.
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 Covalent
Between 0.4 and 1.7 then it is Polar Covalent
Greater than 1.7 then it is Ionic
For C and N,
E.N of Nitrogen = 3.04
E.N of Carbon = 2.55
________
E.N Difference 0.49 (Weakly Polar Covalent)
For N and S,
E.N of Nitrogen = 3.04
E.N of Sulfur = 2.58
________
E.N Difference 0.46 (Weakly Polar Covalent)
For K and F,
E.N of Fluorine = 3.98
E.N of Potassium = 0.28
________
E.N Difference 3.70 (Ionic)
For N and N,
E.N of Nitrogen = 3.04
E.N of Nitrogen = 3.04
________
E.N Difference 0 (Non Polar Covalent)
Answer: electrons are arranged in shells around an atom's nucleus.
Explanation:
Answer:
LiOH (aq) + VCl3(aq) ---> LiCl(aq) + V(OH)3(s) - unbalanced
3LiOH (aq) + VCl3(aq) ---> 3LiCl(aq) + V(OH)3(s) - balanced
3Li+OH- (aq) + V3+(Cl-)3(aq) ---> 3Li+Cl-(aq) + V3+(OH-)3(s) - showing ions
3Li+(aq) + 3OH- (aq) + V3+(aq) + 3Cl-(aq) ---> 3Li+(aq) + 3Cl-(aq) + V3+(OH-)3(s) (some courses don't show the charges in insoluble ionic compounds - so V(OH)3(s)) - Showing soluble ionic compounds as separate ions.
3OH- (aq) + V3+(aq) ---> V3+(OH-)3(s) (or V(OH)3(s)) - without spectator ions
Explanation:
i don't know if this is right ore not but i hope this helps even if it is just a little bit sorry if this does not help i truly apologize
