A polar bond is formed with atoms having different electronegativities. The bonding electrons are attracted more towards the atom with greater electronegativity resulting in unequal sharing of electrons. Therefore the molecule develop partial charges and becomes polar. Polar molecules have dipole moment that is the partial charge on molecules due to differences in electronegativity between atoms.
A non-polar bond is formed with atoms having the same electronegativity, hence the bonded pair of electron is shared equally between atoms. Non-polar molecules have no moment.
Note that: symmetrical molecules having polar bonds are non-polar because the dipoles of the bond exert equal and opposite effect. Hence the dipoles cancel the charges.
Example: HCl
In HCl, Cl is more electronegative therefore Cl atom pulls the electron pair of the covalent bond towards itself and develops a partial negative charge. Consequently H develops a partial positive charge. This therfore leads to the formation of a dipole.
Answer:
0.25M HCl
Explanation:
The reaction of HCl with NaOH is:
HCl + NaOH ⇄ H₂O + NaCl
<em>Where 1 mole of HCl reacts per mole of NaOH</em>
The end point was reached when the student added:
0.0500L × (0.1mol / L) = 0.00500 moles of NaOH
As 1 mole of HCl reacted per mole of NaOH, moles of HCl present are:
<em>0.00500 moles HCl</em>
The volume of the sample of hydrochloric acid was 20.0mL = 0.0200L, and concentration of the sample is:
0.00500 mol HCl / 0.0200L = <em>0.25M HCl</em>
To answer the question above, multiply the given number of moles by the molar masses.
(A) (0.20 mole) x (32 g / 1 mole) = 6.4 grams O2
(B) (0.75 mole) x (62 g / 1 mole) = 46.5 grams H2CO3
(C) (3.42 moles) x (28 g / 1 mole) = 95.7 grams CO
(D) (4.1 moles) x (29.88 g / 1 mole) = 122.508 g Li2O
The answer to the question above is letter D.
Explanation:
White precipitate of silver chloride get dissolves in excess ammonia to formation of complex between silver ions, chloride ions and ammonia molecules.
The chemical reaction is given as:
![AgCl(s)+2NH_3(aq)\rightarrow Ag[(NH_3)_2]^+Cl^-(aq)](https://tex.z-dn.net/?f=AgCl%28s%29%2B2NH_3%28aq%29%5Crightarrow%20Ag%5B%28NH_3%29_2%5D%5E%2BCl%5E-%28aq%29)
When 1 mole of silver chloride is added to 2 mole of an aqueous ammonia it form coordination complex of diaaminesilver(I) chloride.
Answer:
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