Chemistry terms - Help! (1 Viewer)

Trent566

New Member
Joined
Mar 12, 2004
Messages
15
Gender
Undisclosed
HSC
N/A
I have problems with understanding of some chemical terms, can someone please explain to me what they mean please:

1. Anhydrous
2. Polar/Non-Polar

Thanks
 

Affinity

Active Member
Joined
Jun 9, 2003
Messages
2,062
Location
Oslo
Gender
Undisclosed
HSC
2003
anhydrous -> 'without water'
polar -> the molecues have it's positive and negative charges distributed unevenly

non-polar is the opposite of polar
 

santaslayer

Active Member
Joined
May 29, 2003
Messages
7,816
Location
La La Land
Gender
Male
HSC
2010
1. without water; especially water of crystallization

2. Having a pair of equal and opposite charges
 

CM_Tutor

Moderator
Moderator
Joined
Mar 11, 2004
Messages
2,644
Gender
Male
HSC
N/A
Originally posted by santaslayer
1. without water; especially water of crystallization
So, taking an example, anhydrous copper(II) sulfate = CuSO<sub>4</sub>, a white solid
hydrated copper(II) sulfate = CuSO<sub>4</sub>.5H<sub>2</sub>O, a blue solid

As for polar / non-polar, it is used in relation to molecules and bonds. An object is polar if it has an unequal distribution of charge, such that one end is appreciably positive, and the other appreciably negative. A covalent bond will be polar if there is a sufficiently large difference in electronegativity between the two atoms involved in the bond - for example, a C-H bond is non-polar, but an O-H bond is polar - the O slightly negative and the H slightly positive.

A molecule will be polar if it has polar bonds and its geometry is such that the individual bond dipoles reinforce.

For example, CO<sub>2</sub> has polar C=O bonds, but since it is linear, the individual bond dipoles cancel one another's effects, leaving a net non-polar molecule.

By contrast, SO<sub>2</sub> is bent, and so the bond dipoles of the polar S=O bonds reinforce, producing a net polar molecule.

There is no HSC case of a substance being polar in the absence of polar bonds.
 

Users Who Are Viewing This Thread (Users: 0, Guests: 1)

Top