Answer:
a) ammonium ion
b) amide ion
Explanation:
The order of decreasing bond angles of the three nitrogen species; ammonium ion, ammonia and amide ion is NH4+ >NH3> NH2-. Next we need to rationalize this order of decreasing bond angles from the valence shell electron pair repulsion (VSEPR) theory perspective.
First we must realize that all three nitrogen species contain a central sp3 hybridized carbon atom. This means that a tetrahedral geometry is ideally expected. Recall that the presence of lone pairs distorts molecular structures from the expected geometry based on VSEPR theory.
The amide ion contains two lone pairs of electrons. Remember that the presence of lone pairs causes greater repulsion than bond pairs on the outermost shell of the central atom. Hence, the amide ion has the least H-N-H bond angle of about 105°.
The ammonia molecule contains one lone pair, the repulsion caused by one lone pair is definitely bless than that caused by two lone pairs of electrons hence the bond angle of the H-N-H bond in ammonia is 107°.
The ammonium ion contains four bond pairs and no lone pair of electrons on the outermost nitrogen atom. Hence we expect a perfect tetrahedron with bond angle of 109°.
The answer is: <span>supersaturation</span>
Answer:
The indepandent variable is wether its hot or cold water, so the tempurature of the water. The dependant variable is how much sugar disolves.
Explanation:
The indepandant variable is what changes or has a direct affect on the depandant. The depandant is what is being mesured or tested.
apply quantum theory to, especially form into quanta, in particular restrict the number of possible values of (a quantity) or states of (a system) so that certain variables can assume only certain discrete magnitudes.
Answer:
percent composition
Explanation:
the percentage composition had to be mnow in order to work out empirical formula
