I don't understand how to do these questions. Can someone explain to me?:
1.) What are the relative numbers of H3O+ and OH- ions in an acidic, an alkaline, and a neutral solution?
2.) How many times more hydroxide ions are there in a solution with a pH of 9 than in a solution with a pH of 3?
3.) An example of base is...
a. pure water b. vinegar (i think it's this one?)
c. ammonia d. urine
2007-01-18 11:54:27 · 7 answers · asked by PloyCMBrown 2 in Education & Reference ➔ Homework Help
1. ?
2. ?
3. I think urine? Pure water is a neutral, ammonia is more basic than acidic (experiments). I don't know... because vinegar is sour. Sour liquids are acidic, and bitter liquids are basic.
2007-01-18 12:00:56 · answer #1 · answered by chomama 3 · 0⤊ 0⤋
ammonia is the base, water is usually neutral, vinegar is acidic, urine contains ammonia.
2007-01-18 11:57:32 · answer #2 · answered by dasdesperado 2 · 0⤊ 0⤋
Sweet, R .... probably REACTIVE A branched polysaccharide has many terminal glycosidic bonds. These glycosidic bonds can be hydrolysed to release glucose. Many bonds means a rapid release of glucose. Hence animals, more active than plants, store glycogen which is more branched than the amylopectin component of starch. Possibly one could argue that a branched structure fits more glucose residues into a given volume. Storage polysaccharides are always described as "compact" so that "cells can store many molecules".
2016-05-24 05:08:57 · answer #3 · answered by Deborah 4 · 0⤊ 0⤋
1 acidic More H30 Alkaline more OH, neutral balanced
2007-01-18 12:04:21 · answer #4 · answered by Don't look too close! 4 · 0⤊ 0⤋
umm i dont remember bio that well but i only know 3 which is c. ammonia...vinegar is an acid
2007-01-18 11:56:55 · answer #5 · answered by AzN x SPIKER 2 · 0⤊ 0⤋
I believe the last one is C. Ammonia.
2007-01-18 11:59:08 · answer #6 · answered by Muffins 1 · 0⤊ 0⤋
Thats more like chemistry to me!
I thought a base was alkali, so i'd go for ammonia on the last one.......
2007-01-18 11:57:15 · answer #7 · answered by Ben G 1 · 1⤊ 0⤋