I am answering this question with respect to organic molecules in proteins. In case of proteins, the sequence of proteins (order) greatly affects the function of protein.For example, in enzymes the binding site or catalytic site has a catalytic residues (organic molecules = amino acids) at specific positions. And these amino acids together catalyzes the reaction. Like Serine, Histidine and Aspartic in chymotripsin together catalyzes the hydrolysis of protein and they have specific order (i.e serine 195, histidine 57 and aspartic acid 102. So, order is very important in functionality of protein.
Secondly, the shape is very important. Like, in above example the -OH group of Asp protonates the carbonyl groups in substrate, the -OH group of Ser acts as a nucleophile and the -NH group of His acts as a base by abstracting the protons during the reaction. So, it is the shape of these amino acids which are helping in functioning the enzyme.
So, the correct answer is Order and Shape.
The empirical formula is N₂O₅.
The empirical formula is the <em>simplest whole-number ratio of atoms</em> in a compound.
The ratio of atoms is the same as the ratio of moles, so our job is to calculate the <em>molar ratio of N:O</em>.
I like to summarize the calculations in a table.
<u>Element</u> <u>Moles</u> <u>Ratio¹ </u> <u> ×2² </u> <u>Integers</u>³
N 1.85 1 2 2
O 4.63 2.503 5.005 5
¹To get the molar ratio, you divide each number of moles by the smallest number (1.85).
²Multiply these values by a number (2) that makes the numbers in the ratio close to integers.
³Round off the number in the ratio to integers (2 and 5).
The empirical formula is N₂O₅.
Answer:
Explanation:
Mixing Ammonia gas into a solution of Copper(II) Sulfate will give Ammonium Sulfate and a precipitate of Copper(II) Hydroxide (Cu(OH)₂). The Ksp of Cu(OH)₂ is published => 2.2 x 10⁻²². Such gives a solubility* of the Cu(OH)₂ to be ~1.77 x 10⁻⁷M => [Cu⁺²] ~1.77 x 10⁻⁷M and [OH⁻] = 2(1.77 x 10⁻⁷)M = 3.53 x 10⁻⁷M. The reaction of Ammonium Hydroxide and Copper(II) Sulfate will generate 1 x 10⁻⁴ mole Cu(OH)₂ as a precipitate but only 1.77 x 10⁻⁷ mole of the hydroxide will remain in 1 Liter of solution b/c of extreme limited solubility.
*Solubility of 1:2 ionization ratio salts = CubeRt(Ksp/4).
When there are equal number of H+ and OH- ions, the pH of water is 7.