Answer:
amino group
Explanation:
There are twenty (20) amino acids in nature. Generally, each amino acid is structurally made up of a central carbon atom called alpha carbon attached to a hydrogen, carboxylic acid group (-COOH) and an amine group (-NH2). However, one particular amino acid called PROLINE posseses an exception to this.
Proline, which is the only cyclic amino acid, is also the only amino acid that forms a secondary amine group i.e. loss of hydrogen atoms in its amine group when in a protein structure. This means that when in a protein, PROLINE does not have an AMINE GROUP.
Answer: He did not discuss about any of these.
Explanation: Dalton proposed some of the postulates for his atomic theory. They are:
1) Matter is made up of atoms which are not divisible.
2) Atoms of different elements combine in a fixed ratio to form compounds.
3) The atomic properties of given element are same including mass. This states that all the atoms of an element have same mass but the atoms of different elements have different masses.
4) No atoms are either created or destroyed during a chemical reaction.
5) Atoms of an element are identical in mass, size and other chemical and physical properties.
As it is visible from the postulates, he only discussed only about the atoms but not subatomic particles or isotopes.
The answer:
<span>The equation of its dissolution in water is: AgNO3 → Ag + (aq) + NO3- (aq)
and </span>AgNO3 → Ag + (aq) + NO3- (aq)
1 mol 1mol 1mol
? -------- 0.854mo
so for finding the value, it is sufficients to complute 1 x 0.854 mol =0.854 mol
so, 0.854 mol is required for the reaction to form 0.854 mol of Ag
When the concentration is expressed in molality, it is expressed in moles of solute per kilogram of solvent. Since we are given the mass of the solvent, which is water, we can compute for the moles of solute NaNO3.
0.5 m = x mol NaNO3/0.5 kg water
x = 0.25 mol NaNO3
Since the molar mass of NaNO3 is 85 g/mol, the mass is
0.25 mol * 85 g/mol = 21.25 grams NaNO3 needed
The answer would A. Sand grain I got 100 on the sedimentary assignment.
