Answer:
The acid-base reaction produces glycine reduction, and hence the increase of glycine pH.
Explanation:
The glycine is an amino acid with the following chemical formula:
NH₂CH₂COOH
The COOH functional group is what gives the acid properties in the molecule.
Hence, when NaOH is added to glycine an acid-base reaction takes place in which COOH reacts with the NaOH added:
NH₂CH₂COOH + OH⁻ ⇄ NH₂CH₂COO⁻ + H₂O
The glycine concentration starts to shift to its ion form (NH₂CH₂COO⁻) because of the reaction with NaOH, that is why the pH glycine increases when NaOH is added.
Therefore, the acid-base reaction produces glycine reduction, and hence the increase of glycine pH.
I hope it helps you!
Answer:
Decreasing the volume of solvent in the solution of molecule A
Explanation:
We know that one of the factors that affect the rate of reaction is the concentration of the reactants. The greater the concentration of reactants, the faster the rate of reaction (the greater the frequency of collision between reactants).
Hence, when we decrease the volume of solvent in the solution of molecule A, the concentration of the solution increases and consequently more particles of molecule A are available to collide with particles of molecule B resulting in a higher rate of reaction.
Answer:
1. 80g
2. 1.188mole
Explanation:
1. We'll begin by obtaining the molar mass of CH4. This is illustrated below:
Molar Mass of CH4 = 12 + (4x1) = 12 + 4 = 16g/mol
Number of mole of CH4 from the question = 5 moles
Mass of CH4 =?
Mass = number of mole x molar Mass
Mass of CH4 = 5 x 16
Mass of CH4 = 80g
2. Mass of O2 from the question = 38g
Molar Mass of O2 = 16x2 = 32g/mol
Number of mole O2 =?
Number of mole = Mass /Molar Mass
Number of mole of O2 = 38/32
Number of mole of O2 = 1.188mole
Answer:
(a) 5s. n = 5. Sublevel s, l = 0. Number of orbitals = 1
(b) 3p. n = 3. Sublevel p, l = 1. Number of orbitals = 3
(c) 4f. n =4. Sublevel f, l = 3. Number of orbitals = 7
Explanation:
The rules for electron quantum numbers are:
1. Shell number, 1 ≤ n
2. Sublevel number, 0 ≤ l ≤ n − 1
So,
(a) 5s. n = 5, shell number 5. Sublevel s, l = 0. Number of orbitals = 2l +1 = 1
(b) 3p. n = 3, shell number 3. Sublevel p, l = 1. Number of orbitals = 2l +1 = 3
(c) 4f. n =4, shell number 4. Sublevel f, l = 3. Number of orbitals = 2l +1 = 7
The answer///////////b, 5 orbits
