Answer:
(D) (CH3CH2)2NH
Explanation:
In order to decide which base is strongest we need to calculate its PKb
PKb = -log [Kb]
A large Kb value and small PKb value gives the strongest base
Compound Kb PKb
(A) C6H5NH2 - 4 x 10^-10 9.349
(B) NH3 1.76x 10^-5 4.754
(C) CH3NH2 4.4x 10^-4 3.357
(D) (CH3CH2)2NH 8.6x 10^-4 3.066
(E) C5H5N 1.7x10^-9 8.77
Clearly (CH3CH2)2NH is the strongest base.
Bonds formed between atoms can be classified as ionic and covalent
Ionic bonds are formed between atoms that have a high difference in the electronegativity values.
In contrast, bonds formed between atoms that have a difference in electronegativity lower than the ionic counterparts are polar covalent bonds. If the atoms have very similar electronegativities, they form non-polar covalent bonds.
In H2S, the S atom is bonded to 2 H atoms. The electronegativity of H = 2.2 and S= 2.56. Since the difference is not high the bond formed will be covalent (polar covalent).
Answer:
The concentration of O2 will begin decreasing and The concentrations of CO2 and O2 will be equal.
Explanation:
Equilibrium occurs when the velocity of the formation of the products it's equal to the velocity of the formation of the reactants, thus the concentrations of the compounds remain constant.
Analyzing the information and the reaction given, we can notice that in equilibrium the rate (velocity) of formation of O2 (product) is equal to the rate of formation of CO2 (reactant).
As the CO2 and H2O are placed in the reaction, the Le Chateliêr's principle states that the equilibrium must shift to reestablish the equilibrium, thus, they must be consumed, and the concentration of O2 must increase.
As state above, in equilibrium, the concentrations didn't change, thus, the concentrations of CO2 and O2 will not change.
The concentrations of CO2 and O2 depends on the rate of the reaction and the initial quantities presented, so it's not possible to affirm they'll be equal.
Answer:
attached below
Explanation:
Structure of two acyclic compounds with 3 or more carbons that exhibits one singlet in 1H-NMR spectrum
a) Acetone CH₃COCH₃
Attached below is the structure
b) But-2-yne (CH₃C)₂
Attached below is the structure
Answer:
d
Explanation:
sugar molecules are being broken down