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Montano1993 [528]
3 years ago
5

Answer questions a-c about the Bronsted acid-base reaction below using the identifying letters A-D below each structure. The pKa

's for the acids of interest are: phenol (pKa = 9.9), and carbonic acid (pKa = 6.4). a) The stronger acid is b) Its conjugate base is c) The species that predominate at equilibrium are (two letters, e.g. ac)
Chemistry
1 answer:
GuDViN [60]3 years ago
8 0

Answer:

a  H2CO3 b HCO3- and c H+ and HCO3-

Explanation:

As the pKa value of phenol is more than that of carbonic acid(H2CO3), the carbonic acid will have high Ka value than that of phenol.

  The acid that contain high Ka value act as stong acid.From that point of view H2CO3 is a strong acid than phenol as the Ka value of carbonic acid is greater than that of phenol.

  The conjugate base of H2CO3 is bicarbonate ion(HCO3-)

c The species that predorminates at equilibrium are H+ and HCO3-

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Write a structural formula for each of the following: (a) 3-Methylcyclobutene (b) 1-Methylcyclopentene (c) 2,3-Dimethyl-2-penten
Lemur [1.5K]

Answer:

Attached below

Explanation:

(a) 3-Methylcyclobutene

(b) 1-Methylcyclopentene

(c) 2,3-Dimethyl-2-pentene

(d) (Z)-3-Hexene  

(e) (E)-2-Pentene

(f) 3,3,3-Tribromopropene

(g) (Z,4R)-4-Methyl-2-hexene

(h) (E,4S )-4-Chloro-2-pentene

(i) (Z)-1-Cyclopropyl-1-pentene

(j) 5-Cyclobutyl-1-pentene

(k) (R)-4-Chloro-2-pentyne

(l) (E)-4-Methylhex-4-en-1-yne

The structural formula of a compound ( chemical ) is the representation of the molecular structure of the compounds ( i.e. it shows how the atoms in the compound is arranged, also the chemical bonding within the molecules are represented as well.

8 0
3 years ago
How many moles of HCl are needed to react with 0.87 mol Al?
WARRIOR [948]
The answer is ( 2.6 mol HCl )
5 0
3 years ago
The standard free-energy change for this reaction in the direction written is 23.8 kJ/mol. The concentrationsof the three interm
Natali [406]

Answer: The actual free-energy change for the reaction  -8.64 kJ/mol.

Explanation:

The given reaction is as follows.

  Fructose 1,6-bisphosphate \rightleftharpoons Glyceraldehyde 3-phosphate + DHAP

For the given reaction, \Delta G^{o} is 23.8 kJ/mol.

As we know that,

       \Delta G = \Delta G^{o} + RT ln Q

Here,    R = 8.314 J/mol K,       T = 37^{o} C

                                                    = (37 + 273) K

                                                    = 310.15 K

Fructose 1,6-bisphosphate = 1.4 \times 10^{-5} M

Glyceraldehyde 3-phosphate = 3 \times 10^{-6} M

DHAP = 1.6 \times 10^{-5} M

Expression for reaction quotient of this reaction is as follows.

    Reaction quotient = \frac{[DHAP][\text{glyceraldehyde 3-phosphate}]}{[/text{Fructose 1,6-bisphosphate}]}

        Q = \frac{1.6 \times 10^{-5} \times 3 \times 10^{-6}}{1.4 \times 10^{-5}}

            = 3.428 \times 10^{-6}

Now, we will calculate the value of \Delta G as follows.

          \Delta G = \Delta G^{o} + RT ln Q

                      = 23800 + 8.314 \times 310.15 \times ln(3.428 \times 10^{-6})

                      = -8647.73 J/mol

                      = -8.64 kJ/mol

Thus, we can conclude that the actual free-energy change for the reaction  -8.64 kJ/mol.

7 0
3 years ago
Which of the following reasons correctly explains the color changes that take place when ethylenediamine (C2N2H8) is added to th
pickupchik [31]

Answer:

The water ligands surrounding the cobalt metal center are being replaced by ethylenediamine and chloride ligands which results in a different crystal field splitting. Thus, the energy associated with electron transitions between the do-orbitals will differ for the two compounds showing a color change.

Explanation:

The five d-orbitals are usually degenerate. Upon approach of a ligand, the d-orbitals split into two sets of orbitals depending in the nature of the crystal field.

The magnitude of crystal field splitting is affected by the nature of the ligand. Ligands having filled p-π orbitals such as ethylenediamine lead to greater crystal field splitting.

The change in the colour that takes place when ethylenediamine is added to the solution of cobalt(II) chloride occurs due to a different crystal field splitting pattern. Thus, the energy associated with electron transitions between the d-orbitals now differ for the two compounds showing a color change.

7 0
3 years ago
Where are cardiac muscles used? Are they voluntary?
Ghella [55]

Answer:

Heart

Not voluntary.

Explanation:

The heart is made up of special muscle called the cardiac muscle that contains interconnected muscle fibres which are supplied with blood through the coronary artery. Also so that the waves of contraction can travel throughout the mass of the muscle.

6 0
3 years ago
Read 2 more answers
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