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Ierofanga [76]
2 years ago
11

Write chemical equations and corresponding equilibrium expressions for each of the two ionization steps of carbonic acid. Part A

Write chemical equations for first ionization step of carbonic acid. Express your answer as a chemical equation. Identify all of the phases in your answer. nothing Request Answer Part B Complete previous part(s) Part C Write chemical equations for second ionization step of carbonic acid. Express your answer as a chemical equation. Identify all of the phases in your answer. nothing
Chemistry
1 answer:
lesantik [10]2 years ago
6 0

<u>Answer:</u> The chemical equations and equilibrium constant expression for each ionization steps is written below.

<u>Explanation:</u>

The chemical formula of carbonic acid is H_2CO_3. It is a diprotic weak acid which means that it will release two hydrogen ions when dissolved in water

The chemical equation for the first dissociation of carbonic acid follows:

               H_2CO_3(aq.)\rightleftharpoons H^+(aq.)+HCO_3^-(aq.)

The expression of first equilibrium constant equation follows:

Ka_1=\frac{[H^+][HCO_3^{-}]}{[H_2CO_3]}

The chemical equation for the second dissociation of carbonic acid follows:

               HCO_3^-(aq.)\rightarrow H^+(aq.)+CO_3^{2-}(aq.)

The expression of second equilibrium constant equation follows:

Ka_2=\frac{[H^+][CO_3^{2-}]}{[HCO_3^-]}

Hence, the chemical equations and equilibrium constant expression for each ionization steps is written above.

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Explain how your experimental data for Rf values are, or are not, consistent with your predictions of enantiomeric and diastereo
hichkok12 [17]

Answer:

Answer is explained below.

Explanation:

As (+) menthol and (-) menthol are enantiomers whose physical properties are same except optical activity so we can expect they have similar Rf values.

Whereas diastereomers have different physical properties and different Rf values.

For example when the (+) menthol , (-) menthol, isomenthol and neomenthol undergo TLC (thin layer chromatography) the

Rf values of.(+menthol) = .447

Rf (+isomenthol) = .395

Rf (+neomenthol)= .487

Rf (-menthol) = .434

The above data shows that (+) menthol and (-) menthol have almost same Rf values and vary a little i.e 0.447 and 0.437. So we can conclude them as enantiomers

Whereas (+) menthol or (+) neomenthol or (+) isomenthol i.e 0.447 , 0.395 and 0.487 have different Rf values. We can conclude them as diasteromers.

(+) menthol and (-) menthol - enantiomers

(+) menthol and (+) neomenthol- diastereomers

(-) menthol and (+) isomenthol - diastereomers

3 0
2 years ago
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Up to a point, the elongation of a spring is directly proportional to the force applied to it. Once you extend the spring more t
lana [24]

yes 2.5 is correct for plato!

8 0
2 years ago
The factor 0.01 corresponds to which prefix? A) milli B) deci C) deka D) centi
BARSIC [14]
The answer is: D) centi
5 0
2 years ago
Tag all the carbon atoms with pi bonds in this molecule. If there are none, please check the box.
snow_tiger [21]

Answer:

Pi bonds (π bonds) are covalent chemical bonds where two lobes of an orbital involved in the bond overlap with two lobes of the other orbital involved. These orbitals share a nodal plane that passes through the nuclei involved. Are generally weaker than sigma links, because their negatively charged electronic density is further from the positive charge of the atomic nucleus, which requires more energy.

They are frequent components of multiple bonds, as is the molecule indicated in our exercise.

The characteristics that distinguish pi bonds from other kinds of interactions between atomic species are described below, beginning with the fact that this union does not allow the free rotation movement of atoms, such as carbon. For this reason, if there is rotation of the atoms, the bond is broken.

Explanation:

In order to describe the formation of the pi bond, first we must talk about the hybridization process, as this is involved in some important links.

Hybridization is a process where hybrid electronic orbitals are formed; that is, where orbitals of atomic sub-levels s and p can get mixed. This causes the formation of sp, sp2 and sp3 orbitals, which are called hybrids.

In this sense, the formation of pi bonds occurs thanks to the overlapping of a pair of lobes belonging to an atomic orbital over another pair of lobes that are in an orbital that is part of another atom.

This orbital overlap occurs laterally, so the electronic distribution is mostly concentrated above and below the plane formed by the linked atomic nuclei, and causes the pi bonds to be weaker than the sigma bonds.

When talking about the orbital symmetry of this type of junction, it should be mentioned that it is equal to that of the p-type orbitals as long as it is observed through the axis formed by the bond. In addition, these junctions are mostly made up of p orbitals.

Since pi bonds are always accompanied by one or two more links (one sigma or another pi and one sigma), it is relevant to know that the double bond that is formed between two carbon atoms has less bond energy than that corresponding to two Sometimes the sigma link between them.

4 0
2 years ago
What is the percent composition by mass of nitrogen in the compound N2H4 (gram-formula mass = 32 g/mol)?
Alex

Answer:

\large \boxed{93\, \% }

Explanation:

1. Calculate the molar mass of N₂H₄

2N = 2 × 14 = 28

4H = 2 ×  1  = <u>  4</u>

           Tot. = 32

2. Calculate the mass percent of N

\text{\% of element} = \dfrac{\text{mass of element}}{\text{mass of compound}} \times 100 \, \% = \dfrac{\text{28}}{\text{32}} \times 100 \, \% =\mathbf{88 \, \%}\\\\\text{The percentage of N in N$_{2}$H$_{4}$ is $\large \boxed{\mathbf{88\, \% }}$}}

3 0
2 years ago
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