12. The most common factors that cause chemical reactions to occur are all the following except

Which of the compounds above are strong enough acids to react almost completely with a hydroxide ion (pka of h2o = 15.74) or wit

luda_lava [24]

The compounds can react with OH⁻ and HCO₃⁻ only C₅H₆N pyridinium

<h3><em>Further explanation </em></h3>

In an acid-base reaction, it can be determined whether or not a reaction occurs by knowing the value of pKa or Ka from acid and conjugate acid (acid from the reaction)

Acids and bases according to Bronsted-Lowry

Acid = donor (donor) proton (H + ion)

Base = proton (receiver) acceptor (H + ion)

If the acid gives (H +), then the remaining acid is a conjugate base because it accepts protons. Conversely, if a base receives (H +), then the base formed can release protons and is called the conjugate acid from the original base.

From this, it can be seen whether the acid in the product can give its proton to a base (or acid which has a lower Ka value) so that the reaction can go to the right to produce the product.

The step that needs to be done is to know the pKa value of the two acids (one on the left side and one on the right side of the arrow), then just determine the value of the equilibrium constant

Can be formulated:

K acid-base reaction = Ka acid on the left : K acid on the right.

or:

pK = acid pKa on the left - pKa acid on the right

K = equilibrium constant for acid-base reactions

pK = -log K;

$K~=~10^{-pK}$

K value> 1 indicates the reaction can take place, or the position of equilibrium to the right.

There is some data that we need to complete from the problem above, which is the pKa value of some compounds that will react, namely:

pyridinium pKa = 5.25

acetone pKa = 19.3

butan-2-one pKa = 19

Let's look at the K value of each possible reaction:

pka H₂O = 15.74, pka of H₂CO₃ = 6.37)

1. C₅H₆N pyridinium

* with OH⁻

C₅H₆N + OH- ---> C₅H₅N- + H₂O

pK = pKa pyridinium - pKa H₂O

pK = 5.25 - 15.74

pK = -10.49

$K~=~10^{4.9}$

K values> 1 indicate the reaction can take place

* with HCO3⁻

C₅H₆N + HCO₃⁻-- ---> C₅H₅N⁻ + H₂CO₃

pK = 5.25 - 6.37

pK = -1.12

$K`=~10^{1.12]$

Reaction can take place

2. Acetone C₃H₆O

* with OH-

C₃H₆O + OH⁻ ---> C₃H₅O- + H₂O

pK = 19.3 - 15.74

pK = 3.56

$K~=~10^{ -3.56}$

Reaction does not happen

* with HCO₃-

C₃H₆O + HCO₃⁻ ----> C₃H₅O⁻ + H₂CO₃

pK = 19.3 - 6.37

pK = 12.93

$K`=~10 ^{-12.93}$

Reaction does not happen

3. butan-2-one C₄H₇O

* with OH-

C₄H₇O + OH- ---> C₄H₆O- + H₂O

pK = 19 - 15.74

pK = 3.26

$K~=~10^{-3.26}$

Reaction does not happen

* with HCO₃⁻

C₄H₇O + HCO₃⁻ ---> C₄H₆O⁻ + H₂CO₃

pK = 19 - 6.37

pK = 12.63

$K~=~ 10^{-12.63}$

Reaction does not happen

So that can react with OH⁻ and HCO₃⁻ only C₅H₆N pyridinium

<h3><em>Learn more </em></h3>

the lowest ph

brainly.com/question/9875355

the concentrations at equilibrium.

brainly.com/question/8918040

the ph of a solution

brainly.com/question/9560687

Keywords : acid base reaction, the equilibrium constant

5 0

2 years ago

Read 2 more answers

From the following enthalpy of reaction data and data in Appendix C, calculate ΔH∘f for CaC2(s): CaC2(s)+2H2O(l)→Ca(OH)2(s)+C2H2

sashaice [31]

Answer:

From the following enthalpy of reaction data and data in Appendix C, calculate ΔH∘f for CaC2(s): CaC2(s)+2H2O(l)→Ca(OH)2(s)+C2H2(g)ΔH∘=−127.2kJ

ΔHf°(C2H2) = 227.4 kJ/mol

ΔHf°(H2O) = -285.8 kJ/mol and

ΔHf°(Ca(OH)2) = -985.2 kJ/mol

(Ans)

ΔHf° of CaC2 = -59.0 kJ/mol

Explanation:

CaC2(s) + 2 H2O(l) → Ca(OH)2(s) + C2H2 (g) = −127.2kJ

ΔHrxn = −127.2kJ

ΔHrxn = ΔHf°(C2H2) + ΔHf°(Ca(OH)2) - ΔHf°(CaC2)- 2ΔHf°(H2O);

ΔHf°(CaC2) = ΔHf°(C2H2) + ΔHf°(Ca(OH)2) - 2ΔHf°(H2O) – ΔHrxn

Where

ΔHf°(C2H2) = 227.4 kJ/mol

ΔHf°(H2O) = -285.8 kJ/mol and

ΔHf°(Ca(OH)2) = -985.2 kJ/mol

ΔHf°(CaC2) =227.4 - 985.2 + 2x285.8 + 127.2 = -59.0 kJ/mol

ΔHf°(CaC2) = -59.0 kJ/mol

7 0

2 years ago

10.000g of boron (B) combines with hydrogento form 11.554g of a pure compound. What is the empirical formula of this compound?

Citrus2011 [14]

Answer:

B3H5

Explanation:

The law of conservation of mass states that matter in an closed system is neither created nor destroyed by physical transformations or chemical reactions but changes from one form to the other.

That is, the sum of masses of the reactants = The sum of masses of the product

10.00g of Boron + x grams of Hydrogen = 11.55g of the product

Mass of hydrogen = 11.55 - 10.00 = 1.55g

Molar mass of Boron = 10.811g

Molar mass of Hydrogen = 1.00784g

Number of moles of Boron = (mass of Boron)/(molar mass of Boron) = 10/10.811 = 0.9249 mols

Number of moles of Hydrogen = (mass of Hydrogen)/(molar mass of Hydrogen) = 1.55/1.00784 =1.5379mols

0.9249 mols of Boron combines with 1.5379mols of Hydrogen

Dividing both sides mols by 0.9249 gives

1 mole of Boron combines with 1.66266 mols of Hydrogen

Converting 1.66266 to fractions we have 1.66266 approximately 5/3

or 1 mole of Boron combines with 5/3 moles of Hydrogen

Multiplying both sides by 3 we have

3 moles of Boron combines with 5 moles of Hydrogen

Molecular formula of the compound is

B3H5

4 0

2 years ago

A hypothetical substance has a melting point of −10°C and a boiling point of 155°C. If this substance is heated from 2°C to the

igor_vitrenko [27]

Answer:

The specific heat capacity of liquid and the het of vaporization is used.

.

Explanation:

Step 1: Data given

A substance at temperature 2°C.

The substance has a melting point of −10°C and a boiling point of 155°C.

The initial temperature is 2°C which is between the melting point (-10°C) and the boiling point (155°C). At 2°C, the substance is liquid.

At 155°C, the substance changes from liquid to gas.

To calculate the heat gained for the change of 2°C liquid to 155°C liquid, specific heat capacity of the liquid (C) is needed.

To calculate the heat gained for the change of liquid to 155°C gas, heat of vaporization (D) is needed.

The <u>specific heat of the solid is not used</u> because the substance is changed from liquid to gas. it doesn't come in the state of solid.

<u>Heat of fusion is not used</u>, because it's used when there is a change from its state from a solid to a liquid,

<u>The specific heat capacity of the gas is not used</u>, because the substance only formes gas after reaching 155 °C

5 0

2 years ago

How much heat will be absorbed by a 50.3 g piece of aluminum (specific heat = 0.930 J/g･°C) as it changes temperature from 23.0°

shepuryov [24]

Answer:

quantity of heat= <u>mc</u>*they

=50.3*0.930*(67-23)

=46.779*44

=2058.276

quantity of heat =2058.3kJ

8 0

2 years ago