Rank the following amine derivatives from highest acidity (lowest pKa value) to lowest acidity (highest pKa value).
Highest acidity
anilinium ion
aniline
ammonium ion
secondary amine
amide
Lowest acidity

Answers

Answer 1

Answer:

anilinium ion > ammonium ion > amide > aniline > secondary amine

Explanation:

Acidity of amine derivatives can derived from their pKa values.

The rule of thumb for acidity with relation to pKa values is that:

As the pKa decreases the acid strength increases and the conjugate base decreases. Similarly, as the pKa increases, the acid strength decreases and the conjugate base increase.

Hence the stronger the acid , the  lower pKa value  and the weaker the acid , the stronger the pKa value.

So the pKa value for anilinium ion = 4.6

ammonium ion = 9.4

Amide = 15

Similarly, for aniline and secondary amine, in order to determine the derivative with the higher acidity, we will consider the electron withdrawing substituent group.

The more difficult the electron are being withdraw from the electron withdrawing substituent , the more acidic the compound.

In aniline , the stabilized benzene ring attached to NH₂ makes it a less electron withdrawing group compared to the straight chains structure found in secondary amine where electron are easily withdraw by nucleophilic substitution reactions.

Thus, from highest acidity (lowest pKa value) to lowest acidity (highest pKa value).

the amine derivatives ranking is as follows:

anilinium ion > ammonium ion > amide > aniline > secondary amine


Related Questions

Select the true statement concerning voltaic and electrolytic cells. Select one: a. Voltaic cells involve oxidation-reduction reactions while electrolytic cells involve decomposition reactions. b. Voltaic cells require applied electrical current while electrolytic cells do not. . c. all electrochemical cells, voltaic and electrolytic, must have spontaneous reactions. d. Electrical current drives nonspontaneous reactions in electrolytic cells.

Answers

Answer:

Electrical current drives nonspontaneous reactions in electrolytic cells.

Explanation:

Electrochemical cells are cells that produce electrical energy from chemical energy.

There are two types of electrochemical cells; voltaic cells and electrolytic cells.

A voltaic cell is an electrochemical cell in which electrical energy is produced from spontaneous chemical process while an electrolytic cell is an electrochemical cell where electrical energy is produced from nonspontaneous chemical processes. Current is needed to drive these nonspontaneous chemical processes in an electrolytic cell.

Answer:

electrolytic cells generate electricity through a non-spontaneous reaction while voltaic cells absorb electricity to drive a spontaneous reaction.

Explanation:

Answer via Educere/ Founder's Education

What is the ph of 0.36M HNO3 ?

Answers

Answer:

0.44

Explanation:

We know that the pH of any acid solution is given by the negative logarithm of its hydrogen ion concentration. Hence, if I can obtain the hydrogen ion concentration of any acid, I can obtain its pH.

For the acid, HNO3, [H^+] = [NO3^-]= 0.36 M

pH= -log [H^+]

pH= - log[0.36]

pH= 0.44

Which of the following represents six molecules of water? 6HO 2 6H 2O 1 6H 2O H 6O

Answers

Answer:

6H20 represents six molecules of water

Answer:

6H20 represents six molecules of water

Explanation:

suppose you make lemonade with one can lemonade concentrate mixed with four cans of water. What is the fraction of the final product that is water

Answers

Answer:

0.8 part of the product is water

Explanation:

Volume (or parts) of water = 4

Volume (or parts) of lemonade = 1

Total volume = 4 + 1 = 5

Fraction of water = Volume of water / Total volume = 4 / 5 = 0.8

Find the standard enthalpy of formation of iodine atoms. (Round your answer to one decimal place.) Standard enthalpy of formation

Answers

Answer:

Enthalpy of formation is the energy change when one mole of a substance is formed from its constituent atoms under standard conditions

Which of the following types of electromagnetic radiation have higher frequencies than visible light and which have shorter frequencies than visible light? Sort them accordingly. ltems (6 items) (Drag and drop into the appropriate area below)
a. Gamma rays
b. Infrared radiation
c. Ultraviolet liht
d. X-rays
e. Microwaves
f. Radio waves

Answers

Answer:

Higher frequency than visible light -  Ultraviolet light, X-rays, and Gamma rays

Lower frequency than visible light - Infrared radiation, microwaves, and Radio waves

Explanation:

The frequencies of electromagnetic radiations vary according to their wavelengths. The relationship between the frequency and wavelength of the waves is expressed such that:

      λ = c/f, where λ = wavelength, c = speed of light, and f = frequency.

Thus, there is an inverse relationship between the wavelength and the frequency of electromagnetic waves.

The order of the electromagnetic waves based on their frequency from the lowest to the highest is radio waves, microwaves, infrared, visible light, ultraviolet, X-rays, and gamma-rays

Hence, electromagnetic waves with higher frequencies than visible light include ultraviolet light, X-rays, and Gamma rays while those with lower frequencies include Infrared radiation, microwaves, and Radio waves.

Answer:

need points

Explanation:

A compound is found to contain 26.94 % nitrogen and 73.06 % fluorine by weight. To answer the questions, enter the elements in the order presented above. 1. What is the empirical formula for this compound? 2. The molecular weight for this compound is 104.02 g/mol. What is the molecular formula for this compound?

Answers

Answer:

THE EMPIRICAL FORMULA OF THE COMPOUND IS NF2

THE MOLECULAR FORMULA OF THE COMPOUND IS N2F4

Explanation:

To calculate the empirical formula for the compound, we:

1. Write out the percentage weight of each elements

N = 26.94%

F = 73.06 %

2. Divide each by its atomic mass

( N= 14, F = 19)

N = 26.94 / 14 = 1.924

F = 73.06 / 19 = 3.845

3. Divide each by the smaller of the values

N = 1.924 / 1.924 = 1

F = 3.845 / 1.924 = 1,998

4. Round up to a whole number and write the empirical formula

N= 1

F = 2

So the empirical formula of the compound is N F2

To calculate the molecular formula, we:

(N F2 )n = molecular weight

( 14 + 19*2) n = 104.02

52 n = 104.02

n = 2.000

The molecular formula of the compound will be:

(N F2)2 = N2F4

In conclusion, the empirical formula of the compound is NF2 and the molecular formula of the compound is N2F4

1.78 L of gas is at a pressure of 735 torr. What is the volume in liters (L) when the pressure decreases to 700.0 torr

Answers

Explanation:

using boyles law

p1v1=p2v2

735 x 1.76 = 700 x V2

1293.6 = 700 x V2

V2 = 1293.6/700

V2 = 1.85L

If 100-mL of 1.0 M Sr(OH)2 is added to 100 mL of 1.0 M HCl, the pH of the mixture would be _____. Group of answer choices

Answers

Answer:

pH = 13.7

Explanation:

A strong acid (HCl) reacts with a strong base Sr(OH)₂ producing water and a salt, thus:

2HCl + Sr(OH)₂ → 2H₂O + SrCl₂

To solve this problem, we need to find initial moles of both reactants and, with the chemical equation find limiting reactant and moles in excess to find pH as follows:

The initial moles of HCl and Sr(OH)₂ are:

100mL = 0.1L ₓ (1.0mol / L) = 0.100 moles of both HCl and Sr(OH)₂

As 2 moles of HCl reacts per mole of Sr(OH)₂, moles of Sr(OH)₂ that reacts with 0.100 moles of HCl are:

0.100 moles HCl ₓ (1 mol Sr(OH)₂ / 2 mol HCl) = 0.050 moles Sr(OH)₂

That means HCl is limiting reactant and after reaction will remain in solution:

0.100 mol - 0.050mol =

0.050 moles of Sr(OH)₂

Find pH:

1 mole of Sr(OH)₂ contains 2 moles of OH⁻, 0.050 moles contains 0.050×2 = 0.100 moles of OH⁻. In 200mL = 0.2L:, molar concentration of OH⁻ is:

0.100 moles / 0.2L =

[OH⁻] = 0.5M

As pOH of a solution is -log[OH⁻],

pOH = -log 0.5M

pOH = 0.301

And knowing:

pH = 14 - pOH

pH = 14 - 0.301

pH = 13.7

There are 454 grams in one pound. How many pounds are in 700 grams

Answers

Answer:

1.543 pounds = 700 grams

Come up with a definition for density

Answers

Density measures how tightly packed particles are.

If particles are tightly packed together, they will be more dense.

If they are loosely together, they will be less dense.

However, a common mistake is thinking that if something

is more dense it means that it's heavier.

However, that's not the case.

It has to do with how particles are packed in an object.

Measures how tightly packed particles are

CI
Which of the following statements is INCORRECT?
(1)
(2)
the compound contains a o molecular orbital formed by the overlap of one carbon
sp2 hybrid orbital and one hydrogen sp3 hybrid orbital
the compound contains a T molecular orbital formed by the overlap of two
unhybridized carbon p atomic orbitals
the compound contains a polar C-Cl bond
each carbon atom of the C=C bond is sp2 hybridized
(3)
(4)​

Answers

Answer:

The compound contains a o molecular orbital formed by the overlap of one carbon sp2 hybrid orbital and one hydrogen sp3 hybrid orbital.

Explanation:

Molecular orbital is function which describes wave like behavior of an electron in a molecule. The molecular orbital theory describes the electronic structure of molecule using quantum mechanics. Electrons are not assigned to individual bonds between atoms. The compound contains sp2 hybrid orbial which is polar C - CI bond.

How are Math, Physics, Chemistry, and Biology all related?

Answers

Answer:

- you have to do maths in all 3

- atoms make up everything even parts of a cel and theyre studied in chem and physics

- chemistry is used in biology by finding out what different substances are eg cytoplasm in a cell

Draw the major organic product that is expected when cyclopentanecarboxylic acid is treated with each of the following reagents:

a. NaOH
b. [H+]

Answers

Answer:

a. Sodium cyclopentanecarboxylate

b. No reaction

Explanation:

In this case, in the cyclopentanecarboxylic acid we have a carboxylic acid functional group. Therefore we have an "acid". The acids by definition have the ability to produce hydronium ions ([tex]H^+[/tex]).

With this in mind, for molecule a. we will have an acid-base reaction, because NaOH is a base. When we put together an acid and a base we will have as products a salt and water. In this case, the products are  Sodium cyclopentanecarboxylate (the salt) and water.

For the second molecule, we have the hydronium ion  ([tex]H^+[/tex]). This ion can not react with an acid. Because, the acid will produce the hydronium ion also, so a reaction between these compounds is not possible.

See figure 1

I hope it helps!

a boy capable of swimming 2.1m/a in still water is swimming in a river with a 1.8 m/a current. At what angle must he swim in order to end up directly opposite his starting point?

Answers

Answer:

The boy must swim at an angle of 59°northwest to get to a position directly opposite his starting point.

Explanation:

To get to a point directly opposite his starting point, the boy must travel at an angle x, in a direction northwest of his starting point. The speed of the boy and the speed of the river current forms a right-angled triangle with an an opposite side of 1.8 m/a and a hypotenuse of 2.1 m/a having an angle x.

Sin x = opp/ hyp

Sin x = 1.8/2.1

x = sin⁻¹ (1.8/2.10

x = 58.99

x = 59°

Therefore, the boy must swim at an angle of 59° in the northwesterly direction to get to a position directly opposite his starting point.

0.25 L of aqueous solution contains 0.025g of HCLO4 (strong acid) what will be the Ph of the solution g

Answers

Answer:

The pH of the solution will be 3

Explanation:

The strength of acids is determined by their ability to dissociate into ions in aqueous solution. A strong acid is any compound capable of completely and irreversibly releasing protons or hydrogen ions, H⁺. That is, an acid is said to be strong if it is fully dissociated into hydrogen ions and anions in solution.

Being pH=- log [H⁺] or pH= - log [H₃O⁺] and being a strong acid, all the HClO₃ dissociates:

HClO₄      +    H₂O        →      H₃O⁺      +      ClO₄-  

So: [HCLO₄]= [H₃O⁺]

The molar concentration is:

[tex]molar concentration=\frac{number of moles of solute}{volume solution}[/tex]

The molar mass of HClO₄ being 100 g / mole, then if 100 grams of the compound are present in 1 mole, 0.025 grams in how many moles are present?

[tex]moles of HClO_{4} =\frac{0.025 grams*1 mole}{100 grams}[/tex]

moles of HClO₄= 0.00025

Then:

[tex][HClO_{4}]=\frac{0.00025 moles}{0.25 L}[/tex]

[tex][HClO_{4}]=0.001 \frac{ moles}{ L}[/tex]

Being [HCLO₄]= [H₃O⁺]:

pH= - log 0.001

pH= 3

The pH of the solution will be 3

Chlorine monoxide and dichlorine dioxide are involved in the catalytic destruction of stratospheric ozone. They are related by the equation:

2ClO(g) ⇌ Cl2O2(g) for which Kc is 4.96×10^11 at 273 K.

For an equilibrium mixture in which [Cl2O2] is 6.00 x 10^-6M, what is [ClO]?

Answers

Answer:

[ClO] = 3.48×10¯⁹ M.

Explanation:

The following data were obtained from the question:

Equilibrium constant (Kc) = 4.96×10¹¹

Concentration of Cl2O2, [Cl2O2] = 6x10¯⁶ M.

Concentration of ClO, [ClO] =.?

The equation for the reaction is given below:

2ClO(g) ⇌ Cl2O2(g)

The equilibrium constant for a reaction is simply defined as the ratio of the concentration of product raised to their coefficient to the concentration of the reactant raised to their coefficient.

The equilibrium constant, Kc for the reaction is given by:

Kc = [Cl2O2] / [ClO]²

Thus, we can calculate the concentration of ClO, [ClO] as follow:

Kc = [Cl2O2] / [ClO]²

4.96×10¹¹ = 6x10¯⁶ / [ClO]²

Cross multiply

4.96×10¹¹ × [ClO]² = 6x10¯⁶

Divide both side by 4.96×10¹¹

[ClO]² = 6x10¯⁶ / 4.96×10¹¹

[ClO]² = 1.21×10¯¹⁷

Take the square root of both side

[ClO] = √ (1.21×10¯¹⁷)

[ClO] = 3.48×10¯⁹ M

Therefore, the concentration of ClO, [ClO] is 3.48×10¯⁹ M.

Rectangular cube 3.2 m length 1.2 m in height and 5 m in length is split into two parts. The container has a movable airtight divider that divides its length as necessary. Part A has 58 moles of gas and part B has 165 moles of a gas.

Required:
At what length will the divider to equilibrium?

Answers

Answer:

The length the divider is to  equilibrium from Part A = 1.30 m and from Part B = 3.70 m

Explanation:

Given that:

A rectangular cube with 3.2 m breadth, 1.2 m height and 5 m in length is splitted into two parts.

The diagrammatic expression for the above statement can be found in the attached diagram below.

The container has a movable airtight divider that divides its length as necessary.

Part A has 58 moles of gas

Part B has 165 moles of a gas.

Thus, the movable airtight divider will stop at a length where the pressure on it is equal on both sides.

i.e

[tex]\mathtt{P = P_A = P_B}[/tex]

Using the ideal gas equation,

PV = nRT

where, P,R,and  T are constant.

Then :

[tex]\mathsf{\dfrac{V_A}{n_A}= \dfrac{V_B}{n_B}}[/tex]

[tex]\mathsf{\dfrac{L_A \times B \times H}{n_A}= \dfrac{L_B \times B \times H}{n_B}}[/tex] --- (1)

since Volume of a cube = L × B × H

From the question; the L = 5m

i,e

[tex]\mathsf{L_A +L_B}[/tex] = 5

[tex]\mathsf{L_A = 5 - L_B}[/tex]

From equation (1) , we divide both sides by (B × H)

Then :

[tex]\mathsf{\dfrac{L_A }{n_A}= \dfrac{L_B }{n_B}}[/tex]

[tex]\mathsf{\dfrac{5-L_B}{58}= \dfrac{L_B }{165}}[/tex]

By cross multiplying; we have:

165 ( 5 - [tex]\mathsf{L_B}[/tex] )  = 58 (

825 - 165[tex]\mathsf{L_B}[/tex]  = 58

825 = 165[tex]\mathsf{L_B}[/tex] +58

825 = 223[tex]\mathsf{L_B}[/tex]

[tex]\mathsf{L_B}[/tex] = 825/223

[tex]\mathsf{L_B}[/tex]  = 3.70 m

[tex]\mathsf{L_A = 5 - L_B}[/tex]

[tex]\mathsf{L_A = 5 - 3.70}[/tex]

[tex]\mathsf{ L_A}[/tex] = 1.30 m

The length the divider is to  equilibrium from Part A = 1.30 m and from Part B = 3.70 m

Which of the following do we need to know in order to calculate pH during an acid-base titration of a strong monoprotic acid with a strong monoprotic base? Select all that apply

a. the concentration of the acid
b. the concentration of the base titrant
c. the initial volume of the acid solution
d. the volume of the titrant used

Answers

Answer:

the volume of the titrant used

Explanation:

Acid-base titrations are usually depicted on special graphs referred to as titration curve. A titration curve is a graph that contains a plot of the volume of the titrant as the independent variable and the pH of the system as the dependent variable.

Hence, a titration curve is a graphical plot showing the pH of the analyte solution plotted against the volume of the titrant as the reaction is in progress. The titration curve is drawn by plotting data obtained during a titration, that is, volume of the titrant added (plotted on the x-axis) and pH of the system (plotted on the y-axis).

Which response includes all the following processes that are accompanied by an increase in entropy? 1) 2SO 2(g) + O 2(g) → SO 3(g) 2) H 2O(l) → H 2O(s) 3) Br 2(l) → Br 2(g) 4) H 2O 2(l) → H 2O(l) + 1/ 2O 2(g)

Answers

Answer: Reaction (1) , (3) and (4) are accompanied by an increase in entropy.

Explanation:

Entropy is the measure of randomness or disorder of a system. If a system moves from  an ordered arrangement to a disordered arrangement, the entropy is said to decrease and vice versa.

(1) [tex]2SO_2(g)+O_2(g)\rightarrow SO_3(g)[/tex]

3 moles of reactant are changing to 1 mole of product , thus the randomness is increasing. Thus the entropy also increases.

2) [tex]H_2O(l)\rightarrow H_2O(s)[/tex]

1 mole of Liquid reactant is changing to 1 mole of solid product , thus the randomness is decreasing. Thus the entropy also decreases.

3) [tex]Br_2(l)\rightarrow Br_2(g)[/tex]

1 mole of Liquid reactant is changing to 1 mole of gaseous product , thus the randomness is increasing. Thus the entropy also increases.

4)  [tex]H_2O_2(l)\rightarrow H_2O(l)+\frac{1}{2}O_2(g)[/tex]

1 mole of Liquid reactant is changing to half mole of gaseous product and 1 mole of liquid product, thus the randomness is increasing. Thus the entropy also increases.

Gas is contained in a 9.00-L vessel at a temperature of 24.0°C and a pressure of 5.00 atm. (a) Determine the number of moles of gas in the vessel. mol (b) How many molecules are in the vessel? molecules

Answers

Answer:

a. Moles in the vessel = 1.85 moles of the gas

b. 1.11x10²⁴ molecules are in the vessel

Explanation:

a.It is possible to determine moles of a gas using the general law of gases:

PV = nRT

Where P is pressure: 5.00atm; V is volume = 9.00L; R is gas constant: 0.082atmL/molK; T is absolute temperature: 273.15K +24.0 = 297.15K

Computing the values:

PV / RT = n

5.00atm* 9.00L / 0.082atmL/molK*297.15K = n

Moles in the vessel = 1.85 moles of the gas

b. With Avogadro's number we can convert moles of any compound to molecules thus:

Avogadro's number = 6.022x10²³ molecules / mole

1.85moles ₓ (6.022x10²³ molecules / mole) =

1.11x10²⁴ molecules are in the vessel

A sample of ammonia gas was allowed to come to equilibrium at 400 K. 2NH3(g) <----> N2(g) 3H2(g) At equilibrium, it was found that the concentration of H2 was 0.0484 M, the concentration of N2 was 0.0161 M, and the concentration of NH3 was 0.295 M. What was the initial concentration of ammonia

Answers

Answer:

0.327 M

Explanation:

Step 1: Write the balanced equation

2 NH₃(g) ⇄ N₂(g) + 3H₂(g)

Step 2: Make an ICE chart

        2 NH₃(g) ⇄ N₂(g) + 3 H₂(g)

I              x             0            0

C          -2y            +y         +3y

E         x-2y             y           3y

Step 3: Find the value of y

The concentration of N₂ at equilibrium is 0.0161 M. Then,

y = 0.0161

Step 4: Find the value of x

The concentration of NH₃ at equilibrium is 0.295 M. Then,

x-2y = 0.295

x-2(0.0161) = 0.295

x = 0.327

A 1.362 g sample of an iron ore that contained Fe3O4 was dissolved in acid with all of the iron being reduced to iron (II). The solution was acidified with sulfuric acid and titrated with 39.42 mL of 0.0281 M KMnO4, which oxidized the iron (II) to iron (III) while reducing the permanganate to manganese (II). Generate the balanced net ionic equation for the reaction. What is the mass percent of iron in this iron ore sample?

Answers

Answer:

a. MnO₄⁻ + 8H⁺ + 5Fe²⁺ → 5Fe³⁺ + Mn²⁺ + 4H₂O

b. 18.17% of Fe in the sample

Explanation:

a. In the reaction, Fe²⁺ is oxidized to Fe³⁺ and permanganate, MnO₄⁺ reduced to Mn²⁺, thus:

Fe²⁺ → Fe³⁺ + 1e⁻

MnO₄⁻ + 5e⁻ + 8H⁺ → Mn²⁺ + 4H₂O

5 times the iron and suming the manganese reaction:

MnO₄⁻ + 5e⁻ + 8H⁺ + 5Fe²⁺ → 5Fe³⁺ + 5e⁻ + Mn²⁺ + 4H₂O

MnO₄⁻ + 8H⁺ + 5Fe²⁺ → 5Fe³⁺ + Mn²⁺ + 4H₂O

b. Moles of permanganate in the titration are:

0.03942L × (0.0281 moles / L) = 1.108x10⁻³ moles of MnO₄⁻

Based on the reaction, 1 mole of permanganate reacts with 5 moles of iron, if 1.108x10⁻³ moles of MnO₄⁻ reacts, moles of iron are:

1.108x10⁻³ moles of MnO₄⁻ × (5 moles Fe²⁺ / 1 mole MnO₄⁻) =

4.431x10⁻³ moles of Fe²⁺. Molar mass of Fe is 55.845g/mol. 4.431x10⁻³ moles of Fe²⁺ are:

4.431x10⁻³ moles of Fe²⁺ ₓ (55.845g / mol) =

0.2474g of Fe you have in your sample.

Percent mass is:

0.2474g Fe / 1.362g sample ₓ 100 =

18.17% of Fe in the sample

The mass percent of iron in the sample is 22.6%.

The net ionic equation of the reaction is;

5Fe^2+(aq) + 8H^+(aq) + MnO4^-   -----> 5Fe^3+(aq) + Mn^2+(aq) + 4H2O(l)

Number of moles of MnO4^-  = 39.42/1000 L × 0.0281 M = 0.0011 moles

If 5 moles of Fe^2+ reacts with 1 mole of MnO4^-

x moles of Fe^2+ reacts with 0.0011 moles

x =  5 moles × 0.0011 moles/1 mole

x = 0.0055 moles

Mass of Fe^2+ =  0.0055 moles × 56 g/mol = 0.308 g

Mass percent of iron = 0.308 g/ 1.362 g × 100/1

= 22.6%

Learn more: https://brainly.com/question/10643807

A baseball has a mass of 0.145 kilograms. If acceration due to gravity is 9.8m/s,what is the weight of the baseball in newtons?

Answers

Answer:

I hope it works

Explanation:

As we know that

w=m*g

given m=0.145 , g=9.8

hence we get

w= (9.8)*(0.145)

w=1.421 m/sec 2

if its help-full thank hit the stars and brain-list it thank you

Use the standard half-cell potentials listed below to calculate the standard cell potential for the following reaction occurring in an electrochemical cell at 25°C. (The equation is balanced.) 3 Cl2(g) + 2 Fe(s) → 6 Cl-(aq) + 2 Fe3+(aq) Cl2(g) + 2 e- → 2 Cl-(aq) E° = +1.36 V Fe3+(aq) + 3 e- → Fe(s) E° = -0.04 V

Answers

The cell potential for the electrochemical cell has been 1.40 V.

The standard reaction for the cell will be:

[tex]\rm 3\;Cl_2\;+\;2\;Fe\;\rightarrow\;6\;Cl^-\;+\;2\;Fe^3^+[/tex]

The half-reaction of the cells has been:

[tex]\rm Fe^3^+\;+\;3\;e^-\;\rightarrow\;Fe[/tex]

The potential for this reduction has been -0.04 V.

[tex]\rm Cl_2\;+\;2\;e^-\;\rightarrow\;2\;Cl^-[/tex]

The potential for the reduction has been 1.36 V.

The cell potential has been: Potential of reduction - Potential of oxidation

Cell potential = 1.36 - (-0.04) V

Cell potential = 1.40 V.

The cell potential for the electrochemical cell has been 1.40 V.

For more information about the electrochemical cell, refer to the link:

https://brainly.com/question/22550969

What class of organic product results when 1-heptyne is treated with a mixture of mercuric acetate in aqueous sulfuric acid, and then HOCH2CH2OH with catalytic sulfuric acid

Answers

Answer:

2-methyl-2-pentyl-1,3-dioxolane

Explanation:

In this case, we have two reactions:

First reaction:

1-heptyne + mercuric acetate -------> Compound A

Second reaction:

Compound A + HOCH2CH2OH -------> Compound C

First reaction

In the first reaction, we have as a main functional group a triple bond. We have to remember that mercuric acetate in sulfuric acid will produce a ketone. The carbonyl group (C=O) would be placed in the most substituted carbon of the triplet bond (in this case, carbon 2). With this in mind, we will have as a product: heptan-2-one. (See figure 1).

Second reaction

In this reaction, we have as reagents:

-) Heptan-2-one

-) Ethylene-glycol [tex]HOCH_2CH_2OH[/tex]

-) Sulfuric acid [tex]H_2SO_4[/tex]

When we put ethylene-glycol with a ketone or an aldehyde we will form a cyclic acetal. In this case, this structure would be formed on carbon 2 forming 2-methyl-2-pentyl-1,3-dioxolane. (See figure 2).

I hope it helps!

For a particular reaction at 235.8 °C, ΔG=−936.92 kJ/mol , and ΔS=513.79 J/(mol⋅K) . Calculate ΔG for this reaction at −9.9 °C.

Answers

Answer:

-138.9 kJ/mol

Explanation:

Step 1: Convert 235.8°C to the Kelvin scale

We will use the following expression.

K = °C + 273.15 = 235.8°C + 273.15 = 509.0 K

Step 2: Calculate the standard enthalpy of reaction (ΔH°)

We will use the following expression.

ΔG° = ΔH° - T.ΔS°

ΔH° = ΔG° / T.ΔS°

ΔH° = (-936.92kJ/mol) / 509.0K × 0.51379 kJ/mol.K

ΔH° = -3.583 kJ (for 1 mole of balanced reaction)

Step 3: Convert -9.9°C to the Kelvin scale

K = °C + 273.15 = -9.9°C + 273.15 = 263.3 K

Step 4: Calculate ΔG° at 263.3 K

ΔG° = ΔH° - T.ΔS°

ΔG° = -3.583 kJ/mol - 263.3 K × 0.51379 kJ/mol.K

ΔG° = -138.9 kJ/mol

Calculate the pH of a buffer that is 0.13 M in lactic acid and 0.10 M in sodium lactate. Express your answer using two decimal places.

Answers

Answer:

pH of the buffer is 3.75

Explanation:

It is possible to find pH of a buffer using Hendersson-Hasselbalch equation:

pH = pKa + log [A⁻] / [HA]

Where [A⁻] is molar concentration of the conjugate base and [HA] concentration of the weak acid

In the lactic buffer, pKa = 3.86. Lactic acid is the weak acid and its conjugate base is tha lactate salt. H-H equation for this buffer is:

pH = 3.86 + log [Lactate] / [Lactic acid]

Replacing with the concentrations of the problem:

pH = 3.86 + log [0.10M] / [0.13M]

pH = 3.75

pH of the buffer is 3.75

9
What might happen if acidic chemicals were emitted into
the air by factories? Choose the best answer.
A
The acid would destroy metallic elements in the air
B
The acid would be neutralized by bases within clouds
C
Acid rain might destroy ecosystems and farmland
D
Violent chemical reactions would take place within the
atmosphere
co search
O
BI

Answers

Your answer is B. Acid rain might destroy ecosystems and farmland

The complex ion Fe(CN)63- is paramagnetic with one unpaired electron. The complex ion Fe(SCN)63- has five unpaired electrons. Where does SCN- lie in the spectrochemical series with respect to CN-?

Answers

Answer:

SCN- is a weak field ligand while CN- is a strong field ligand

Explanation:

The spectrochemical series is an arrangement of ligands according to their magnitude of crystal field splitting. Ligands that cause only a small degree of crystal field splitting are called weak field ligands while ligands that cause large crystal field splitting are called strong field ligands.

Strong field ligands often lead to the formation of low spin complexes with the least number of unpaired electrons while high spin complexes are formed by weak field ligands.

CN- is a strong field ligand as it lies towards the right hand side of the spectrochemical series.

SCN- is a weak field ligand hence it forms a high spin complex having the maximum number of unpaired electrons for Fe^3+, hence the answer.

SCN⁻ lies in the lower (weak field) region of the spectrochemical series while CN⁻ lies in the higher (stronger field) region.

CN⁻ is a strong field ligand with a large splitting constant, and it is high up in the spectrochemical series.

Conversely, SCN⁻ is a weak field ligand with a low splitting constant, and it is lower in the spectrochemical series.

Hence, SCN⁻ lies in the lower (weak field) region of the spectrochemical series while CN⁻ lies in the higher (stronger field) region.

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