An alkyne with the molecular formula C6H10 was treated with ozone followed by water to produce only one type of carboxylic acid. Draw the structure of the starting alkyne and the product of ozonolysis.

The balanced equation is given as,

2Al³⁺ + 3H₂O + Br⁻ → 2 Al + BrO³⁻ + 6H⁺

Explanation:

H₂O + Br⁻ + Al³⁺ → Al + BrO₃⁻ + H⁺

Here the half reactions are:

Al³⁺→ Al     [reduction]

Br⁻ → BrO₃⁻ [oxidation]

Now we have to balance the half reactions as,

Al³⁺ + 3e⁻ → Al

To balance H atoms we have to add water and electrons.

3H₂O + Br⁻→ BrO³⁻ + 6H⁺ + 6e⁻

Now we have to balance the electrons as,

2Al³⁺ + 6e⁻ → 2 Al

3H₂O + Br⁻→ BrO³⁻ + 6H⁺ + 6e⁻

Now we have to add  both the equations as,

2Al³⁺ + 6e⁻ + 3H₂O + Br⁻ → 2 Al + BrO³⁻ + 6H⁺ + 6e⁻

6 electrons on both sides of the equation gets cancelled and so the balanced reaction can be written as,

2Al³⁺ + 3H₂O + Br⁻ → 2 Al + BrO³⁻ + 6H⁺

Answer:

rate = k [Br⁻][H⁺]²[BrO₃⁻]

Explanation:

Here we are going to determine the rate law for the reaction

5 Br-(aq) + BrO3-(aq) + H+(aq) → 3 Br2(aq) + 3 H2O(l)

by performing experiments in which we vary concentration of the reactants and determining the effect this has on the initial rate.

Exp     [  Br− ]             [BrO3− ]                [ H+]                 (Δ[BrO3− ]/Δt)0 M·s−1

I            0.10                0.10                     0.10                      6.8 x 10-4

II           0.15                0.10                      1.0                          10-3

III          0.10                0.20                    0.10                       1.4 x 10-3

IV         0.10                0.10                     0.25                      4.3 10-3

The way to do the comparison is by taking experiments in which we keep constant the concentration of two of the reactants and vary the third and study the effect this change has on the initial rate of the reaction.

Comparing experiments I and III we see that the initial reaction doubled when we doubled the  [BrO3− ]    while keeping the other two the same. Thus the reaction rate is of order 1 respect to  [BrO3− ].

In experiments I and IV we increased the concentration of H⁺ by 2.5 times, and the rate of the reaction increased by a factor of 6.3 which is 2.5 squared . If you do not see it, lets try using logarithms

(0.25/ 0.10)^ x = 4.3 x 10⁻³ / 6.8 x 10⁻⁴

2.5 ^ x  = 6.3235

x log 2.5 = log 6.3235

 0.40 x = 0.80 ∴ x = 2

Therefore the rate is second order respect to [H⁺].

Comparing experiments I and II we see that increasing the concentration of Br⁻ by a factor of 1.5, the initial rate also went up by a factor of 1.5 ( 1.0 x 10⁻³ / 6.8 x 10⁻⁴ =1.5. Thus the rate is first order respect to [ Br⁻ ].

Then our rate law is

rate = k [Br⁻][H⁺]²[BrO₃⁻]

The rate law for the reaction 5 Br-(aq) + BrO3-(aq) + H+(aq) → 3 Br2(aq) + 3 H2O(l) is determined to be rate = k[Br-]^1[BrO3-]^1[H+]^0 by analyzing how the rate changes with varying initial concentrations of the reactants.

To determine the rate law for the reaction: 5 Br-(aq) + BrO3-(aq) + H+(aq) → 3 Br2(aq) + 3 H2O(l), we need to examine how the rate changes with varying initial concentrations of the reactants. The change in rates as concentrations are altered provides the order of reaction with respect to each reactant. Our data from experiments I, II, and III indicate that when the Br- concentration increases by a factor of 1.5 (from 0.10 to 0.15), the rate also increases by a similar factor (from 6.8 x 10-4 to 1.0 x 10-3), suggesting a first-order dependence: [Br-]^1. However, when we double the initial concentration of BrO3 (experiment III vs I), the rate also doubles indicating first-order dependence on BrO3 as well: [BrO3]^1. The experiment IV suggests a zero-order dependence on [H+]. Hence, the rate law for this reaction is rate = k[Br-]^1[BrO3-]^1[H+]^0, where k is the rate constant.

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Answer:

pH = 11.38

Explanation:

The pH of a solution is given as:

pH = pKw − pOH = 14.00 + log[OH-]

Since the concentration of Ca(OH)2 is 0.0012 M, the [OH−] is twice that, or 0.0024 M, resulting in pOH = 2.62 and pH = 11.38.

Final answer:

The pH of a solution of 0.0012M calcium hydroxide at 25.0°C is 11.38.

Explanation:

The pH of a solution of 0.0012M calcium hydroxide at 25.0°C can be calculated using the concentration of hydroxide ions ([OH-]).

Calcium hydroxide is a strong base that dissociates completely in water to form two hydroxide ions for every formula unit dissolved.

The concentration of the solute is 0.0012M, but because Ca(OH)2 is a strong base, the actual concentration of hydroxide ions ([OH-]) is two times this, or 2 × 0.0012M = 0.0024M.

Since [OH-] = 0.0024M, the pOH can be calculated as pOH = -log([OH-]) = -log(0.0024) = 2.62. The pH can then be calculated using the expression pH = 14 - pOH = 14 - 2.62 = 11.38.

The solubility of Cu(OH)2(s) increases in the presence of a strong acid due to the dissolution of the salt and formation of Cu2+ ions and water. The balanced net ionic equation for this process is Cu(OH)2(s) + 2H+(aq) → Cu2+(aq) + 2H2O(l). The equilibrium constant (K) for the reaction can be calculated using the concentrations of the reactants and products at equilibrium.

The balanced net ionic equation to show why the solubility of Cu(OH)2(s) increases in the presence of a strong acid is:

Cu(OH)2(s) + 2H+(aq) → Cu2+(aq) + 2H2O(l)

Cu(OH)2 is a sparingly soluble salt, meaning it has low solubility in water. However, in the presence of a strong acid, the concentration of H+ ions increases, leading to the dissolution of Cu(OH)2 and the formation of Cu2+ ions and water.

The equilibrium constant for this reaction can be calculated by using the concentrations of the reactants and products at equilibrium. The equation for calculating the equilibrium constant (K) is:

K = [Cu2+][H2O2]/[Cu(OH)2][H+]2

Note: The square brackets indicate the concentrations of the species in the solution.

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Answer:

Bond Order = ½ [# bonded e--# antibonded e-]

C2= ½ [6-2] = 2

C2+= ½ [5-2] = 1.5

C2- = ½ [7-2] = 2.5

Bond Length: C2+> C2> C2-

Bond Energy: C2-> C2> C2+

Explanation:

Bond order is a measurement of the number of electrons involved in bonds between two atoms in a molecule.

How to calculate bond order

1. Draw the Lewis structure.

2. Count the total number of bonds.

3. Count the number of bond groups between individual atoms.

4. Divide the number of bonds between atoms by the total number of bond groups in the molecule.

Bond energy (E) is defined as the amount of energy required to break apart a mole of molecules into its component atoms. It is a measure of the strength of a chemical bond.

The higher the bond order, the shorter the bond length and the greater the bond energy.

The molecular orbital configuration of a chemical specie shows the arrangement of electrons in such a specie into molecular orbitals in accordance with the Aufbau principle.

For C2;

σ1s2 σ*1s2 σ2s2 σ*1s2 π2py2 π2pz2

Bond order = 1/2(8 - 4) = 2

For C2^-;

σ1s2 σ*1s2 σ2s2 σ*1s2 π2py2 π2pz2 σ2px1

Bond order = 1/2(9 - 4) = 2.5

For C2^+;

σ1s2 σ*1s2 σ2s2 σ*1s2 π2py2 π2pz1

Bond order = 1/2(7 - 4) = 1.5

We must recall that the higher the bond order, the shorter the bond length and the greater the bond energy. Hence;

a) In order of increasing bond energy; C2^+ <  C2 < C2^-

b) In order of increasing bond length; C2^- < C2 < C2^+

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In iodine tribromide, the I-Br sigma bond is formed through the straight-on overlap of sp3d hybrid atomic orbitals on iodine with p orbitals on bromine.

In iodine tribromide (IBr3), the central iodine (I) atom is surrounded by three bromine (Br) atoms. The I-Br bond that forms is a sigma (σ) bond, which is the result of a straight-on overlap of atomic orbitals. Sigma bonding typically involves the hybridized orbitals. Specifically, in IBr3, the I atom undergoes hybridization and forms two lone pairs and three σ bonds with Br atoms using sp3d hybrid orbitals. So, the hybrid orbital on the iodine atom that makes up the σ bond with a bromine atom is the sp3d hybrid orbital.

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The sigma bond in IBr3 between the central I atom and an outer Br atom is formed by the overlap of a sp3d hybrid orbital from iodine with a p orbital from bromine to accommodate the T-shaped geometry of the molecule.

In iodine tribromide (IBr3), the sigma bond between the central iodine (I) atom and an outer bromine (Br) atom is typically formed by the overlap of a hybrid orbital from the iodine with a p orbital from the bromine. When we consider the valence shell electronic configuration of iodine (5s2 5p5), upon forming IBr3, three of the p orbitals would be used to form sigma bonds with the bromine atoms, and this would likely involve hybridization to accommodate the molecular geometry of IBr3. The exact nature of the hybrid orbitals would depend on the geometry of the molecule, which, due to the presence of only three bond pairs and two lone pairs, adopts a T-shaped geometry, pointing towards sp3d hybridization. However, the exact details of this hybridization can only be confirmed with molecular orbital calculations

Answer:

The correct answer is option 3. Run a test reaction of crude oil with ocean water over time with Oil Spill Eater present

Explanation:

In any laboratory experiment, all the apparatus needed to carry out a particular experiment must be provided. In this case, our apparatus will be crude oil with ocean water and oil spill eater which is the enzyme used.

We can then run a test reaction of crude oil with ocean water over time with Oil Spill Eater present.

In the oxidation of HBr by O2, the elementary reactions add up to give the overall reaction and thus confirm Hess's Law. The rate determining step is the first one, and the intermediates in the reaction are HOOBr and HOBr. The inability to detect these among the final products does not disprove this mechanism.

The overall reaction for the gas-phase oxidation of HBr by O2 is written as: 4 HBr(g) + O2(g) → 2 H2O(g) + 2 Br2(g)

First, we need to confirm if the elementary reactions add up to the overall reaction. Based on Hess's Law, we can see that if we add the elementary reactions: HBr(g) + O2(g) → HOOBr(g), HOOBr(g) + HBr(g) → 2 HOBr(g), and HOBr(g) + HBr(g) → H2O(g) + Br2(g), they give the same overall reaction, thus confirming Hess's Law.

Second, the experimentally determined rate law is first order with respect to both HBr and O2, which suggests the rate determining step is the first one: HBr(g) + O2(g) → HOOBr(g). This is because the rate-determining step usually determines the order of the reaction.

The intermediates in this reaction are the species that are produced in one step and consumed in another, in this case HOOBr(g) and HOBr(g).

Lastly, the inability to detect HOBr or HOOBr among the final products does not necessarily disprove the mechanism. This is because these are intermediates and are typically used up in subsequent reaction steps, leading them to not appear in the final product of the reaction.

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The given elementary steps add up to the overall reaction. The rate-determining step involves HBr and O₂ as indicated by the rate law. HOBr and HOOBr are intermediates, and their absence among products does not disprove the mechanism.

let's analyze and confirm each part.

a. Confirming the Elementary Reactions:

The given elementary steps are:

Adding these reactions together:

Combining the species on both sides gives us:

4 HBr(g) + O₂(g) → 2 H₂O(g) + 2 Br₂(g)

This confirms that the elementary steps add up to the overall reaction.

b. Rate-Determining Step:

The given rate law is first order with respect to both HBr and O₂, indicating that the rate-determining step involves one molecule each of HBr and O₂. Hence, the first step, HBr(g) + O₂(g) → HOOBr(g), is the rate-determining step.

c. Intermediates:

Intermediates are species that appear in the reaction mechanism but not in the overall reaction. Here, HOOBr and HOBr are intermediates.

d. Detecting Intermediates:

Not detecting HOBr or HOOBr among the products does not disprove the mechanism. Intermediates typically have short lifespans and do not accumulate significantly; hence they might not be detected in significant amounts in the product mixture.

Answer:

a. 86.7KJ/mol

b. 2.5 * 10^30

c. 1.42 * 10^55

d. Lightning produces more amount of NO from nitrogen and oxygen thus giving a better crop as NO is essential for crop growth

Explanation:

Complete question is as follows;

Consider the reaction N2(g)+ O2(g) —-> 2NO(g)

Given that ΔG0 for the reaction at 25 degrees celsius is 173.4 kJ/mol a) calculate the standard freeenergy of formation of NO and B) calculate KP ofthe reaction. C) One of the starting substances in smog formationis NO. Assuming that the temperature in a running automobile engine is 1100 degreesC, estimate KP of the above reaction.D) As farmers know, lighting helps to produce a better crop.Why?

solution

Please check attachment for complete solution and step by step explanation

The mass of (NH4)3PO4 is 111.75grams.

HOW TO CALCULATE THE MASS:

mass (g) = moles (mol) × molar mass (g/mol)

{14 + 1(4)}3 + 31 + 16(4)

= (14+4)3 + 31 + 64

= 54 + 31 + 64

= 149g/mol

mass in grams = 149 × 0.75

mass in grams = 111.75g

Therefore, the mass of (NH4)3PO4 is 111.75grams.

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Answer:

A solution that is 0.10 M HCN and 0.10 M LiCN.

Explanation:

A buffer solution is a solution of a weak acid and its conjugate base or a solution of a weak base and its conjugate acid, so the pH of the solution changes in a very little range when an acid or base is added to the solution.

Let's evaluate each statement:

(a) A solution that is 0.10 M LiOH and 0.10 M HNO₃

This is not a buffer solution since the HNO₃ is not the acid conjugate of the LiOH, and also, the LiOH is a strong base and the HNO₃ is a strong acid.

(b) A solution that is 0.10 M HCN and 0.10 M LiCN

This is a buffer solution, with the following reaction:

HCN + H₂O ⇄ CN⁻Li⁺ + H₃O⁺  

The acid HCN is a weak acid and the LiCN is its conjugate base. The fact that the concentrations are equal for both is appropriate for the buffer solution.

(c) A solution that is 0.10 M NaCl and 0.10 M HCl

This is not a buffer solution since the HCl is a strong acid. It dissociates in water to form Cl⁻ and H⁺.

(d) A solution that is 0.10 M Li and 0.10 M HNO₃

This is not a buffer solution since the HNO₃ is not the conjugate base of Li, the HNO₃ is a strong acid that dissociates in water to form H⁺ and NO₃⁻.

(e) A solution that is 0.10 M HCN and 0.10 M LiCl

This is not a buffer solution since the LiCl is not the conjugate base of the acid HCN.

Therefore, the correct option is: A solution that is 0.10 M HCN and 0.10 M LiCN.

I hope it helps you!  

Answer:

are you asking if it's true or false?

Answer:

no

Explanation:

The question is incomplete, the complete question is:

Standard reduction potentials for zinc(II) and copper(II)

The standard reduction potential for a substance indicates how readily that substance gains electrons relative to other substances at standard conditions. The more positive the reduction potential, the more easily the substance gains electrons. Consider the following:

Zn2+(aq)+2e−→Zn(s),Cu2+(aq)+2e−→Cu(s), E∘red=−0.763 V E∘red=+0.337 V

Part B

What is the standard potential, E∘cell, for this galvanic cell? Use the given standard reduction potentials in your calculation as appropriate.

Express your answer to three decimal places and include the appropriate units.

Answer:

1.100 V

Explanation:

E∘cell= E∘cathode - E∘anode

E∘cathode= +0.337 V

E∘anode= −0.763 V

E∘cell= 0.337-(-0.763)

E∘cell= 1.1V

Answer:

[tex]T_{2,H_2O}=89^oC[/tex]

Explanation:

Hello,

In this case, we notice that the energy gained by the first sample of water is lost by the second sample of water as they are heated and cooled respectively, therefore, in terms of heats:

[tex]Q_{1,H_2O}=-Q_{2,H_2O}[/tex]

Which in terms of masses, heat capacities and temperatures is:

[tex]m_{1,H_2O}*Cp_{1,H_2O}*(T_{EQ}-T_{1,H_2O})=-m_{2,H_2O}*Cp_{2,H_2O}*(T_{EQ}-T_{2,H_2O})[/tex]

In such a way, as the heat capacity is the same and the initial temperature of the cold water is required, we solve for it via:

[tex]m_{1,H_2O}*(T_{EQ}-T_{1,H_2O})=-m_{2,H_2O}*(T_{EQ}-T_{2,H_2O})\\T_{EQ}-T_{2,H_2O}=\frac{m_{1,H_2O}*(T_{EQ}-T_{1,H_2O})}{-m_{2,H_2O}} \\\\T_{2,H_2O}=T_{EQ}+\frac{m_{1,H_2O}*(T_{EQ}-T_{1,H_2O})}{m_{2,H_2O}}[/tex]

With the given data we obtain:

[tex]T_{2,H_2O}=T_{EQ}+\frac{m_{1,H_2O}*(T_{EQ}-T_{1,H_2O})}{m_{2,H_2O}}\\\\T_{2,H_2O}=48.9^oC+\frac{102g*(48.9^oC-22.6^oC)}{66.9g} =48.9^oC+40.1^oC\\T_{2,H_2O}=89^oC[/tex]

Which means the second sample was hot.

Regards.

Answer:

The initial temperature of the second sample of water is 8.8 °C

Explanation:

Step 1 :Data given

Mass of water = 102 grams

Initial temperature of water = 22.6 °C =

Mass of other water = 66.9 grams

The final temperature is 48.9 °C

The specific heat capacity of liquid water is 4.184 J/g *K = 4.184 J/g°C

Step 2: Calculate the initial temperature of the second sample water

Heat gained = heat lost

Qgained = -Q lost

Q = m* C* ΔT

m(water1)*C(water)*ΔT(water1) = -m(water2) * C(water) *ΔT(water2)

⇒with m(water1) = the mass of the water at 22.6 °C = 102 grams

⇒with C(water) = the specific heat of water = 4.184 J/g°C

⇒with ΔT = the change of water of the first sample = 48.9 - 22.6 = 26.3 °C

⇒with m(water2) = the mass of the other unknown sample water = 66.9 grams

⇒with with C(water) = the specific heat of water = 4.184 J/g°C

⇒with ΔT = the change of water of the second sample = 48.9 - T1

102 * 4.184 * 26.3 °C = 66.9 * 4.184 * (48.9 - T1)

102 * 26.3 = 66.9 * (48.9 - T1)

2682.6 = 3271.4 - 66.9 T1

-588.8 = -66.9 T1

T1 = 8.8 °C

The initial temperature of the second sample of water is 8.8 °C

Answer:

The equivalent circuit for the electrode while the electrolyte gel is fresh

From the uploaded diagram the part A is the electrolyte, the part part B is the electrolyte gel when is fresh and the part C is the surface of the skin

Now as the electrolyte gel start to dry out the resistance [tex]R_s[/tex] of the gel begins to increase and this starts to limit the flow of current . Now when the gel is then completely dried out  the resistance of the gel [tex]R_s[/tex] then increases to infinity  and this in turn cut off flow of current.

The diagram illustrating this is shown on the second uploaded image

Explanation:

Find figure in the attachment

Answer:

Explanation:

solution to the question is found below

Answer:

Explanation:

see answer below in the attached file.

The carnitine shuttle system is an important system that helps in the fatty acid oxidation.

The functions of the enzymes and compounds of the carnitine shuttle system include the following:

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Answer:

Step 1, formation of the iminium ion

2) enol reaction with iminium ion to form beta-amino enone

Explanation:

Please find attached the detailed reaction mechanism and the structure of iminium ion separately

The Mannich reaction is a three-component reaction that forms a ß-amino ketone or aldehyde. It involves protonation, nucleophilic attack, formation of an iminium ion, enol attack, and deprotonation.

The Mannich reaction is a three-component organic reaction involving a ketone, an aldehyde, and an amine under acid-catalyzed conditions to form a ß-amino ketone or ß-amino aldehyde. The mechanism of the reaction can be broken down into several steps:

Given these steps, the structure of iminium ion 2 involves a nitrogen atom double-bonded to a carbon atom derived from the original carbonyl.

Answer:

Option 4. loss of electrons, resulting in an increased oxidation number.

Explanation:

Oxidation is a process involving loss of electron(s). When this happens the oxidation number of the atom being oxidised increases. This can be seen when calcium (Ca) reacts with chlorine (Cl2) to form calcium chloride (CaCl2) according to the equation given below:

Ca + Cl2 —> CaCl2

The oxidation number of calcium increases from 0 to +2. This implies that calcium is being oxidised as it loses its electrons. The oxidation number of chlorine decreases from 0 to - 1 as it gains electron.

Now, we can see that the oxidation of calcium i.e lose of electrons increased its oxidation number from 0 to +2.

From the simple illustrations above, we can see clearly that oxidation involves loss of electrons, resulting in an increased oxidation number.

Answer:

See the attached file for the structure

Explanation:

See the attached file

Answer:

1. NaI     Y     AgI

2. K2S     Y   CuS

3. K2CO3      N

Explanation:

1. When we add NaI to the mixture, the reaction that takes place is:

Such a reaction does not happen with Cu⁺².

2. When we add K₂S to the mixture, the reaction that takes place is:

Such a reaction does not happen with Ag⁺.

3. When we add K₂CO₃ to the mixture, the reactions that take place are:

This means both Cu⁺² and Ag⁺ would precipitate, thus they would not be separated.

Answer:

Explanation:

The question is not completely clear because some missing parts or grammar (syntax) errors.

Interpreting it, the question is what is the Kelvin temperatue of the air in a tire, when the pressure in the tire has increased to 225 kPa, if the initial conditions of the air inside the tire were 188kPa of pressure and a temperature of 32ºC.

To solve this, you can assume constant volume and use the law of Gay-Lussac for ideal gases:

          [tex]\dfrac{P_1}{T_{1}}=\dfrac{P_2}{T_2}[/tex]

That equation works with absolute temperatures, i.e. Kelvin.

Then solve for T₂, substitute and compute:

        [tex]T_2=P_2\times \dfrac{T_1}{P_{1}}=225kPa\times \dfrac{305.15K}{188kPa}[/tex]

        [tex]T_2=413.90K\approx 414K\longleftarrow answer[/tex]

Answer:I disagree, the molarity is meant to be higher if all other factors remain constant.

Explanation:

Molarity=amount of substance/volume of the liquid

So,the lesser,the volume, the higher the molarity and vice versa.

The molarity calculation of Na2CO3 could indeed be lower if the precipitate was not adequately dried. Excess moisture adds to the mass, misrepresenting the actual quantity of Na2CO3, which may lead to a lower calculated molarity.

In the field of chemistry, the student's claim about the molarity of Na2CO3 (sodium carbonate) being inaccurately low due to incomplete drying of the precipitate can be valid. As the drying process helps in removing water molecules from the precipitate, any remaining moisture can add extra mass, causing the quantity of pure Na2CO3 to be undervalued. While calculating the molarity, this would imply more volume per mole, hence a lower molarity. This is a common issue in stoichiometry where proper handling and processing of chemical substances significantly affect the outcome of calculations.

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Answer:

Explanation:

question solved.

The rate law is defined as the rate of a chemical reaction, which is dependent on the concentration of the reactants.

It can be calculated by the formula:

[tex]\text v_o &= \text k [\text A]^x [\text B]^y[/tex]

where,

In the given reaction:

[tex]\begin{aligned}2\;\text{ClO}_2\left(aq \right )+2\text{OH}^-\left(aq \right )\rightarrow\text{ClO}_3^-\left(aq \right )+\text{ClO}_2^-+\text{H}_2\text{O} \end{aligned}[/tex]

Using the rate law formula, we get:

Rate = K [Cl]ⁿ  [OH⁻]ⁿ

0.0248 =  K [0.060]ⁿ  [0.030 ]ⁿ       ........(1)

0.00276 = K  [0.020]ⁿ  [0.030 ]ⁿ   .........(2)

0.00828 = K  [0.020]ⁿ  [0.090]ⁿ     ..........(3)

Dividing (1) equation by (2), we get:

[tex]\begin {aligned} \dfrac {0.0248}{0.00276}&= \dfrac{\text K (0.06)^n (0.03)^n} {\text K (0.02)^n (0.03)^n}\\\\9 &= \dfrac{0.06}{0.02}^n\\\\3 ^2 &= 3 ^ m\\\text m &=2\\\end {aligned}[/tex]

Dividing the equation (2) by (3), we get:

[tex]\begin {aligned} \dfrac {0.00276}{0.00828 }&= \dfrac{\text K (0.02)^n (0.03)^n} {\text K (0.02)^n (0.09)^n}\\\\\dfrac{1}{3} &= \dfrac{0.03}{0.09}^n\\\\ \dfrac{1}{3} ^1 &= \dfrac{1}{3} ^n\\\\\text n &=2\\\end {aligned}[/tex]

Based on the rate law:

a. Rate = K [ClO₂]²⁻ [OH⁻]¹

b. 0.0248 = K [0.06]² [0.03]¹

   K = [tex]\dfrac {0.0248}{(0.06)^2 \times 0.03}[/tex]  = 229.63 m⁻²s⁻¹

c. Rate = K [ClO]₂ [OH⁻]ⁿ

  Rate = 229.62 x (0.2)² (0.05)¹

Therefore, the rate of the reaction is 0.45926 m/sec.

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The volume of Potassium chloride solution is 4.8 L.

Explanation:

Molarity is found by dividing the number of moles of the given substance by its volume in litres, and so the unit of molarity is mol/L.

Here number of moles is given as 3.34 mol

Molarity = 0.7 M

Volume = ? L

Molarity = [tex]$\frac{moles}{Volume (L) }[/tex]

To find the volume we have to rearrange the equation as,

Volume (L) = [tex]$\frac{moles}{molarity}[/tex]

Now, we have to plug in the values as,

Volume (L) = [tex]$\frac{3.34 mol}{0.7 mol/L}[/tex]

                  = 4.77 ≈ 4.8 L

So the volume of Potassium chloride solution is 4.8 L.

Answer:

See explaination andd attachment

Explanation:

Treatment of aldehydes and ketones with a suitable base can lead to the formation of a nucleophilic species called an enolate that reacts with electrophiles. These C nucleophiles are useful for making new carbon-carbon bonds.

Please kindly see attachment for the drawings.

Answer:

A.) ΔG = 107.8 kJ/mol

ΔG rxn is positive implies that the reaction is non spontaneous.

B) P (NO) = 5.0 x 10^-7 atm

C) T > 923.4 K, therefore, the temperature should be 924K in order to make the reaction spontaneous.

Explanation:

N2O(g) + NO2(g) - 3NO(g)

A) ΔG = ΔG products - ΔG reactants

 ΔGf (N2O) = 103.7 kJ/mol

ΔGf (NO2) = 51.3 kJ/mol

ΔGf (NO) = 87.6 kJ/mol

ΔG rxn = 3*87.60 - (103.7 + 51.3)

= 107.8 kJ/mol

ΔG rxn is positive implies that the reaction is non spontaneous.

(B). P (N2O) = P (NO2) = 1 atm

ΔG rxn = ΔG°rxn + RTln(K)

ΔG rxn = ΔG°rxn + RT×ln (P NO)^3 / [P(N2O) × P(NO2)]

When the reaction ceases to be spontaneous, then ΔG rxn = 0.

0 = 107.8 × 10^3 + 8.314 × 298 × ln (P NO)^3 / (1 * 1)

107.8 × 10^3 = - 8.314 × 298 × ln (P NO)^3

ln (P NO)^3 = - 43.51

3 × ln (P NO) = - 43.51

ln (P NO) = -14.50

P (NO) = e^(-14.50)

P (NO) = 5.0 x 10^-7 atm

(C). For a reaction to be spontaneous, ΔG rxn should be negative.

It is known that:

ΔG= ΔH - TΔS

ΔG= ΔH - TΔS < 0

ΔH< TΔS

T < ΔH / ΔD

For the reaction:

ΔH = 3×NO - (N2O + NO2)

= 3×(91.3) - (81.6 + 33.2)

= 159.1 kJ/mol

ΔS = 3×210.8 - (220.0 + 240.1)

= 172.3 J/mol.K

= 0.1723 kJ/mol.K

From the above expression.

T > 159.1 / 0.1723

T > 923.4 K

Therefore, the temperature should be 924K in order to make the reaction spontaneous.

Answer:

Explanation:

Substance A

Answer: The new pressure will be 1.42 atm

Explanation:

To calculate the final pressure of the system, we use the equation given by Gay-Lussac Law. This law states that pressure of the gas is directly proportional to the temperature of the gas at constant pressure.

Mathematically,

[tex]\frac{P_1}{T_1}=\frac{P_2}{T_2}[/tex]

where,

[tex]P_1\text{ and }T_1[/tex] are the initial pressure and temperature of the gas.

[tex]P_2\text{ and }T_2[/tex] are the final pressure and temperature of the gas.

We are given:

[tex]P_1=1.18atm\\T_1=4^0C=(4+273)K=277K\\P_2=?\\T_2=60^0C=(60+273)K=333K[/tex]

Putting values in above equation, we get:

[tex]\frac{1.18}{277}=\frac{P_2}{333}\\\\P_2=1.42[/tex]

Hence, the new pressure will be 1.42 atm

Answer:

The correct option is: provide a source of counterions to prevent the build-up of charge at both the cathode to the anode.

Explanation:

A salt bridge is a U-shaped glass tube that is used in a voltaic cell or galvanic cell to connect the oxidation and reduction half-cells and complete the electric circuit.

It allows the ions to pass through it, thus preventing the accumulation of charge on the anode and cathode as the chemical reaction proceeds.

Therefore, the correct option is: provide a source of counterions to prevent the build-up of charge at both the cathode to the anode.