The beta-ketoester Claisen product is generated from the product of this final step by addition of dilute HCl. Write the final step of the Claisen condensation using curved arrows to show electron reorganization.

Answer:

Explanation:

An electron-donating heteroatom substituent at position-2 of a furan promotes regiospecific opening of the 7-oxa bridge of the Diels-Alder cycloadduct with hexafluoro-2-butyne, producing a 4-heterosubstituted 2,3-di(trifluoromethyl)phenol building block in a single step. The phenol and heteroatom substituent are easily transformed to the corresponding iodide or triflate that readily undergoes Heck, Suzuki, and Stille reactions to install a variety of substituents in high yields. This methodology provides a facile and general synthesis of 1,4-disubsituted 2, 3-di(trifluoromethyl)benzenes.

Final answer:

The Diels-Alder reaction of 1,3-butadiene with hexafluoro-2-butyne would form a six-membered ring with a hexafluoroalkyl side group, yielding a bicyclic compound.

Explanation:

The Diels-Alder reaction between 1,3-butadiene and hexafluoro-2-butyne would yield a six-membered ring product, specifically a bicyclic compound due to the formation of a cyclohexene derivative where the double bond is part of the ring and the hexafluoroalkyl group is attached to two adjacent carbons in the ring. It involves the 1,3-butadiene acting as the diene and the hexafluoro-2-butyne acting as the dienophile in an intermolecular reaction. The Diels-Alder reaction creates two new sigma bonds and a pi bond across the alkene and the alkyne to form the new cyclohexene ring.

The given isotopes of carbon, i.e., 12C and 13C, have the same number of protons but the number of neutrons differs. So, the statement 12C and 13C have different numbers of neutrons, is correct.

Isotopes are members of the same element's family but have variable numbers of neutrons despite having the same number of protons.

The element carbon can have different isotopes, based on the number of neutrons.

The isotopes of carbon include, C10, C11, C12, C13, C14, C15, and C16.

The given isotope, 12C and 13C, possess similar chemical properties and the same atomic number.

But the mass of the 13C isotope of carbon is more, because it possesses, one extra neutron, as compared to the 12C isotope of carbon.

Thus, the statement about the isotopes, “12C and 13C have different numbers of neutrons”, is correct.

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The statement that 13C has one more electron than 12C is incorrect. Both carbon-12 (12C) and carbon-13 (13C) isotopes have the same number of electrons due to both having the same atomic number. They only differ in the number of neutrons they possess.

The statement '13C has one more electron than 12C' is false. Both isotopes, carbon-12 (12C) and carbon-13 (13C) have the same atomic number '6', which represents the number of protons and, under stable conditions, also views it as the same number of electrons.

So, while 12C and 13C indeed differ in the number of neutrons (12C has 6 neutrons and 13C has 7), they possess the same number of protons and electrons. Furthermore, their atomic weights are different as a result of the difference in the number of neutrons. The atomic weight of 12C is closer to 12 and that of 13C is closer to 13.

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Answer:

The partial pressure of the other gases is 0.009 atm

Explanation:

Step 1: Data given

Air is about 78.0% nitrogen molecules and 21.0% oxygen molecules and 1% of other gases.

The atmospheric pressure = 0.90 atm

Step 2: Calculate mol fraction

If wehave  100  moles of air, 78 moles will be nitrogen,

21  moles will be  oxygen, and  1  mol will be other gases.

Mol fraction = 1/100 = 0.01

Step 3: Calculate the partial pressure of the other gases

Pgas = Xgas * Ptotal

⇒ Pgas = the partial pressure = ?

⇒ Xgas = the mol fraction of the gas = 0.01

⇒Ptotal = the total pressure of the pressure = 0.90 atm

Pgas = 0.01 * 0.90 atm

Pgas = 0.009 atm

The partial pressure of the other gases is 0.009 atm

Answer:

The answer is B

Explanation:

Answer:

[tex]BaI_2\ ^.6H_2O[/tex]

Explanation:

Hello,

In this case, by knowing the volume and the molarity of the barium iodide, is it possible to compute the residue's mass as shown below:

[tex]m_{BaI_2}=0.500L*0.1133\frac{molBaI_2}{L}*\frac{391.136gBaI_2}{1molBaI_2} =22.16gBaI_2[/tex]

Nevertheless, the obtained value is lower than the obtained by 6.133 g which means that mass corresponds to water forming a hydrate. In such a way, one could know how many waters in form of hydrate remain with the residue by a trial-error procedure as shown below:

[tex]m=28.28g=0.500L*0.1133\frac{molBaI_2}{L}*\frac{(391.136+18)gBaI_2\ ^.H_2O}{1molBaI_2} =23.17g\rightarrow No\\m=28.28g=0.500L*0.1133\frac{molBaI_2}{L}*\frac{(391.136+2*18)gBaI_2\ ^.2H_2O}{1molBaI_2} =24.20g\rightarrow No\\m=28.28g=0.500L*0.1133\frac{molBaI_2}{L}*\frac{(391.136+3*18)gBaI_2\ ^.3H_2O}{1molBaI_2} =25.22g\rightarrow No\\m=28.28g=0.500L*0.1133\frac{molBaI_2}{L}*\frac{(391.136+4*18)gBaI_2\ ^.4H_2O}{1molBaI_2} =26.24g\rightarrow No\\[/tex]

[tex]m=28.28g=0.500L*0.1133\frac{molBaI_2}{L}*\frac{(391.136+5*18)gBaI_2\ ^.5H_2O}{1molBaI_2} =27.26g\rightarrow No\\m=28.28g=0.500L*0.1133\frac{molBaI_2}{L}*\frac{(391.136+6*18)gBaI_2\ ^.6H_2O}{1molBaI_2} =28.28g\rightarrow Yes\\[/tex]

Therefore, the formula is barium iodide hexahydrate:

[tex]BaI_2\ ^.6H_2O[/tex]

Best regards.

The question is incomplete, here is the complete question:

A chemist adds 180.0 mL of a 1.42 M sodium carbonate solution to a reaction flask. Calculate the mass in grams of sodium carbonate the chemist has added to the flask. Be sure your answer has the correct number of significant digits.

Answer: The mass of sodium carbonate that must be added are 40.9 grams

Explanation:

To calculate the mass of solute, we use the equation used to calculate the molarity of solution:

[tex]\text{Molarity of the solution}=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (in mL)}}[/tex]

Molar mass of sodium carbonate = 106 g/mol

Molarity of solution = 1.42 M

Volume of solution = 180.0 mL

Putting values in above equation, we get:

[tex]1.42mol/L=\frac{\text{Mass of sodium carbonate}\times 1000}{160g/mol\times 180}\\\\\text{Mass of sodium carbonate}=\frac{160\times 1.42\times 180}{1000}=40.9g[/tex]

Hence, the mass of sodium carbonate that must be added are 40.9 grams

Explanation:

(a)  Formula for critical stress is as follows.

         [tex]\sigma_{c} = \frac{k_{IC}}{\tau \sqrt{\pi \times a}}[/tex]

Here,  [tex]K_{IC}[/tex] = 54.8

          [tex]\tau[/tex] = 0.99

            a = 0.8 mm = [tex]0.8 \times 10^{-3}[/tex] m

Putting the given values into the above formula as follows.

          [tex]\sigma_{c} = \frac{k_{IC}}{\tau \sqrt{\pi \times a}}[/tex]

                       = [tex]\frac{54.8}{0.99 \times \sqrt{3.14 \times 0.8 \times 10^{-3}}}[/tex]

                       = 1107 MPa

Hence, value of critical stress is 1107 MPa.

(b)    Applied stress value is given as 1205 MPa and since it is more than the critical stress (1107 MPa) as a result, a fracture will occur.

Answer:

Germanium is an element in the same group with Carbon and Silicon. The atomic number is 32. The relative atomic mass is usually measured with the Sample of an isotope. In this case Germanium has a relative atomic mass of 72.63

Answer:

The relative atomic mass of the sample of Germanium is 72.89 u

Explanation:

For a sample of germanium we have the following compositions:

m70Ge = 70 u

%70Ge = 22.6%

m72Ge = 72 u

%72Ge = 25.45%

m74Ge = 74 u

%74Ge = 36.73%

m76Ge = 76 u

%76Ge = 15.22%

To calculate the atomic mass of Germanium we must add all the atomic masses of its isotopes multiplied by its percentage of abundance and all this divided by 100, in this way:

mGe = [(m70Gex%70Ge)+(m72Gex%72Ge)+(m74Gex%74Ge)+(m76Gex%76Ge)]/100

replacing values:

mGe = [(70x22.6)+(72x25.45)+(74x36.73)+(76x15.22)]/100 = 72.89 u

Answer: It is a Lewis acid/base reaction, and [tex]Fe^{3+}[/tex] is the Lewis acid.

Explanation:

According to the Lewis concept, an acid is defined as a substance that accepts electron pairs and base is defined as a substance which donates electron pairs.

[tex]Fe^{3+}[/tex] can readily accept electrons and thus act as a lewis acid which is short of electrons.

[tex]CN^-[/tex] can readily lose electrons and thus act as a lewis base which has excess of electrons.

It is a Lewis acid/base reaction, and [tex]Fe^{3+}[/tex] is the Lewis acid.

Answer: = D

Explanation:

The atomic mass increases from Ne to O2 to Cl2 hence the boiling point also increases, therefore

Ne < O2 < Cl2

Answer:

1. 2C2H6 + 7O2 —> 4CO2 + 6H2O

2. 56moles of O2

Explanation:

Ethane undergo Combustion to produce CO2 and H20 according the equation below:

C2H6 + O2 —> CO2 + H2O

Let us balance the equation. There are 6 atoms of H on the left side and 2 atoms on the right side. It can be balanced by putting 3 in front of H2O as shown below:

C2H6 + O2 —> CO2 + 3H2O

There are 2 atoms of C on left and 1atom on the right. It can be balanced by putting 2 in front of CO2 as shown below:

C2H6 + O2 —> 2CO2 + 3H2O

There are a total of 7 atoms of O on the right and 2 atoms on the left. It can be balanced by putting 7/2 in front of O2 as shown below:

C2H6 + 7/2O2 —> 2CO2 + 3H2O

Now we multiply through by 2 to remove the fraction as shown below

2C2H6 + 7O2 —> 4CO2 + 6H2O

Now the equation is balanced

2. 2C2H6 + 7O2 —> 4CO2 + 6H2O

From the equation above,

2 moles of ethane(C2H6) combined with 7moles of O2.

Therefore, 16moles of ethane(C2H6) will combine with = (16x7)/2 = 56moles of O2

Explanation:

Grey Goose vodka has an alcohol content of 40.0 % (v/v).

Volume of vodka = V = 100 mL

This means that 40.0 mL of alcohol is present 100 mL of vodka.

Volume of ethanol=V' = 40.0 mL

Mass of ethanol = m

Density of the ethanol = d = 0.789 g/mL

[tex]m=d\times V' = 0.789 g/ml\times 40.0 mL=31.56 g[/tex]

Volume of water = V''= 100 ml - 40.0 mL = 60.0 mL

Mass of water = m'

Density of the water = d' = 1.00 g/mL

[tex]m'=d'\times V'' = 1.00 g/ml\times 60.0 mL=60.0 g[/tex]

a.)

Moles of ethanol = n= [tex]\frac{31.56 g}{46g/mol}=0.6861 mol[/tex]

Volume of vodka = V = 100 mL = 0.100 L ( 1mL=0.001 L)

Molarity of the ethanol:

[tex]=\frac{0.6861 mol}{0.100 L}=6.861 M[/tex]

6.861 M the molarity of ethanol in this vodka.

b) Mass of ethanol = 31.56 g

Moles of ethanol = n= [tex]\frac{31.56 g}{46g/mol}=0.6861 mol[/tex]

Volume of vodka = V = 100 mL

Mass of vodka = m

Density of the water = D = 0.935 g/mL

[tex]M=D\times V=0.935 g/ml\times 100 ml=93.5 g[/tex]

The percent by mass of ethanol % (m/m):

[tex]\frac{31.56 g}{93.5 g}\times 100=33.75\%[/tex]

33.75% is the percent by mass of ethanol % (m/m) in this vodka.

c)

Moles of ethanol = n= [tex]\frac{31.56 g}{46g/mol}=0.6861 mol[/tex]

Mass of solvent that is water = 60.0 g = 0.060 kg ( 1g = 0.001 kg)

Molality of ethanol in vodka :

[tex]m=\frac{0.6861 mol}{0.060 kg}=11.435 m[/tex]

11.435 m is the molality of ethanol in this vodka.

d)

Moles of ethanol = [tex]n_1=\frac{31.56 g}{46g/mol}=0.6861 mol[/tex]

Moles of water = [tex]n_2=\frac{60.0 g}{18 g/mol}=3.333 mol[/tex]

Mole fraction of ethanol = [tex]\chi_1[/tex]

[tex]\chi_1=\frac{n_1}{n_1+n_2}=\frac{0.6861 mol}{0.6861 mol+3.333 mol}[/tex]

= 0.1707

Mole fraction of water = [tex]\chi_2[/tex]

[tex]\chi_2=\frac{n_2}{n_1+n_2}=\frac{3.3333 mol}{0.6861 mol+3.333 mol}[/tex]

= 0.8290

e)

The vapor pressure of vodka = P

Mole fraction of ethanol = [tex]\chi_1=0.1707[/tex]

Mole fraction of water = [tex]\chi_2=0.8290[/tex]

The vapor pressures of ethanol  = [tex]p_1=45.0 Torr[/tex]

The vapor pressures of pure water = [tex]p_2=23.8Torr[/tex]

[tex]P=\chi_1\times p_1+\chi_2\times p_2[/tex]

[tex]P=0.1707\times 45.0torr+0.8290\times 23.8 Torr=27.41 torr[/tex]

The vapor pressure of vodka is 27.41 Torr.

The change in enthalpy for the dissolution of the compound is 344.11 J/g.

In a coffee-cup calorimeter experiment, the change in enthalpy for the dissolution of a compound can be determined using the equation q = mCΔT, where q is the amount of heat absorbed or released, m is the mass of the compound, C is the specific heat of the solution, and ΔT is the change in temperature. Given that the initial temperature of the water is 23.2°C and the final temperature is 31.8°C, the change in temperature is ΔT = 31.8°C - 23.2°C = 8.6°C.

Since 10.00 g of the compound was added to 75.0 g of water, the total mass of the solution is 75.0 g + 10.00 g = 85.0 g.

Using the equation q = mCΔT, where C is the specific heat of water (4.18 J/(g·°C)), the amount of heat involved in the dissolution can be calculated as q = (85.0 g)(4.18 J/(g·°C))(8.6°C) = 3,441.14 J.

The change in enthalpy for the dissolution of the compound is thus 3,441.14 J/10.00 g = 344.11 J/g of the compound.

Explanation:

The integrated first law is given by :

[tex][A]=[A]_o\times e^{-k\times t}[/tex]

Where:

[tex][A]_o[/tex] = initial concentration of reactant

[A] = concentration of reactant after t time

k = rate constant

a)

[tex][A_o]=x[/tex]

[tex][A]=\frac{x}{2}[/tex]

t = 1000 s

[tex]\frac{x}{2}=x\times e^{-k\times 1000 s}[/tex]

Solving for k:

[tex]k=0.06934 s^{-1}[/tex]

The rate constant for this reaction is [tex]0.06934 s^{-1}[/tex].

b)

[tex][A_o]=0.67 mol/L[/tex]

[tex][A]=0.53 mol/L[/tex]

t = 25 s

[tex]0.53 mol/L=0.67 mol/L\times e^{-k\times 25s}[/tex]

Solving for k:

[tex]k=0.009376 s^{-1}[/tex]

The rate constant for this reaction is [tex]0.009376 s^{-1}[/tex].

c) 2 A → B +C

[tex][A_o]=0.153 mol/L[/tex]

[tex][A]=?[/tex]

[tex][B]=0.034 mol/L[/tex]

According to reaction, 1 mole of B is obtained from 2 moles of A.

Then 0.034 mole of B will be obtained from:

[tex]\frac{2}{1}\times 0.034 mol= 0.068 mol[/tex] of A

So, the concentration left after 115 seconds:

[tex][A]=0.153 mol/L-0.068 mol/L=0.085 mol/L[/tex]

t = 115 s

[tex]0.085mol/L=0.53 mol/L\times e^{-k\times 115 s}[/tex]

Solving for k:

[tex]k=0.01592 s^{-1}[/tex]

The rate constant for this reaction is [tex]0.01592 s^{-1}[/tex].

Answer:

There are 0.183 grams of bicarbonate in their tablet

Explanation:

Step 1: Data given

0.003 moles of bicarbonate was present in their portion of an antacid tablet

Bicarbonate = HCO3-

Molar mass of bicarbonate = 61.02 g/mol

Step 2: Calculate mass bicarbonate

Mass bicarbonate = moles bicarbonate * molar mass bicarbonate

Mass bicarbonate = 0.003 moles * 61.02 g/mol

Mass bicarbonate = 0.183 grams

There are 0.183 grams of bicarbonate in their tablet

Answer: Option (c) is the correct answer.

Explanation:

It is known that when Gibb's free energy, that is, [tex]\Delta G[/tex] has a negative value then the reaction will be spontaneous and the formation of products is favored more rapidly.  

Activation energy is defined as the minimum amount of energy required to initiate a chemical reaction.

So, when reactants of a chemical reaction are unable to reach towards its activation energy then a catalyst is added to lower the activation energy barrier so the reaction can take place rapidly.

Since, the given reaction has low activation energy. Therefore, there is no need to add a catalyst.

And, when value of [tex]\Delta G[/tex] is positive then the reaction is spontaneous in nature and formation of products is less favored.

Thus, we can conclude that for the given situation positive delta G is the reason that a reaction might form products very slowly, or not at all.

Answer:

Explanation:

First PCL to form a scaffold, combined with gelatin.

They are made by first three forms A) made by just PCL.B) made by gelatin ratio PCL is 3:1 and last is C) made by gelatin PCL.

The decoration rate is 1%, 25% and 50% respectively.

A) Growth factor is stuck in 21 days, and can not spread. In this case in vivo experiment for 7 days used highly degradable scaffold use the ability to break down due to decomposition in vivo degradation rate depends on the scaffold's acid byproduct impact.

B) the amount of scaffolded degradation.

First, with scaffold A.

10 mg scaffold weight A= 1 per cent degradation.

Following degradation wt is 9.967=? Degradation per cent.

So, degradation (9.967* 1)/10= 0.9967 per cent.

Likewise for C) scaffold c 10 mg wt. Loss of 50 per cent.

After wt 8.33=? Degradation per cent.

Degradation (8.33* 50)/10=41.65 per cent.

Scaffold c degrades significantly, since the loss of wt is even greater.

Answer: The coefficient of the solid in the balanced equation is 1.

Explanation:

A double displacement reaction is one in which exchange of ions take place. The salts which are soluble in water are designated by symbol (aq) and those which are insoluble in water and remain in solid form are represented by (s) after their chemical formulas.

The balanced chemical equation is:

[tex]Na_2SO_4(aq)+Ba(NO_3)_2(aq)\rightarrow BaSO_4(s)+2NaNO_3(aq)[/tex]

Thus a coefficient of 1 is placed in front of the solid.

The coefficient of the solid in the balanced equation is 1.

The balanced chemical equation is:

            Ba(NO3)2 + Na2SO4 → BaSO4 + NaNO3

Thus, the coefficient of the solid in the balanced equation is 1.

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The volume V2 is 0.400 ml

Explanation:

The dilution equation is the product of initial values of molarity and volume which is equal to the product of final values of molarity and volume.

The dilution equation is given by

                          M1 [tex]\times[/tex] V1 = M2 [tex]\times[/tex] V2

where,

M represents the molarity of the solution  

V represents the volume of the solution

M1 and V1 are the initial values  of the solution

M2 and V2 are the final values of the solution

                            M1 [tex]\times[/tex] V1 = M2 [tex]\times[/tex] V2

               (2.5 [tex]\times[/tex] [tex]10^{-5}[/tex]) [tex]\times[/tex] 10 = (6.25 [tex]\times 10^{-4}[/tex]) [tex]\times[/tex] V2  

                                    V2 = 2.5 [tex]\times 10^{-4}[/tex] / (6.25 [tex]\times 10^{-4}[/tex])  

                                    V2 = 0.4

    The volume V2 is 0.400 ml

The volume of solution required is 0.400 mL.

In this case, we have to use the dilution formula;

C1V1 = C2 V2

In this case;

C1 =  6.25 x 10-4 M

V1 = ?

C2 = 2.50 x 10-5 M

V2 = 10.0 mL

Making V1 the subject of the formula;

V1 = C2V2/C1

V1 = 2.50 x 10-5 M x 10.0 mL/6.25 x 10-4 M

V1 = 0.400 mL

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Answer:  0.821 moles

Explanation:

Moles of electron = 1 mole

According to mole concept:

1 mole of an atom contains [tex]6.022\times 10^{23}[/tex] number of particles.

We know that:

Charge on 1 electron = [tex]1.6\times 10^{-19}C[/tex]

Charge on 1 mole of electrons = [tex]1.6\times 10^{-19}\times 6.022\times 10^{23}=96500C[/tex]

To calculate the charge passed, we use the equation:

[tex]I=\frac{q}{t}[/tex]

where,

I = current passed = 60 A

q = total charge = ?

t = time required = 22.0 minutes =[tex]22.0\times 60=1320sec[/tex]

Putting values in above equation, we get:

[tex]60A=\frac{q}{1320}\\\\q={60A}\times {1320s}=79200C[/tex]

As 96500 C contains = 1 mole of electrons

79200 C contains = [tex]\frac{1}{96500}\times 79200 =0.821[/tex] mole of electrons

Thus 0.821 moles of electrons travel through the wire.

Answer:

4.5g/ml

Explanation:

metal sample has a mass = 7.56 g

cylinder previously filled with water = 20.00 mL

final volume in the cylinder =  21.68 mL

[tex]density = \frac{mass}{V_f - V_i}[/tex]

[tex]density = \frac{7.56}{(21.68 - 20.00)}[/tex]

density = 4.5g/ml

The solubility product constant (Ksp) is calculated as 2.392.

To solve this problem, we can use the Nernst equation:

Ecell = E°cell - (RT/nF) * ln(Q)

where:

Ecell is the measured potential difference between the rod and the SHE (0.5812 V)

E°cell is the standard cell potential

R is the ideal gas constant (8.314 J/(mol·K))

T is the temperature in Kelvin (25°C = 298 K)

n is the number of electrons transferred in the balanced redox reaction

F is Faraday's constant (96485 C/mol)

Q is the reaction quotient

In this case, the balanced redox reaction is:

Ag2C2O4(s) ⇌ 2Ag+(aq) + C2O4^2-(aq)

The solubility product constant (Ksp) for silver oxalate can be expressed as:

Ksp = [Ag+]^2 * [C2O4^2-]

Since the rod is positive, it means that Ag+ ions are being reduced to Ag(s) on the rod, so we can write:

E°cell = E°red,cathode - E°red,anode

The standard reduction potential for Ag+ to Ag(s) is 0.7996 V, and the standard reduction potential for C2O4^2- to CO2(g) is 0.1936 V.

Therefore, E°cell = 0.7996 V - 0.1936 V = 0.606 V

Now we can substitute all the values into the Nernst equation:

0.5812 V = 0.606 V - (8.314 J/(mol·K) * 298 K / (2 * 96485 C/mol)) * ln(Q)

Simplifying:

0.5812 V = 0.606 V - (0.0257 V) * ln(Q)

0.0257 V * ln(Q) = 0.606 V - 0.5812 V

0.0257 V * ln(Q) = 0.0248 V

ln(Q) = 0.0248 V / 0.0257 V

ln(Q) = 0.964

Q = e^(0.964)

Q ≈ 2.622

Since Q = [Ag+]^2 * [C2O4^2-], we can write:

[Ag+]^2 * [C2O4^2-] ≈ 2.622

The solubility product constant (Ksp) is the product of the concentrations of the ions at equilibrium, so we can assume that the concentrations are equal to each other:

[Ag+] ≈ [C2O4^2-]

Therefore, we can substitute [Ag+] for [C2O4^2-]:

[Ag+]^2 * [Ag+] ≈ 2.622

[Ag+]^3 ≈ 2.622

Taking the cube root of both sides:

[Ag+] ≈ (2.622)^(1/3)

[Ag+] ≈ 1.378

Therefore, the approximate solubility of silver oxalate is 1.378 M.

The solubility product constant (Ksp) can now be calculated as:

Ksp = [Ag+]^2 * [C2O4^2-] ≈ (1.378)^2 * (1.378) ≈ 2.392

The solubility product constant for silver oxalate at [tex]25\°C[/tex] is approximately [tex]\( 10^{-22.125} \)[/tex].

The solubility product constant [tex](\( K_{sp} \))[/tex] for silver oxalate can be calculated using the Nernst equation and the measured potential difference between the silver rod and the standard hydrogen electrode (SHE). The reaction occurring at the silver rod can be represented as:

[tex]\[ \text{Ag}_2\text{C}_2\text{O}_4(s) \rightleftharpoons 2\text{Ag}^+(aq) + \text{C}_2\text{O}_4^{2-}(aq) \][/tex]

The Nernst equation for this reaction at [tex]25\°C[/tex] is given by:

[tex]\[ E = E^{\circ} - \frac{0.0592}{n} \log \frac{1}{[\text{Ag}^+]^2[\text{C}_2\text{O}_4^{2-}]} \][/tex]

where [tex]\( E \)[/tex] is the measured potential (0.5812 V), [tex]\( E^{\circ} \)[/tex] is the standard reduction potential for the [tex]\( \text{Ag}^+ \)[/tex] / Ag couple, [tex]\( n \)[/tex] is the number of electrons transferred in the half-reaction (which is 2 for this reaction), and the concentrations of [tex]\( \text{Ag}^+ \)[/tex] and [tex]\( \text{C}_2\text{O}_4^{2-} \)[/tex] are equal to the solubility of silver oxalate since they come from its dissolution.

 First, we need to find the standard reduction potential [tex]\( E^{\circ} \)[/tex] for the [tex]\( \text{Ag}^+ \)[/tex] / Ag couple, which is 0.7996 V.

 Rearranging the Nernst equation to solve for the solubility \( S \), we get:

[tex]\[ S = [\text{Ag}^+] = [\text{C}_2\text{O}_4^{2-}] = 10^{\frac{(E^{\circ} - E)n}{0.0592}} \][/tex]

Plugging in the values:

[tex]\[ S = 10^{\frac{(0.7996 - 0.5812) \times 2}{0.0592}} \][/tex]

[tex]\[ S = 10^{\frac{0.2184 \times 2}{0.0592}} \][/tex]

[tex]\[ S = 10^{\frac{0.4368}{0.0592}} \][/tex]

[tex]\[ S = 10^{7.375} \][/tex]

[tex]\[ S \approx 10^{-7.375} \][/tex]

The solubility product constant [tex]\( K_{sp} \)[/tex] is given by:

[tex]\[ K_{sp} = [\text{Ag}^+]^2[\text{C}_2\text{O}_4^{2-}] \][/tex]

[tex]\[ K_{sp} = (S)^2(S) \][/tex]

[tex]\[ K_{sp} = (10^{-7.375})^2(10^{-7.375}) \][/tex]

[tex]\[ K_{sp} = 10^{-7.375 \times 3} \][/tex]

[tex]\[ K_{sp} = 10^{-22.125} \][/tex]

Answer:

A. Increased entropy of the water molecules.

Explanation:

Entropy is the quantitative measure of disorder or randomness in a system or element. In thermodynamics, entropy or hydrophobic effect is the free energy change of water enclosing a solute. Hence the existing negative free charges enhances the effect of hydrophilicty hence the aggregation of non-polar molecules.

The aggregation of nonpolar molecules in water is primarily due to the increased entropy of the water molecules, as they rearrange themselves to accommodate nonpolar molecules.

The aggregation of nonpolar molecules or groups in water is thermodynamically due to the increased entropy of the water molecules. When nonpolar molecules or groups are added to water, water molecules arrange themselves in order to minimize the disruption of hydrogen bonds, which increases the entropy of water. In other words, the introduction of nonpolar substances leads to a more disordered and random arrangement of water molecules, increasing the system's overall entropy.

Options B, C, and D although relevant, are not the primary reason. While enthalpy may decrease due to aggregation and Van der Waals forces might influence nonpolar association, the primary factor contributing to the phenomenon is the change in water's entropy.

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Answer:

1. The concentration of [tex]NO_2[/tex] is equal to the concentration of [tex]N_2O_4[/tex] : False

2. The rate of the dissociation of [tex]N_2O_4[/tex] is equal to the rate of formation of [tex]N_2O_4[/tex]: True

3. The rate constant for the forward reaction is equal to the rate constant of the reverse reaction: False

4. The concentration of [tex]NO_2[/tex] divided by the concentration of [tex]N_2O_4[/tex] is equal to a constant : False

Explanation:

[tex]N_2O_4(g)\rightleftharpoons 2NO_2(g)[/tex]

The concentrations of reactant and product is constant and not equal.

The rate of forward and backward reactions are equal. Thus rate of the dissociation of [tex]N_2O_4[/tex] is equal to the rate of formation of [tex]N_2O_4[/tex]

For a reversible reaction, the equilibrium constant for the forward reaction is inverse of the equilibrium constant for the backward reaction and not equal.

Equilibrium constant is the ratio of the concentration of products to the concentration of reactants each term raised to its stochiometric coefficients.

[tex]K_c=\frac{[NO_2]^2}{[N_2O_4]}[/tex]

Final answer:

Statements 1 and 3 are False, and 2 and 4 are True.  At equilibrium, the rate of formation and dissociation of N₂O₄ are equal but their concentrations are not necessarily the same. Rate constants for forward and reverse reactions are different yet define the equilibrium constant. The concentration ratio of NO₂ to N₂O₄ is constant at a given temperature.

Explanation:

When dinitrogen tetroxide (N₂O₄) is in equilibrium with nitrogen dioxide (NO₂), the following statements can be made:

Explanation:

It is known that standard entropies for [tex]N_{2}(g)[/tex] is 191.6 J/mol K, [tex]O_{2}(g)[/tex] = 205 J/mol K, and [tex]NO_{2}(g)[/tex] is 239.7 J/mol K at 298 K.

Therefore, we will calculate the value of [tex]\Delta S^{o}_{reaction}[/tex] from standard absolute entropies as follows.

     [tex]\Delta S^{o}_{reaction} = \sum \Delta S^{o}_{products} - \sum \Delta S^{o}_{reactants}[/tex]

                   = 2 mole of [tex]NO_{2}(g)[/tex] - 1 mole of [tex]N_{2}(g)[/tex] + 2 mole of [tex]O_{2}(g)[/tex]

                   = [tex]2 \times 239.7 J/mol K - 1 \times 191.6 J/mol K + 2 \times 205 J/mol K[/tex]

                   = -122.2 J/K

The entropy change for 1.90 moles of [tex]N_{2}(g)[/tex] reacting is as follows.

        [tex]\Delta S^{o}_{system}[/tex] = 1.90 moles of [tex]N_{2}(g) \times 122.2 J/K/ 1 \text{mol of} N_{2}(g)[/tex]

                           = 232.18 J/K

Thus, we can conclude that the entropy change for the given system is 232.18 J/K.

If an extra amount of a product is added to a system at equilibrium, the system will shift towards the reactants to counteract this change. This shift can cause changes in the amounts of all substances involved, including the added product, the other product, and the reactants.

When an extra amount of a product is added to a system at equilibrium, the system shifts to counteract this change and re-establish equilibrium, as described by Le Châtelier's principle. Specifically, the system will shift in the direction that reduces the excess amount of the product, i.e., towards the reactants. With this shift, the quantities of all substances in the system can change.

Therefore, the appropriate answer is d. all of the above. Adding more of one product will indeed increase the amount of that product initially (a), but the system will shift to remove the added product by moving towards the reactants. In this process, the amount of the other product may also change as it is consumed or generated (b), and the amounts of reactants will also adjust as they are generated or consumed (c).

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Explanation:

Let us assume that volume of acetic acid added is V ml.

So,   [tex][CH_{3}COOH] = \frac{0.05 \times 100}{100 + V}[/tex]

and,   [tex][CH_{3}COONa] = \frac{0.05 \times V}{100 + V}[/tex]

Expression for the buffer solution is as follows.

     pH = [tex]pK_{a} + log \frac{[CH_{3}COONa]}{[CH_{3}COOH]}[/tex]

     5.1 = [tex]4.76 + log \frac{0.05 \times V}{0.05 \times 100}[/tex]

         0.34 = log V - 2

         log V = 2.34

or,        V = 218.77 ml

Now, we will calculate the molarity of the buffer with respect to acetate as follows.

  = [tex][CH_{3}COO^{-}] + [CH_{3}COOH][/tex]

  = [tex]\frac{0.05 \times 218.77}{318.77} + \frac{0.05 \times 100}{318.77}[/tex]

  = 0.0499 M

or,  = 0.05 M (approx)

Thus, we can conclude that molarity of the resulting buffer with respect to acetate is 0.05 M.

Final answer:

To make a buffer of pH 5.1 using 0.05 M acetic acid, the Henderson-Hasselbalch equation is used to calculate the amount of 0.05 M sodium acetate needed. The result is based on achieving the correct ratio of acetate ion to acetic acid. The molarity of the resulting buffer depends on the total moles of acetate and acid in the final solution volume.

Explanation:

To determine how many mL of 0.05 M sodium acetate should be added to 100 mL of 0.05M acetic acid to make a buffer of pH 5.1, we can use the Henderson-Hasselbalch equation, which is pH = pKa + log([A-]/[HA]), where [A-] is the concentration of the acetate ion and [HA] is the concentration of acetic acid. Given that the pKa of acetic acid is 4.76, plugging in the pH 5.1 gives us the equation 5.1 = 4.76 + log([A-]/[0.05]).

Solving for [A-], we find that the ratio of [A-] to [HA] needed is approximately 2.2. Since the volume of the acetic acid solution is 100 mL and its concentration is 0.05M, to achieve the desired ratio, the amount of sodium acetate needed can be calculated based on the molarity and the final volume of the solution. The resulting molarity of the buffer with respect to acetate (acetate + acetic acid) will be determined by the total moles of acetate ions and acetic acid divided by the total volume of the solution after the addition of sodium acetate.

Answer: The nuclear fission reaction of U-234.043 is written below.

Explanation:

In a nuclear reaction, the total mass and total atomic number remains the same.

For the given fission reaction:

[tex]^{234.043}_{92}\textrm{U}\rightarrow ^{137.159}_{54}\textrm{Xe}+^{A}_{38}\textrm{Sr}+3^{1.0087}_0\textrm{n}+180MeV[/tex]

To calculate A:

Total mass on reactant side = total mass on product side

234.043 = 137.159 + A + 3(1.0087)

A = 93.858

Now, the chemical equation becomes:

[tex]^{234.043}_{92}\textrm{U}\rightarrow ^{137.159}_{54}\textrm{Xe}+^{93.858}_{38}\textrm{Sr}+3^{1.0087}_0\textrm{n}+180MeV[/tex]

Hence, the nuclear fission reaction of U-234.043 is written above.

Answer:

[tex]1.86*10^{20[/tex]

Explanation:

The equation for the reaction is:

[tex]Zn^{2+}_{(aq)} + 4OH^-_{(aq)}[/tex]     ⇄     [tex]Zn(OH)^{2-}_{4(aq)}[/tex]

Oxidation can be defined as the addition of oxygen, removal of hydrogen and/or loss of electron during an electron transfer. Oxidation process occurs at the anode.

On the other hand; reduction is the removal of oxygen, addition of hydrogen and/ or the process of electron gain during an electron transfer. This process occurs at  the cathode.

The oxidation-reduction process with its standard reduction potential is as follows:

[tex]Zn(OH)^{2-}_{4(aq)} + 2e^- ----->Zn_{(s)} +4OH^-_{(aq)}[/tex]            [tex]E^0_{anode} = -1.36 V[/tex]

At the zinc electrode (cathode); the reduction process of the reaction with its standard reduction potential is :

[tex]Zn^{2+}_{(aq)} +2e^- -----> Zn_{(s)}[/tex]                  [tex]E^0_{cathode} = -0.76 V[/tex]

The standard cell potential [tex]E^0_{cell}[/tex] is given as:

[tex]E^0_{cell}=E^0_{cathode}-E^0_{anode}[/tex]

[tex]E^0_{cell}[/tex] = -0.76 V - (- 1.36 V)

[tex]E^0_{cell}[/tex] = -0.76 V + 1.36 V

[tex]E^0_{cell}[/tex] = +0.60 V

Now to determine the formation constant [tex]k_f[/tex] of the [tex]E^0_{cell}[/tex] ; we use the expression:

[tex]E^0_{cell}[/tex] = [tex]\frac{RT}{nF}Ink_f[/tex]

where;

[tex]E^0_{cell}[/tex] = +0.60 V

R = universal gas constant = 8.314 J/mol.K

T = Temperature @ 25° C = (25+273)K = 298 K

n = numbers of moles of electron transfer = 2

F = Faraday's constant = 96500 J/V.mol

[tex]+ 0.60 V = \frac{(8.314)(298)}{n(96500)} Ink_f[/tex]

[tex]+0.60 V = \frac{(0.0257)}{n}Ink_f[/tex]

[tex]+0.60V = \frac{0.0592}{n}log k_f[/tex]

[tex]logk_f = \frac{+0.60V*n}{0.0592}[/tex]

[tex]logk_f = \frac{+0.60V*2}{0.0592}[/tex]

[tex]logk_f = 20.27[/tex]

[tex]k_f= 10^{20.27}[/tex]

[tex]k_f = 1.86*10^{20}[/tex]

Therefore, the formation constant [tex]k_f[/tex] for the reaction is = [tex]1.86*10^{20[/tex]