Pyridine is a weak base that is used in the manufacture of pesticides and plastic resins. It is also a component of cigarette smoke. Pyridine ionizes in water as follows: C5H5N+H2O⇌C5H5NH++OH− The pKb of pyridine is 8.75. What is the pH of a 0.460 M solution of pyridine? (Assume that the temperature is 25 ∘C.) Express the pH numerically to two decimal places.

Answer: Concentration of the dye in solution is 1.159 M

Explanation:

According to Beer-Lambert law, the absorbance is directly proportional to the concentration of an absorbing species.

Formula used :

[tex]A=\epsilon \times C\times l[/tex]

where,

A = absorbance of solution = 1.657

C = concentration of solution = ?

l = length of the cell = 0.701 cm

[tex]\epsilon[/tex] = molar absorptivity of this solution =

[tex]1.657=2.038Lmol^{-1}cm^{-1}\times C\times 0.701cm[/tex]

[tex]C=1.159mol/L[/tex]

Thus the concentration of the dye in solution is 1.159 M

Concentration of the dye in solution is 1.159 M

According to Beer-Lambert law, the absorbance is directly proportional to the concentration of an absorbing species. The concentration can be calculated by using this law which is given as:

[tex]A=E*l*c[/tex]

where,

A = absorbance of solution = 1.657

c = concentration of solution = ?

l = length of the cell = 0.701 cm

E= molar absorptivity of this solution = [tex]2.038 L mol^{-1} cm^{-1}[/tex]

On substituting the values:

[tex]A=E*l*c\\\\1.657=2.038*0.701*c\\\\c=\frac{1.657}{2.038*0.701}\\\\ c=1.159mol/L[/tex]

Thus, the concentration of the dye in solution is 1.159 M.

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Answer:

There will be no reaction between the product of hydrolysis and I2-KI (-ve). When the product of hydrolysis is tested with Benedict reagent, a brick-red precipitate is observed.

Explanation:

Benedict's reagent as a chemical reagent is a mixture of sodium carbonate, sodium citrate and copper(II) sulfate pentahydrate. It is often used in place of Fehling's solution to detect the presence of reducing sugars. A positive test with Benedict's reagent is shown by a color change from clear blue to a brick-red precipitate.

Glycogen and starch are both complex structures containing repeating units of glucose(a reducing sugar). The polysaccharides have non redusing ends and so cannot react with benedict reagent.

When they are hydrolysed, glucose which is a reducing sugar can then test positive with Benedict reagent.

Final answer:

After hydrolysis, glycogen and starch would no longer give a positive iodine test but would give a positive Benedict's test, as they break down into glucose monomers, which are reducing sugars capable of reacting with Benedict's reagent.

Explanation:

If you hydrolyzed glycogen and starch and then tested the solutions with I₂-KI (iodine test) and Benedict's solution, you would get different results compared to the tests performed on the non-hydrolyzed polysaccharides. Glycogen and starch are both polysaccharides that give a positive reaction with the iodine test due to their coiled structures, which trap iodine molecules and create a bluish-black color. However, they do not give a positive Benedict's test result in their non-hydrolyzed form because they are not free-reducing sugars and therefore do not react with the Benedict's reagent.

Once hydrolyzed, glycogen and starch break down into glucose monomers. Glucose is a reducing sugar, which will react with Benedict's reagent. Upon heating with Benedict's solution, the glucose will reduce the copper(II) ions in the reagent to copper(I) oxide, resulting in a color change from blue to greenish or orangish precipitate, depending on the amount of glucose present. Therefore, after hydrolysis, the solutions of glycogen and starch would give a negative result with the iodine test (no blue-black color) and a positive result with Benedict's test (color change indicating the presence of reducing sugar).

Answer:

The answer to your question is there will be 20.5 g left of HCl

Explanation:

Data

HCl = 31.4 g

NaOH = 12 g

Excess of HCl = ?

Balanced chemical reaction

                HCl  +  NaOH  ⇒   NaCl  +  H₂O

Molar weight of HCl = 1 + 35.5 = 36.5 g

Molar weight of NaOH = 23 + 16+ 1 = 40 g

Calculate the limiting reactant

theoretical yield HCl/NaOH = 36.5/40 = 0.9125

experimental yield HCl/NaOH = 31.4/12 = 2.62

Conclusion

The limiting reactant is NaOH because the experimental yield increases.

Calculate the mass of Excess reactant

               36.5 g of HCl ---------------- 40 g of NaOH

                x                     ---------------- 12 g of NaOH

                x = (12 x 36.5) / 40

                x = 438 / 40

                x = 10.95 g of HCl

Excess HCl = 31.4 - 10.95

                   = 20.5 g

Answer:

There will remain 20.5 grams of hydrochloric acid

Explanation:

Step 1: Data given

Mass of hydrochloric acid = 31.4 grams

Mass of sodium hydroxide = 12.0 grams

Molar mass hydrochloric acid (HCl) = 36.46 g/mol

Molar mass sodium hydroxide (NaOH) = 40.0 g/mol

Step 2: The balanced equation

HCl + NaOH → NaCl + H2O

Step 3: Calculate moles

Moles = mass / molar mass

Moles HCl = 31.4 grams / 36.46 g/mol

Moles HCl = 0.861 moles

Moles NaOH = 12.0 grams / 40.0 g/mol

Moles NaOH = 0.3 moles

Step 4: Calculate the limiting reactant

The limiting reactant is NaOH. It will completely be consumed (0.3 moles). Hcl is in excess. There will react 0.3 moles. There will remain 0.861 - 0.3 = 0.561 moles

Step 5: Calculate mass HCl

Mass HCl = moles HCl * molar mass

Mass HCl = 0.561 moles * 36.46 g/mol

Mass HCl = 20.5 grams

There will remain 20.5 grams of hydrochloric acid

Answer:

ΔrxnHº  =-1,124.3 kJ

Explanation:

To solve this question we first need to search  for the enthalpies of formation of reactants and products and calculate the change in enthalpy of reaction utilizing the equation

ΔrxnHº = ∑ νΔfHº reactants - Σ νΔfHº products

where ν is the stoichiometric coefficient of the compound  in the balanced equation .

ΔfHº H₂S(g) = -20.50 kJmol⁻¹

ΔfHº 0₂(g)  = 0  ( O₂ is in standard state )

ΔfHº H₂O(l) = -285.8 kJmol⁻¹

ΔfHº SO₂(g) = -296.84 kJmol⁻¹

ΔrxnHº =  [2 x  -285.8  + 2 x  -296.84] - [2 x -20.50]

           =    - 1,124.3 kJ

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: 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: 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:

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: [tex]K_a[/tex] of PABA is 0.000022

Explanation:

[tex]NH_2C_6H_5COOH\rightarrow H^++NH_2C_6H_5COO^-[/tex]

  cM              0             0

[tex]c-c\alpha[/tex]        [tex]c\alpha[/tex]          [tex]c\alpha[/tex]

So dissociation constant will be:

[tex]K_a=\frac{(c\alpha)^{2}}{c-c\alpha}[/tex]

Give c= 0.055 M and [tex]\alpha[/tex] = ?

[tex]K_a=?[/tex]

[tex]pH=-log[H^+][/tex]

[tex]2.96=-log[H^+][/tex]

[tex][H^+]=1.09\times 10^{-3}[/tex]

[tex][H^+]=c\times \alpha[/tex]

[tex]1.09\times 10^{-3}=0.055\times \alpha[/tex]

[tex]\alpha=0.02[/tex]

Putting in the values we get:

[tex]K_a=\frac{(0.055\times 0.02)^2}{(0.055-0.055\times 0.02)}[/tex]

[tex]K_a=0.000022[/tex]

Thus [tex]K_a[/tex] of PABA is 0.000022

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

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:

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

Explanation:

Below is an attachment containing the solution.

Answer:

[tex]r=k[UO_2^+]^2[H^+][/tex]

Explanation:

The given reaction is :-

[tex]2UO_2^++4H^+\rightarrow U^{4+}+UO_2^{2+}+2H_2O[/tex]

According to the law of mass action:-

The rate of the reaction is directly proportional to the active concentration of the reactant which each are raised to the experimentally determined coefficients which are known as orders. The rate is determined by the slowest step in the reaction mechanics.

Order of in the mass action law is the coefficient which is raised to the active concentration of the reactants. It is experimentally determined and can be zero, positive negative or fractional.

The order of the whole reaction is the sum of the order of each reactant which is raised to its power in the rate law.  

m = 2 =  is order with respect to [tex]UO_2^+[/tex]

n = order with respect to H+

overall order = m+n = 3

n = 3 - m = 3 - 2 = 1

Rate law is:-

[tex]r=k[UO_2^+]^2[H^+][/tex]

Answer: The value of [tex]\Delta S^o[/tex] for the surrounding when given amount of hydrogen gas is reacted is -31.02 J/K

Explanation:

Entropy change is defined as the difference in entropy of all the product and the reactants each multiplied with their respective number of moles.

The equation used to calculate entropy change is of a reaction is:

[tex]\Delta S^o_{rxn}=\sum [n\times \Delta S^o_{(product)}]-\sum [n\times \Delta S^o_{(reactant)}][/tex]

For the given chemical reaction:

[tex]H_2(g)+F_2(g)\rightarrow 2HF(g)[/tex]

The equation for the entropy change of the above reaction is:

[tex]\Delta S^o_{rxn}=[(2\times \Delta S^o_{(HF(g))})]-[(1\times \Delta S^o_{(H_2(g))})+(1\times \Delta S^o_{(F_2(g))})][/tex]

We are given:

[tex]\Delta S^o_{(HF(g))}=173.78J/K.mol\\\Delta S^o_{(H_2)}=130.68J/K.mol\\\Delta S^o_{(F_2)}=202.78J/K.mol[/tex]

Putting values in above equation, we get:

[tex]\Delta S^o_{rxn}=[(2\times (173.78))]-[(1\times (130.68))+(1\times (202.78))]\\\\\Delta S^o_{rxn}=14.1J/K[/tex]

Entropy change of the surrounding = - (Entropy change of the system) = -(14.1) J/K = -14.1 J/K

We are given:

Moles of hydrogen gas reacted = 2.20 moles

By Stoichiometry of the reaction:

When 1 mole of hydrogen gas is reacted, the entropy change of the surrounding will be -14.1 J/K

So, when 2.20 moles of hydrogen gas is reacted, the entropy change of the surrounding will be = [tex]\frac{-14.1}{1}\times 2.20=-31.02J/K[/tex]

Hence, the value of [tex]\Delta S^o[/tex] for the surrounding when given amount of hydrogen gas is reacted is -31.02 J/K

A 55-gram serving of cereal supplies 270 mg of sodium, which is 11% of the recommended daily allowance. The recommended daily allowance contains 2454.55 mg of sodium and 106.82 mol of sodium. It also contains 6.43 x 10²⁵ atoms of sodium.

To find the moles of sodium in the recommended daily allowance, we need to know the amount of sodium in the daily allowance. The problem states that a 55-gram serving of the cereal supplies 270 mg of sodium, which is 11% of the recommended daily allowance.

So, the total amount of sodium in the daily allowance is

270 mg / 0.11 = 2454.55 mg.

To convert this to moles, we use the molar mass of sodium, which is 22.99 g/mol.

Therefore, the number of moles of sodium in the daily allowance is 2454.55 mg / 22.99 g/mol = 106.82 mol.

To find the number of atoms of sodium in the daily allowance, we use Avogadro's number (6.022 x 10²³ atoms/mol). Therefore, the number of atoms of sodium in the daily allowance is 106.82 mol x (6.022 x 10²³ atoms/mol) = 6.43 x 10²⁵ atoms.

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To find the moles and atoms of sodium in the recommended daily allowance, divide the mass of sodium in mg by its molecular weight, then multiply the resulting moles by Avogadro's number to get atoms. There are roughly 0.107 moles and 6.44 x 1022 atoms of sodium in the allowance.

One 55-gram serving of cereal provides 270 mg of sodium, which is 11% of the recommended daily allowance. To calculate the total recommended daily allowance, we divide 270 mg by 0.11, getting approximately 2455 mg of sodium. The molecular weight of sodium is about 22.99 g/mol, so we can find the number of moles of sodium in the daily allowance by dividing the mass (in grams) by the molecular weight. This results in approximately 0.107 moles of sodium.

To find the number of atoms of sodium, we use Avogadro's number, which is approximately 6.022 x 1023 atoms/mole. Multiplying the number of moles by Avogadro's number gives us the number of atoms. So, there are roughly 6.44 x 1022 sodium atoms in the recommended daily allowance of sodium.

Answer:

At physiological pH,the carboxylic acid group of an amino acid will be deprotonated while the amino group will be protonated,yielding the zwitter ion form.

Explanation:

Deprotonated means removal of protons in an acid base reaction and protonated means addition of protons in an acid base reaction.

Both protonated and de protonated reaction takes place in catalytic acid bade reaction by changing either it's mass or it's charge.

During formation of zwitter ion, the carboxylic acid will be deprotonated by donating the H+ ion while the amino acid is protonated by taking the H+ ion.

R-CH-COOH                               R-CH-COO-

   I                                ⇒                 I

  NH₂                                              NH₃               (Zwitter ion)

At physiological pH, the carboxylic acid group of an amino acid will be negatively charged, while the amino group will be positively charged, yielding the zwitterion form.

Amino acids can exist as zwitterions at physiological pH. The carboxylic acid group of an amino acid will be negatively charged as a carboxylate ion (-COO-) at physiological pH. On the other hand, the amino group of an amino acid will be positively charged as an ammonium ion (-NH3+) at physiological pH. This combination of positive and negative charges gives rise to the zwitterion form of an amino acid.

Answer: Friction (Fk) = 46.2N.

Explanation: In the attachment, there is the drawing of the forces acting on the skiing person. As the person is in movement, the formula to calculate the friction is Fk = μk.N, where μk is the coefficient of kinetic friction and N is normal force. Normal force is the force necessary to keep a person "above" the ground. To find the normal force as suggested, we use the representation of the forces in the attachment. With it, it's shown that to calculate N:

cos 15° = [tex]\frac{N}{Fg}[/tex]

N = Fg·cos 15°

N = 608·(-0.76)

N = - 461.9N

Now, with N and knowing the coefficient, which can be found in Physics Text Books and it is 0.1, we have:

Fk = 0.1·(-461.9)

Fk = - 46.2N

The friction is negative because it is pointing to the opposite side of the reference.

Answer:

Part A K´p = 3.4 x 10⁴

Part B K´p = 2.4 x 10⁻¹⁴

Part C K´p = 1.2 x 10⁹

Explanation:

The methodolgy to answer this question is to realize that the equilibrium in part A is the reverse of the the equilibrium given:

2PH₃(g) + As₂(g)  ⇌ 2 AsH₃(g) + P₂(g)  Kp = 2.9 x 10⁻⁵

Kp = [AsH₃]²[P₂]/[PH₃]²[As]  = 2.9 x 10⁻⁵

Likewise, part B is this equilibrium where the coefficients have been multiplied by 3,  and part C is  is the  reverse equilibrium multipled by 2.

Given the way that the equilibrium constant, Kp, is expressed mathematically, we can see that if we reserse the equilibrium its constant will be the inverse of the old Kp.

Similarly if we multiply the coefficient by a factor we have to raise the expression for Kp to that factor.

With that in mind lets answer the three parts of the questiion.

Part A

Lets call the new equilibrium K´p :

K´p = 1/ Kp = 1/2.9 x 10⁻⁵ = 3.4 x 10⁴

Part B

K´p = ( Kp )³ = ( 2.9 x 10⁻⁵ )³ = 2.4 x 10⁻¹⁴

Part C

K´p = ( 1/Kp)² = ( 1/ 2.9 x 10⁻⁵ )² = 1.2 x 10⁹

If you can not see why this is so, lets show it for part C

K´p = [As₂]²[PH₃]⁴ / [P₂]²[AsH₃]⁴

rearranging this equation:

K´p = (1 / {  [AsH₃]⁴[P]² / [As₂]²[PH₃]⁴ } =) 1 / Kp²

Final answer:

The Kp for Part A (the reverse reaction) is the reciprocal of the original Kp, for Part B (the original times 3) is the original Kp to the third power, and for Part C (exchanged reactants/products with doubled coefficients) is the reciprocal of original Kp squared.

Explanation:

The original reaction is 2PH₃(g) + As₂(g) ⇌ 2AsH3(g) + P₂(g) and has Kp=2.9×10⁻⁵ at 873 K.

Part A is the reverse of the original reaction. For the reverse reaction, the equilibrium constant Kp is the reciprocal of the original reaction. Thus, Kp for Part A is 1/(2.9×10⁻⁵).

Part B has the original reaction multiplied by 3. To find the new Kp, we raise the original Kp to the power of 3. So, Kp for Part B is (2.9×10⁻⁵)³.

Part C is the original reaction with products and reactants exchanged and coefficients doubled. This combines the effects of reversing the reaction and squaring its coefficients. Therefore, Kp for Part C is (1/(2.9×10⁻⁵))².

The question is incomplete, here is the complete question:

At 25°C Henry's Law constant for carbon dioxide gas in water is 0.031 M/atm . Calculate the mass in grams of gas that can be dissolved in 425. mL of water at 25°C and at a partial pressure of 2.92 atm. Round your answer to 2 significant digits.

Answer: The mass of carbon dioxide that can be dissolved is 1.7 grams

Explanation:

To calculate the molar solubility, we use the equation given by Henry's law, which is:

[tex]C_{CO_2}=K_H\times p_{CO_2}[/tex]

where,

[tex]K_H[/tex] = Henry's constant = [tex]0.031M/atm[/tex]

[tex]C_{CO_2}[/tex] = molar solubility of carbon dioxide gas

[tex]p_{CO_2}[/tex] = partial pressure of carbon dioxide gas = 2.92 atm

Putting values in above equation, we get:

[tex]C_{CO_2}=0.031M/atm\times 2.92 atm\\\\C_{CO_2}=0.0905M[/tex]

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]

Given mass of carbon dioxide = ? g

Molar mass of carbon dioxide = 44 g/mol

Molarity of solution = [tex]0.0905mol/L[/tex]

Volume of solution = 425 mL

Putting values in above equation, we get:

[tex]0.0905mol/L=\frac{\text{Mass of carbon dioxide}\times 1000}{44g/mol\times 425}\\\\\text{Mass of solute}=\frac{44\times 425\times 0.0905}{1000}=1.7g[/tex]

Hence, the mass of carbon dioxide that can be dissolved is 1.7 grams

Answer : The molecular of the compound is, [tex]C_8H_8[/tex]

Solution :

If percentage are given then we are taking total mass is 100 grams.

So, the mass of each element is equal to the percentage given.

Mass of C = 92.26 g

Mass of H = 100 - 92.26 = 7.74 g

Molar mass of C = 12 g/mole

Molar mass of H = 1 g/mole

Step 1 : convert given masses into moles.

Moles of C = [tex]\frac{\text{ given mass of C}}{\text{ molar mass of C}}= \frac{92.26g}{12g/mole}=7.688moles[/tex]

Moles of H = [tex]\frac{\text{ given mass of H}}{\text{ molar mass of H}}= \frac{7.74g}{1g/mole}=7.74moles[/tex]

Step 2 : For the mole ratio, divide each value of moles by the smallest number of moles calculated.

For C = [tex]\frac{7.688}{7.74}=0.99\approx 1[/tex]

For H = [tex]\frac{7.74}{7.74}=1[/tex]

The ratio of C : H = 1 : 1

The mole ratio of the element is represented by subscripts in empirical formula.

The Empirical formula = [tex]C_1H_1[/tex]

The empirical formula weight = 1(12) + 1(1) = 13 gram/eq

Now we have to calculate the molecular mass of polymer.

As, the mass of polymer of [tex]7.02\times 10^{18}[/tex] molecules = 1.22 mg = 0.00122 g

So, the mass of polymer of [tex]6.022\times 10^{23}[/tex] molecules = [tex]\frac{6.022\times 10^{23}}{7.02\times 10^{18}}\times 0.00122g=104.6g/mol[/tex]

Now we have to calculate the molecular formula of the compound.

Formula used :

[tex]n=\frac{\text{Molecular formula}}{\text{Empirical formula weight}}[/tex]

[tex]n=\frac{104.6}{13}=8[/tex]

Molecular formula = [tex](C_1H_1)_n=(C_1H_1)_8=C_8H_8[/tex]

Therefore, the molecular of the compound is, [tex]C_8H_8[/tex]

The molecular formula for styrene is C8H8.

The molecular formula for styrene can be determined using the given information. Firstly, we need to find the mass of 7.02 * 10^18 molecules of styrene. Given that 7.02 * 10^18 molecules weigh 1.22 mg, we can calculate the molecular weight of the styrene as follows:

Therefore, the molecular formula of styrene is C8H8.

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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]

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.

Explanation:

        [tex]{\displaystyle {\ce {HA[/tex]  ⇄  [tex]{H^+}+{A^{-}}} }}[/tex]

Here HA is a protonic acid such as acetic acid, [tex]CH_3COOH[/tex]

        [tex]K_a[/tex] = [tex]\dfrac{[H^{+}] [A^{-}]}{[HA]}[/tex]

         [tex]\\{\displaystyle {\ce {{HA}+ H_2O}[/tex] ⇄  [tex][H_3O^+] [A^-][/tex]

        [tex]K_a[/tex] = [tex]\dfrac{[H_3O^{+}] [A^{-}]}{[HA]}[/tex]

Here, [tex]K_a[/tex] is the dissociation constant of an acid.

Answer:

Explanation:

To calculate the acid dissociation constant (Ka) of the acid, we need to know the concentration of the acid and the concentration of the conjugate base at equilibrium, as well as the concentration of H3O+ or OH- ions in the solution. The general equation for the dissociation of a weak acid, HA, is:

HA + H2O ⇌ H3O+ + A-

The acid dissociation constant can be written as:

Ka = [H3O+][A-]/[HA]

We can assume that the initial concentration of the acid is equal to the concentration of HA at equilibrium, since the acid is completely dissociated in solution. Therefore, we can write the equilibrium concentration of HA as [HA]0 - [H3O+], where [HA]0 is the initial concentration of the acid.

We are not given the values of [HA], [H3O+], or [A-], but we can use the pH of the solution to calculate the concentration of H3O+:

pH = -log[H3O+]

Solving for [H3O+], we get:

[H3O+] = 10^-pH

Substituting this expression for [H3O+] into the equation for Ka, we get:

Ka = [H3O+][A-]/([HA]0 - [H3O+])

Ka = (10^-pH)[A-]/([HA]0 - 10^-pH)

If we know the concentration of the conjugate base, [A-], we can substitute that value into the equation. If we do not know the concentration of the conjugate base, we can assume that it is small compared to [HA]0 and therefore can be neglected. In that case, the equation simplifies to:

Ka ≈ 10^-pH

Therefore, the acid dissociation constant of the acid is approximately equal to 10^-pH. We would need more information about the acid and the solution to calculate a more exact value for Ka.

Explanation:

It is known that for high concentration of [tex]M^{2+}[/tex], reduction will take place. As, cathode has a positive charge and it will be placed on left hand side.

Now, [tex]E^{o}_{cell}[/tex] = 0 and the general reaction equation is as follows.

         [tex]M^{2+} + M \rightarrow M + M^{2+}[/tex]

                3.00 M        n = 2       30 mM

         E = [tex]0 - \frac{0.0591}{2} log \frac{50 \times 10^{-3}}{1}[/tex]

            = [tex]-\frac{0.0591}{2} log (5 \times 10^{-2})[/tex]

            = 0.038 V

Therefore, we can conclude that voltage shown by the voltmeter is 0.038 V.

The question involves calculation of equilibrium concentrations in a complex formation reaction in a solution, where the reaction is between a metal ion M²⁺ and cyanide ions CN⁻ to form a complex [M(CN)4]²⁻. The stated formation constant for the reaction allows formulating an equation for calculating the equilibrium concentration of M²⁺.

To determine the concentration of M²⁺ ions at equilibrium, we need to consider the process of complex formation here, which is M²⁺ + 4CN⁻  → [M(CN)4]²⁻. The formation constant for this reaction is given as 7.7 x 10¹⁶. Given a 0.150 mole quantity of M(NO3)2 is added to a liter of 0.820 M NaCN solution, initially we will have [M²⁺] = 0.150 M and [CN⁻] = 0.820 M.

As the reaction proceeds to equilibrium, [M²⁺] drops by x amount reacting with 4x amount of [CN⁻]. At equilibrium, [M²⁺] = 0.150 - x, [CN-] = 0.820 - 4x and [M(CN)4]²⁻ = x. Applying the formation constant: 7.7 x 10¹⁶ = [M(CN)4]²⁻ / ([M²⁺][CN-]⁴) = x / (0.150 - x)(0.820 - 4x)⁴

This is a difficult equation to solve directly, but if we make the assumption that x, which relates to the equilibrium concentration of [M²⁺], is much less than initial concentrations of M²⁺ and CN⁻, this simplifies the equation and allows determination of the equilibrium concentration of M²⁺.

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To find the concentration of M2+ ions at equilibrium, use the formation constant and initial concentrations of M(NO3)2 and NaCN.

The concentration of M2+ ions at equilibrium can be determined using the formation constant and the initial concentrations of M(NO3)2 and NaCN. First, calculate the initial concentration of M2+ ions by multiplying the concentration of M(NO3)2 (0.150 M) by its stoichiometric coefficient (2). Then, use the formation constant (7.70 × 10^16) to calculate the concentration of [M(CN)4]2- ions at equilibrium. Since the stoichiometric coefficient of M2+ ions is also 1, the concentration of M2+ ions at equilibrium will be equal to the concentration of [M(CN)4]2- ions.

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The sum of all of the coefficients in the balanced chemical equation of AH of Fe2O3 is 9.

A chemical equation is said to be balanced when the number of atoms of each element on both sides of the equation are the same.

According to this question, the balanced equation for the enthalpy change of Fe2O3 is given as follows:

4Fe + 3O2 → 2Fe2O3

The sum of all the coefficients in this balanced equation is as follows: 4 + 3 + 2 = 9.

Therefore, the sum of all of the coefficients in the balanced chemical equation of AH of Fe2O3 is 9.

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Final answer:

The balanced chemical equation for the formation of iron(III) oxide (Fe₂O₃) from iron (Fe) and oxygen (O₂) is 4 Fe (s) + 3 O₂ (g) → 2 Fe₂O₃ (s), resulting in a sum of 9 for all coefficients in the equation.

Explanation:

The question asks for the writing and balancing of the chemical equation associated with the standard enthalpy of formation (ΔHf) of iron(III) oxide (Fe₂O₃). The standard enthalpy of formation refers to the change in enthalpy when one mole of a compound is formed from its constituent elements in their standard states. For Fe₂O₃, this reaction involves the combination of iron (Fe) and oxygen (O₂) to form iron(III) oxide.

The balanced chemical equation for this reaction is:

4 Fe (s) + 3 O₂ (g) → 2 Fe₂O₃ (s)

Here, we ensure that the atom counts for Fe and O are equal on both sides of the equation: 4 Fe atoms on the left match the 4 Fe atoms on the right (within 2 molecules of Fe₂O₃), and 6 O atoms on the left (3 O₂) match the 6 O atoms on the right (within 2 molecules of Fe₂O₃). Thus, the equation is balanced, and the sum of all coefficients is 4 + 3 + 2 = 9.

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.

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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