30.0 mL of 0.20 M AgNO, are added to 100.0 mL of 0.10 M HCI in a thermally nsulated vessel. The following reaction takes place: Ag (aq)+ Cl (aq)AgCI (s) The two solutions were initially at 22.00°C and the final temperature was 22.80 C Calculate the heat of this reaction in k.Jimol of AgCI formed. Assume a combined mass of 120 g and a specific heat capacity of 4.18 JK-'g for the reaction mixture.

Answer:-ΔG=-101.5KJ

Explanation:We have to calculate ΔG for the reaction  so using the formula given in the equation we can calculate the \Delta G for the reaction.

We need to convert the unit ofΔS in terms of KJ/Kelvin as its value is given in terms of J/Kelvin

Also we need to convert the temperature in Kelvin as it is given in degree celsius.

[tex]\Delta H=-137.5\\ \Delta S=-120J/K\\ \Delta S=-0.120KJ/K\\ T=25^{.C}\\ T=273+25=298 K\\ \Delta G=?\\ \Delta G=\Delta H-T\Delta S\\\Delta G=-137.5KJ-(278\times -0.120)\\ \Delta G=-137.5+35.76\\\Delta G=-101.74\\\Delta G=Negative[/tex]

After calculating forΔG we found that the value ofΔG is negative and its value is -101.74KJ

 For a  reaction to be spontaneous the value of \Delta G \ must be negative .

As the ΔG for the given reaction is is negative so the reaction will be spontaneous in nature.

In this reaction since the entropy of reaction is positive and hence when we increase the temperature term then the overall term TΔS would become more positive  and hence the value of ΔG would be less negative .

Hence the value of ΔG  would become more positive with the increase in temperature.

So we found the value of ΔG to be -101.74KJ

Answer:

ΔG = -101.591 KJ

Explanation:

Gibbs free energy -

It is a thermodynamic quantity , which is given by the change in enthalpy minus the product of the change in entropy and absolute temperature.

i.e.,

ΔG is given as the change in gibbs free energy ( KJ )

ΔS is given as the change in entropy ( KJ /K )

ΔH is given as the change in ethalphy ( KJ )

T = temperature ( Kelvin ( K ))

ΔG  =  ΔH - TΔS

The sign of ΔG determines the reaction spontaneity , as

ΔG = negative , the reaction is spontaneous and

If ΔG = positive , the reaction is non spontaneous .

Given -

For the reaction ,

C₂H₄ (g) + H₂(g) ---> C₂H₆(g)

ΔH = - 137.5 KJ

ΔS = - 120.5 J /K

Since ,

1 KJ = 1000 J

1 J = 1 / 1000KJ

ΔS = - 120.5 / 1000 KJ /K

ΔS = -0.1205 KJ /K

T = 25°C

(adding 273 To °C to convert it to K)

T = 25 + 273 = 298 K

Putting the values on the above equation ,

ΔG = ΔH - TΔS

ΔG = -137.5 KJ - 298 * (-0.1205 KJ / K)

ΔG = -137.5 KJ + 35.909 KJ

ΔG = -101.591 KJ

Since,

the value of ΔG is negative ,

hence, the reaction is spontaneous.

For the above reaction ,

If the temperature is increased ,

ΔG = ΔH - TΔS

From the above equation ,

the value of TΔS will increase ,

As a result the value of ΔG will be more positive , by increasing the temperature.

Answer :

(a) The work done by the gas on the surroundings is, 17537.016 J

(b) The entropy change of the gas is, 73.0709 J/K

(c) The entropy change of the gas is equal to zero.

Explanation:

(a) The expression used for work done in reversible isothermal expansion will be,

[tex]w=nRT\ln (\frac{V_2}{V_1})[/tex]

where,

w = work done = ?

n = number of moles of gas  = 4 mole

R = gas constant = 8.314 J/mole K

T = temperature of gas  = 240 K

[tex]V_1[/tex] = initial volume of gas  = [tex]V_1[/tex]

[tex]V_2[/tex] = final volume of gas  = [tex]9V_1[/tex]

Now put all the given values in the above formula, we get:

[tex]w=4mole\times 8.314J/moleK\times 240K\times \ln (\frac{9V_1}{V_1})[/tex]

[tex]w=17537.016J[/tex]

The work done by the gas on the surroundings is, 17537.016 J

(b) Now we have to calculate the entropy change of the gas.

As per first law of thermodynamic,

[tex]\Delta U=q-w[/tex]

where,

[tex]\Delta U[/tex] = internal energy

q = heat

w = work done

As we know that, the term internal energy is the depend on the temperature and the process is isothermal that means at constant temperature.

So, at constant temperature the internal energy is equal to zero.

[tex]q=w[/tex]

Thus, w = q = 17537.016 J

Formula used for entropy change:

[tex]\Delta S=\frac{q}{T}[/tex]

[tex]\Delta S=\frac{17537.016J}{240K}=73.0709J/K[/tex]

The entropy change of the gas is, 73.0709 J/K

(c) Now we have to calculate the entropy change of the gas when the expansion is reversible and adiabatic instead of isothermal.

As we know that, in adiabatic process there is no heat exchange between the system and surroundings. That means, q = constant = 0

So, from this we conclude that the entropy change of the gas must also be equal to zero.

Answer:4-nitrobenzoic acid is more acidic than Benzoic acid

4-chlorobenzoic acid is more acididc than 4-methyl benzoic acid

P-Nitrophenol is more acidic than meta-nitrophenol

Explanation:Acidity of an acid can be explained in terms of the stability of conjugate base formed.

1. 4-nitrobenzoic acid is more acidic as compared to benzoic acid because of the presence of nitro group at 4-position that is para position of the benzene ring. Nitro group is an electron withdrawing group and it withdraws the electron density through resonance effect.

Here the conjugate base would be benzoate anion which has a carboxylate anion attached with the benzene ring . So any group which can withdraw the electron density from benzoate anion will stabilise the benzoate anion and subsequently it would increase the acidity .

In case of  benzoic acid there is no extra withdrawl of electron density whereas in case of 4-nitrobenzoic acid the nitro group stabilises the benzoate anion by withdrawing electron density thereby stabilising the benzoate anion and increasing the acidity.

2. 4-chlorobenzoic acid is more acidic than 4-methyl benzoic acid because 4-chlorobenzoic acid has Cl group which is a good electron withdrawing group through inductive effect so the benzoate anion formed can be stabilised by the electron withdrawing Cl atom  which would increase the acidity of 4-chlorobenzoic acid.

4-methylbenzoic acid has an electron donating Methyl group which donates electron density through inductive effect hence a methyl group would intensify the negative charge on the benzoate anion through electron donation and subsequently it would destabilise the benzoate anion thereby decreasing its acidity.

3. In case of phenols the conjugate base formed is phenoxide anion and the negative charge that is its electron density is delocalised over the whole phenol ring. The negative charge electron density is more prominent at ortho and para position rather than the meta position. p-nitrophenol is more acidic than m-nitrophenol because p-nitrophenol has the nitro group at para position where it can stabilise the phenoxide anion through electron withdrawl via  resonance whereas in case of m-nitrophenol as the nitro group is present at meta position so it can not stabilize prominently through electron withdrawl via resonance.

In each pair, 4-nitrobenzoic acid, 4-chlorobenzoic acid, and p-nitrophenol are more acidic due to their electron withdrawing groups that can better stabilize the negative charge after ionization.

The acidity of a compound can be determined by its ability to donate a proton (hydrogen ion), and this ability is often influenced by other groups present in the compound. In each pair, the compound with the group that is more electron withdrawing will generally be more acidic because they can better stabilize the negative charge of the carboxylate ion after ionization occurs.

Benzoic acid vs. 4-nitrobenzoic acid: The presence of the nitro group in the 4-nitrobenzoic acid makes it more electron withdrawing compared to benzoic acid, making it more acidic.

4-Methylbenzoic acid vs. 4-chlorobenzoic acid: The chlorine atom in 4-chlorobenzoic acid is more electron withdrawing than the methyl group in 4-methylbenzoic acid, making 4-chlorobenzoic acid more acidic.

p-Nitrophenol vs. m-nitrophenol: In this case, p-nitrophenol would be more acidic. Although the nitro group is meta to the phenol group in m-nitrophenol, and para in p-nitrophenol, the para position allows for more effective resonance stabilization post-ionization, enhancing its acidity.

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Answer:

The pH of the final solution is 0.16 .

Explanation:

The pH of the solution is defined as negative logarithm of hydrogen ion concentration in a solution.

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

Concentration of HCl = 0.16 M

[tex]HCl(aq)\rightarrow H^+(aq)+Cl^-(aq)[/tex]

HCl is a string acid .1 molar of HCl gives 1 molar of of hydrogen ions.

[tex][H^+]=0.16 M[/tex]

Concentration of [tex]HNO_3[/tex] = 0.52 M

[tex]HNO_3(aq)\rightarrow H^+(aq)+NO_{3}^-(aq)[/tex]

Nitric is a string acid .1 molar of nitric acid gives 1 molar of of hydrogen ions.

[tex][H^+]'=0.52 M[/tex]

Total hydrogen ion concentration:

[tex][H^+]''=[H^+]+[H^+]'[/tex]

=0.16 M+0.52 M=0.68 M

The pH of the solution:

[tex]\pH=-\log[H^+]''=-\log[0.68 M][/tex]

pH = 0.16

The pH of the final solution is 0.16 .

Answer:

Diethyl ether

Explanation:

vapor pressure - the pressure exerted by the gaseous molecules on the walls of the container , is called its vapor pressure.

The compound with higher boiling point , will have lower vapor pressure ,

and the compound with lower boiling point , will have higher vapor pressure.

Hence, Boiling point and vapor pressure have inverse relation.

The vapor pressure and boiling point both, depends on the inter molecular interactions , i,e, the interaction between the molecules.  

Since, the compound with stronger inter molecular interactions, will not easily convert to gas, hence will have higher boiling point between , therefore , its vapor pressure would be less.

But the compound with less inter molecular interactions , can easily vaporize to convert to gaseous state and hence will have lower boiling point, therefore, its vapor pressure would be higher .

Among all the options , diethyl ether have lowest boiling point , hence, will have highest vapor pressure , at room temperature.

Final answer:

The value of the equilibrium constant, Kp, for the given reaction can be calculated based on the partial pressures of the reactants and products at equilibrium. In this case, the values are 1.425 atm for X₂ and 5.7 atm for X. Using the formula Kp = (p(X)²) / p(X₂), we find that Kp = 22.8.

Explanation:

The value of the equilibrium constant, Kp, for the reaction X₂(g) ⇌ 2X(g), can be calculated using the partial pressures of the reactants and products at equilibrium. In this case, the initial pressure of X₂ is 1.55 atm, and the total pressure at equilibrium is 2.85 atm.

Let's assume that the partial pressure of X₂ at equilibrium is p(X₂), and the partial pressure of X at equilibrium is p(X).

According to the equation, the number of moles of X₂ decreases by 1 (from 2 to 1) and the number of moles of X increases by 2 (from 0 to 2). This means that at equilibrium, the equilibrium partial pressure of X₂ is half of the total pressure, and the equilibrium partial pressure of X is twice the total pressure.

Therefore, p(X₂) = 2.85/2 = 1.425 atm and p(X) = 2 * 2.85 = 5.7 atm.

To calculate the value of Kp, we use the formula:

Kp = (p(X)²) / p(X₂)

Substituting the values we obtained, we get:

Kp = (5.7²) / 1.425 = 22.8

Answer:

The enthalpy of the reaction is -123.5 kJ.

Explanation:

[tex]P4 (s) + 6 Cl_2 (g)\rightarrow 4 PCl_3 (l) ,\Delta H_1 =-1280 kJ[/tex]..(1)

[tex]P4 (s) + 10 Cl_2 (g)\rightarrow 4 PCl_5 (l) ,\Delta H_2 =-1774 kJ[/tex]..(2)

[tex]PCl_3 (l) + Cl_2(g)\rightarrow PCl_5(s),\Delta H_{rxn}=x[/tex]...(3)

(2) - (1)

[tex]4PCl_3 (l) + 4Cl_2(g)\rightarrow 4PCl_5(s),\Delta H_{rxn}=y[/tex]

Dividing equation by 4 we get (3)

[tex]PCl_3 (l) + Cl_2(g)\rightarrow PCl_5(s),\Delta H_{rxn}=\frac{y}{4}[/tex]...(3)

[tex]\Delta H_{rxn}=y=(-1774 kJ)-(-1280 kJ)=-494 kJ[/tex]

[tex]\Delta H_{rxn}=x=\frac{y}{4}={-494 kJ}{4}=-123.5 kJ[/tex]

The enthalpy of the reaction is -123.5 kJ.

Answer:

chlorous acid HClO₂

Explanation:

The Ka is the acidity constant and it tells you about the acidity strength of a compound. If the value of Ka is high the compound is a strong acid. If the value of Ka is low the compound is a weak acid.  

The problem gives us the following compounds with the Ka values:

HF, 3.5 × 10⁻⁴

HCOOH, 1.8 × 10⁻⁴  

HClO₂, 1.1 × 10⁻²  

HCN, 4.9 × 10⁻¹⁰

HNO₂, 4.6 × 10⁻4

The chlorous acid HClO₂ have the highest Ka, 1.1 × 10⁻², so this one is the strongest acid.

The strength of an acid is determined by its Ka value. Among the given options: HF, HCOOH, HClO2, HCN, and HNO2, the strongest acid is HClO2 as it has the highest Ka value of 1.1 × 10-2.

The acidity of a substance is determined by its Ka value, which is the acid dissociation constant. A higher Ka value indicates a stronger acid because it shows the acid dissociates more completely in solution. The substances you've listed are all acids and their respective Ka values are given. Here, the strongest acid would have the highest Ka value.

Looking at the given options: HF (Ka = 3.5 × 10-4), HCOOH (Ka = 1.8 × 10-4), HClO2 (Ka = 1.1 × 10-2), HCN (Ka = 4.9 × 10-10), and HNO2 (Ka = 4.6 × 10-4), the strongest acid based on their Ka values would be HClO2 as it has the highest Ka value of 1.1 × 10-2.

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Answer:

Rank the resulting by neutrons, from most to least

Answer:

The correct answer is: X is nitrogen dioxide, and Y is a metal oxide

Explanation:

Combustion of compound of containing nitrogen and metal will give nitrogen  dioxide and metal oxide as product. During combustion reaction a compound reacts with oxygen in order to yield oxides of elements present in the compound.

The general equation is given as:

[tex]4M_3N_x+7xO_2\rightarrow 4xNO_2+6M_2O_x[/tex]

Hence, the correct answer is :X is nitrogen dioxide, and Y is a metal oxide.

In a combustion reaction, a compound containing nitrogen and a metal typically forms nitrogen dioxide (compound X) from the nitrogen atoms and a metal oxide (compound Y) from the metal atoms.

In the context of your question about how a compound containing nitrogen and a metal reacts in a combustion reaction, the outcome depends on the specific reactant. Usually, compounds containing nitrogen atoms tend to form nitrogen oxides under high heat or combustion conditions, with nitrogen dioxide (NO2) being a common example. This would be compound X.

Regarding the metal component, in a combustion reaction, metals commonly react with oxygen in the environment to produce metal oxides. This would make compound Y a metal oxide. Therefore, the correct pair of products according to your question tends to be: X is nitrogen dioxide, and Y is a metal oxide.

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Answer: The mass of EDTA that would be needed is 24.3 grams.

Explanation:

We are given:

Concentration of [tex]Ca^{2+}[/tex] ions = 20 mg/L

Converting this into grams/ Liter, we use the conversion factor:

1 g = 1000 mg

So, [tex]\Rightarrow \frac{20mg}{L}\times {1g}{1000mg}=0.02g/L[/tex]

Now, we need to calculate the mass of calcium present in 44 gallons of drum.

Conversion factor used:  1 gallon = 3.785 L

So, 44 gallons = (44 × 3.785)L = 166.54 L

Calculating the mass of calcium ions in given amount of volume, we get:

In 1L of volume, the mass of calcium ions present are 0.02 g.

Thus, in 166.54 L of volume, the mass of calcium ions present will be = [tex]\frac{0.02g}{1L}\times 166.54L=3.3308g[/tex]

The chemical equation for the reaction of calcium ion with EDTA to form Ca[EDTA] complex follows:

[tex]EDTA+Ca^{2+}\rightarrow Ca[EDTA][/tex]

Molar mass of EDTA = 292.24 g/mol

Molar mass of [tex]Ca^{2+}[/tex] ion = 40 g/mol

By Stoichiometry of the reaction:

40 grams of calcium ions reacts with 292.24 grams of EDTA.

So, 3.3308 grams of calcium ions will react with = [tex]\frac{292.24g}{40g}\times 3.3308g=24.33g[/tex] of EDTA.

Hence, the mass of EDTA that would be needed is 24.3 grams.

Answer : The pH of 0.001 M [tex]H_2SO_4[/tex] is, 2.69 and the pH after mixing the solution is, 3.30.

Explanation :

First we have to calculate the concentration of hydrogen ion.

The balanced dissociation reaction will be,

[tex]H_2SO_4\rightarrow 2H^++SO_4^{2-}[/tex]

The concentration of [tex]H_2SO_4[/tex] = x = 0.001 M

The concentration of [tex]H^+[/tex] ion = 2x = 2 × 0.001 M = 0.002 M

The concentration of [tex]SO_4^{2-}[/tex] = x = 0.001 M

Now we have to calculate the pH of 0.001 M [tex]H_2SO_4[/tex].

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

[tex]pH=-\log (0.002)[/tex]

[tex]pH=2.69[/tex]

Now we have to calculate the molarity after mixing the solution.

[tex]M_1V_1=M_2V_2[/tex]

where,

[tex]M_1[/tex] = molarity of [tex]H_2SO_4[/tex] solution = 0.001 M

[tex]V_1[/tex] = volume of [tex]H_2SO_4[/tex] solution = 250 ml

[tex]M_2[/tex] = molarity of after mixing = ?

[tex]V_2[/tex] = volume of after mixing = 250 + 750 = 1000 ml

Now put all the given values in the above formula, we get the molarity after mixing the solution.

[tex](0.001M)\times 250ml=M_2\times (1000ml)[/tex]

[tex]M_2=2.5\times 10^{-4}M[/tex]

The concentration of [tex]H^+[/tex] ion = [tex]2\times (2.5\times 10^{-4}M)=5\times 10^{-4}M[/tex]

Now we have to calculate the pH after mixing the solution.

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

[tex]pH=-\log (5\times 10^{-4})[/tex]

[tex]pH=3.30[/tex]

Therefore, the pH of 0.001 M [tex]H_2SO_4[/tex] is, 2.69 and the pH after mixing the solution is, 3.30.

Answer:

True

Explanation:

A disaccharide is a sugar and the general molecular formula of a disaccharide is C₁₂H₂₂O₁₁.

A disaccharide is formed when two monosachharide units are joined by a covalent bond called the glycosidic bond.

The glycosidic bond in a disaccharide is formed by dehydration reaction between the two monosachharide units. The removal of the water molecule results in the formation of the glycosidic linkage.

For example: maltose a disaccharide, is formed when two molecules of glucose are joined by a (1→4) glycosidic bond. As, the glycosidic bond is formed between the carbon 1 of one glucose unit and carbon 4 of another glucose unit.

Therefore, in a disaccharide the two monosaccharide units are joined by a glycosidic bond or linkage.

Therefore, the given statement is TRUE.

Final answer:

To calculate the pH of a 0.289 M ammonium chloride solution, first determine the Ka of the ammonium ion from the Kb of ammonia and set up an ICE table to find the hydronium ion concentration. Then calculate the pH using the negative logarithm of the hydronium ion concentration.

Explanation:

The pH of a solution of ammonium chloride can be calculated by first determining the Ka (acid dissociation constant) of the ammonium ion (NH4+), which is the conjugate acid of ammonia (NH3). Given that the Kb for ammonia is 1.8×10−5, the Ka for ammonium can be calculated using the relationship Ka = Kw/Kb, where Kw is the ion-product constant for water (1.0×10−14 at 25°C). In this case, Ka = 1.0×10−14 / 1.8×10−5 = 5.6×10−10.

Knowing the Ka, we can set up an ICE (Initial, Change, Equilibrium) table to find the concentration of hydronium ions, H3O+, produced. Since ammonium chloride is a strong electrolyte, it completely dissociates in water, thus initial [NH4+] is 0.289 M, and initial [H3O+] is 0. After the equilibrium is established, we calculate the concentration of H3O+ and subsequently find the pH of the solution. The pH is determined using the formula pH = -log[H3O+]. For a solution of ammonium chloride, this results in an acidic pH due to the formation of H3O+ ions.

The pH of a 0.289 M solution of ammonium chloride is approximately 4.89

This is calculated by determining the dissociation constant (Ka) for NH₄⁺ and using the concentration of H₃O⁺ ions to find the pH.

To determine the pH of a 0.289 M solution of ammonium chloride (NH₄Cl), we need to consider the dissociation of the ammonium ion (NH₄⁺) in water:

The equilibrium constant for this reaction is the acid dissociation constant (Ka) of the ammonium ion.

We can calculate Ka using the relation Ka = Kw / Kb.

Given Kw = 1.0 × 10⁻¹⁴ and Kb for ammonia (NH₃) as 1.8 × 10⁻⁵, we find:

Assuming the dissociation of NH₄⁺ is small, the concentration of H₃O⁺ is 'x', and using the initial concentration of NH₄⁺ (0.289 M) in the expression for Ka:

Simplifying for 'x', we get:

The pH is then calculated as:

Hence, the pH of a 0.289 M solution of ammonium chloride is approximately 4.89

Answer:

The concentration of acetic acid in vinegar of that trial would be greater than the actual concentration.

Explanation:

"The titrator" contains the base solution (NaOH) with which the soution of vinegar (acetic acid) is being titrated.

Under the assumption that the tip of the syringe was not filled before the initial volume reading was recorded, part of the volume of the base that you release will be retained in the tip of the syringe, and, consequently, the actual volume of base added to the acetic acid will be less than what you will calculate by the difference of readings.

So,  in your calculations you will use a larger volume of the base than what was actually used, yielding a fake larger number of moles of base than the actual amount added.

So, as at the neutralization point the number of equivalents of the base equals the number of acid equivalents, you will be reporting a greater number of acid equivalents, which in turn will result in a greater concentration than the actual one. This means that the concentration of acetic acid in vinegar of that trial would be greater than the actual concentration.

If the syringe tip was not filled with NaOH before recording the initial volume, the concentration of acetic acid calculated would be less than the actual concentration.

If the tip of the syringe was not filled with NaOH before the initial volume reading was recorded, the concentration of acetic acid in vinegar calculated from that trial would be less than the actual concentration. This is because the actual volume of NaOH dispensed during titration would be over-reported.

When the titration is performed, the volume of NaOH required to reach the equivalence point appears larger than it truly is, leading to a miscalculation of the moles of NaOH used. Consequently, this will result in the calculated concentration of acetic acid being lower than its true value.

In titration, ensuring that the titrant (in this case, NaOH) is ready and properly measured is crucial for accurate results. The initial volume reading must be correct to avoid errors in determining the volume of NaOH added, which directly affects the accuracy of the acetic acid concentration estimation.

Answer:

  Because it uses the residual energy of the fluid used by the first engine.

Explanation:

  A combined cycle power generation counts with two heat engines that work in tandem from the same source of heat. The engines turn the energy into mechanical energy.

  The cycle is much more efficient than the other, almost 60% more.

  I hope this answer helps you.

Answer:

The energy for vacancy formation in silver is 1.1 ev/atom

Explanation:

The total number of sites is equal to:

[tex]N=\frac{N_{A} \rho }{A}[/tex]

Where

NA = Avogadro´s number = 6.023x10²³atom/mol

A = atomic weight of silver = 107.9 g/mol

ρ = density of silver = 9.5 g/cm³

Replacing:

[tex]N=\frac{6.023x10^{23}*9.5 }{107.9} =5.3x10^{22} atom/cm^{3} =5.3x10^{28} m^{-3}[/tex]

The energy for vacancy is equal:

[tex]Q=-RTln(\frac{N_{v} }{N} )[/tex]

Where

R = 8.314 J/mol K = 8.614x10⁻⁵ev/atom K

T = 800°C = 1073 K

Nv = number of vacancy = 3.6x10²³m⁻³

Replacing:

[tex]Q=-8.614x10^{-5} *1073*ln(\frac{3.6x10^{23} }{5.3x10^{28} } )=1.1ev/atom[/tex]

Final answer:

The energy for vacancy formation in silver is 2.87 x 10^-19 J.

Explanation:

To calculate the energy for vacancy formation in silver, we can use the equation: E = kTln(N/V), where E is the energy, k is the Boltzmann constant (1.38 x 10^-23 J/K), T is the temperature in Kelvin, N is the number of vacancies, and V is the volume.

Given that the equilibrium number of vacancies is 3.6 x 10^23 m^-3, we can use the density of silver to calculate the volume. The density of silver at 800˚C is 9.5 g/cm^3, which is equivalent to 9.5 x 10^6 kg/m^3.

Therefore, the energy for vacancy formation in silver is:

E = (1.38 x 10^-23 J/K) x (1073 K) x ln(3.6 x 10^23 / (9.5 x 10^6)) = 2.87 x 10^-19 J.

Answer: The pH of the solution is 1.136

Explanation:

To calculate the moles from molarity, we use the equation:

[tex]\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}[/tex]

Molarity of ammonia = 0.3764 M

Volume of ammonia = 47.41 mL = 0.04741 L   (Conversion factor: 1 L = 1000 mL)

Putting values in above equation, we get:

[tex]0.3764mol/L=\frac{\text{Moles of ammonia}}{0.04741L}\\\\\text{Moles of ammonia}=0.01784mol[/tex]

Molarity of nitric acid = 0.3838 M

Volume of ammonia = (47.41 + 10.00) mL = 57.41 mL= 0.05741 L

Putting values in above equation, we get:

[tex]0.3838mol/L=\frac{\text{Moles of nitric acid}}{0.05741L}\\\\\text{Moles of nitric acid}=0.02203mol[/tex]

After the completion of reaction, amount of nitric acid remained = 0.022 - 0.0178 = 0.0042 mol

For the reaction of ammonia with nitric acid, the equation follows:

                       [tex]NH_3+HNO_3\rightarrow NH_4NO_3[/tex]

At [tex]t=0[/tex]             0.0178   0.022

Completion        0     0.0042        0.0178

As, the solution of the reaction is made from strong acid which is nitric acid and the conjugate acid of weak base which is ammonia. So, the pH of the reaction will be based totally on the concentration of nitric acid.

To calculate the pH of the reaction, we use the equation:

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

where,

[tex][H^+]=\frac{0.0042mol}{0.05741L}=0.0731M[/tex]

Putting values in above equation, we get:

[tex]pH=-\log(0.0731)\\\\pH=1.136[/tex]

Hence, the pH of the solution is 1.136

Answer : The frequency of a photon of radiation is, [tex]4.47\times 10^9s^{-1}[/tex]

Explanation : Given,

Wavelength of the radiation = 670.8 nm

First we have to convert wavelength form 'nm' to 'm'.

Conversion used : [tex](1nm=10^{-9}m)[/tex]

So, the wavelength of the radiation = 670.8 nm = [tex]670.8\times 10^{-9}m[/tex]

Now we have to calculate the frequency of a photon of radiation.

Formula used : [tex]\nu =\frac{c}{\lambda}[/tex]

where,

[tex]\nu[/tex] = frequency of a photon of radiation

[tex]\lambda[/tex] = wavelength of the radiation

c = speed of light = [tex]3\times 10^8m/s[/tex]

Now put all the given values in the above formula, we get the frequency of a photon of radiation.

[tex]\nu =\frac{3\times 10^8m/s}{670.8\times 10^{-9}m}[/tex]

[tex]\nu =4.47\times 10^9s^{-1}[/tex]

Therefore, the frequency of a photon of radiation is, [tex]4.47\times 10^9s^{-1}[/tex]

Answer:

[tex]f=4.47x10^5GHz[/tex]

Explanation:

Hello,

In this case, we relate the speed of light, wavelength and frequency via the shown below equation expressed in the proper SI system of units:

[tex]f=\frac{c}{\lambda } =\frac{3x10^8m/s}{670.8nm*\frac{1x10^{-9}m}{1nm} } =4.47x10^{14}Hz*\frac{1GHz}{1x10^9Hz}\\ f=4.47x10^5GHz[/tex]

Best regards.

Answer: The [tex]\Delta H_f[/tex] for [tex]H_2O(l)[/tex] in the reaction is 297.9 kJ/mol.

Explanation:

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

The equation that is used to calculate enthalpy change of a reaction is:

[tex]\Delta H_{rxn}=\sum [n\times \Delta H_f(product)]-\sum [n\times \Delta H_f(reactant)][/tex]

For the given chemical reaction:

[tex]2H_2O(l)\rightarrow 2H_2(g)+O_2(g)[/tex]

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

[tex]\Delta H_{rxn}=[(2\times \Delta H_f_{(H_2)})+(1\times \Delta H_f_{(O_2)})]-[(2\times \Delta H_f_{(H_2O)})][/tex]

We are given:

Enthalpy of substances present in their standard form are taken to be 0.

[tex]\Delta H_f_{(H_2)}=0kJ/mol\\\Delta H_f_{(O_2)}=0kJ/mol\\\Delta H_{rxn}=595.8kJ[/tex]

Putting values in above equation, we get:

[tex]595.8=[(2\times (0))+(1\times (0))]-[2\times (\Delta H_f_{H_2O})]\\\\\Delta H_f_{H_2O}=297.9kJ/mol[/tex]

Hence, the [tex]\Delta H_f[/tex] for [tex]H_2O(l)[/tex] in the reaction is 297.9 kJ/mol.

Answer:

95.8 kg

Explanation:

At the end of the feeding process, there is steam in the tank at 15 MPa and 400ºC because the process is adiabatic. So, use the steam tables (In this case I use data from van Wylen Six Edition, table B.13) in order to get the specific volume of superheated steam.

The specific volume data reported is [tex]v=0.01565\frac{m^{3}}{kg}[/tex]

The mass can be calculated with the definition of specific volume:

[tex]v=\frac{V}{m}\\m=\frac{V}{v}=\frac{1.5m^{3}}{0.01565\frac{m^{3}}{kg}} =95.8kg[/tex]

Answer:

In the first test precipitates AgCl and PbCl2. In the second one there is SnCl4 and SnS2 that are very soluble, and there ir more SnCl4 that SnS2.

Explanation:

This problem is about the cualitity studies about ions. The acidity is a factor for this studies. The chlorides and sulfides groups are mostly solubles except Pb2+, Ag+ and Hg2+ for chlorides and Sr+2, Ba+2, Pb+2 y Hg+2 for sulfides.

In the first case we have a high concentration of HCl. It means that all ions reaction with HCl. In the second one there is no reaction because in the solution we have SnCl4 that is very soluble and SnS2 is very soluble too. There is more SnCl4 because for Le Chatelier if we add more reactive the balance tends to reactive.

Answer: The amount of sulfur present in the original material is [tex]1.34\times 10^7tons[/tex]

Explanation:

Converting given amount of mass in tons to grams, we use the conversion factor:

1 ton = 907185 g            .......(1)

So, [tex]2.68\times 10^7=2.431\times 10^{13}g[/tex]

To calculate the number of moles, we use the equation:

[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex]     ......(2)

Given mass of sulfur dioxide = [tex]2.431\times 10^{13}g[/tex]

Molar mass of sulfur dioxide = 64 g/mol

Putting values in above equation, we get:

[tex]\text{Moles of sulfur dioxide}=\frac{2.431\times 10^{13}g}{64g/mol}=3.79\times 10^9mol[/tex]

For the given chemical reaction:

[tex]S(s)+O_2(g)\rightarrow SO_2(g)[/tex]

By Stoichiometry of the reaction:

1 mole of sulfur dioxide is produced from 1 mole of sulfur

So, [tex]3.79\times 10^9[/tex] moles of sulfur dioxide will be produced from = [tex]\frac{1}{1}\times 3.79\times 10^9=3.79\times 10^9[/tex] moles of sulfur.

Now, calculating the mass of sulfur using equation 2:

Moles of sulfur = [tex]3.79\times 10^9mol[/tex]

Molar mass of sulfur = 32 g/mol

Putting values in equation 2, we get:

[tex]3.79\times 10^9mol=\frac{\text{Mass of sulfur}}{32g/mol}\\\\\text{Moles of sulfur}=121.54\times 10^{11}g[/tex]

Converting this value in tons using conversion factor 1, we get:

[tex]\Rightarrow (\frac{1ton}{907185g})\times 121.54\times 10^{11}g\\\\\Rightarrow 13397491.6tons=1.34\times 10^7tons[/tex]

Hence, the amount of sulfur present in the original material is [tex]1.34\times 10^7tons[/tex]

Answer:

Average molecular weight of air is 28.84 g/mol.

Explanation:

The average molecular weight of a mixture is determined from their molar composition and molecular weight.

Average molecular weight :[tex]\sum (\chi_i\times m_i)[/tex]

[tex]\chi_1[/tex] : mole fraction of the 'i' component.

[tex]m_i[/tex] = Molecular weight of i component

Average molecular weight of air with approximate molar composition of 79% nitrogen gas and 21% of oxygen gas can be calculated as:

Average molecular weight of air:

[tex]79\%\times 28 g/mol+21\%\times 32 g/mol[/tex]

[tex]=0.79\times 28 g/mol+0.21\times 32 g/mol=28.84 g/mol[/tex]

Final answer:

The average molecular weight of air can be calculated using its molar composition of 79% N2 and 21% O2.

Explanation:

The average molecular weight of air can be calculated using its molar composition of 79% N2 and 21% O2. The molar weight of N2 is 28.01 g/mol and the molar weight of O2 is 32.00 g/mol. We can calculate the average molecular weight using the formula:

Average Molecular Weight = (0.79 * 28.01 g/mol) + (0.21 * 32.00 g/mol)

Average Molecular Weight = 28.89 g/mol

Answer : The value of [tex]K_{goal}[/tex] for the final reaction is, [tex]1.238\times 10^{-7}[/tex]

Explanation :

The following equilibrium reactions are :

(1) [tex]2CO_2(g)+2H_2O(l)\rightleftharpoons CH_3COOH(l)+2O_2[/tex] [tex]K_1=5.40\times 10^{-16}[/tex]

(2) [tex]2H_2(g)+O_2(g)\rightleftharpoons 2H_2O(l)[/tex] [tex]K_2=1.06\times 10^{10}[/tex]

(3) [tex]CH_3COOH(l)\rightleftharpoons 2C(s)+O_2(g)[/tex] [tex]K_3=2.68\times 10^{-9}[/tex]

The final equilibrium reaction is :

[tex]CO_2(g)\rightleftharpoons C(s)+O_2(g)[/tex] [tex]K_{goal}=?[/tex]

Now we have to calculate the value of [tex]K_{goal}[/tex] for the final reaction.

First half the equation 1, 2 and 3 that means we are taking square root of equilibrium constant and then add all the equation 1, 2 and 3 that means we are multiplying all the equilibrium constant, we get the final equilibrium reaction and the expression of final equilibrium constant is:

[tex]K_{goal}=\sqrt{K_1\times K_2\times K_3}[/tex]

Now put all the given values in this expression, we get :

[tex]K_{goal}=\sqrt{(5.40\times 10^{-16})\times (1.06\times 10^{10})\times (2.68\times 10^{-9})}[/tex]

[tex]K_{goal}=1.238\times 10^{-7}[/tex]

Therefore, the value of [tex]K_{goal}[/tex] for the final reaction is, [tex]1.238\times 10^{-7}[/tex]

The value of the equilibrium constant, Kgoal, for the reaction CO2(g)⇌C(s)+O2(g) can be calculated by multiplying the equilibrium constants of the individual reactions involved.

The value of the equilibrium constant, Kgoal, for the reaction CO2(g) ⇌ C(s) + O2(g) can be determined using the given information:

Since reaction 3 is the sum of reactions 1 and 2, we can use the equations to calculate the value of Kgoal:

Kgoal = K1 × K2 × K3

Substituting the values:

Kgoal = (5.40×10-16) × (1.06×1010) × (2.68×10-9) = 1.47×10-14

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The equilibrium constants K'eq for the given reactions can be calculated using the formula K'eq = exp(-∆Go'/RT), where ∆Go' is the standard Gibbs free energy change, R is the gas constant, and T is the temperature in Kelvin.

For the given reactions the equilibrium constants K'eq can be calculated using the formula: K'eq = exp(-∆Go'/RT), where ∆Go' is the standard Gibbs free energy change, R is the gas constant (8.314 J/mol K at 25°C or 298.15 K) and T is the temperature in Kelvin.

Note: In the formulas above, the Gibbs free energy change is converted to J/mol (from kJ/mol) by multiplying by 10^3.

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Answer :

true

Explanation:

when titanium react with oxygen it form TiO₂ Titanium oxide which is passive in nature it forms layer of  TiO₂ when large amount of oxygen is passed through titanium at very high temperature it does not react with oxygen impurities for this reason titanium mostly used in aerospace and chemical industries

Answer:The product formed on reaction with hydroxide ion as nucleophile is 2R-hexane-2-ol.

The product formed on reaction with water would be a 50:50 mixture of

2S-hexane-2-ol. and 2R-hexane-2-ol.

Explanation:

2S-iodohexane on reactiong with hydroxide ion would undergo SN² substitution reaction that is substitution bimolecular. Hydroxide ion has a negative charge and hence it is a quite good  nucleophile .

The rate of a SN² reaction depends on both the substrate and nucleophile . Here the substrate is a secondary carbon center having Iodine as a leaving group.SN² reaction  takes place here as hydroxide ion is a good nucleophile and it can attack the secondary carbon center from the back side leading to the formation of 2R-hexane-2-ol.

In a SN² reaction since the the nucleophile attacks from the back-side so the product formation takes place with the inversion of configuration.

When the same substrate S-2-iodohexane undergoes a substitution reaction with water as a nucleophile then the reaction occurs through (SN¹) substitution nucleophilic unimolecular mechanism .

The rate of a SN¹ reaction depends only on the nature of substrate and is independent of the nature of nucleophile.

The SN¹ reaction is a 2 step reaction , in  the first step leaving group leaves leading to the formation of a carbocation and once the carbocation is formed then any weaker nucleophile or even solvent molecules can attack leading the formation of products.

In this case a secondary carbocation would be generated in the first step and then water will attack this carbocation to form the product in the second step.

The product formed on using water as a nucleophile would be a racemic mixture of R and S isomers of hexane -2-ol in 50:50 ratio. The two products formed would be 2R-hexane-2-ol and 2S-hexane-2-ol.

Kindly refer the attachment for reaction mechanism and structure of products.

Answer :

Lewis-Randall rule : Lewis-Randall rule states that, the fugacity of a component is directly proportional to the mole fraction of the component in the solution in an ideal solution.

The equation used for the Lewis-Randall rule for fugacity of species in an ideal solution is :

[tex]f_i=X_i\times f_i^o[/tex]

where,

[tex]f_i[/tex] = fugacity in the solution

[tex]f_i^o[/tex] = fugacity of a pure component

[tex]X_1[/tex] = mole fraction of component

Final answer:

The Lewis-Randall rule indicates that the fugacity of a component in an ideal solution equals the product of the component's mole fraction and the fugacity of the pure component, central to thermodynamics and mixture behavior understanding.

Explanation:

The Lewis-Randall rule provides a foundation for understanding the fugacity of species in an ideal solution. It states that the fugacity of a component in an ideal mix is equal to the product of the mole fraction of that component in the mixture and the fugacity of the pure component at the same temperature and pressure. In mathematical terms, for a component i in an ideal solution, this can be expressed as fi = xi × fi°, where fi is the fugacity of the component in the mixture, xi is the mole fraction of the component, and fi° is the fugacity of the pure component.

This rule is pivotal in the field of thermodynamics and is particularly relevant when dealing with ideal solutions where the interactions between different species are similar to those present in pure substances. It provides a basis for calculating the fugacity coefficient, which is used to assess how the real behavior of a gas differs from the ideal predicted by Raoult's law in mixtures. The Lewis-Randall rule, alongside Raoult's and Henry's laws, forms a fundamental part of the theoretical framework for understanding phase equilibria and the behavior of mixtures.

Answer:

25.275 moles of oxygen gas will be required to completely react with all the samarium metal.

Explanation:

[tex]4Sm+3O_2\rightarrow 2Sm_2O_3[/tex]

Number of moles samarium metal = 33.7 moles

According to reaction, 4 moles of  samarium reacts with 3 moles of oxygen gas.

Then 33.7 moles of samarium will react with:

[tex]\frac{3}{4}\times 33.7 mol=25.275 mol[/tex]of oxygen gas.

25.275 moles of oxygen gas will be required to completely react with all the samarium metal.

Answer:

Moles of oxygen gas required to react completely with 33.7 moles of samarium metal is [tex]\fbox{25.3 \text{ mol}}[/tex].

Explanation:

A chemical equation in which the number of atoms of each element is the same on the reactant and product side is called a balanced chemical equation.

The balanced chemical equation can be used to determine the stoichiometric ratio between the reactant and the product. The stoichiometric ratio thus enables us to calculate:

1. Amount of one reactant required to react completely with the other reactant.

2. Amount of the product that can be produced from the given amount of the reactant.

Step 1: Write the chemical equation for the reaction between samarium metal and oxygen gas.

The chemical formula for oxygen gas is [tex]\text{O}_{2}[/tex].

Samarium has +3 oxidation state within the samarium/oxygen compound. So, the chemical formula of the samarium oxygen compound is [tex]\text{Sm}_{2}\text{O}_{3}[/tex].

The chemical equation is as follows:

[tex]\fbox{\text{Sm}+\text{O}_{2} \rightarrow \text{Sm}_{2}\text{O}_{3}\\\end{minipage}}[/tex]

Step 2: Balance the chemical equation for the reaction between oxygen and samarium metal.

The number of oxygen atoms in the product side is 3 and in the reactant side is 2. Put coefficient 2 in front of [tex]\text{Sm}_{2}\text{O}_{3}[/tex] and 3 in front of [tex]\text{O}_{2}[/tex] to balance the oxygen atoms.

[tex]\fbox{\text{Sm}+3\text{O}_{2} \rightarrow 2 \text{Sm}_{2}\text{O}_{3}\\\end{minipage}}[/tex]

The number of samarium atoms in the product side is 4 and in the reactant side is 1. Put coefficient 4 in front of Sm in the reactant side.

[tex]\fbox{\\4\text{Sm}+3\text{O}_{2} \rightarrow 2 \text{Sm}_{2}\text{O}_{3}\\\end{minipage}}[/tex]

Step 3: Determine the stoichiometric ratio between samarium and oxygen from the above balanced chemical equation.

According to the balanced chemical equation, we can see that the stoichiometric ratio between samarium and oxygen is 4:3.

Step 4: Use unitary method and calculate the moles of oxygen required to completely react with the given moles of samarium metal as follows:

[tex]\text{moles of O}_{2} = \left( \text{moles of Sm} \right)\left( \frac{3 \text{ mol of O}_{2}}{4 \text{ mol of Sm}} \right)[/tex]                               ...... (1)

Step 5: Substitute 33.7 mol for moles of Sm in equation (1).

[tex]\text{Moles of O}_{2} = \left( \text{33.7 mol} \right) \left( \frac{3 \text{ mol of O}_{2}}{4 \text{ mol of Sm}} \right)\\\text{Moles of O}_{2}= 25.275 \text{ mol}\\\text{Moles of O}_{2}= 25.3 \text{ mol}[/tex]

Note:

Do not forgot to balance the reaction. The reaction must be balanced in order to calculate the amount (mol) of oxygen required to completely react with the given amount of samarium.

Learn more:

1. Balanced chemical equation https://brainly.com/question/1405182

2. Learn more about how to calculate moles of the base in given volume https://brainly.com/question/4283309

Answer details:

Grade: Senior School

Subject: Chemistry

Chapter: Some basic concept of chemistry

Keywords: samarium, oxygen gas, samarium/oxygen compound, 33.7 moles, 25.3 mol, balanced equation, stoichiometric ratio, coefficient, balance, moles, completely react.