Consider the following reaction: 2CH3OH(g)  2CH4(g) + O2(g) ΔH = +252.8 kJ a) Calculate the amount of heat transferred when 24.0 g of CH3OH(g) is decomposed by this reaction at constant pressure. b) For a given sample of CH3OH, the enthalpy change during the reaction is 82.1 kJ. How many grams of methane gas are produced?

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:

  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.

Final answer:

The radius of a gallium atom in a primitive cubic unit cell, with an edge length of 362 pm, is calculated to be 181 picometers by dividing the edge length by 2.

Explanation:

The question asks for the radius of a gallium atom given that gallium crystallizes in a primitive cubic unit cell with an edge length of 362 pm. In a primitive cubic unit cell, the atoms are located at the corners of the cube, and the length of the edge of the cube is equal to twice the atomic radius. Therefore, to find the radius of the gallium atom, we divide the edge length by 2.

Radius of gallium atom = edge length / 2 = 362 pm / 2 = 181 pm.

This calculation reveals that the radius of a gallium atom is 181 picometers in a primitive cubic unit cell structure.

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:

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.

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:

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:

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 molecular formula of the oxide of metal be [tex]X_2O_3[/tex]. The balanced equation for the reaction is given by:

[tex]X_2O_3+3CO\rightarrow 3CO_2+2X[/tex]

Explanation:

Let the molecular formula of the oxide of metal be [tex]X_2O_y[/tex]

[tex]X_2O_y+yCO\rightarrrow yCO_2+2X[/tex]

Mass of metal product = 1.68 g

Moles of metal X =[tex]\frac{1.68 g}{55.9 g/mol}=0.03005 mol[/tex]

1 mol of metal oxide produces 2 moles of metal X.

Then 0.03005 moles of metal X will be produced by:

[tex]\frac{1}{2}\times 0.03005 mol=0.01502 mol[/tex] of metal oxide

Mass of 0.01502 mol of metal oxide = 2.40 g (given)

[tex]0.01502 mol\times (2\times 55.9 g/mol+y\times 16 g/mol)=2.40 g[/tex]

y = 2.999 ≈ 3

The molecular formula of the oxide of metal be [tex]X_2O_3[/tex]. The balanced equation for the reaction is given by:

[tex]X_2O_3+3CO\rightarrow 3CO_2+2X[/tex]

Final answer:

To show the simplest formula of the oxide is X2O3, we calculate the moles of metal (X) and oxygen from given masses, find their ratio, and deduce the empirical formula. The balanced equation for the reaction with carbon monoxide is X2O3(s) + 3CO(g) → 2X(s) + 3CO2(g).

Explanation:

To prove that the simplest formula of the oxide is X2O3, first we need to calculate the moles of metal X produced. Since the molar mass of X is given as 55.9 g/mol, we divide the mass of metal product (1.68 g) by the molar mass of X to obtain the number of moles:

moles of X = 1.68 g / 55.9 g/mol = 0.03005 mol

We know that the initial mass of the oxide is 2.40 g and the product (X) is 1.68 g, so the mass of oxygen in the oxide is:

mass of O = 2.40 g - 1.68 g = 0.72 g

rationalizing the ratio, we get approximately 2:3

Thus, the empirical formula of the oxide is X2O3.

Balanced Equation for the Reaction

The balanced equation for the reaction of metal X's oxide with carbon monoxide to obtain metal X and carbon dioxide is:

X2O3(s) + 3CO(g) → 2X(s) + 3CO2(g)

This equation shows that the oxide of metal X reacts with carbon monoxide in a 1:3 mole ratio to produce the pure metal and carbon dioxide in a 2:3 mole ratio.

Answer:

Rank the resulting by neutrons, from most to least

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.

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 :

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: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 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 :

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 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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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:

q = 40.57 kJ; w = -3.10 kJ; strong H-bonds must be broken.

Explanation:

1. Heat absorbed

q = nΔH = 1 mol × (40.57 kJ/1 mol) = 40.57 kJ

2. Change in volume

V(water) = 0.018 L

pV = nRT

1 atm × V = 1 mol × 0.082 06 L·atm·K⁻¹mol⁻¹ × 373.15 K

V = 30.62 L

ΔV = V(steam) - V(water) = 30.62 L - 0.018 L = 30.60 L

3. Work done

w = -pΔV = - 1 atm × 30.60 L = -30.60 L·atm

w = -30.60 L·atm × (101.325 J/1 L·atm) = -3100 J = -3.10 kJ

4. Why the difference?

Every gas does 3.10 kJ of work when it expands at 100 °C and 1 atm.

The difference is in the heat of vaporization. Water molecules are strongly hydrogen bonded to each other, so it takes a large amount of energy to convert water from the liquid phase to the vapour phase.

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:

A) 5-methylhex-2-yne

B) (2Z)-5-methylhex-2-ene

C) (2E)-5-methylhex-2-ene

Explanation:

The given compound must be alkyne as it is undergoing reduction with two equivalents of hydrogen molecule.

Also as it is giving acetic acid on oxidative ozonolysis, it must have triple bond after two carbons in the chain.

The structure of hydrocarbon formed after reduction will give us the structure of alkyne by these information.

Reaction with hydrogen molecule in presence of Lindlar's catalyst gives cis alkene.

Reaction with hydrogen molecule in presence of Na, ammonia gives trans alkene.

The structure of compound is shown in the figure

The achiral hydrocarbon (A) with a molecular formula C7H12 is identified as Hept-1-yne. Upon reacting with a Lindlar catalyst and H2, it forms Hept-1-ene (B). When Hept-1-yne is treated with Sodium in ammonia (Na, NH3), it forms 1-heptyne (C).

The hydrocarbon A that has a molecular formula of C7H12 should be Hept-1-yne in light of the fact that Hept-1-yne upon hydrogenation, utilizing Pd-C as a catalyst and two equivalents of H2, forms 4-methyl hexane, which is exact to what was stated in the question. When Hept-1-yne is treated with ozone (O3), it gives two oxidative cleavage products, one being CH3COOH (acetic acid). The reaction gives us a clue about the presence of a triple bond at the end of the heptane chain. Hence, the structure of compound A (Hept-1-yne) is identified as CH3-(CH2)4-C≡CH.

Compound B can be identified as Hept-1-ene. This is due to the fact that Hept-1-yne, upon reacting with a Lindlar catalyst and H2, forms Hept-1-ene. This conversion is a result of the partial reduction of the triple bond to a double bond.Lastly, compound C can be identified as 1-heptyne. This is because when we treat Hept-1-yne with Sodium in ammonia (Na, NH3, it selectively reduces the triple bond to a trans double bond, a process known as dissolving metal reduction.

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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

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:

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.

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:

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: 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 :

(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.

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:

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