The 1995 Nobel Prize in Chemistry was shared by Paul Crutzen, F. Sherwood Rowland, and Mario Molina for their work concerning the formation and decomposition of ozone in the stratosphere. Rowland and Molina hypothesized that chlorofluorocarbons (CFCs) in the stratosphere break down upon exposure to UV radiation, producing chlorine atoms. Chlorine was previously identified as a catalyst in the breakdown of ozone into oxygen gas. Using the enthalpy of reaction for two reactions with ozone, determine the enthalpy of reaction for the reaction of chlorine with ozone. ( 1 ) ClO ( g ) + O 3 ( g ) ⟶ Cl ( g ) + 2 O 2 ( g ) Δ H ∘ rxn = − 122.8 kJ ( 2 ) 2 O 3 ( g ) ⟶ 3 O 2 ( g ) Δ H ∘ rxn = − 285.3 kJ ( 3 ) O 3 ( g ) + Cl ( g ) ⟶ ClO ( g ) + O 2 ( g ) Δ H ∘ rxn = ?

Explanation:

The given data is as follows.

Molecular weight of azulene = 128 g/mol

Hence, calculate the number of moles as follows.

      No. of moles = [tex]\frac{mass}{\text{molecular weight}}[/tex]

                            = [tex]\frac{0.392 g}{128 g/mol}[/tex]

                            = 0.0030625 mol of azulene

Also,    [tex]-Q_{rxn} = Q_{solution} + Q_{cal}[/tex]

       [tex]Q_{rxn} = n \times dE[/tex]

         [tex]Q_{solution} = m \times C \times (T_{f} - T_{i})[/tex]

              [tex]Q_{cal} = C_{cal} \times (T_{f} - T_{i})[/tex]

Now, putting the given values as follows.    

     [tex]Q_{solution} = 1.17 \times 10^{3} g \times 10^{3} \times 4.184 J/g^{o}C \times (27.60 - 25.20)^{o}C[/tex]

                   = 11748.67  J

So,  [tex]Q_{cal} = 786 J/^{o}C \times (27.60 - 25.20)^{o}C[/tex]

                    = 1886.4 J

Therefore, heat of reaction will be calculated as follows.

        [tex]-Q_{rxn}[/tex] = (11748.67 + 1886.4) J

                      = 13635.07 J

As,  [tex]Q_{rxn} = n \times dE[/tex]

          13635.07 J = [tex]-n \times dE[/tex]

                dE = [tex]\frac{13635.07 J}{0.0030625 mol}[/tex]

                     = 4452267.75 J/mol

or,                 = 4452.26 kJ/mol       (as 1 kJ = 1000 J)

Thus, we can conclude that [tex]\Delta E[/tex] for the given combustion reaction per mole of azulene burned is 4452.26 kJ/mol.

Answer:

A) More NH3(g) will form.

Explanation:

According to the Le Chatellier Principle if we remove product from a reaction which is in equilibrium, then the effect will to produce more product in order to replace what has been removed.

In this reaction also , on spraying liquid water, Ammonia being water soluble will be consumed. Hence, The reaction will shift in forward direction leading to more formation of Ammonia (NH3).

Hence option A will be correct .

Final answer:

Spraying water into the equilibrium involving N2, H₂, and NH₃ will cause more NH₃ to form because NH₃ is more soluble in water than N₂ or H₂, and removing NH₃ by dissolving it in water shifts the equilibrium towards producing more NH₄. the correct option is A.

Explanation:

The question asks what effect spraying liquid water into the equilibrium N2(g) + 3H2(g) ⇒ 2NH3(g) will have, given that NH₃ is far more soluble in water than N2 or H2. According to Le Chatelier's Principle, if a change is applied to a reaction at equilibrium, the system will adjust to partially counteract that change. In this case, spraying water will remove NH₃ by dissolving it. This reduction in the concentration of NH3 will cause the system to shift towards the product side to replace the removed NH₃, thus more NH₃ will form. Therefore, the correct answer is A) MoreNH₃(g) will form.

Answer:

When the solution (with phenolphthalein) changes to colorless

Explanation:

When titrating with HCl is common to add phenolphthalein as an acid-base indicator.

Phenolphthalein is pink or fucsia when added into a basic solution. On the other hand when it is in acid solutions, is colorless.

So, when titrating, the NaOH solution will be initialy pink due to the phenolphthalein and when reaching the equivalence point, that color will fade out into colorless. This is how you know you hace reached the equivalent point.

Answer:

The standard molar enthalpy of formation of 1 mole of NO gas is 90.25 kJ/mol.

Explanation:

We have :

[tex]N_2(g) + 2O_2 \rightarrow 2NO_2(g), \Delta H^o_{1} = 66.4 kJ[/tex]

[tex]2NO(g) + O_2\rightarrow 2NO_2(g),\Delta H^o_{2} = -114.1 kJ[/tex]

To calculate the standard molar enthalpy of formation

[tex]N_2+O_2\rightarrow 2NO(g),\Delta H^o_{3} = ?[/tex]...[3]

[1] - [2] = [3] (Hess's law)

[tex]N_2+O_2\rightarrow 2NO(g),\Delta H^o_{3} = ?[/tex]

[tex]\Delta H^o_{3} =\Delta H^o_{1} - \Delta H^o_{2} [/tex]

[tex]\Delta H^o_{3}=66.4 kJ - [ -114.1 kJ] = 180.5 kJ[/tex]

According to reaction [3], 1 mole of nitrogen gas and 1 mole of oxygen gas gives 2 mole of nitrogen monoxide.

So, the standard molar enthalpy of formation of 1 mole of NO gas :

[tex]\Delta H^o_{f,NO}=\frac{\Delta H^o_{3}}{2 mol}[/tex]

[tex]\Delta H^o_{f,NO}=\frac{180.5 kJ}{2 mol}=90.25 kJ/mol[/tex]

To calculate the standard molar enthalpy of formation of NO(g), the given reactions are used by reversing the second reaction and adding its enthalpy change to the first reaction's ΔH. Dividing the total by 2 gives 90.25 kJ/mol for the formation of 1 mole of NO(g) from its elements.

To calculate the standard molar enthalpy of formation of NO(g), we can use the given enthalpy changes for the reactions provided. The two reactions are:

We want to find the enthalpy change for the formation of NO(g) from its elements, which is N2(g) and O2(g). This is the standard molar enthalpy of formation for NO(g). We can set up an enthalpy diagram to help visualize the process.

Working through the math:
ΔHf[NO] = (66.4 kJ + 114.1 kJ) / 2
ΔHf[NO] = 180.5 kJ / 2
ΔHf[NO] = 90.25 kJ/mol

Therefore, the standard molar enthalpy of formation of NO(g) is 90.25 kJ/mol.

Answer:

Strong electrolyte

Explanation:

Bulb doesn't light when water is used because water doesn't have any appreciable amount of ions for the conduction of electricity.

Strong electrolytes ionize almost completely in solution, thus providing a huge amount of ions to conduct electricity. More the number of ions, more charge is carried and hence greater current and thus leading to bulb glowing brighter.

Weak electrolytes ionize partially in solution, thus providing some amount of ions for carrying charge. This results in a small amount of current going to the bulb and hence the bulb lights up with very low intensity.

Non electrolytes doesn't ionize either and hence is similar to water. These have almost zero conductivity and hence the bulb doesn't light at all.

Answer: The concentration of weak acid is [tex]3.674\times 10^{-2}M[/tex]

Explanation:

To calculate the concentration of acid, we use the equation given by neutralization reaction:

[tex]n_1M_1V_1=n_2M_2V_2[/tex]

where,

[tex]n_1,M_1\text{ and }V_1[/tex] are the n-factor, molarity and volume of acid which is [tex]H_2A[/tex]

[tex]n_2,M_2\text{ and }V_2[/tex] are the n-factor, molarity and volume of base which is NaOH.

We are given:

[tex]n_1=2\\M_1=?M\\V_1=25.00mL\\n_2=1\\M_2=0.1036M\\V_2=17.73mL[/tex]

Putting values in above equation, we get:

[tex]2\times M_1\times 25.00=1\times 0.1036\times 17.73\\\\M_1=\frac{1\times 0.1036\times 17.73}{2\times 25.00}=3.674\times 10^{-2}M[/tex]

Hence, the concentration of weak acid is [tex]3.674\times 10^{-2}M[/tex]

Answer: The nitrogen atom is getting reduced.

Explanation:

Oxidation reaction is defined as the reaction in which an atom looses its electrons. The oxidation number of the atom gets increased during this reaction.

[tex]X\rightarrow X^{n+}+ne^-[/tex]

Reduction reaction is defined as the reaction in which an atom gains electrons. The oxidation number of the atom gets reduced during this reaction.

[tex]X^{n+}+ne^-\rightarrow X[/tex]

The given chemical equation follows:

[tex]Fe_2S_3+12HNO_3\rightarrow 2Fe(NO_3)_3+3S+6NO_2+6H_2O[/tex]

On reactant side:

Oxidation state of iron atom = +3

Oxidation state of nitrogen atom = +5

Oxidation state of sulfur atom = -2

On product side:

Oxidation state of iron atom = +3

Oxidation state of sulfur atom = 0

Oxidation state of nitrogen atom in [tex]NO_2[/tex] = +4

Oxidation state of nitrogen atom in [tex]Fe(NO_3)_3[/tex] = +5

As, the oxidation state of nitrogen atom in [tex]NO_2[/tex] is decreasing from +5 to +4. So, it is getting reduced.

And, oxidation state of sulfur atom is increasing from -2 to 0. So, it is getting oxidized.

Hence, the nitrogen atom is getting reduced.

The element that gets reduced in the chemical reaction is nitrogen.

Reduction in chemistry means when an atom accepts electrons to become more negatively charged.

According to this question, the following chemical reaction is given:

Fe2S3 + 12HNO3 → 2Fe(NO3)3 + 3S + 6NO2 + 6H2O

On reactant side:

On product side:

As indicated above, the oxidation number of nitrogen reduces from +5 to +4, hence, it is the element that gets reduced.

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The molecule acetonitrile or CH3CN, indeed has 16 valence electrons and a triple bond follows its Lewis structure. However, the statement that acetonitrile has a pair of nonbonding electrons is incorrect because all 16 valence electrons are used in the molecule's bonds.

The molecule in question is acetonitrile, or CH3CN. To analyze these statements, it's crucial to consider the Lewis structure of this molecule: the carbon atom bonded to three hydrogen atoms, that carbon bonded to another carbon atom, and that second carbon bonded to a nitrogen through a triple bond. Hydrogen has 1 valence electron, carbon has 4, and nitrogen has 5.

a. The combined total of valence electrons for acetonitrile is indeed 16 - (3x1 for hydrogen, 2x4 for carbon, and 1x5 for nitrogen).
b. There is a triple bond between the carbon and nitrogen atoms.
c. The entirety of the 16 valence electrons are used in the molecule's bonds, leaving no nonbonding electrons. This statement is false.
d. All atoms in the structure satisfy the octet rule, each having a complete outermost shell.
e. Neither carbon atom nor the nitrogen atom has a nonzero formal charge; all atoms in acetonitrile have a formal charge of zero.

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The false statement about acetonitrile (CH3CN) is that it has one pair of nonbonding electrons. All the outer shell electrons in acetonitrile are involved in bonding, hence, there are no non-bonding electrons in the molecule.

Based on your question about acetonitrile (CH3CN), which is an important industrial chemical, the statement that is FALSE is: Acetonitrile has one pair of nonbonding electrons. When drawing the Lewis structure for acetonitrile, we find that all the outer shell electrons are involved in bonding, hence, there are no non-bonding electrons in acetonitrile. The molecule's other properties mentioned in the options are correct: it has 16 valence electrons, a triple bond between the carbon and nitrogen atoms, all atoms satisfy the octet rule, and the carbon attached to the nitrogen, as well as the nitrogen atom itself, have nonzero formal charges due to the triple bond and unequal sharing of electrons.

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Explanation:

Entropy -

In a system, the randomness is measured by the term entropy .

Randomness basically refers as a form of energy that can not be used for any work.

The change in entropy is given by amount heat per change in temperature.

When solid is converted to liquid or gas entropy increases,

As the molecules in solid state are tightly packed and has more force of attraction between the molecules, but as it is converted to liquid or gas, the force of attraction between the molecule decreases and hence entropy increases.

So,

The particles of the substance , if are tightly held by strong force of attraction will decrease the entropy ,

And

If the particles are loosely held , the entropy will increase , i.e. , when gas is converted to liquid or solid .

A. boiling water to form steam ,

change of state from liquid to gas ,

and , hence entropy increases .

B. dissolution of solid KCl in water  ,

the number of particles increases ,

and hence , entropy increases .

C. mixing of two gases into one container  ,

the number of particles increases ,

and hence , entropy increases .

D. freezing water to form ice  ,

change of state from liquid to solid ,

and hence , entropy decreases .

E. melting ice to form water .

change of solid to liquid ,

and hence , entropy increases .

Final answer:

The process that results in a decrease in entropy is freezing water to form ice, as it involves water transitioning from a disordered liquid state to an ordered solid state.

Explanation:

The process that produces a decrease in the entropy of the system is freezing water to form ice (Option D). During freezing, water molecules move from a higher state of disorder in the liquid phase to a more ordered solid phase, resulting in reduced entropy. Freezing involves the transition of water from the liquid phase, where the molecules are more disordered, to the solid phase, where they are arranged in a structured pattern. This structured pattern, characterized by a fixed position of the molecules in the crystal lattice of ice, embodies a lower state of entropy.

Even though freezing results in a decrease in the system's entropy, this does not violate the second law of thermodynamics because the surrounding environment's entropy increases when heat is released during the freezing process, ensuring that the total entropy of the system plus its surroundings does not decrease.

Answer:

A polar molecule may have one or more lone pairs.

A polar molecule has an uneven distribution of electron density.

Explanation:

A molecule in which the dipole bonds will not cancel is a polar molecule. It depends on the geometry of the molecule, and the direction of the dipole. So, to be polar, the molecule must have at least one polar bond. It may have nonpolar bonds, but the total dipole must be different from 0.

A polar bond is formed between elements that have large differents in electronegativities, such as chlorine and hydrogen. When an atom has a large electronegativity, it has lone pairs of electrons in a bond because it has a small size and a great number of electrons in the valence shell. So, a polar molecule may have on or more lone pairs.

Because of the lone pairs presented, and because of the dipole different from 0, it will be partial charges in the atoms, so a polar molecule has an uneven distribution of electrons density.

A polar molecule is a molecule that has opposite charges at both ends of the molecule.

The true statements about polar molecules are;

To say that a molecule is polar means that it has opposite charges at either end of the molecule.

These charges result from uneven distribution of electrons in the molecule. This uneven distribution of electrons in the molecule is also caused by the large difference in electronegativity between the bonding atoms. This electronegativity difference causes the electron density of the bond to draw closer one of bonding atoms thereby conferring a partial negative charge on that atom and a resulting partial positive charge on the other atom.

Many polar molecules have lone pairs on atoms in the molecule such as NH3, H2O, H2S, etc.

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Answer:

(C) arsenious acid, Ka = 6 x 10⁻¹⁰

Explanation:

A buffer is prepared by a weak acid and the conjugate base coming from its salt. Its function is to resist abrupt changes in pH when an acid or a base are added. The best working range of a buffer is in the range of pKa ± 1. Let's consider the 5 options and their pKa (pKa = -log Ka).

(A) phthalic acid, K1 = 1.3 x 10⁻³ (1st ionization)     pKa = 2.9

(B) hydrogen phthalate, K2 = 3.9 x 10⁻⁵                pKa = 4.4

(C) arsenious acid, Ka = 6 x 10⁻¹⁰                           pKa = 9

(D) formic acid, Ka = 1.8 x 10⁻⁵                               pKa = 4.7

(E) phenol, Ka = 1.3 x 10⁻¹⁰                                      pKa = 9.8

The acid whose pKa is closer to the desired pH is arsenious acid. Its working range of pH is 8 - 10. In the second place, phenol could work as a buffer system since the working pH range is 8.8 - 10.8.

Answer:

1. False

2. False

3. True

4. False

5. False

6. False

Explanation:

1. Because HNO₃ is corrosive when handling it is important to use gloves and also protection glasses. If it touches the skin, then it's important to wash the area affected.

2. The dilute of acid is exothermic, so it releases a huge amount of heat. For precaution, the acid must be added at the water slowly stirring, so it can be controlled.

3. Fe(NO₃)₃ is an oxidizer and in contact with the skin it can cause irritation, so it must be handled carefully and with protection.

4. KSC can irritate the eyes, so it's toxic.

5. KSCN is not inflammable and it's not combustible, so when heated it will only change the state for gas, but it will not cause the evolution of cyanide gas.

6. If KSCN reacts with a strong base, it will dissociate, and the ions SCN⁻, which can't react with the OH⁻ ions of the base. So, it will only cause the evolution of cyanide gas if it reacts with a strong acid.

The answers have been given according to which is True or False. It explores the science of chemicals, especially Nitric Acid.

What is a Chemical Reaction?

Two substances will achieve a reaction has occurs when the molecular or ionic structure of a substance is rearranged to create a new form.

The statements thus are:

1) True.

2) False.

3) True.

4) False

5) False

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Answer:

D. The behavior of an atom's electrons can be described by circular orbits around a nucleus.

Explanation:

Which of the following statements is INCORRECT?

A. It is not possible to know the exact location of an electron and its exact energy simultaneously. CORRECT. This is known as the Heisenberg's uncertainty principle.

B. The energies of an hydrogen atom's electrons are quantized. CORRECT. According to the modern atomic model, the energy levels are quantized, that is, they have a discrete amount of energy.

C. The wave mechanical model correctly predicts all the energy states and the orbitals available to an electron in an hydrogen atom. CORRECT. The characteristics of the energy levels are explained by Schrödinger's wave equation.

D. The behavior of an atom's electrons can be described by circular orbits around a nucleus. INCORRECT. This was postulated by Bohr's atomic model but it is now considered to be incorrect.

E. Electrons have both wave and particle properties. CORRECT. This is what De Broglie called wave-particle duality.

Statement D is incorrect because Niels Bohr's model of electrons moving in circular orbits around a nucleus has been replaced by the quantum mechanical model, which describes electrons as occupying 'orbitals' rather than specific paths.

The statement D, 'The behavior of an atom's electrons can be described by circular orbits around a nucleus', is incorrect. This is a description of a model proposed by Niels Bohr to explain the behavior of electrons in an atom, known as the Bohr model. However, this model has been superseded by the wave mechanical model (also known as the quantum mechanical model) which states that we can't know the exact location of an electron, but we can predict where it's likely to be—an area known as an 'orbital'. Electrons do not travel in defined circular orbits as originally proposed by Bohr.

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Answer: The value of [tex]K_c[/tex] is coming out to be 0.412

Explanation:

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

[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex]     .....(1)

Given mass of [tex]Sb_2S_3[/tex] = 1.00 kg = 1000 g   (Conversion factor: 1 kg = 1000 g)

Molar mass of [tex]Sb_2S_3[/tex] = 339.7 g/mol

Putting values in equation 1, we get:

[tex]\text{Moles of }Sb_2S_3=\frac{1000g}{339.7g/mol}=2.944mol[/tex]

Given mass of hydrogen gas = 10.0 g

Molar mass of hydrogen gas = 2 g/mol

Putting values in equation 1, we get:

[tex]\text{Moles of hydrogen gas}=\frac{10.0g}{2g/mol}=5mol[/tex]

Given mass of hydrogen sulfide = 72.6 g

Molar mass of hydrogen sulfide = 34 g/mol

Putting values in equation 1, we get:

[tex]\text{Moles of hydrogen sulfide}=\frac{72.6g}{34g/mol}=2.135mol[/tex]

The chemical equation for the reaction of antimony sulfide and hydrogen gas follows:

                  [tex]Sb_2S_3(s)+3H_2(g)\rightarrow 2Sb(s)+3H_2S(g)[/tex]

Initial:            2.944      5

At eqllm:      2.944-x     5-3x         2x        3x

We are given:

Equilibrium moles of hydrogen sulfide = 2.135 moles

Calculating for 'x', we get:

[tex]\Rightarrow 3x=2.135\\\\\Rightarrow x=\frac{2.135}{3}=0.712[/tex]

Equilibrium moles of hydrogen gas = (5 - 3x) = (5 - 3(0.712)) = 2.868 moles

Volume of the container = 25.0 L

Molarity of a solution is calculated by using the formula:

[tex]\text{Molarity}=\frac{\text{Moles}}{\text{Volume}}[/tex]

The expression of [tex]K_c[/tex] for above equation, we get:

[tex]K_c=\frac{[H_2S]^3}{[H_2]^3}[/tex]

The concentration of solids and liquids are not taken in the expression of equilibrium constant.

[tex]K_c=\frac{(\frac{2.135}{25})^3}{(\frac{2.868}{25})^3}\\\\K_c=0.412[/tex]

Hence, the value of [tex]K_c[/tex] is coming out to be 0.412

The estimated value of K for the reaction Sb2S3 (s) + 3 H2 (g) <--> 2 Sb (s) + 3 H2S (g) at 713K is approximately 1. This estimate is based on initial and equilibrium moles and concentrations of H2 and H2S.

The reaction is Sb2S3 (s) + 3 H2 (g) <--> 2 Sb (s) + 3 H2S (g). This type of equation is a chemical equilibrium reaction. Because Sb2S3 and Sb are solids, they are not included in the equilibrium constant expression. Only H2 (g) and H2S (g) are included. To solve the problem, the initial moles of H2 and H2S need to be calculated, and the moles of H2S at equilibrium can be found. After finding the moles, they can be converted into concentrations by dividing by the volume of the container (25.0 L).

Initially, we have 1.00 kg of Sb2S3, which is excess and doesn't appear in the equilibrium equation. So, it can be ignored. Similarly, for hydrogen, 10 g converts to approximately 5 moles. Since no H2S is present initially, let's denote its moles as 0. When the equilibrium is established we know that only 72.6g of H2S is present and that is approximately 3 moles. By stoichiometry, 1 mole of H2 will form 1 mole of H2S. Therefore, 3 moles of H2 have been used to form 3 moles of H2S. Subtracting the used quantity from the total, there is approximately 2 moles of H2 left. Plugging all these into the equilibrium equation, we get K = [H2S]^3 / [H2]^3 which is approximately equal to 1.

This process illustrates the concept of chemical equilibrium and how to calculate the equilibrium constant, K, for a reaction. It emphasizes the importance of stoichiometry, the mole concept, and concentration when dealing with equilibrium reactions.

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The complete question is here:

A 1.00 kg sample of Sb2S3 (s) and a 10.0 g sample of H2 (g) are allowed to react in 25.0L container at 713K. At equilibrium, 72.6 g H2S (g) is present? What is the value of K at 713K for the following reaction? The value of Kp?

Sb2S3 (s) + 3 H2 (g) <--> 2 Sb (s) + 3 H2S (g)

Answer:

The reaction will be spontaneous

Explanation:

To determine if the reaction will be spontaneous or not at this temperature, we need to calculate the Gibbs's energy using the following formula:

[tex]\Delta G= \Delta H - T * \Delta S [/tex]

If the Gibbs's energy is negative, the reaction will be spontaneous, but if it's positive it will not.

Calculating the [tex]\Delta G= -1267 - 473 K* \Delta S [/tex] :

[tex]\Delta G= -1267 - 473 K* \Delta S [/tex]

Now, other factor we need to determine is the sign of the S variation. When talking about gases, the more moles you have in your system the more enthropic it is.

In this reaction you go from 7 moles to 8 moles of gas, so you can say that you are going from one enthropy to another higher than the first one. This results in: [tex]\Delta S>0[/tex}

Back to this expression:

[tex]\Delta G= -1267 - 473 K* \Delta S [/tex]

If the variation of S is positive, the Gibbs's energy will be negative always and the reaction will be spontaneous.

The minimum standard reduction potential at the cathode in this galvanic cell should be at least 1.23V to provide the needed power. There is no defined maximum limit for this value. A suitable half-reaction for the cathode could involve the reduction of silver ions to silver metal.

For a galvanic cell to work, the potential at the cathode must be greater than that at the anode. In this specific case, the reduction potential at the anode, E0anode, is given as +0.13V. The difference in potentials between the cathode and anode corresponds to the provided voltage of the cell, which is 1.10V in this particular scenario.

Using this information, the reduction potential at the cathode can be calculated using E0cell = E0cathode - E0anode formula. Substituting the known values, 1.10V = E0cathode - (+0.13V), we find that E0cathode should be at least 1.23V to provide the necessary power.

There is technically no maximum limit for E0cathode as increasing this value only increases the power provided by the cell. As for a suitable half-reaction, a reduction of silver ions (Ag+) to silver metal (Ag) having a reduction potential of +0.80V could be used:

Ag+ + e- --> Ag

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There is a minimum standard reduction potential for the cathode: 0.97 V. The maximum standard reduction potential is generally assumed to be less than +2.87 V. An example of a balanced cathode half-reaction is 2H⁺ (aq) + 2e⁻ → H₂ (g).

Yes, there is a minimum standard reduction potential for the half-reaction at the cathode. The cell potential, E°cell, is calculated as:

Given that E°anode = -0.13 V (since it’s the reduction potential, but we use it as an oxidation potential here), and the cell must provide at least 1.10 V, we have:

Solve for E°cathode:

Yes, there is a maximum standard reduction potential. Practically, the maximum standard reduction potential of the cathode is determined by the potential of the reference electrode (often chosen to be +2.87 V for fluorine). Therefore, any standard reduction potential must be less than this value.

The question is already answered from one point of view but I'll add BeH2 has sp hybridization and H2O has sp3 hybridization getting a different structure type cause where the electron pairs are located around the central atom, that means that you will see the next structures from these molecules.

Answer:

Water (HoH) has

2

bonding electrons and 2 lone pairs. The placement of these electrons is going to be a tetrahedral electron-pair geometry. The HBeH molecule has 2 bonding electron pairs and 0 lone pairs. The bonding pairs must be as far from one another as possible.

Explanation:

I hope that helps any one in the future who needs it <3

Answer:

The order of solubility is AgBr <   Ag₂CO₃ < AgCl

Explanation:

The solubility constant give us the molar solubilty of ionic compounds. In general for a compound AB the ksp will be given by:

Ksp = (A) (B) where A and B are the molar solubilities = s²  (for compounds with 1:1  ratio).

It follows then  that the higher the value of Ksp the greater solubilty of the compound if we are comparing compounds with the same ionic ratios:

Comparing AgBr: Ksp = 5.4 x 10⁻¹³ with AgCl: Ksp = 1.8 x 10⁻¹⁰, AgCl will be more soluble.

Comparing Ag2CO3: Ksp = 8.0 x 10⁻¹²  with AgCl Ksp = AgCl: Ksp = 1.8 x 10⁻¹⁰ we have the complication of  the ratio of ions 2:1 in Ag2CO3,  so the answer is not obvious. But since we know that

Ag2CO3 ⇄ 2 Ag⁺ + CO₃²₋

Ksp Ag2CO3  = 2s x s = 2 s² =  8.0 x 10-12

s = 4 x 10⁻12 ∴ s= 2 x 10⁻⁶

And for AgCl

AgCl  ⇄ Ag⁺ + Cl⁻

Ksp = s² = 1.8 x 10⁻¹⁰  ∴ s = √ 1.8 x 10⁻¹⁰   = 1.3 x 10⁻⁵

Therefore, AgCl is more soluble than Ag₂CO₃

The order of solubility is AgBr <   Ag₂CO₃ < AgCl

Answer:

5.4 x 10^-13

Explanation:

Edmentum says it is correct

Answer:

B, because that was the answer on quizlet.

Explanation:

Answer:

False statement is

At a given temperature, for a given gas, every molecule has the same speed

Explanation:

While checking each and every statement given

The mean free of a molecule actually depends on the size of the molecule because the mean free path is defined as the average distance between two successive collisions of the gas molecules

If the size of the molecule is more, the average distance between two successive collisions decrease and as a result the mean free path of the molecule decreases

Dalton's law of partial pressures is applicable for only ideal gases which means we are assuming that the size of the molecule of a gas is negligible and there are no intermolecular forces of attraction

These two assumptions gets applied at high temperature and low pressure

So Dalton's law of partial pressures tells us that total pressure of the gas is equal to the sum of the partial pressures of the individual gas components

∴ It explains the independent nature of the gas molecule

At a given temperature, for a given gas, all molecules of the gas do not have same speed but overall the average speed of the gas remains same because speed of each molecule of a gas depends on the collision with other molecules of the gas and as the collisions can't be the same therefore molecules of a gas have different speeds

Actually pressure is generated by the collisions of molecules with the container walls because when the gas molecules collide with the container they generate a force which in turn produce the pressure

At high pressure gas do not tend to behave ideally as there will be intermolecular forces and we will write the ratio of PV/nRT as Z which is the compressibility factor of a gas and it will be different for different gases as different gases has different intermolecular forces of attraction

Answer:

D.) Products are weakly favored

Explanation:

For the reaction:

2SO₃ ⇄ O₂ + 2SO₂ + 198kJ/mol

The kc is defined as:

kc = [O₂] [SO₂]² / [SO₃]²

As the kc is 8,1:

8,1 [SO₃]² =  [O₂] [SO₂]²

The products are favored 8,1 times. This is a weakly favored because the usual kc are in the order of 1x10⁴. Thus, right answer is:

D.) Products are weakly favored

I hope it helps!

Final answer:

For the reaction 2SO3 → O2 + 2SO2 with an equilibrium constant Kc of 8.1 at 25 °C, the products are weakly favored since Kc is greater than 1 but much less than 10^3.  So the correct answer is D.

Explanation:

The equilibrium constant Kc indicates whether the reactants or products are favored in a chemical reaction at equilibrium. For the reaction 2SO3 → O2 + 2SO2, the equilibrium constant Kc is given as 8.1 at 25 °C. Since the value of Kc is greater than 1, this means that the reaction favors the formation of products over reactants. If the Kc value was less than 1, the reactants would be favored.

According to the table provided, values of Kc greater than 103 indicate a strong tendency for the reaction to favor the formation of products, suggesting a strongly favored product side for such large Kc values. In this case, with a Kc of 8.1, which is greater than 1 but significantly less than 103, the products are favored to a lesser extent. Therefore, we could conclude that the products are weakly favored, making option D the correct answer.

Answer:

The principle quantum number determines the overall size and energy of an orbital.

Explanation:

The principle quantum number (n) is an integer that has possible values of 1, 2, 3.... and determines the overall size and energy of an orbital.

Angular momentum quantum number (L) determines the shape of atomic orbital.

The location of the electron in the cell is given by quantum mechanics. The principal quantum number determines the size of the atom and the energy of the orbital.

The quantum mechanism helps in the determination of the location of electrons in an atom with the help of four quantum numbers.

The principal quantum number is the first quantum number that helps in the analysis of the energy of the shell to which the atom belongs.

The principal number determines the size of an atom and the energy of the orbital. Thus, option F is correct.

Learn more about principal quantum number, here:

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Answer:

molarity of I3^-(aq) in the solution = 0.21 M

Explanation:

We are given the balanced chemical reaction and the volume and molarity of the  Na₂S₂O₃ so we can calculate the moles of the thiosulfate that were required, and then can calculate the molarity the  I₃⁻ by the definition of molarity.

First lets  convert the volume of Na₂S₂O₃ to liters:

35.8 mL x 1 L/1000 mL = 0.0358 L

# moles Na₂S₂O₃ = 0.350 mol/L x 0.0358 L = 0.0125 mol

From the stoichiometry of the reaction we know 2 mol Na₂S₂O₃ s react with 1 mol I₃⁻ , therefore mol of I₃⁻  will be given by

1 mol I₃⁻ / 2 mol Na₂S₂O₃  x 0.0125 mol Na₂S₂O₃ = 0.0063 mol  I₃⁻

and its molarity is:

0.0063 mol  I₃⁻ / 0.030 L = 0.21 M

Answer:

U⁴⁺  

Explanation:

Fluorine stabilizes metals in higher oxidation states with high M:F ratios.

Hard acids prefer to bind to hard bases.

U⁴⁺ is the hardest acid in the list and F⁻ is a hard base, so an eight-coordinate complex of U⁴⁺ is the most stable.

The metal ion that can most adequately exist in the form of 8-coordinate complex would be:

d). U⁴⁺  

Thus, option d is the correct answer.

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Answer:

a)  ν = 2.7 x 10¹³  s⁻¹

b)  λ=  5.7 x 10⁻⁷ m

c)  Neither will be visible

d)  d= 13,500 m

Explanation:

The relationship ν = c/λ  where

ν = frequency of the radiation

c = speed of light = 3 x 10^8 m/s

λ = wavelength of the radiation

will be used to solve parts a), b) and c).

For part d. we know that all electromagnetic radiation travel at the speed of light so d= speed x time

a) ν = 3 x 10^8 m/s / 11 x 10 ⁻⁶ m       ( 1 μm = 10⁻6 m )

      = 2.7 x 10¹³  s⁻¹

b)  λ= c/ν = 3 x 10^8 m/s / 5.30 x 10¹⁴ s⁻¹ = 5.7 x 10⁻⁷ m

c)  The visible spectrum range is 380 to 470 nm

in a. converting to nm:

11 x 10⁻⁶ m x  10⁹ nm/m = 11,000 nm

in b. converting to nm:

5.7 x 10⁻⁷ m x  10⁹ nm/ m = 570 nm

Neither of these radiations will be visible to the human eye.

d) d= 3 x 10^8 m/s x 45 x 10⁻⁶ s = 13,500 m

The frequency of the radiation is 2.73 x 10¹³ Hz.

The wavelength of the radiation is 566 nm

The distance traveled by the electromagnetic radiation is 13,500 m.

The given parameters;

The frequency of the radiation is calculated as follows;

c = fλ

[tex]f = \frac{c}{\lambda} \\\\f = \frac{3\times 10^{8}}{11 \times 10^{-6}} \\\\f = 2.73 \times 10^{13} \ Hz[/tex]

The wavelength of the radiation is calculated as follows;

[tex]\lambda = \frac{c}{f} \\\\\lambda = \frac{3\times 10^{8}}{5.3\times 10^{14}} \\\\\lambda = 5.66 \times 10^{-7} \ m\\\\\lambda = 566 \ \times 10^{-9} \ m\\\\\lambda = 566 \ nm[/tex]

The distance traveled by the electromagnetic radiation is calculated as;

d = vt

d = 3 x 10⁸ x 45 x 10⁻⁶

d = 13,500 m

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Answer: The energy released for the the given amount of hydrogen -1 atom is [tex]1.2474\times 10^{11}J[/tex]

Explanation:

First we have to calculate the mass defect [tex](\Delta m)[/tex].

The given equation follows:

[tex]4_{1}^{1}\textrm{H}\rightarrow _{2}^{4}\textrm{He}+2_0^{1}\textrm{e}[/tex]

To calculate the mass defect, we use the equation:

Mass defect = Sum of mass of product - Sum of mass of reactant

[tex]\Delta m=(2m_{e}+m_{He})-(4m_{H})[/tex]

We know that:

[tex]m_e=0.00054858g/mol\\m_{H}=1.00782g/mol\\m_{He}=4.00260g/mol[/tex]

Putting values in above equation, we get:

[tex]\Delta m=((2\times 0.00054858)+4.00260)-(4\times 1.00782)=-0.027583g=-2.7583\times 10^{-5}kg[/tex]

(Conversion factor: 1 kg = 1000 g )

To calculate the energy released, we use Einstein equation, which is:

[tex]E=\Delta mc^2[/tex]

[tex]E=(-2.7583\times 10^{-5}kg)\times (3\times 10^8m/s)^2[/tex]

[tex]E=-2.4825\times 10^{11}J[/tex]

The energy released for 4 moles of hydrogen atom is [tex]2.4825\times 10^{11}J[/tex]

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

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

Given mass of hydrogen atom = 2.01 g

Molar mass of hydrogen atom = 1 g/mol

Putting values in above equation, we get:

[tex]\text{Moles of hydrogen atom}=\frac{2.01g}{1g/mol}=2.01mol[/tex]

We need to calculate the energy released for the fusion of given amount of hydrogen atom. By applying unitary method, we get:

As, 4 moles of hydrogen atom releases energy of = [tex]2.4825\times 10^{11}J[/tex]

Then, 2.01 moles of hydrogen atom will release energy of = [tex]\frac{2.4825\times 10^{11}}{4}\times 2.01=1.2474\times 10^{11}J[/tex]

Hence, the energy released for the the given amount of hydrogen -1 atom is [tex]1.2474\times 10^{11}J[/tex]

Answer:

The correct option is: A) N₂

Explanation:

The Bond length of a chemical bond is the length of a chemical bond formed between two given atoms.

Bond length is inversely proportional to the bond order of the chemical bond, which is the total number of bonds between two atoms. Thus as the bond order increases, the bond length decreases.

A) N₂: The nitrogen-nitrogen bond in dinitrogen is a triple bond (N≡N).

Thus the bond order = 3.

B) O₂: The oxygen-oxygen bond in dioxygen is a triple bond (O=O).

Thus the bond order = 2.

C) SO₂: Sulfur dioxide is a resonance stabilized molecule and its resonance hybrid shows that the sulfur-oxygen bond in sulfur dioxide is a partial double bond.

Thus the bond order = 1.5

D) SO₃: Sulfur trioxide is a resonance stabilized molecule and its resonance hybrid shows that the sulfur-oxygen bond in sulfur trioxide is a partial double bond.

Thus the bond order = 1.33

Since the bond order of N₂ is the largest, therefore, the N-N bond length is the shortest.

Answer: [tex]5.81\times 10^6J/mol[/tex]

Explanation:

Heat of combustion is the amount of heat released when 1 mole of the compound is completely burnt in the presence of oxygen.

[tex]C_{12}H_{22}O_{11}(s)+12O_2\rightarrow 12CO_2(g)+11H_2O(g)[/tex]

To calculate the moles, we use the equation:

[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}=\frac{0.05392g}{342g/mol}=1.577\times 10^{-4}moles[/tex]

Thus [tex]1.577\times 10^{-4}moles[/tex] of sucrose releases =  916.6 J of heat

1 mole of sucrose releases =[tex]\frac{916.6}{1.577\times 10^{-4}}\times 1=5.81\times 10^6J[/tex] of heat

Thus ∆H value for the combustion reaction is [tex]5.81\times 10^6J/mol[/tex]

The ∆H for the combustion reaction of 0.05392 g of sucrose, given that the qrxn is -916.6 J, is calculated to be approximately -58364.2 kJ. This value is negative indicating an exothermic reaction where heat is released.

In this problem, we are tasked with finding the enthalpy change, or ∆H, for the combustion reaction of sucrose. The heat of the reaction, qrxn, is given as -916.6 J for a 0.05392 g pellet of sucrose.

Firstly, we have to convert the grams of sucrose to moles. The molecular weight of sucrose, C12H22O11, is about 342.3 g/mol so the moles of sucrose burned would be 0.05392 g ÷ 342.3 g/mol = 0.000157 mol.

Next, the change in enthalpy per mole of sucrose can be calculated using ∆H = qrxn / moles of sucrose which gives us -916.6 J / 0.000157 mol = -58364152.2293 J/mol. Since the question requires the answer in kJ, we must divide this by 1000 to convert J to kJ. Hence, ∆H = -58364.1523 kJ/mol.

It is important to note that ∆H is negative because the reaction is exothermic, meaning heat is released during the combustion of sucrose. Therefore, the ∆H for the combustion reaction of sucrose is approximately -58364.2 kJ.

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Answer:

See the explanation

Explanation:

In this case, in order to get an elimination reaction we need to have a strong base. In this case, the base is the phenoxide ion produced the phenol (see figure 1).  

Due to the resonance, we will have a more stable anion therefore we will have a less strong base because the negative charge is moving around the molecule (see figure 2).

Finally, the phenoxide will attack the primary carbon attached to the Cl. The C-Cl bond would be broken and the C-O would be produced at the same time to get a substitution (see figure 1).

Final answer:

The phenoxide anion prefers acting as a nucleophile in an SN2 mechanism on 3-chloro-1,2-propanediol because the electronegative chloride creates an electrophilic carbon that is more reactive towards nucleophilic attack than the deprotonation of an alcohol.

Explanation:

The phenoxide anion does not act as a base to deprotonate one of the alcohols on 3-chloro-1,2-propanediol because the presence of the chloride makes that carbon a more electrophilic center, which is highly susceptible to nucleophilic attack.

In an SN2 mechanism, the nucleophile favors attacking an electrophilic carbon, here significantly activated by the chloride leaving group, rather than deprotonating an alcohol which is a less electrophilic and less favorable process.

This is especially true when considering that alcohols are not particularly acidic, and thus their protons are not as easily abstracted by a base as compared to more acidic hydrogens (e.g., hydrogens adjacent to carbonyl groups).

The SN2 reaction is characterized by the simultaneous bond formation by the nucleophile and bond breaking by the leaving group, typically observed in primary alkyl halides. The phenoxide anion is a good nucleophile due to its negative charge, making it highly reactive towards electrophilic carbons, particularly against an atom that bears a good leaving group like chloride.