Part A Describe the electrodes in this nickel-copper galvanic cell. Drag the appropriate items to their respective bins. View Available Hint(s) ResetHelp Nickel Copper Standard reduction potentials for nickel(II) and copper(II) The standard reduction potential for a substance indicates how readily that substance gains electrons relative to other substances at standard conditions. The more positive the reduction potential, the more easily the substance gains electrons. Consider the following: Ni2+(aq)+2e−→Ni(s),Cu2+(aq)+2e−→Cu(s), E∘red=−0.230 V E∘red=+0.337 V Part B What is the standard potential, E∘cell, for this galvanic cell? Use the given standard reduction potentials in your calculation as appropriate.

Answer:

Steps

i) In thee mitochondrion, pyruvate carboxylase converts pyruvate to oxloacetate.

ii) Malate dehydrogenase in the mitochondrion reduces oxaloacete to to malate.

iii) Malate dehydrogenase in the cytoplasm oxidizes malate to oxaloacetate.

iv) Phosphoenolpyruvate carboxykinase decarboxylates and phosphorylates oxaloacetate.forming phosphoenolpyruvate.  

Explanation:

Gluconeogenesis is the synthesis of glucose from non - carbohydrate compounds. The substrates for gluconeogenesis are lactate, pyruvate, amino acids, propionate and glycerol.

Gluconeogenesis occurs only in cytosol but the precursor is produced in mitochondria. In the conversion of pyruvte to phosphoenolpyruvate occur in mitochondria and cytosol.

Step -1:

Pyruvate carboxylase is a biotin dependent enzyme located in mitochondria. It converts pyruvate to oxlaoacetate and carbondioxide in the presence of ATP.Oxlaocetate synthesized in mitochondrial matrix has to be transported to cytosol for gluconeogenesis. Oxaloacetate is impermeble, cannot be sent out of mitochondria. So it has to be converted to malate.

Step -2:

Malate dehydrogenase in mitochondria converts oxaloacetate synthesized in mitochondrial matrix to malate. And then  it is transported to cytosol.

Step 3:

Malate dehydrogenase responsible for reversible reaction in cytosol converts malate to oxaloacetate.

Step -4

Phosphoenolpyruvate carboxy-kinase in cytosol converts oxaloacetate to PEP. The enzyme transfer high energy phosphate bond from GTP to oxaloacetate to from PEP and liberated carbondioxide.

Therefore, the steps of glycolysis converts phosphoenolpyruvate to pyruvate are as follows.

i) In thee mitochondrion, pyruvate carboxylase converts pyruvate to oxloacetate.

ii) Malate dehydrogenase in the mitochondrion reduces oxaloacete to to malate.

iii) Malate dehydrogenase in the cytoplasm oxidizes malate to oxaloacetate.

iv) Phosphoenolpyruvate carboxykinase decarboxylates and phosphorylates oxaloacetate.forming phosphoenolpyruvate.  

Gluconeogenesis uses pyruvate as a starting point to generate glucose, going through several steps that largely mirror glycolysis in reverse order. However, the first step in gluconeogenesis varies, converting pyruvate into oxaloacetate before continuing with the conversion into phosphoenolpyruvate. It's a complex process that allows the cell to regulate the production and use of glucose efficiently.

The process of converting pyruvate to phosphoenolpyruvate via gluconeogenesis is not a simple reversal of the glycolysis steps. It starts with the conversion of pyruvate to oxaloacetate. Oxaloacetate serves as a substrate for the enzyme phosphoenolpyruvate carboxykinase (PEPCK), which transforms oxaloacetate into phosphoenolpyruvate (PEP). From this step, gluconeogenesis nearly mimics glycolysis, but in reverse. PEP is converted back into 2-phosphoglycerate, then into 3-phosphoglycerate. Then, 3-phosphoglycerate is converted into 1,3 bisphosphoglycerate and then into glyceraldehyde-3-phosphate. Next, two molecules of glyceraldehyde-3-phosphate combine to form fructose-1-6-bisphosphate, which is then converted into fructose 6-phosphate and then into glucose-6-phosphate. Finally, a series of reactions generates glucose. This sequence enables glycolysis and gluconeogenesis to be regulated independently of each other.

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Answer:

B) 0.32 %

Explanation:

Given that:

[tex]K_{a}=1.8\times 10^{-5}[/tex]

Concentration = 1.8 M

Considering the ICE table for the dissociation of acid as:-

[tex]\begin{matrix}&CH_3COOH&\rightleftharpoons &CH_3COOH&+&H^+\\ At\ time, t = 0 &1.8&&0&&0\\At\ time, t=t_{eq}&-x&&+x&&+x\\ ----------------&-----&-&-----&-&-----\\Concentration\ at\ equilibrium:-&1.8-x&&x&&x\end{matrix}[/tex]

The expression for dissociation constant of acid is:

[tex]K_{a}=\frac {\left [ H^{+} \right ]\left [ {CH_3COO}^- \right ]}{[CH_3COOH]}[/tex]

[tex]1.8\times 10^{-5}=\frac{x^2}{1.8-x}[/tex]

[tex]1.8\left(1.8-x\right)=100000x^2[/tex]

Solving for x, we get:

x = 0.00568  M

Percentage ionization = [tex]\frac{0.00568}{1.8}\times 100=0.32 \%[/tex]

Option B is correct.

Final answer:

After setting up the equilibrium expression for acetic acid dissociation and solving the equation, the percent ionization of a 1.8 M HC2H3O2 solution is determined to be approximately 0.1%. The closest answer option given is D) 0.18%.

Explanation:

To calculate the percent ionization for a 1.8 M HC2H3O2 solution, we need to set up an equilibrium expression based on the acid dissociation constant (Ka) and initial concentration of acetic acid. We start with the dissociation reaction:

HC2H3O2 \<=> H+ + C2H3O2-

Let's assume that the concentration of H+ (x) is equal to the concentration of C2H3O2- (x) at equilibrium due to stoichiometry:

Ka = [H+][C2H3O2-]/[HC2H3O2]

If we let x represent the molarity of H+ and C2H3O2- ions at equilibrium and assuming that x is small enough that (1.8 M - x) ≈ 1.8 M, we can simplify the expression to:

Ka = x^2/1.8 M

The equation becomes:

1.8 \ 10^-5 = x^2/1.8

Solving for x gives us:

x ≈ \ 10^-5 \ 1.8)}\> ≈ 1.8 \ 10^-3 M

The percent ionization is then calculated by:

Percent ionization = (x / [HC2H3O2] initial) \ 100%

Percent ionization = (1.8 \ 10^-3 M / 1.8 M) \ 100% = 0.1%

Since none of the options given matches the calculated value strictly, and provided that we've made an approximation, we'll have to choose the closest answer, which is:

D) 0.18%

Answer:

The formula of Estrone = [tex]C_{18}H_{22}O_2[/tex]

Explanation:

Mass of water obtained = 1.388 g

Molar mass of water = 18 g/mol

Moles of [tex]H_2O[/tex] = 1.388 g /18 g/mol = 0.07711 moles

2 moles of hydrogen atoms are present in 1 mole of water. So,

Moles of H = 2 x 0.07711 = 0.1542 moles

Molar mass of H atom = 1.008 g/mol

Mass of H in molecule = 0.1542 x 1.008 = 0.1555 g

Mass of carbon dioxide obtained = 5.543 g

Molar mass of carbon dioxide = 44.01 g/mol

Moles of [tex]CO_2[/tex] = 5.543 g  /44.01 g/mol = 0.126 moles

1 mole of carbon atoms are present in 1 mole of carbon dioxide. So,

Moles of C = 0.126 moles

Molar mass of C atom = 12.0107 g/mol

Mass of C in molecule = 0.126 x 12.0107 = 1.5133 g

Given that the Estrone only contains hydrogen, oxygen and carbon. So,

Mass of O in the sample = Total mass - Mass of C  - Mass of H

Mass of the sample = 1.893 g

Mass of O in sample =  1.893 - 1.5133 - 0.1555 = 0.2242 g  

Molar mass of O = 15.999 g/mol

Moles of O  = 0.2242  / 15.999  = 0.01401 moles

Taking the simplest ratio for H, O and C as:

0.1542 : 0.01401 : 0.126

= 11 :1 : 9

The empirical formula is = [tex]C_9H_{11}O[/tex]

Molecular formulas is the actual number of atoms of each element in the compound while empirical formulas is the simplest or reduced ratio of the elements in the compound.

Thus,  

Molecular mass = n × Empirical mass

Where, n is any positive number from 1, 2, 3...

Mass from the Empirical formula = 9×12 + 11×1 + 16= 135 g/mol

Given, Molar mass = 270.36 g/mol

So,  

Molecular mass = n × Empirical mass

270.36 = n × 135

⇒ n = 2

The formula of Estrone = [tex]C_{18}H_{22}O_2[/tex]

Final answer:

To find the molecular formula of estrone, calculate the empirical formula first by finding the moles of carbon and hydrogen. Then, determine the mole ratio between carbon and hydrogen and round to the nearest whole number to get the empirical formula. Finally, divide the molar mass of estrone by the molar mass of the empirical formula to find the ratio, and round to the nearest whole number to get the molecular formula.

Explanation:

To find the molecular formula for estrone, we need to determine the empirical formula first. The empirical formula represents the simplest whole-number ratio of the elements in a compound. We can use the information from the combustion analysis to calculate the empirical formula.

First, we need to determine the moles of carbon and hydrogen in the estrone sample.

Moles of carbon = mass of CO₂ / molar mass of CO₂

Moles of carbon = 5.543 g / 44.01 g/mol = 0.1259 mol

Moles of hydrogen = mass of H₂O / molar mass of H₂O

Moles of hydrogen = 1.388 g / 18.015 g/mol = 0.0770 mol

Next, we need to find the mole ratio between carbon and hydrogen by dividing the moles of each element by the smallest mole value.

Dividing both moles of carbon and hydrogen by 0.0770 mol, we get the ratio 1.6367 : 1.

Finally, we round the ratio to the nearest whole number to obtain the empirical formula. In this case, the empirical formula is C2H2O.

To find the molecular formula, we need to know the molar mass of estrone. The molar mass of estrone is given as 270.36 g/mol. Dividing the molar mass of estrone by the molar mass of the empirical formula (42.08 g/mol), we get a ratio of approximately 6.42.

Rounding the ratio to the nearest whole number, we find that the molecular formula for estrone is C6H6O3.

The bond angles are determined by their molecular structures - SO2 is 120°, BF3 is 180°, SCl2 is 120°, CO2 is 180°, PF3 is slightly less than 109.5°, and CH4 is 109.5°.

The bond angles in various compounds are determined by the molecule's electron-pair geometry and molecular structure. For the SO2 compound, O-S-O angle corresponds to a bent molecular structure with a bond angle of 120° (electron-pair geometry: trigonal planar). The F-B-F angle in BF3 has a linear molecular structure that leads to a bond angle of 180° (electron-pair geometry: linear). In SCl2, the Cl-S-Cl angle is 120° due to its bent structure (electron-pair geometry: trigonal planar). For CO2, the O-C-O angle is 180° because of its linear structure (electron-pair geometry: linear). In PF3, the F-P-F angle is slightly less than 109.5° because of its trigonal pyramidal structure (electron-pair geometry: tetrahedral). Finally, in CH4, the H-C-H is 109.5° as it has a tetrahedral structure (electron-pair geometry: tetrahedral).

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Answer:

(a) 4

(b) 7

(c) 8

(d) 8

(e) 4

Explanation:

In order to draw the Lewis structure, we need to take into account the octet rule: atoms will share electrons to have 8 electrons in their valence shell.

(a) How many valence electrons does Si have initially?

Si has 4 valence electrons. It has to share 4 pairs of electrons to reach the octet.

(b) How many valence electrons does each Cl have initially?

Cl has 7 valence electrons. Each Cl has to share 1 pair of electrons to reach the octet.

(c) How many valence electrons surround the Si in the SiCl₄ molecule?

Si is surrounded by 8 electrons in SiCl₄.

(d) How many valence electrons surround each Cl in the SiCl₄ molecule?

Each Cl is surrounded by 8 electrons in SiCl₄.

(e) How many bonding pairs of electrons are in the SiCl₄ molecule?

There are 4 bonding pairs between Si and Cl.

Silicon (Si) forms SiCl₄ by bonding with four chlorine (Cl) atoms. Si initially has 4 valence electrons, while each Cl has 7. In SiCl₄, Si has 8 valence electrons, and each Cl has 8. There are four bonding pairs of electrons in the molecule, formed by single bonds between Si and Cl.

Lewis Structure for the Formation of SiCl₄:

(a) **Valence Electrons in Si (Silicon):** Silicon is in Group 14 of the periodic table, so it has 4 valence electrons.

(b) **Valence Electrons in Each Cl (Chlorine) Atom:** Chlorine is in Group 17, so each Cl atom has 7 valence electrons.

(c) **Valence Electrons Surrounding Si in SiCl₄ Molecule:** In SiCl₄, Silicon forms four single bonds with four Cl atoms. Each bond involves two electrons. Therefore, around Silicon, there are 4 * 2 = 8 valence electrons.

(d) **Valence Electrons Surrounding Each Cl in SiCl₄ Molecule:** Each Cl atom is involved in one bond with Si, contributing one pair of electrons. Therefore, each Cl has a total of 8 valence electrons around it (7 from its own valence shell plus 1 from the shared bond).

(e) **Bonding Pairs of Electrons in SiCl₄ Molecule:** Since Si forms four single bonds with four Cl atoms, there are four bonding pairs of electrons in the SiCl₄ molecule.

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Final answer:

The conditions that can alter the value of ΔG for the reaction A + B ⇌ C + D are explained. Adding a catalyst does not change the value of ΔG. Increasing the concentrations of products decreases ΔG. Coupling with ATP hydrolysis decreases ΔG. Increasing the concentrations of reactants also decreases ΔG.

Explanation:

The reaction A + B ⇌ C + D has a positive standard change in free energy (ΔG°). Changing the conditions of the reaction can alter the value of ΔG. Let's classify the conditions:

a. Adding a catalyst: Adding a catalyst can decrease the activation energy and increase the rate of the reaction. However, it does not change the value of ΔG.

b. Increasing [C] and [D]: Increasing the concentrations of products (C and D) will shift the equilibrium towards the reactants and decrease the value of ΔG.

c. Coupling with ATP hydrolysis: Coupling the reaction with ATP hydrolysis, which is an exergonic reaction, can decrease the value of ΔG and make the overall reaction more spontaneous.

d. Increasing [A] and [B]: Increasing the concentrations of reactants (A and B) will shift the equilibrium towards the products and decrease the value of ΔG.

Answer: 656.6 nm.

Explanation:

Using Rydberg's Equation for hydrogen atom:

[tex]\frac{1}{\lambda}=R_H\left(\frac{1}{n_i^2}-\frac{1}{n_f^2} \right )[/tex]

Where,

[tex]\lambda[/tex] = Wavelength of radiation

[tex]R_H[/tex] = Rydberg's Constant

[tex]n_f[/tex] = Higher energy level  = 3  (least energetic for visible series)

[tex]n_i[/tex]= Lower energy level  = 2

We have:

[tex]n_f=3, n_i=2[/tex]

[tex]R_H=1.097\times 10^7 m^{-1}[/tex]

[tex]\frac{1}{\lambda}=1.097\times 10^7 m^{-1}\times \left(\frac{1}{2^2}-\frac{1}{3^2} \right )[/tex]

[tex]\frac{1}{\lambda}=1.097\times 10^7 m^{-1}\times \frac{5}{36}[/tex]

[tex]\frac{1}{\lambda}=0.1523\times 10^{7} m[/tex]

[tex]\lambda=6.566\times 10^{-7}m=656.6nm[/tex]

([tex]1 m= 10^9nm[/tex])

The wavelength of the photon emitted when the hydrogen atom undergoes a transition from n = 2 to n = 3 is 656.6 nm.

Answer: The value of equilibrium constant for reverse reaction is [tex](\frac{1}{95})^2[/tex]

Explanation:

The given chemical equation follows:

[tex]N_2O_5(g)\rightarrow 2NO_2(g)+\frac{1}{2}O_2(g)[/tex]

The equilibrium constant for the above equation is 95.

We need to calculate the equilibrium constant for the reverse equation of above chemical equation, which is:

[tex]O_2(g)+4NO_2(g)\rightarrow 2N_2O_5(g)[/tex]

The equilibrium constant for the reverse reaction will be the reciprocal of the initial reaction.

If the equation is multiplied by a factor of '2', the equilibrium constant of the reverse reaction will be the square of the equilibrium constant  of initial reaction.

The value of equilibrium constant for reverse reaction is:

[tex]K_{eq}'=(\frac{1}{95})^2[/tex]

Hence, the value of equilibrium constant for reverse reaction is [tex](\frac{1}{95})^2[/tex]

The equilibrium constant for the reaction O2(g) + 4 NO2(g) --> 2 N2O5(g) at 25ºC is (1/95)^2. This is due to Le Chatelier's principle stating that K for the reverse reaction is the reciprocal of the original one and it's squared when the reaction is doubled.

The second reaction in your question is the reverse reaction of the first one, but also multiplied by two. According to Le Chatelier's principle, the equilibrium constant (K) of the reverse reaction is the reciprocal of the original one. So the K of the reverse of 1: N2O5(g) --> 2 NO2(g) + ½ O2(g) would be 1/95. However, the actual reaction to the question is twice that.

When you double a reaction, you square its equilibrium constant. Thus, the K for the reaction: O2(g) + 4 NO2(g) --> 2 N2O5(g) would be (1/95)^2.

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Answer:

The correct answer is D) 7.081

Explanation:

The water equilibrium is the following:

H₂O ⇄  H⁺ + OH⁻      Kw

Where Kw= [H⁺] x [OH⁻]

In pure water: [H⁺] = [OH⁻]= C

If we introduce C in the expression for Kw:

Kw= [H⁺] x [OH⁻]= C x C= C²

⇒ C= [tex]\sqrt{Kw}[/tex]= [tex]\sqrt{6.88 x 10^{-15} }[/tex]= 8.29 x 10⁻⁸

pH= -log [H⁺] = -log C = -log (8.29 x 10⁻⁸) = 7.081

Final answer:

At 20°C, the pH of pure water is 7.000.

Explanation:

The ion-product constant of water (Kw) is the product of the concentrations of hydrogen ions (H3O+) and hydroxide ions (OH-) in water. At 20°C, the value of Kw is 6.88 x 10^-15. Since pure water is neutral, the concentration of hydrogen ions and hydroxide ions in pure water are equal. Therefore, at 20°C, the concentration of hydrogen ions and hydroxide ions in pure water is 6.88 x 10^-15 M.

The pH of a solution is a measure of the concentration of hydrogen ions. It is calculated as the negative logarithm (base 10) of the hydrogen ion concentration. Therefore, at 20°C, the pH of pure water is 7.000 (approximately) since the hydrogen ion concentration and the hydroxide ion concentration in pure water are both 6.88 x 10^-15 M.

Answer:

one-quarter as strong

Explanation:

Coulomb's law gives the mathematical expression to calculate the electrical force (F) between the charge of the nucleus (q₊) and the charge of an electron (q₋) separated by a distance (r).

[tex]F=\frac{k.q_{+}.q_{-}}{r^{2} }[/tex]

where,

k is the Coulomb's constant

The force between the nucleus and an electron in the level 1 is:

[tex]F_{1}=\frac{k.q_{+}.q_{-}}{r_{1}^{2} }[/tex]

Considering the distance to an electron from n = 2 is twice as great as the distance to an electron from n = 1, the force between the nucleus and an electron in the level 2 is:

[tex]F_{2}=\frac{k.q_{+}.q_{-}}{r_{2}^{2} }=\frac{k.q_{+}.q_{-}}{(2r_{1})^{2} }=\frac{1}{4} \frac{k.q_{+}.q_{-}}{(r_{1})^{2} }=\frac{1}{4}F_{1}[/tex]

The equilibrium constant for the reaction at 25 °C, [tex]K_{eq}[/tex] is 11.2.

ΔG for the reaction at body temperature (37.0 °C) is -9.04 kJ/mol.

[tex]\triangle G^o= -RT*lnK_{eq}[/tex]

where,

[tex]\triangle G ^o[/tex] = standard Gibbs free energy  = -5.980 kJ/mol = -5980 J/mol

R = gas constant = 8.314 J/K.mol

T = temperature = 298K

[tex]K_{eq}[/tex]   = equilibrium constant = ?

→ Calculation for  [tex]K_{eq}[/tex] :

Now, substituting the values in the above formula:

[tex]\triangle G^o= -RT*lnK_{eq}\\\\-5980J/mol=-(8.314J/K.mol)*(298K)* lnK_{eq}\\\\K_{eq}=11.2[/tex]

Thus, the value of [tex]K_{eq}[/tex] is 11.2.

→ Calculation for [tex]\triangle G_{rxn}[/tex]:

Chemical reaction:

A(aq) ⇌B(aq)

The formula used is:

[tex]\triangle G_{rxn}=\traingle G^o+RT lnQ\\\\\triangle G_{rxn}=\traingle G^o+RT ln\frac{[B]}{[A]}[/tex]

where,

On substituting the values in the above formula:

[tex]\triangle G_{rxn}=\traingle G^o+RT ln\frac{[B]}{[A]}\\\\\triangle G_{rxn}=(-5980J/mol)+[(8.314J/mole.K)*(310K)*ln\frac{0.55}{1.8}] \\\\\triangle G_{rxn}=-9035.75J/mole=-9.04kJ/mol[/tex]

Therefore, the value of [tex]\triangle G_{rxn}[/tex] is -9.04 kJ/mol.

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Answer:

The coefficients found are: 2,3,10,4,3,2,5

The sum of the coefficients is: 2+3+10 + 4 + 3 + 2 + 5 = 29

Explanation:

K2CrO4+Na2SO3+HCl→KCl+Na2SO4+CrCl3+H2O

On the left side we 2 times K , on the right side we have 1 time K

So we have to multiply KCl by 2

K2CrO4+Na2SO3+HCl→2KCl+Na2SO4+CrCl3+H2O

On the left side wehave 1 Cl, on the right side we have 5 Cl

So the left side we have to multiply by 5

K2CrO4+Na2SO3+5 HCl→2KCl+Na2SO4+CrCl3+H2O

On the left side we have 5 times H, on the right side, we have 2 times H

So on the left side the HCl we should multiply by 10 (instead of 5)

On the right side, we should multiply H2O by 5

K2CrO4+Na2SO3+ 10HCl→2KCl+Na2SO4+CrCl3+ 5H2O

So on the left side we have 10 times Cl, on the right side KCl and CrCL3 should be doubled

K2CrO4+Na2SO3+ 10HCl→4 KCl+Na2SO4+ 2CrCl3+ 5H2O

On the right side we have 4 times K, and 2 times Cr. This mean on the left side K2CrO4 should be multiplied by 2

2K2CrO4+Na2SO3+ 10HCl→4 KCl+Na2SO4+ 2CrCl3+ 5H2O

On the left side we have 8 times O plus 3x times O

On the right side we have 5 times O plus 4x times O

To equal this we should multiply on the left side Na2SO3 by 3

and Na2SO4 on the right side also by 3

Now the entire equation is balanced

2 K2CrO4 + 3 Na2SO3 + 10 HCl → 4 KCl + 3 Na2SO4 + 2 CrCl3 + 5 H2O

The coefficients found are: 2,3,10,4,3,2,5

The sum of the coefficients is: 2+3+10 + 4 + 3 + 2 + 5 = 29

Final answer:

To balance the chemical equation, coefficients are placed in front of the compounds to ensure the same number and types of atoms on both sides. The balanced equation is: 3K2CrO4 + 3Na2SO3 + 14HCl → 3KCl + 3Na2SO4 + 2CrCl3 + 7H2O and the coefficients are 3,3,14,3,3,2,7.

Explanation:

To balance the given chemical equation, we must ensure that the number of atoms of each element is the same on both sides of the equation. Let's add coefficients to balance:

Start with Chrome atoms: There is 1 Cr on the left side and 3 on the right side, so place a coefficient 3 in front of K2CrO4.

Next, look at sulfur atoms: There are 3 S on the left and 3 S on the right, so we are now balanced for sulfur.

For potassium atoms, put a coefficient 3 in front of KCl on the product side.

Balance sodium by having a coefficient 3 in front of Na2SO4 since there are 6 Na on the reactant side.

Hydrogen atoms are balanced by placing a coefficient 6 in front of HCl to get 6 H atoms on both sides.

Now, balance the chloride ions: We have 6 Cl from HCl, but we need 9 Cl on the products side (3 from KCl and 6 from CrCl3). So the final coefficient for HCl is 14.

Finally, place a coefficient 7 in front of H2O to balance out the oxygen atoms.

The balanced equation is therefore: 3K2CrO4 + 3Na2SO3 + 14HCl → 3KCl + 3Na2SO4 + 2CrCl3 + 7H2O. The coefficients in the order that they appear in the equation are: 3,3,14,3,3,2,7.

Answer: The mole fraction of hydrogen gas at 20°C is 0.975  

Explanation:

We are given:

Vapor pressure of water vapor at 20°C = 17.5 torr

Total pressure at 20°C = 700.0 torr

Vapor pressure of hydrogen gas at 20°C = (700.0 - 17.5) torr = 682.5 torr

To calculate the mole fraction of hydrogen gas at 20°C, we use the equation given by Raoult's law, which is:

[tex]p_{H_2}=p_T\times \chi_{H_2}[/tex]

where,

[tex]p_{H_2}[/tex] = partial pressure of hydrogen gas = 682.5 torr

[tex]p_T[/tex] = total pressure = 700.0 torr

[tex]\chi_{H_2}[/tex] = mole fraction of hydrogen gas = ?

Putting values in above equation, we get:

[tex]682.5torr=700.0torr\times \chi_{H_2}\\\\\chi_{H_2}=\frac{682.5}{700.0}=0.975[/tex]

Hence, the mole fraction of hydrogen gas at 20°C is 0.975

Answer: 0.975

Explanation:

First, calculate the partial pressure of hydrogen gas. The partial pressure of hydrogen gas, PH2, is the difference between the total pressure and the partial pressure of water vapor.

PH2=700.0 torr−17.5torr=682.5torr

Now, use the following equation to calculate the mole fraction of hydrogen gas.

PH2 = XH2 × Ptotal

XH2 = PH2 / Ptotal

= 682.5 torr / 700.0 torr = 0.975

Answer:

[tex][H_{3}O^{+}]=x M = 2.5\times 10^{-5}M[/tex] and pH = 4.6

Explanation:

Construct an ICE table to calculate changes in concentration at equilibrium.

[tex]C_{6}H_{5}COOH+H_{2}O\rightleftharpoons C_{6}H_{5}COO^{-}+H_{3}O^{+}[/tex]

I(M): 0.17                                    0.42                0

C(M): -x                                        +x                  +x

E(M): 0.17-x                                0.42+x             x

So, [tex]\frac{[C_{6}H_{5}COO^{-}][H_{3}O^{+}]}{[C_{6}H_{5}COOH]}=K_{a}(C_{6}H_{5}COOH)[/tex]

or, [tex]\frac{(0.42+x)x}{(0.17-x)}=6.3\times 10^{-5}[/tex]

or, [tex]x^{2}+0.4201x-(1.071\times 10^{-5})=0[/tex]

So, [tex]x=\frac{-0.4201+\sqrt{(0.4201)^{2}+(4\times 1\times 1.071\times 10^{-5})}}{(2\times 1)}M[/tex]

([tex]ax^{2}+bx+c=0\Rightarrow x=\frac{-b\pm \sqrt{b^{2}-4ac}}{2a};x< 0.17M[/tex])

So, [tex]x=2.5\times 10^{-5}[/tex]M

Hence [tex][H_{3}O^{+}]=x M = 2.5\times 10^{-5}M[/tex]

[tex]pH=-log[H_{3}O^{+}]=-logx=-log(2.5\times 10^{-5})=4.6[/tex]

The pH is 4.44, and the [H3O+] is 3.63 × 10^-5 M.

The question asks about the calculation of the hydronium ion concentration ([H3O+]) and the pH of a buffer solution consisting of benzoic acid (C6H5COOH) and sodium benzoate (C6H5COONa).

Write the equilibrium expression for benzoic acid dissociation:
C6H5COOH ↔ C6H5COO- + H3O+.

Use the Henderson-Hasselbalch equation:
pH = pKa + log ([A-]/[HA]),
where [A-] = concentration of benzoate ion and [HA] = concentration of benzoic acid.

Insert the values into the Henderson-Hasselbalch equation:
pH = 4.20 + log (0.42/0.17).
Calculate the pH.

Determine [H3O+] from pH:
[H3O+] = 10-pH.

Using the provided concentrations and the pKa of benzoic acid (4.20), the pH and [H3O+] can be calculated with high precision. For this specific buffer:

The pH is calculated to be 4.44.

The [H3O+] can then be found as 3.63 × 10-5 M.

Answer:

The correct answer is: 'Nitrogen is relatively inert'.

Explanation:

As, atomic number of nitrogen is 7 and its electronic distribution is 2, 5. Hence, there are 5 valence electrons present in a nitrogen atom. And, to attain stability it will gain three electrons from a donor atom. Hence, it will make a triple bond.

[tex][N]=1s^22s^22p^3[/tex]

Therefore, nitrogen has 5 valence electrons and makes 3 bonds in neutral compounds.Thus nitrogen will combine with another nitrogen atom to completes its octet to form [tex]N_2[/tex] gas molecule. Due to the presence of this triple bond, the gas molecule is almost inert as its bond dissociation energy is very high.

Hence, the correct option is:- nitrogen is relatively inert.

Answer: option d

Explanation:

Project teams must be proactive. Each member of the team should empowered and each member can share ideas and experience in order to make the WBS more precise.

Answer: 0.069 mmol/L

Explanation:

Molarity of a solution is defined as the number of moles of solute dissolved per Liter of the solution.

[tex]Molarity=\frac{n\times 1000}{V_s}[/tex]

where,

n= moles of solute  = [tex]3.46\mu mol[/tex] =  [tex]0.001\times 3.46=3.46\times 10^{-3}mmol[/tex]

1micro (µ) = 0.001 milli (m)

[tex]V_s[/tex] = volume of solution in ml= 50 ml

[tex]Molarity=\frac{3.46\times 10^{-3}\times 1000}{50ml}=0.069mmol/L[/tex]

Thus concentration in mmol/L of the chemist's potassium dichromate solution is 0.069.

Final answer:

To calculate the concentration of the potassium dichromate solution in mmol/L, convert 3.46μmol to mmol, then divide by the volume in liters, resulting in a concentration of 0.069 mmol/L.

Explanation:

The question asks to calculate the concentration of a potassium dichromate solution in millimoles per liter (mmol/L) given that 3.46μmol of potassium dichromate is dissolved in a 50 mL volumetric flask.

First, convert micromoles (μmol) to millimoles (mmol). Recall that 1μmol = 0.001 mmol:

3.46μmol = 3.46 x 0.001 = 0.00346 mmol

Next, we need to find the concentration in mmol/L. To do this, we use the formula for concentration:

Concentration (mmol/L) = (amount of solute in mmol) / (volume of solution in L)

Since the volume of the solution is 50 mL, we convert this to liters (L) because concentration is expressed in mmol/L. Recall that 1000 mL = 1 L.

50 mL = 0.05 L

Now, we can calculate the concentration:

After rounding to two significant digits, the concentration is 0.069 mmol/L.

Answer:

B. Strong attractive forces hold the gas molecules together.

Explanation:

Kinetic molecular theory postulates:-

Hence, The correct option which is not a postulate of kinetic molecular theory is:- B. Strong attractive forces hold the gas molecules together.

Answer: The mass of hydrogen sulfide that can be dissolved is 2.86 grams.

Explanation:

Henry's law states that the amount of gas dissolved or molar solubility of gas is directly proportional to the partial pressure of the gas.

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

[tex]C_{H_2S}=K_H\times p_{liquid}[/tex]

where,

[tex]K_H[/tex] = Henry's constant = [tex]0.087M/atm[/tex]

[tex]p_{H_2S}[/tex] = partial pressure of hydrogen sulfide gas = 2.42 atm

Putting values in above equation, we get:

[tex]C_{H_2S}=0.087M/atm\times 2.42atm\\\\C_{H_2S}=0.2105M[/tex]

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

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

We are given:

Molarity of solution = 0.2105 M

Molar mass of hydrogen sulfide = 34 g/mol

Volume of solution = 400.0 mL

Putting values in above equation, we get:

[tex]0.2105M=\frac{\text{Mass of hydrogen sulfide}\times 1000}{34g/mol\times 400.0mL}\\\\\text{Mass of }H_2S=\frac{0.2105\times 34\times 400}{1000}=2.86g[/tex]

Hence, the mass of hydrogen sulfide that can be dissolved is 2.86 grams.

Answer:

Cr₂O₇⁻²(aq) and ClO₃⁻(aq)

Explanation:

At a redox reaction, one substance must be reduced (gain electrons) and others must be oxidized (lose electrons). To evaluate the potential of the substance to be reduced, it's placed a reaction, in standards conditions, with H₂.

The potential reduction is quantified by E°, and as higher is the value of E°, as easy is to the compound to be reduced. So, at a redox reaction, the compound with the greatest E° will be reduced, and the other will be oxidized, in a spontaneous reaction. The values of E° are:

RuO₄⁻(aq) to RuO₄²⁻(aq) E° = + 0.59 V (the reduction reaction is the opposite of the oxidation reaction).

Ni⁺²(aq) E° = -0.257 V

I₂(s) E° = +0.535 V

Cr₂O₇⁻²(aq) E° = +1.33 V

ClO₃⁻(aq) E° = +0.890 V

Pb²⁺(aq) E° = -0.125 V

So, the substances that have E° higher than the E° of the RuO₄⁻²(aq) are Cr₂O₇⁻²(aq) and ClO₃⁻(aq), which are the substances that can oxidize RuO₄⁻(aq) to RuO₄²⁻(aq).

The correct substances that can oxidize RuO42(aq) to RuO4(aq) under standard conditions are Cr2O72(aq) and ClO3(aq).

To determine which substances can act as oxidizing agents for RuO42−(aq), we need to look at the standard reduction potentials (E°) for the given half-reactions. An oxidizing agent is a substance that accepts electrons, and for a redox reaction to occur spontaneously, the substance with the higher reduction potential will oxidize the substance with the lower reduction potential.

 The half-reaction for the reduction of RuO4’(aq) to RuO4(aq) is as follows:

[tex]\[ \text{RuO}_4(aq) + e^- \rightarrow \text{RuO}_4^{2-}(aq) \][/tex]

 We do not have the standard reduction potential for this half-reaction, but we can compare the potentials of the given substances to determine if they can act as oxidizing agents.

 From Appendix E, we can find the standard reduction potentials for the given substances:

1.[tex]\( \text{Ni}^{2+}(aq) + 2e^- \rightarrow \text{Ni}(s) \) with \( E^\circ = -0.25 \) V[/tex]

2.[tex]\( \text{I}_2(s) + 2e^- \rightarrow 2\text{I}^-(aq) \) with \( E^\circ = +0.54 \) V[/tex]

3.[tex]\( \text{Cr}_2\text{O}_7^{2-}(aq) + 14\text{H}^+(aq) + 6e^- \rightarrow 2\text{Cr}^{3+}(aq) + 7\text{H}_2\text{O}(l) \) with \( E^\circ = +1.33 \) V[/tex]

4. [tex]\( \text{ClO}_3^-(aq) + 6\text{H}^+(aq) + 6e^- \rightarrow \text{Cl}^-(aq) + 3\text{H}_2\text{O}(l) \) with \( E^\circ = +1.45 \) V[/tex]

5.[tex]\( \text{Pb}^{2+}(aq) + 2e^- \rightarrow \text{Pb}(s) \) with \( E^\circ = -0.13 \) V[/tex]

 Comparing these potentials with the half-reaction for RuO42(aq), we can see that Ni2+(aq) and Pb2+(aq) have negative reduction potentials, which means they cannot oxidize RuO42(aq) because they are stronger reducing agents than RuO42(aq).

I2(s) has a positive reduction potential, but it is lower than the reduction potentials of Cr2O72(aq) and ClO3(aq). Therefore, I2(s) can oxidize RuO42’(aq), but there are stronger oxidizing agents available.

Cr2O72(aq) and ClO3(aq) have the highest reduction potentials in the list, which means they are the strongest oxidizing agents and can oxidize RuO4(aq) to RuO4(aq) under standard conditions.

 Thus, the correct substances that can oxidize RuO42(aq) to RuO4(aq) under standard conditions are Cr2O72(aq) and ClO3(aq)

Final answer:

To calculate the mass of dry ice needed to completely sublime in water, we can use the concept of stoichiometry and the ideal gas law equation. The moles of CO2 can be determined using the ideal gas law equation, and then converted to moles of water using the stoichiometry of the balanced equation. Finally, the mass of dry ice can be calculated using the moles of CO2 and the molar mass of CO2.

Explanation:

In order to calculate the mass of dry ice that should be added to the water, we need to use the concept of stoichiometry. The equation for the sublimation of carbon dioxide is CO2(s) -> CO2(g). We can use the molar ratio between CO2(s) and CO2(g) to convert the volume of water to moles of CO2. From there, we can use the molar mass of CO2 to calculate the mass of dry ice needed.

First, we need to calculate the moles of CO2 that can be produced using the ideal gas law equation:

n = PV / RT

where P is the pressure, V is the volume, R is the ideal gas constant, and T is the temperature. Since temperature and pressure are not given, we cannot directly solve for moles of CO2. However, we can assume that the pressure and temperature remain constant throughout the process, so the ideal gas equation can still be used to find the moles of CO2.

The next step is to use stoichiometry to convert moles of CO2 to moles of water. The balanced equation is:

CO2(s) + H2O(l) -> H2CO3(aq)

From the balanced equation, we can see that the mole ratio between CO2 and water is 1:1. Therefore, if we know the moles of CO2, we also know the moles of water. The molar mass of CO2 is 44.01 g/mol, so we can calculate the mass of dry ice needed using the equation:

mass = moles * molar mass

Since we know the volume of water (15.0 L) and the final temperature (28°C), we can plug those values into the ideal gas equation and calculate the moles of CO2. Then, we can use the moles of CO2 to calculate the mass of dry ice needed. The final mass will depend on the pressure and temperature at which the dry ice is added, as well as the final temperature of the water.

Final answer:

The student's question involves calculating the mass of dry ice needed to produce fog in water until the temperature drops to 28°C. To perform this calculation, details such as the heat capacity of water, the heat of sublimation of CO2, and the complete temperature value are critical, but these are not provided in the question.

Explanation:

Dry ice, known chemically as solid carbon dioxide (CO2), sublimes at room temperature and pressure, meaning it converts directly from solid to gas. In the context of the question, when dry ice is added to warm water, the heat transfer from the water to the dry ice accelerates this sublimation process. The interaction cools the water down while providing a fog effect until either all the dry ice sublimes or the water temperature drops too much to sustain rapid sublimation.

To calculate the mass of dry ice required to create these special effects until the water cools to 28°C, one would need to determine the amount of heat the water can provide before reaching this temperature, and then use the heat of sublimation for CO2 to find the corresponding mass of dry ice. However, since the question doesn't provide specific heat capacity, heat of sublimation values, or the completion temperature value (presumably 28°C), an exact calculation cannot be performed without this information.

Handling dry ice requires caution due to its extremely cold temperature, and it is essential to be in a well-ventilated area since the resulting CO2 gas is heavier than air and can displace oxygen.

Answer:

For a: The reaction is not spontaneous.

For b: The reaction is spontaneous.

For c: The reaction is not spontaneous.

For d: The reaction is spontaneous.

Explanation:

For the reaction to be spontaneous, the Gibbs free energy of the reaction must come out to be negative.

Relationship between standard Gibbs free energy and standard electrode potential follows:

[tex]\Delta G^o=-nFE^o_{cell}[/tex]

For a reaction to be spontaneous, the standard electrode potential must be positive.

To calculate the [tex]E^o_{cell}[/tex] of the reaction, we use the equation:

[tex]E^o_{cell}=E^o_{cathode}-E^o_{anode}[/tex]       .......(1)

Substance getting oxidized always act as anode and the one getting reduced always act as cathode.

The chemical reaction follows:

[tex]Fe(s)+Mn^{2+}(aq.)\rightarrow Fe^{2+}(aq.)+Mn(s)[/tex]

We know that:

[tex]E^o_{Fe^{2+}/Fe}=-0.44V\\E^o_{Mn^{2+}/Mn}=-1.18V[/tex]

Calculating the [tex]E^o_{cell}[/tex] using equation 1, we get:

[tex]E^o_{cell}=-1.18-(-0.44)=-0.74V[/tex]

As, the standard electrode potential is coming out to be negative. So, the reaction is not spontaneous.

The chemical reaction follows:

[tex]Fe(s)+2Ag^{+}(aq.)\rightarrow Fe^{2+}(aq.)+2Ag(s)[/tex]

We know that:

[tex]E^o_{Fe^{2+}/Fe}=-0.44V\\E^o_{Ag^{+}/Ag}=0.80V[/tex]

Calculating the [tex]E^o_{cell}[/tex] using equation 1, we get:

[tex]E^o_{cell}=0.80-(-0.44)=1.24V[/tex]

As, the standard electrode potential is coming out to be positive. So, the reaction is spontaneous.

The chemical reaction follows:

[tex]Zn(s)+Mg^{2+}(aq.)\rightarrow Zn^{2+}(aq.)+Mg(s)[/tex]

We know that:

[tex]E^o_{Zn^{2+}/Zn}=-0.76V\\E^o_{Mg^{2+}/Mg}=-2.37V[/tex]

Calculating the [tex]E^o_{cell}[/tex] using equation 1, we get:

[tex]E^o_{cell}=-2.37-(-0.76)=-1.61V[/tex]

As, the standard electrode potential is coming out to be negative. So, the reaction is not spontaneous.

The chemical reaction follows:

[tex]2Al(s)+3Pb^{2+}(aq.)\rightarrow 2Al^{3+}(aq.)+3Pb(s)[/tex]

We know that:

[tex]E^o_{Al^{3+}/Al}=-1.66V\\E^o_{Pb^{2+}/Pb}=-0.13V[/tex]

Calculating the [tex]E^o_{cell}[/tex] using equation 1, we get:

[tex]E^o_{cell}=-0.13-(-1.66)=1.53V[/tex]

As, the standard electrode potential is coming out to be positive. So, the reaction is spontaneous.

Answer:

D) Fe3+ + 3e− → Fe (s) Eo = −0.036V

Explanation:

In electrolysis we have to add  electrical energy for redox reactions non-spontaneous because ΔºG is positive.

These  4 reduction reactions  are all non-spontaneous because their reduction potentials are negative   (ΔGº = - nFεº , ΔGº will be positive). So the ion most easily reduced is the least negative, in this case  Fe3+ + 3e− → Fe (s) Eo = −0.036V.

Explanation:

It'd be better to use cyclohexane.  The possible explanation is that the freezing temperature will change by 20.1 degrees for each mole of substance added to 1 kg of cyclohexane, although the same amount added to naphthalene will change its freezing point just by 6.94 degrees.

It is so much easier to identify a larger change more adequately than a smaller one.  You would actually not have a 1 molal solution in operation, so the variations in freezing points would be even smaller than the ones already described.

Final answer:

Cyclohexane, with its higher freezing point depression constant, would give a more accurate determination of molar mass than naphthalene because it results in more significant temperature changes, making precise measurements easier.

Explanation:

The determination of the molar mass of a substance using freezing point depression will be more accurate with a solvent that has a higher freezing point depression constant (Kƒ). In this case, cyclohexane, with a Kƒ of 20.1°C/m, would give a more accurate determination of molar mass compared with naphthalene, which has a Kƒ of 6.94°C/m, assuming that the substance is equally soluble in both. This is because a higher Kƒ value indicates a greater change in freezing point per mole of solute, making the temperature change easier to measure precisely.

Answer:

It’s either 34.2 OR 34.3 either would be correct!

Explanation:

The correct answer is (E) [tex]34.17\text{ mL}[/tex].

To solve this problem, we will use stoichiometry to relate the moles of [tex]AgNO_3[/tex] needed to the moles of Ag2SO4 formed. Here is the step-by-step solution:

1. First, we need to write down the balanced chemical equation:

[tex]\[ 2\text{AgNO}_3(aq) + \text{H}_2\text{SO}_4(aq) \rightarrow \text{Ag}_2\text{SO}_4(s) + 2\text{H}_2\text{O}(l) \][/tex]

2. From the equation, we can see that [tex]2[/tex] moles of [tex]AgNO_3[/tex] produce 1 mole of [tex]Ag_2SO_4[/tex] This gives us the mole ratio:

[tex]\[ \frac{2\text{ moles AgNO}_3}{1\text{ mole Ag}_2\text{SO}_4} \][/tex]

3. Calculate the molar mass of [tex]Ag_2SO_4[/tex] using the atomic masses of Ag [tex](107.87 g/mol)[/tex] and [tex]S (32.07 g/mol)[/tex]

[tex]\[ \text{Molar mass of Ag}_2\text{SO}_4 = 2 \times 107.87\text{ g/mol (Ag)} + 32.07\text{ g/mol (S)} + 4 \times 16.00\text{ g/mol (O)} \][/tex]

[tex]\[ \text{Molar mass of Ag}_2\text{SO}_4 = 2 \times 107.87 + 32.07 + 4 \times 16.00 \][/tex]

[tex]\[ \text{Molar mass of Ag}_2\text{SO}_4 = 215.74 + 32.07 + 64.00 \][/tex]

[tex]\[ \text{Molar mass of Ag}_2\text{SO}_4 = 311.81\text{ g/mol} \][/tex]

4. Use the given mass of [tex]Ag_2SO_4[/tex] to find the moles of [tex]Ag_2SO_4[/tex] produced:

[tex]\[ \text{Moles of Ag}_2\text{SO}_4 = \frac{\text{mass of Ag}_2\text{SO}_4}{\text{molar mass of Ag}_2\text{SO}_4} \][/tex]

[tex]\[ \text{Moles of Ag}_2\text{SO}_4 = \frac{0.657\text{ g}}{311.81\text{ g/mol}} \][/tex]

[tex]\[ \text{Moles of Ag}_2\text{SO}_4 \approx 0.002097\text{ moles} \][/tex]

5. Using the mole ratio from step 2, calculate the moles of AgNO3 needed:

[tex]\[ \text{Moles of AgNO}_3 = 2 \times \text{Moles of Ag}_2\text{SO}_4 \][/tex]

[tex]\[ \text{Moles of AgNO}_3 = 2 \times 0.002097 \][/tex]

[tex]\[ \text{Moles of AgNO}_3 = 0.004194\text{ moles} \][/tex]

6. Now, we can use the molarity of AgNO3 to find the volume needed:

[tex]\[ \text{Molarity (M)} = \frac{\text{moles of solute (mol)}}{\text{volume of solution (L)}} \][/tex]

[tex]\[ \text{Volume of AgNO}_3 = \frac{\text{moles of AgNO}_3}{\text{Molarity of AgNO}_3} \][/tex]

[tex]\[ \text{Volume of AgNO}_3 = \frac{0.004194\text{ moles}}{0.123\text{ M}} \][/tex]

[tex]\[ \text{Volume of AgNO}_3 \approx 0.03417\text{ L} \][/tex]

7. Convert the volume from liters to milliliters (since 1 L = 1000 mL):

[tex]\[ \text{Volume of AgNO}_3 = 0.03417\text{ L} \times 1000\text{ mL/L} \][/tex]

[tex]\[ \text{Volume of AgNO}_3 = 34.17\text{ mL} \][/tex]

Answer:

A.) Atoms contain electrons

Explanation:

J.J. Thomson conducted an experiment in which he took a gas at low pressure in a discharge tube and applied high voltage current. When he did so, he noticed that there are some particles emitting from cathode going towards the anode. Then he concluded that they are negatively charged particles and coined them electrons.

Hence, out of the options:- A.) Atoms contain electrons is correct.

Answer:

598.74 mL

Explanation:

Using Ideal gas equation for same mole of gas as

[tex] \frac {{P_1}\times {V_1}}{T_1}=\frac {{P_2}\times {V_2}}{T_2}[/tex]

Given ,  

V₁ = 530 mL

V₂ = ?

P₁ = 1.2 atm

P₂ = 1 atm

T₁ = 17 ºC

T₂ = 0 ºC

The conversion of T( °C) to T(K) is shown below:

T(K) = T( °C) + 273.15  

So,  

T₁ = (17 + 273.15) K = 290.15 K  

T₂ = (0 + 273.15) K = 273.15 K  

Using above equation as:

[tex] \frac {{P_1}\times {V_1}}{T_1}=\frac {{P_2}\times {V_2}}{T_2}[/tex]

[tex] \frac {{1.2}\times {530}}{290.15}=\frac {{1}\times {V_2}}{273.15}[/tex]

Solving for V₂ , we get:

V₂ = 598.74 mL

Answer:

The enthalpy of reaction for the reaction of chlorine with ozone is -162.5 kJ.

Explanation:

[tex]ClO ( g ) + O_3 ( g )\rightarrow Cl ( g ) + 2 O_2 ( g ),\Delta H^o_{1,rxn} =-122.8 kJ [/tex]..[1]

[tex]2 O_3 ( g )\rightarrow 3O_2 ( g ),\Delta H^o_{2,rxn} = -285.3 kJ [/tex]..[2]

[tex]O_3(g) + Cl(g)\rightarrow ClO (g)+O_2(g),\Delta H^o_{3,rxn}=?[/tex]..[3]

The enthalpy of reaction for the reaction of chlorine with ozone can be calculated by using Hess's law:

[2] - [1] = [3]

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

[tex]=-285.3 kJ-(-122.8 kJ)=162.5 kJ[/tex]

The enthalpy of reaction for the reaction of chlorine with ozone is -162.5 kJ.

The entropy change for the surroundings in the given reaction can be calculated using the formula ΔS° surroundings = -ΔH° reaction / T.

The entropy change for the surroundings in the given reaction can be calculated using the formula: ΔS° surroundings = -ΔH° reaction / T. In this case, the enthalpy change of the reaction is -802 kJ mol⁻¹ and the temperature is 298 K. Plugging in these values, we can calculate the entropy change for the surroundings.

ΔS° surroundings = -(-802 kJ mol⁻¹) / 298 K

ΔS° surroundings = 2.69 kJ K⁻¹ mol⁻¹

Therefore, the entropy change for the surroundings is 2.69 kJ K⁻¹ mol⁻¹.

Answer:

a. 2.041 V

b. 2.116 V

c. 2.043 V

Explanation:

a.

In an electrochemical cell;

reduction occurs at the cathode & oxidation at the anode.

Thus, from the question; reduction at cathode=

[tex]PbO_{2(s)} + 3H_{(aq)}^{+} +HSO_{4}_{(aq)} +2e^{-}[/tex] → [tex]PbSO_{4(s)} + 2H_{2} O_{(l)} E^{0}_{(cathode)} = 1.685V[/tex]

Oxidation at anode;

[tex]Pb_{s} + HSO^{-}_{4}[/tex]  →  [tex]PbSO_{4(s)}+H^{+}+2e^{-} E^{0}_{anode}  =0.356V[/tex]

The net process in the Cell is obtained by the algebraic sum of the two half-cell reactions:

Overall reaction:

[tex]PbO_{2(s)}+Pb_{s}+2HSO^{-}_{4(aq)}+ 2 H^{+}_{(aq)}[/tex]  →  [tex]PbSO_{4(s)} + 2H_{2} O_{(l)}[/tex]

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

= 1.685V  -   (- 0.356V)

= 2.041V

b.      

To determine the initial value of E-cell :

In the electrochemical cell; both the anode and the cathode were submerged into 4.30M of Sulfuric acid; therefore:

[tex]H_{2} SO_{4}[/tex] →     [tex]H^{+} + HSO_{4}^{-}[/tex]

Assuming:

[tex]H^{+} = HSO_{4}^{-}[/tex]   such that (;)  [tex]H_{2} SO_{4}[/tex]   = 4.30M

let  : a=  [tex]H^{+}[/tex] & b=  [tex]HSO_{4}^{-}[/tex]

[tex]E^{i} _{cell}[/tex] =  

[tex]E^{0} _{cell}[/tex]  -  [tex]\frac{0.0591}{2}[/tex] ×  log   [tex]\frac{1}{(a)^{2} (b)^{2} }[/tex]

=2.041V -  [tex]\frac{0.0591}{2}[/tex]  ×  log  [tex]\frac{1}{(4.30)^{2} (4.30)^{2} }[/tex]

= 2.041V  -  0.02955  ×  log  [tex]\frac{1}{341.8801}[/tex]

= 2.041V  -  0.02955  ×  log  [tex]{0.0029250026}[/tex]

= 2.041V  -  0.02955  ×  (-2.5339)

= 2.041V   +  0.075V

=2.116V

c.

If concentration of [tex]H^{+}[/tex] is dropped by 76.00%

[tex]H^{+} = HSO_{4}^{-}[/tex]  =  4.30M

= 4.30M  -  4.30M   [tex]\frac{76}{100}[/tex]

= 1.032M

[tex]E_{cell}[/tex]  =  2.041V -  [tex]\frac{0.0591}{2}[/tex]  ×  log  [tex]\frac{1}{(1.032)^{2} (1.032)^{2} }[/tex]

=2.041V  -  0.02955  ×  log  [tex]\frac{1}{1.13427612058}[/tex]

=2.041V  -  0.02955  ×  log  (0.88161954735)

=2.041V  -  0.02955  (- 0.0547)

=2.041V  +   (0.0020)

= 2.043V

The standard electrode potential (E°) of the cell in the lead-acid battery is 2.041 V. The initial value of E_cell in the same condition is also 2.041 V. As [H+] falls by 76%, E_cell can be calculated with the Nernst Equation.

The electrochemical cell in a lead-acid battery consists of an anode made of lead and a cathode made of lead(IV) oxide, both submerged in sulfuric acid. The half reactions are as follows:

(a) You can calculate the cell's standard electrode potential (E°) by subtracting the anode potential from the cathode potential, as per the formula E°cell = E°cathode - E°anode. Therefore, E° = 1.685 V - (-0.356 V) = 2.041 V.

(b) The initial value of E_cell is the same as the standard electrode potential, since initially [H+] = [HSO_4-] = 1M as per the standard conditions, which means E_cell = E° = 2.041 V.

(c) If the H+ concentration drops by 76%, it becomes 0.24M. For calculating the new E_cell we have to utilize the Nernst Equation, which is E=E°-(0.059/n)log(Q), where n is the number of electrons transferred (in this case 2), and Q is the reaction quotient ([Pb][HSO4-])/([PbSO4][H+]^2). Plugging in the values, we can determine the new E_cell.

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