Problem PageQuestion A chemist fills a reaction vessel with nitrogen monoxide gas, chlorine gas, and nitrosyl chloride gas at a temperature of . Under these conditions, calculate the reaction free energy for the following chemical reaction: Use the thermodynamic information in the ALEKS Data tab. Round your answer to the nearest kilojoule.

The change in entropy when 95 g of ethyl acetate freezes at -84.0°C is -59.8 J/K, rounded to two significant digits.

The question involves calculating the change in entropy when 95 g of ethyl acetate freezes at -84.0℃. Given the heat of fusion ( Hf) for ethyl acetate is 10.5 kJ/mol, we first need to convert grams to moles using the molar mass of ethyl acetate (C4H8O2) which is 88.11 g/mol.

Next, we use the formula:

moles = mass (g) / molar mass (g/mol)

For 95 g ethyl acetate, moles = 95 g / 88.11 g/mol = 1.078 moles.

The change in entropy ( S) can be found using the formula:
S =  H / T

where T is the temperature in Kelvin. To convert -84.0℃ to Kelvin, we add 273.15:

T = -84.0 + 273.15 = 189.15 K

The entropy change then is:
S = 10.5 kJ/mol / 189.15 K
S per mole = 0.0555 kJ/K∙mol

Finally, for 1.078 moles, the total entropy change is:
S total = 0.0555 kJ/K∙mol × 1.078 mol
S total = 0.0598 kJ/K

Expressed in J/K (as 1 kJ = 1000 J),
S total = 59.8 J/K.

Since we are calculating the entropy change for freezing, the sign should be negative because entropy decreases during freezing.

Therefore, the change in entropy when 95 g of ethyl acetate freezes at -84.0℃ is -59.8 J/K, rounded to two significant digits.

Answer:

Kp = 0.16

Explanation:

Step 1: Data given

Initial pressure of Cl2 = 4.92 atm

Initial pressure of Br2 = 4.65 atm

The equilibrium pressure of BrCl is found to be 1.597 atm

Step 2: The balanced equation

Cl2(g) + Br2(g) ⇔ 2 BrCl(g

Step 3: The initial pressures

pCl2 = 4.92 atm

pBr2 = 4.65 atm

pBrCl = 0 atm

Step 4: The pressure at the equilibrium

For 1 mol Cl2 we need 1 mol Br2 to produce 2 moles BrCl

pCl2 = 4.92 - X atm

pBr2 = 4.65 - Xatm

pBrCl = 2X atm = 1.597 atm

X = 1.597/2 = 0.7985 atm

pCl2 = 4.92 - X atm = 4.92 - 0.7985 = 4.1215 atm

pBr2 = 4.65 - Xatm = 3.8515 atm

Step 5: Calculate Kp

Kp = (BrCl)² / (Cl2)*(Br2)

Kp = 1.597² / (4.1215*3.8515)

Kp = 0.16

Final answer:

To calculate Kp for the given reaction, start with the initial pressures of Cl2 and Br2, calculate the change in pressure due to the formation of BrCl, and apply these values in the Kp equation. The result is Kp = 0.084.

Explanation:

The student asked how to calculate Kp for the reaction Cl2(g) + Br2(g) ⇌ 2 BrCl(g) at a given temperature, given initial pressures for Cl2 and Br2, and the equilibrium pressure of BrCl. To solve for Kp, we utilize the change in concentration according to the reaction's stoichiometry and apply it to the equation for the equilibrium constant in terms of pressure (Kp).

Initial pressures: Cl2 = 4.92 atm, Br2 = 4.65 atm. At equilibrium, BrCl = 1.597 atm. The change in pressure for Cl2 and Br2 to form 2 BrCl is equal to the pressure of BrCl divided by 2, since the stoichiometry of the reaction dictates twice the amount of BrCl for each reactant consumed. Thus, the change (δ) is 1.597 atm / 2 = 0.7985 atm. Therefore, the pressures of Cl2 and Br2 at equilibrium are 4.92 - 0.7985 atm and 4.65 - 0.7985 atm, respectively.

To find Kp, the equation is Kp = (PBrCl)2 / (PCl2 × PBr2). Substituting the equilibrium pressures into this equation gives Kp = (1.5972) / ((4.92 - 0.7985) × (4.65 - 0.7985)). Solving this yields Kp = 0.084 (to two decimal places).

Answer:

a), b), c) & f)

Explanation:

d) does not apply because Ms value can be either +½ or -½

e) does not apply because Ml - values range from -l to +l, hence l= 2 doesn't exist when l= 1

The four quantum numbers that govern the possible states of an electron in an atom are the principal quantum number (n), the angular momentum quantum number (l), the magnetic quantum number (ml), and the spin quantum number (ms). Allowable combinations of quantum numbers follow specific rules.

The four quantum numbers that govern the possible states of an electron in an atom are the principal quantum number (n), the angular momentum quantum number (l), the magnetic quantum number (ml), and the spin quantum number (ms).

Allowable combinations of quantum numbers for an electron can be determined according to the following rules:

Answer:

[tex]2.55*10^{11[/tex]

Explanation:

Equation for the heterogeneous system is given as:

[tex]2A_{(aq)} + 3 B_{(g)} + C_{(l)}[/tex]      ⇄      [tex]2D_{(s)}[/tex]    [tex]+[/tex]     [tex]3E_{(g)}[/tex]

The concentrations and pressures at equilibrium  are:

[tex][A] = 9.68*10^{-2}M[/tex]

[tex]P_B = 9.54*10^3Pa[/tex]

[tex][C]=14.64M[/tex]

[tex][D]=10.11M[/tex]

[tex]P_E=9.56*10^4torr[/tex]

If we convert both pressure into bar; we have:

[tex]P_B = 9.54*10^3Pa[/tex]

[tex]P_B = (9.54*10^3)*\frac{1}{10^5} bar[/tex]

[tex]P_B=9.54*10^{-2}bar[/tex]

[tex]P_E=9.56*10^4torr[/tex]

1 torr = 0.001333 bar

[tex]9.54*10^4 *0.001333 = 127.5 bar[/tex]

[tex]K=\frac{[P_E]^3}{[A]^2[P_B]^3}[/tex]

[tex]K=\frac{(127.5)^3}{(9.68*10^{-2})^2(9.54*10^{-2})^3}[/tex]

[tex]K=2.55*10^{11[/tex]

Answer:

HF > H2O > NH3 > H2S > CBr4=CO2=0

Explanation:

Dipole moment is a vector quantity. Its a measure of polarity of a bond in a molecule and also a meaure of separation of positive and negative charge in a system.It occurs due to electronegativity difference between the atoms in a molecule.

In order for a molecule to have dipole moment, a molecule must exhibit high electronegativity difference and the shape of the moloecule must be asymmetry

HF has the highest electronegativity difference among all the molecules listed above hence its dipole moment is the greatest.

[tex]H_{2}O[/tex] has a bent structure. There are two O-H bonds hence more charge dipoles. The dipole moment is less than HF molecule because of the net dipole moments of two O-H bonds.

[tex]CO_{2}[/tex] is a linear molecule.However it has polar bonds.But because of the shape of the molecule, the two C-O bond dipoles cancel out each other hence the overall dipole moment will be zero.

Similarly in [tex]CBr_{4}[/tex], ,since the molecule is symmetry, the bond dipole cancels each other out hence the overall dipole moment will be zero.  

The increasing order of the dipole moment will be:

[tex]\rm CO_2[/tex] < [tex]\rm CBr_4[/tex] < [tex]\rm H_2S[/tex] < [tex]\rm NH_3[/tex] < [tex]\rm H_2O[/tex] < HF.

Dipole moment can be described as the measure of the polarity of the molecule. The higher the electronegativity difference, the more polar the bond.

In the given molecules,

The increasing order of the dipole moment will be:

[tex]\rm CO_2[/tex] < [tex]\rm CBr_4[/tex] < [tex]\rm H_2S[/tex] < [tex]\rm NH_3[/tex] < [tex]\rm H_2O[/tex] < HF.

The molecule with the highest dipole moment is HF.

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Final answer:

The molar mass of the unknown protein is 5.79 g/mol.

Explanation:

To find the molar mass of the unknown protein, we can use the formula for osmotic pressure:

π = MRT

Where π is the osmotic pressure, M is the molarity of the solution, R is the ideal gas constant, and T is the temperature in Kelvin. Rearranging the formula, we have:

M = π / (RT)

Substituting the given values:

M = 2.45 torr / (0.0821 L.atm/(mol.K) * 298 K)

M = 0.0996 mol/L

To convert the molarity to grams per liter, we need to multiply it by the molar mass

Molar mass = (0.0996 mol/L) * (10 mL/1 L) * (5.87 mg/1 mL)

Molar mass = 5.79 g/mol

Answer:

The activation energy is lowered by +85.35 KJ/mol

Explanation:

Rate constant is related to the activation energy and the temperature of the reaction through

K = Ae⁻ᴱᵃ/ᴿᵀ

In K = (In A) - (Ea/RT)

In K = (-Ea/RT) + In A

where k = the activity constant

Ea = activation energy

R = molar gas constant

T = absolute temperature in Kelvin

₁₂

At point 1, without the catalyst

In K₁ = (-Ea₁/RT) + In A (eqn 1)

At point 2 with the catalyst

In K₂ = (-Ea₂/RT) + In A (eqn 2)

Note that the molar gas constant, the absolute temperature in Kelvin and the activity constant are all the same for both points.

Subtract (eqn 1) from (eqn 2)

In K₂ - In K₁ = (-Ea₂/RT) - (-Ea₁/RT) + In A - In A

In (K₂/K₁) = (1/RT) (Ea₁ - Ea₂)

We were told that K₂/K₁ = 2.3 × 10¹⁴, then find the difference in Ea

R = 8.314 J/mol.K, T = 37°C = 310.15 K

In (2.3 × 10¹⁴) = (1/(8.314×310.15) (Ea₁ - Ea₂)

33.1 = (1/2578.5871) (Ea₁ - Ea₂)

(Ea₁ - Ea₂) = 33.1 × 2578.5871 = 85351.23 J/mol

Therefore, the activation energy is lowered by +85.35 KJ/mol

The activation energy is lowered by +85.35 KJ/mol

[tex]K = Ae^{\frac{-E_a}{RT} }\\\\In K = (In A) -( \frac{-E_a}{RT})\\\\In K = (\frac{-E_a}{RT}) + In A[/tex]

where

1. At point 1, without the catalyst

[tex]In K_1 = (-Ea_1/RT) + In A[/tex]............(i)

2. At point 2 with the catalyst

[tex]In K_ = (-Ea_2/RT) + In A[/tex]...........(ii)

On subtracting equation (i) from (ii)

[tex]In K_2 - In K_1 = (-Ea_2/RT) - (-Ea_1/RT) + In A - In A\\\\In (K_2/K_2) = (1/RT) (Ea_1 - Ea_2)[/tex]

Given:

On solving:

[tex]In (2.3 * 10^{14}) = (1/(8.314*310.15) (Ea_1 - Ea_2)\\\\33.1 = (1/2578.5871) (Ea_1 - Ea_2)\\\\(Ea_1 - Ea_2) = 33.1 * 2578.5871\\\\ (Ea_1 - Ea_2)= 85351.23 J/mol[/tex]

Thus, the activation energy is lowered by +85.35 KJ/mol.

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Answer:

152 mL is the volume of KOH required to reach the equivalence point.

Explanation:

[tex]HNO_3(aq)+KOH(aq)\rightarrow KNO_3(aq)+H_2O(l)[/tex]

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]HNO_3[/tex]

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

We are given:

[tex]n_1=1\\M_1=0.590 M\\V_1=90.0 mL\\n_2=1\\M_2=0.350 M\\V_2=?[/tex]

Putting values in above equation, we get:

[tex]1\times 0.590 M\times 90.00=1\times 0.350 M\times V_2[/tex]

[tex]V_2=\frac{1\times 0.590 M\times 90.0 mL}{1\times 0.350 M}=151 .7 mL\approx 152 mL[/tex]

152 mL is the volume of KOH required to reach the equivalence point.

Answer:

152 ml.

Explanation:

Given:

Volume of HNO3 = 90 ml

Molar concentration of HNO3 = 0.59 M

Molar concentration of KOH = 0.35 M

Equation of the reaction

KOH + HNO3 --> KNO3 + H2O

Number of moles of HNO3 = molar concentration × volume

= 0.59 × 0.09

= 0.0531 moles.

By stoichiometry, 1 mole of HNO3 reacts with 1 mole of KOH. Therefore,

Number of moles of KOH = 0.0531 moles.

Volume = 0.0531 ÷ 0.350

= 0.152 l

= 152 ml.

Answer:

p(N2O4) = 0.318 atm

p(NO2) = 7.17 atm

Explanation:

Kc for the reaction N2O4 <=> 2NO2 is 0.619 at 45 degrees C If 50.0g of N2O4 is introduced into an empty 2.10L container, what are the partial pressures of NO2 and N2O4 after equilibrium has been achieved at 45 degrees C?

Step 1: Data given

Kc = 0.619

Temperature = 45.0 °C

Mass of N2O4 = 50.0 grams

Volume = 2.10 L

Molar mass N2O4 = 92.01 g/mol

Step 2: The balanced equation

N2O4 ⇔ 2NO2

Step 3: Calculate moles N2O4

Moles N2O4 = 50.0 grams / 92.01 g/mol

Moles N2O4 = 0.543 moles

Step 4: The initial concentration

[N2O4] = 0.543 moles/2.10 L = 0.259 M

[NO2]= 0 M

Step 5: Calculate concentration at the equilibrium

For 1 mol N2O4 we'll have 2 moles NO2

[N2O4] = (0.259 -x)M

[NO2]= 2x

Step 6: Calculate Kc

Kc = 0.619=  [NO2]² / [N2O4]

0.619 = (2x)² / (0.259-x)

0.619 = 4x² / (0.259 -x)

x = 0.1373  

Step 7: Calculate concentrations

[N2O4] = (0.259 -x)M = 0.1217 M

[NO2]= 2x = 0.2746 M

Step 8: The moles

Moles = molarity * volume

Moles N2O4 = 0.1217 M * 2.10  = 0.0256 moles

Moles NO2 = 0.2746 M * 2.10 = 0.577 moles

Step 9: Calculate partial pressure

p*V = n*R*T

⇒ with p = the partial pressure

⇒ with V = the volume = 2.10 L

⇒ with n = the number of moles

⇒ with R = the gas constant = 0.08206 L*atm/mol*K

⇒ with T = the temperature = 45 °C = 318 K

p = (nRT)/V

p(N2O4) = (0.0256 *0.08206 * 318)/ 2.10

p(N2O4) = 0.318 atm

p(NO2) = (0.577 *0.08206 * 318)/ 2.10

p(NO2) = 7.17 atm

In a titration involving a diprotic acid, the equivalence points occur when the acid has lost its protons. If the first equivalence point occurs at 100 mL, the second should occur at 200 mL. resultant species in the solution will vary based on the volume of NaOH added.

A diprotic acid has two protons to donate in a reaction. During titration, protons are removed one at a time, thus presenting two titration curves or equivalence points. If the first equivalence point occurs at 100.0 mL, the second equivalence point typically occurs at twice that volume because the second proton is just as readily removed as the first. Therefore, the second equivalence point will be at 200.0 mL.

Different volumes of added NaOH will result in different major species present. For example, before the equivalence point is reached (0 mL < V< 25 mL), the pH increases gradually as the diprotic acid reacts with the added NaOH to form its conjugate base. At the equivalence point (V = 25 mL), pH increases abruptly as the reaction transitions from acidic to either neutral or basic, depending on whether the diprotic acid is strong or weak, respectively. After the equivalence point (V > 25 mL), the pH is determined by the amount of added NaOH.

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Final answer:

To calculate the entropy change for the surroundings during the reaction of N2(g) and O2(g) to form NO2(g), the standard enthalpy change of the reaction must be known. This value can be used with the equation ΔS° = -ΔH°/T to find the entropy change of the surroundings.

Explanation:

The student is asking to calculate the entropy change for the surroundings (ΔS°surroundings) when 1.90 moles of N2(g) react with O2(g) to form NO2(g) according to the reaction N2(g) + 2O2(g) → 2NO2(g) at standard conditions of 298 K.

To find this, we'll first need the standard enthalpy change (ΔH°) for the reaction, which can be obtained from standard thermodynamic tables. We then apply the equation ΔS° = -ΔH°/T, which relates the entropy change of the surroundings to the enthalpy change of the system at a constant temperature (T).

Given that the standard enthalpy change for the formation of NO2(g) is 33.2 kJ/mol, and the reaction produces 2 moles of NO2 for 1 mole of N2, the standard enthalpy change for the reaction when 1.90 moles of N2 react is (1.90 moles * 33.2 kJ/mol * 2). We'll convert kJ to J by multiplying by 1,000 and then calculate ΔS°surroundings.

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Final answer:

The standard entropy change for the reaction N2(g) + 2O2(g) → 2NO2(g) at 298K can be calculated using the equation: ΔS° = 2∙S°(NO2) - [S°(N2) + 2∙S°(O2)]. The entropy change is -198.3 J/mol K.

Explanation:

The standard entropy change for the reaction N2(g) + 2O2(g) → 2NO2(g) at 298K can be calculated using the equation: ΔS° = 2∙S°(NO2) - [S°(N2) + 2∙S°(O2)].

Using the standard entropy values at 298K from the reference table, we can substitute the values and calculate the entropy change:

ΔS° = 2∙192.5 - [191.5 + 2∙130.6]

ΔS° = -198.3 J/mol K

Answer:

The volume of water needed to recrystallize 0.700 g of benzoic acid is 10.29 mL.

Explanation:

For the complete recrystalization,

the amount of hot water should be such that, the benzoic acid is completely soluble in it.

As we are given that the solubility of benzoic acid in hot water is 68.0 g/L. Now we have to determine the volume of water is needed to recrystallize 0.700 g of benzoic acid.

we conclude that,

As, 68.0 grams of benzoic acid soluble in 1 L of water.

So, 0.700 grams of benzoic acid soluble in  of water.

The volume of water needed = 0.01029 L = 10.29 mL

conversion used : (1 L = 1000 mL)

Therefore, the volume of water needed to recrystallize 0.700 g of benzoic acid is 10.29 mL.

The balanced equation for the combustion of isooctane (C8H18) is C8H18 + 12.5O2 -> 8CO2 + 9H2O.

The balanced equation for the combustion of isooctane (C8H18) to produce carbon dioxide (CO2) and water (H2O) is:

C8H18 + 12.5O2 -> 8CO2 + 9H2O

In this equation, the coefficients have been adjusted to balance the number of atoms on both sides of the equation. There are 8 carbon atoms, 18 hydrogen atoms, and 25 oxygen atoms on both sides of the equation.

Final answer:

The balanced chemical equation for the complete combustion of isooctane (C₈H₁₈) is 2 C₈H₁₈(l) + 25 O₂(g) → 16 CO₂(g) + 18 H₂O(g), comprising a step-by-step process of first balancing the carbon, then hydrogen, and finally the oxygen atoms.

Explanation:

The balanced chemical equation for the combustion of isooctane (C₈H₁₈) to produce carbon dioxide (CO₂) and water (H₂O) can be represented as follows:

2 C₈H₁₈(l) + 25 O₂(g) → 16 CO₂(g) + 18 H₂O(g)

Here is a step-by-step explanation:

After balancing, you end up with the smallest possible integers that balance the equation.

Answer:

0.8 x10^-1 mol/(L s)

Explanation:

2A + B -> 2C + D

For the reaction above, the differential rate is usually expressed as;

Rate = - (1 / 2) Δ[A] / Δt = - Δ[B] / Δt

The negative sign denotes disappearance.

Upon comparing ;  (1 / 2) Δ[A] / Δt = Δ[B] / Δt

If  initial rate of disappearance of A is 1.6x10^-1 mol/(L s);

That means

Δ[B] / Δt = 1.6x10^-1 mol/(L s) / 2

Δ[B] / Δt = 0.8 x10^-1 mol/(L s)

The initial rate of disappearance of B = 1.6x10^-1 mol/(L s)

Answer:

using higher concentration of the nucleophile

Explanation:

In SN2 reaction, the attack of the nucleophile on the substrate occurs simultaneously as the leaving group departs. The entering group normally attacks through the back side of the molecule. The reaction is concerted and bimolecular. This implies that the concentration of the nucleophile is important in the rate equation for the reaction. Hence increasing the concentration of the nucleophile will increase the rate of SN2 reaction.

Answer: The equilibrium constant for the reaction is 2.47

Explanation:

We are given:

Initial moles of ammonia gas = 29.0 moles

Equilibrium moles of nitrogen gas = 13.0 moles

Volume of the tank = 7.50 L

Molarity is calculated by using the formula:

[tex]\text{Molarity}=\frac{\text{Number of moles}}{\text{Volume of tank}}[/tex]

[tex]\text{Initial molarity of ammonia}=\frac{29.0}{7.50}=3.87M[/tex]

[tex]\text{Equilibrium molarity of nitrogen gas}=\frac{13.0}{7.50}=1.73M[/tex]

The chemical equation for the decomposition of ammonia follows:

                      [tex]2NH_3\rightleftharpoons N_2+3H_2[/tex]

Initial:             3.87

At eqllm:       3.87-2x    x      3x

Evaluating the value of 'x'

[tex]\Rightarrow 3x=1.73\\\\x=\frac{1.73}{3}=0.577[/tex]

So, equilibrium concentration of ammonia = (3.87 - 2x) = [3.87 - 2(0.577)] = 2.716 M

Equilibrium concentration of nitrogen gas = x = 0.577 M

The expression of [tex]K_{eq}[/tex] for above equation follows:

[tex]K_{eq}=\frac{[N_2][H_2]^3}{[NH_3]^2}[/tex]

Putting values in above equation, we get:

[tex]K_{eq}=\frac{(2.716)^2}{0.577\times (1.73)^3}\\\\K_{eq}=2.47[/tex]

Hence, the equilibrium constant for the reaction is 2.47

Answer:

The dissociation constant for the acid ( experimental ) is 1.45 lit/mol

Explanation:

The value of dissociation constant can be calculated as,

  [tex]K_{a}[/tex] = C × ∝²

Where, C = concentration of the solution = 0.329M

          ∝ = Degree of dissociation

again , Degree of dissociation can be obtained form :

                      [tex]p_{H}[/tex] = C × ∝

                         ∝ = [tex]\frac{p_{H} }{C}[/tex]

                        ∝ = [tex]\frac{2.327}{0.329}[/tex] = 7.072

So, now [tex]K_{a}[/tex] = C × ∝²

                     = 0.329 ×( 7.072)²

                     = 1.45 lit/ mol

Answer:

The Ka = 6.74 * 10^-5

Explanation:

Step 1: Data given

Concentration of benzoic acid = 0.329 M

pH = 2.327

Step 2: Calculate the Ka

pH = -log (√([HA]*Ka))

2.327 = -log (√(0.329*Ka))

10 ^ - 2.327 = √(0.329*Ka))

0.0047098 = √(0.329*Ka))

2.218 * 10^-5 = 0.329 * Ka

Ka = 6.74 * 10^-5

The Ka = 6.74 * 10^-5

Explanation:

Reaction equations for the given species is as follows.

   [tex]H_{3}PO_{4} + 2NaOH \righleftharpoons Na_{2}HPO_{4} + 2H_{2}O[/tex]

   [tex]NaH_{2}PO_{4} + NaOH \rightleftharpoons Na_{2}HPO_{4} + H_{2}O[/tex]

At the first equivalence point we need 2 × 10 mmol NaOH.

At the second equivalence point we need 5 mmol of NaOH.

Hence, total moles of NaOH required is as follows.

               (20 + 5) mmol = 25 mmol

We assume that volume of NaOH required is V.

      [tex]25 mmol NaOH \times \frac{10^{-3} mol NaOH}{1 mmol NaOH} \times \frac{1000 ml V}{0.10 \text{mol NaOH}}[/tex]

      = 250 ml V

Thus, we can conclude that 250 ml of 0.10 M NaOH must be added to reach second equivalence point of the titration of the [tex]H_{3}PO_{4}[/tex] and NaOH.

In the given case, 250 ml of 0.10 M NaOH must be added to attain the second equivalence point.

The reactions taking place in the given case are:

Based on the given information, a solution comprises 10 mmol of H₃PO₄ and 5 mmol of NaH₂PO₄.

For the first equivalence point, there is a need of 2 × 10 mmol NaOH.

For the second equivalence point, there is a need of 5 mmol NaOH.

Now the total moles of NaOH needed is,

= 20 + 5 mmol

= 25 mmol

Now let the volume of NaOH required be V. Now putting the values we get,

[tex]= 25 mmol NaOH * \frac{10^{-3} mole NaOH}{1 mmol NaOH} * \frac{1000 mLV}{0.10 mol NaOH} \\= 250 ml[/tex]

Thus, 250 ml of 0.10 M NaOH is need to be added to attain the second equivalence point of the titration.

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Answer:

1) 71.9 g/L

2) 0.221 m olal

3)  7.05% by mass

Explanation:

Step 1: Data given

Concentration of sucrose = 0.210 M

Molar weight of sucrose = 342.3 g/mol

Density of solution = 1.02 g/mL

Mass of water = 948.1 grams

Step 2: Convert this value into terms of g/L

(0.210 mol/L) * (342.3 g/mol) = 71.9 g/L

Calculate the molality

Step 1: Calculate mass water

Suppose we have a volume of 1.00L

Mass of the solution = 1000 mL * 1.02 g/mL = 1020 g solution

We know that there are 71.9 g of solute in a liter of solution from the first calculation. This means

(1020 grams solution) - (71.9 g solute) = 948.1 g = 0.9481 kg water

Step 2: Calculate molality

Molality = moles sucrose / mass water

(0.210 mol) / (0.9481 kg) = 0.221 mol/kg = 0.221 m olal

Mass %

% MAss = (mass solute / mass solution)*100%

(71.9 g) / (1020 g) *100% = 7.05% by mass

Answer:

This shown on the second uploaded image

Explanation:

What is occurring in this reaction is the further deprotonation of the base and this would now react with HCl to give the final product

Answer:

0.66 mol

Explanation:

Zero Gauge pressure = 14.7 psi

Pressure read = 173 psi

Actual pressure = 173 psi - 14.7 psi = 158.3 psi

P (psi) = 1/14.696  P(atm)

So, Pressure = 10.77 atm

Given that:

Temperature = 20 °C

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

T(K) = T( °C) + 273.15  

So,  

T₁ = (20+ 273.15) K = 298.15 K

V = 1.50 L

Using ideal gas equation as:

[tex]PV=nRT[/tex]

where,  

P is the pressure

V is the volume

n is the number of moles

T is the temperature  

R is Gas constant having value = 0.0821 L atm/ K mol  

Applying the equation as:

10.77 atm × 1.50 L = n ×0.0821 L atm/ K mol  × 298.15 K

⇒n = 0.66 mol

The vapor pressures of chloroform and acetone in the solution are 52.02 torr and 273.50 torr, respectively, and the total vapor pressure above the solution is 325.52 torr.

To calculate the vapor pressures of chloroform and acetone above the solution, we should make use of Raoult's law. According to Raoult's law, the partial pressure of each component of a mixture is the product of the vapor pressure of the pure component and its mole fraction in the mixture.

First, calculate the mole fractions of each substance. The molar mass of chloroform (CHCl3) is about 119.38 g/mol, and the molar mass of acetone (CH3COCH3) is about 58.08 g/mol. Therefore, the mole fractions of chloroform and acetone are 4.08 g CHCl3 * (1 mol / 119.38 g) = 0.0342 mol and 9.29 g CH3COCH3 * (1 mol / 58.08 g) = 0.1599 mol, respectively. The total moles are 0.0342 mol + 0.1599 mol = 0.1941 mol, so the mole fraction of chloroform is 0.0342/0.1941 = 0.176 and that of acetone is 0.1599/0.1941 = 0.824.

Next, apply Raoult's Law to find the partial pressure of each component in the mixture. The vapor pressure of chloroform is 0.176 * 295 torr = 52.02 torr, and the vapor pressure of acetone is 0.824 * 332 torr = 273.50 torr.

The total vapor pressure above the solution is the sum of the vapor pressures of chloroform and acetone, that is, 52.02 torr + 273.50 torr = 325.52 torr.

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Answer:

[H₂]  = 0.178 M

[I₂]    = 0.184 M

[HI]   = 1.415 M

Explanation:

For the equilibrium:

H₂(g) + I₂(g) ⇄ 2 HI(g)

the equilibrium constant is given by the equation:

Kc = [ HI]² / [H₂][I₂]

Lets use first the reaction quotient which has the same expression as the equilibrium constant to predict the direction the reaction will take, i.e towards reactants or product side.

Q =( 0.895)²/(0.438)(0.444) = 4.12

Q is less than Kc so the reaction will favor the product side.

We can set up the following table to account for all the species at equilibrium:

                                     H₂             I₂                HI

initial                        0.438        0.444          0.895

change                        -x               -x                +2x

equilibrium              0.438 - x    0.444 - x     0.895 + 2x

Now we are in position to express these concentrations  in terms of the equilibrium conctant, Kc

54.3 = (0.895 + 2x)² / (0.438 -x)(0.444 - x)

performing the calculatiopns will result in a quadratic equation:

0.801 + 3.580x +4x² = (0.194 - 0.882x + x²)x 54.3

Upon rearrangement and some algebra, we have

0.801 + 3.580 x + 4x² = 10.534 - 47.893x + 54.3 x²

0 = 9.733 - 51.473 x + 54.3 x²

This equation has two roots X₁ = 0.687 and X₂ = 0.26

The first is physically impossible since it will imply that more 0.687 will make the quantity at equilibrium for both H₂ and I₂ negative.

Therefore the concentrations at equilibrium of each  gas are:

[H₂] = (0.438 - 0.260)              = 0.178 M

[I₂]   = (0.444 - 0.260) M          = 0.184  M

[HI] = [0.895 + 2x(0.260)] M    = 1.415   M

Note if we plug these values into the equilibrium expression we get 61 which is due to the rounding errors propagating in the quadratic equation.

The equilibrium concentrations of H2, I2, and HI at 430 °C are calculated using an expression derived from the reaction quotient equation, plugged into the Kc equation, which is then solved for 'x'. The solutions found are the changes in molarities which applied to the initial molarities give the equilibrium concentrations

Let's denote the change in molarity of H2, I2, and HI as 'x'. At equilibrium, the molarities of H2, I2, and HI will be 0.438+x, 0.444+x, and 0.895-2x respectively. We know the equilibrium constant, Kc = 54.3. Thus, (0.895-2x)2/(0.438+x)(0.444+x) = 54.3. This is a quadratic equation in 'x' and needs to be solved to get the value of 'x'.

After finding 'x', put this value back into the equilibrium concentrations of the gases, i.e., 0.438+x for H2, 0.444+x for I2, and 0.895-2x for HI. These will give you the equilibrium concentrations of the gases at 430 °C.

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The molarity and molality of a solution of 14. g of benzoic acid dissolved in 425.mL of a solvent with a density of 0.92 g/mL are 0.26 M and 0.28 m respectively.

To calculate the molarity and molality of the solution, we'll first need to know how many moles of benzoic acid (C7H6O2) are present. Simply converting grams to moles using the molar mass of benzoic acid (122.12 g/mol), we get 14.g ÷ 122.12 g/mol = 0.11 mol.

Molarity is defined as moles of solute divided by liters of solution. Therefore, the molarity of the solution would be 0.11 mol ÷ 0.425 L = 0.26 M.

On the other hand, molality is calculated as moles of solute divided by kilograms of solvent. To obtain the mass of the solvent in kg, we need to multiply the volume by the density, 425.mL x 0.92 g/mL = 391 g = 0.391 kg. Consequently, the molality of the solution would be 0.11 mol ÷ 0.391 kg = 0.28 m.

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Answer:

H2O<en<phen

Explanation:

The degree of d- splitting is observed from the intensity of colour. The order of d splitting from least to greatest is H2O<en<phen. Phen shows the greatest d-splitting. The degree of splitting of d- orbitals by ligands depends on their relative positions in the spectrochemical series. The spectrochemical series is an experimentally determined series. The series separates the ligands into strong field and weak field ligands. Strong field ligands are found towards the end of the series. Strong field ligands such as en and phen can participate in metal to ligand or ligand to metal pi-bonding. Hence they cause more d-splitting. Ethylendiamine and phenanthroline occur towards the end of the spectrochemical series hence the higher order of d-splitting.

The d-splitting for each ligand is as follows:

H2O (weak field ligand): smallest d-splitting

en (stronger field ligand): medium d-splitting

phen (strongest field ligand): largest d-splitting

The complexes from least to greatest d-splitting are as follows:

[Co(H2O)6]3+

[Co(en)3]3+

[Co(phen)3]3+

The reason for this ordering is that stronger field ligands cause a larger d-splitting. This is because stronger field ligands interact more strongly with the metal's d orbitals, which splits the d orbitals into two sets of orbitals with different energies.

The d-splitting of a transition metal complex is the energy difference between the two sets of d orbitals that are split by the ligand field. The magnitude of the d-splitting depends on the strength of the ligand field. Stronger field ligands cause a larger d-splitting.

The following table shows the spectrochemical series of ligands, which is a ranking of ligands from weakest to strongest field:

| Ligand | Spectrochemical series |

|---|---|---|

| H2O | Weak field |

| en | Medium field |

| phen | Strong field |

Based on the spectrochemical series, we can predict that the d-splitting for each ligand is as follows:

H2O (weak field ligand): smallest d-splitting

en (stronger field ligand): medium d-splitting

phen (strongest field ligand): largest d-splitting

The following table shows the expected d-splitting of the [Co(H2O)6]3+, [Co(en)3]3+, and [Co(phen)3]3+ complexes:

Complex Ligand D-splitting

[Co(H2O)6]3+ H2O Smallest

[Co(en)3]3+ en Medium

[Co(phen)3]3+ phen Largest

The d-splitting of a transition metal complex affects the color of the complex. Complexes with a larger d-splitting absorb higher energy light, which is in the visible region of the spectrum. This is why complexes with strong field ligands tend to be colored.

The following table shows the observed colors of the [Co(H2O)6]3+, [Co(en)3]3+, and [Co(phen)3]3+ complexes:

Complex Ligand Color

[Co(H2O)6]3+ H2O Purple

[Co(en)3]3+ en Purple-pink

[Co(phen)3]3+ phen Pink

The observed colors of the complexes are consistent with the predicted d-splitting. The [Co(phen)3]3+ complex has the largest d-splitting and absorbs the highest energy light, which is why it is pink. The [Co(H2O)6]3+ complex has the smallest d-splitting and absorbs the lowest energy light, which is why it is purple.

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Answer:

Molten solder flows between the two base metals being soldered because of capillary action.

Explanation:

Capillary molten solder takes place when a metal and a melt is brought into contact with the tube and an accessory after heating. Due to the phenomenon of capillary action, the molten metal rises and extends in any direction, due to the small space that remains between the wall of the tube and that of the fitting; With this, when cooling, a completely hermetic union is achieved.

Molten solder flows between two base metals due to capillary action, which allows the solder to spread evenly and adhere to the bases, metal ensuring a strong bond.

Molten solder flows between the two base metals being soldered mainly because of a process known as capillary action. Capillary action is a natural occurrence where a liquid, in this case the molten solder, moves along a narrow space, such as between the two pieces of metal, against the force of gravity. The solder is drawn into the space and spread evenly between the two base metals because of the adhesive forces between the liquid and the surrounding materials are stronger than the cohesive forces within the liquid.

This phenomenon plays a fundamental role in the soldering process, as it allows the solder to spread evenly and adhere to the base metals, ensuring a strong and stable bond between them. This principle is used not only in soldering but also in various other fields such as plant biology, painting, and inkjet printing.

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Answer:

[tex]K_{b}[/tex] is 0.000438 and [tex]pK_{b}[/tex] is 3.36

Explanation:

Methylamine is a monoprotic base.

For a monoprotic base, [tex]K_{b}=\frac{ca^{2}}{(1-a)}[/tex]

where, c is concentration of base in molarity and a is it's degree ionization

Here [tex]a=\frac{6.4}{100}=0.064[/tex] and c = 0.100 M

So, [tex]K_{b}=\frac{(0.100)\times (0.064)^{2}}{(1-0.064)}=0.000438[/tex]

We know, [tex]pK_{b}=-logK_{b}[/tex]

Hence, [tex]pK_{b}=-log(0.000438)=3.36[/tex]

Answer:

a) The Venn diagram is presented in the attached image to this answer.

b) Check Explanation.

c) 0.3333

d) 0.1429

e) 0.6

Explanation:

Let the probability of a lake having high BOD be P(B) = 30% = 0.3

Probability of a lake having low pH = P(P) = 20% = 0.2

Probability that a lake has high BOD and low pH = P(B n P) = 10% = 0.1

Then, probability that a lake has normal BOD = P(B') = 1 - P(B) = 1 - 0.3 = 0.7

Probability that a lake has normal pH = P(P') = 1 - P(P) = 1 - 0.2 = 0.8

Total probability = P(U) = 100% = 1

a) The Venn diagram is presented in the attached image to this answer.

b) Two events are independent if and only if, P(A|B) = P(A) or P(B|A) = P(B).

For this question,

P(B|P) = P(B n P)/P(P) = 0.1/0.2 = 0.5 ≠ P(B) (which is 0.3)

And P(P|B) = P(B n P)/P(B) = 0.1/0.3 = 0.333 ≠ P(P) (which is 0.2).

It is evident that the two events aren't independent of each other.

c) If a stream has high BOD, what is the probability it will also have low pH?

This probability is given as P(P|B) meaning that, the probability of a lake having low pH given that it has high BOD.

Mathematically, this conditional probability is given by

P(P|B) = P(B n P)/P(B) = 0.1/0.3 = (1/3) = 0.3333

d) If a stream has normal levels of BOD, what is the probability it will also have low pH.

This probability is given as P(P|B'); that is, the probability of a lake having low pH given that it has normal BOD.

Mathematically,

P(P|B') = P(B' n P)/P(B')

P(B') = 0.7 (already found above)

But P(B' n P) = ?

Mathematically,

P(B' n P) = P(P) - P(B n P) = 0.2 - 0.1 = 0.1

P(P|B') = 0.1/0.7 = 0.1429

e) What is the probability that a stream will not exhibit either pollution indicator, i.e., will have normal BOD and pH levels?

This is given as P(B' n P')

Mathematically, this represents the region in the Venn diagram outside of the circles representing P(B) and P(P) and it's given mathematically as,

P(B' n P') = P(U) - [P(B n P') + P(B' n P) + P(B n P)] = 1 - (0.2 + 0.1 + 0.1) = 1 - 0.4 = 0.6 or 60%

When the stream has high BOD the probability of low pH is 0.33 and 0.142 when it has normal BOD. 0.6 is the probability that a stream will exhibit neither indicator.

BOD is the biological oxidation demand, that tells about the dissolved oxygen amount in the water body. BOD along with pH are the indicator of pollution.

The Venn diagram is attached in the image below.

The two indicators, high BOD and low pH are dependent on each other. It can be shown as:

[tex]\rm P(A|B) = P(A) \;or \;P(B|A) = P(B)[/tex]

But,

[tex]\begin{aligned}\rm P(B|P) &= \rm \dfrac{P(B \;n \;P)}{P(P)} \\\\&= \dfrac{0.1}{0.2} \\\\&= 0.5 \neq \rm P(B)\end{aligned}[/tex]

And  

[tex]\begin{aligned}\rm P(P|B) &=\rm \dfrac{P(B \;n \;P)}{P(B)} \\\\&= \dfrac{0.1}{0.3} \\\\&= 0.333 \neq \rm P(P) \end{aligned}[/tex]

Hence they are not independent of each other.

The probability of low pH at high BOD is given as P(P|B).

[tex]\begin{aligned}\rm P(P|B) &= \rm \dfrac{P(B \;n \;P)}{P(B)}\\\\ &= \dfrac{0.1}{0.3}\\\\&= 0.3333\end{aligned}[/tex]

Hence, 0.33 is the probability of low pH at high BOD.

The probability of low pH at normal BOD is given as P(P|B').

[tex]\begin{aligned}\rm P(P|B') &= \rm \dfrac{P(B' \;n\; P)}{P(B')}\\\\\rm P(P|B') &= \dfrac{0.1}{0.7} \\\\&= 0.1429\end{aligned}[/tex]

Hence,  0.1429 is the probability of low pH at normal BOD.

The probability that a stream will not exhibit any of the indicators is given by, P(B' n P').

[tex]\begin{aligned}\rm P(B' n P') &= \rm P(U) - [P(B n P') + P(B' n P) + P(B n P)]\\\\&= 1 - (0.2 + 0.1 + 0.1) \\\\&= 0.6\end{aligned}[/tex]

Hence, 0.6 is the probability that neither of the indicators will be expressed.

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Answer:

Rank from most acidic to most basic is

Explanation:

This question is missing options.

Options are

To rank these solutions first calculate either pH or pOH of these solutions.

We will use pH as an indicator to rank these.

Relation between pH and pOH is

    pH + pOH = 14    ................... Eq (A)

For pOH = 8.55

     use equation A

     pH + 8.55 = 14

     pH = 5.45

For pH = 5.45

     pH = 5.45

For 0.0023 M  HCl

HCl is a strong acid. It will ionize 100% in aqueous solution and produce Hydronium Ion. Formula to calculate pH of 0.0023 M  HCl is,

     pH = - log (Molarity)

     pH = - log (0.0023)

     pH = - (-2.64)

     pH = 2.64

For 0.0018 M  KOH

KOH is a base. It will produce Hydroxide Ion in aqueous solution. First calculate pOH and convert it into pH.

    pOH = - log (Molarity)

    pOH = - log (0.0018)

    pOH = 2.74

   use equation A

   pH + 2.74 = 14

   pH = 11.26

Lower the pH more acidic the solution is. Rank from most acidic to most basic is

The solutions, arranged from most acidic to most basic, are H₂SO₄, HCl, NH₄NO₃, NaCl, NaOH, and NaCN.

The arrangement of the given solutions in order of decreasing acidity is as follows:

Strong acids completely dissociate in water, producing a large number of hydronium ions (H₃O⁺) and making the solution highly acidic. Neutral salts, such as NH₄NO₃ and NaCl, do not affect the pH of the solution. Strong bases, like NaOH, ionize completely and produce a high concentration of hydroxide ions (OH⁻), resulting in a basic solution. Weak bases, such as NaCN, only partially ionize, resulting in a lower concentration of OH⁻ ions and a slightly basic solution.

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The complete question is here:

Arrange the following aqueous solutions, all at 25 ∘C, in order of decreasing acidity. Rank from most acidic to most basic. To rank items as equivalent, overlap them.

H₂SO₄

HCl

NH₄NO₃

NaCl

NaOH

NaCN

Answer: The theoretical yield of barium sulfate is 50.9 grams

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 sodium sulfate = 32.4 g

Molar mass of sodium sulfate = 142 g/mol

Putting values in equation 1, we get:

[tex]\text{Moles of sodium sulfate}=\frac{32.4g}{142g/mol}=0.228mol[/tex]

Given mass of barium chloride = 65.3 g

Molar mass of barium chloride = 208.23 g/mol

Putting values in equation 1, we get:

[tex]\text{Moles of barium chloride}=\frac{65.3g}{208.23g/mol}=0.314mol[/tex]

The chemical equation for the reaction of barium chloride and sodium sulfate follows:

[tex]Na_2SO_4+BaCl_2\rightarrow BaSO_4+2NaCl[/tex]

By Stoichiometry of the reaction:

1 mole of sodium sulfate reacts with 1 mole of barium chloride

So, 0.228 moles of sodium sulfate will react with = [tex]\frac{1}{1}\times 0.228=0.228mol[/tex] of barium chloride

As, given amount of barium chloride is more than the required amount. So, it is considered as an excess reagent.

Thus, sodium sulfate is considered as a limiting reagent because it limits the formation of product.

By Stoichiometry of the reaction:

1 mole of sodium sulfate produces 1 mole of barium sulfate.

So, 0.228 moles of sodium sulfate will produce = [tex]\frac{1}{1}\times 0.228=0.228moles[/tex] of barium sulfate

Now, calculating the mass of barium sulfate from equation 1, we get:

Molar mass of barium sulfate = 233.4 g/mol

Moles of barium sulfate = 0.228 moles

Putting values in equation 1, we get:

[tex]0.228mol=\frac{\text{Mass of barium sulfate}}{223.4g/mol}\\\\\text{Mass of barium sulfate}=(0.228mol\times 223.4g/mol)=50.9g[/tex]

Hence, the theoretical yield of barium sulfate is 50.9 grams