The decomposition of SO2Cl2 is first order in SO2Cl2 and has a rate constant of 1.42 x 10-4 s-1 at a certain temperature. How long will it take for the concentration of SO2Cl2 to decrease to 25% of its initial concentration

The question involves calculation of equilibrium concentrations in a complex formation reaction in a solution, where the reaction is between a metal ion M²⁺ and cyanide ions CN⁻ to form a complex [M(CN)4]²⁻. The stated formation constant for the reaction allows formulating an equation for calculating the equilibrium concentration of M²⁺.

To determine the concentration of M²⁺ ions at equilibrium, we need to consider the process of complex formation here, which is M²⁺ + 4CN⁻  → [M(CN)4]²⁻. The formation constant for this reaction is given as 7.7 x 10¹⁶. Given a 0.150 mole quantity of M(NO3)2 is added to a liter of 0.820 M NaCN solution, initially we will have [M²⁺] = 0.150 M and [CN⁻] = 0.820 M.

As the reaction proceeds to equilibrium, [M²⁺] drops by x amount reacting with 4x amount of [CN⁻]. At equilibrium, [M²⁺] = 0.150 - x, [CN-] = 0.820 - 4x and [M(CN)4]²⁻ = x. Applying the formation constant: 7.7 x 10¹⁶ = [M(CN)4]²⁻ / ([M²⁺][CN-]⁴) = x / (0.150 - x)(0.820 - 4x)⁴

This is a difficult equation to solve directly, but if we make the assumption that x, which relates to the equilibrium concentration of [M²⁺], is much less than initial concentrations of M²⁺ and CN⁻, this simplifies the equation and allows determination of the equilibrium concentration of M²⁺.

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To find the concentration of M2+ ions at equilibrium, use the formation constant and initial concentrations of M(NO3)2 and NaCN.

The concentration of M2+ ions at equilibrium can be determined using the formation constant and the initial concentrations of M(NO3)2 and NaCN. First, calculate the initial concentration of M2+ ions by multiplying the concentration of M(NO3)2 (0.150 M) by its stoichiometric coefficient (2). Then, use the formation constant (7.70 × 10^16) to calculate the concentration of [M(CN)4]2- ions at equilibrium. Since the stoichiometric coefficient of M2+ ions is also 1, the concentration of M2+ ions at equilibrium will be equal to the concentration of [M(CN)4]2- ions.

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A 55-gram serving of cereal supplies 270 mg of sodium, which is 11% of the recommended daily allowance. The recommended daily allowance contains 2454.55 mg of sodium and 106.82 mol of sodium. It also contains 6.43 x 10²⁵ atoms of sodium.

To find the moles of sodium in the recommended daily allowance, we need to know the amount of sodium in the daily allowance. The problem states that a 55-gram serving of the cereal supplies 270 mg of sodium, which is 11% of the recommended daily allowance.

So, the total amount of sodium in the daily allowance is

270 mg / 0.11 = 2454.55 mg.

To convert this to moles, we use the molar mass of sodium, which is 22.99 g/mol.

Therefore, the number of moles of sodium in the daily allowance is 2454.55 mg / 22.99 g/mol = 106.82 mol.

To find the number of atoms of sodium in the daily allowance, we use Avogadro's number (6.022 x 10²³ atoms/mol). Therefore, the number of atoms of sodium in the daily allowance is 106.82 mol x (6.022 x 10²³ atoms/mol) = 6.43 x 10²⁵ atoms.

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To find the moles and atoms of sodium in the recommended daily allowance, divide the mass of sodium in mg by its molecular weight, then multiply the resulting moles by Avogadro's number to get atoms. There are roughly 0.107 moles and 6.44 x 1022 atoms of sodium in the allowance.

One 55-gram serving of cereal provides 270 mg of sodium, which is 11% of the recommended daily allowance. To calculate the total recommended daily allowance, we divide 270 mg by 0.11, getting approximately 2455 mg of sodium. The molecular weight of sodium is about 22.99 g/mol, so we can find the number of moles of sodium in the daily allowance by dividing the mass (in grams) by the molecular weight. This results in approximately 0.107 moles of sodium.

To find the number of atoms of sodium, we use Avogadro's number, which is approximately 6.022 x 1023 atoms/mole. Multiplying the number of moles by Avogadro's number gives us the number of atoms. So, there are roughly 6.44 x 1022 sodium atoms in the recommended daily allowance of sodium.

Answer: Part A answer is 180.15588. Part B answer is the moles of HCL required to react with 10 grams of calcium hydroxide.

Explanation:

I just calculated my equation and got my answer.

Answer:

a. 0.076mole

b. 230.88g

Explanation:

a. We'll begin by calculating the molar mass of C6H12O6. This is illustrated below:

Molar Mass of C6H12O6 = (12x6) + (12x1) + (16x6) = 72 + 12 + 96 = 180g/mol

Mass of C6H12O6 = 13.6g

Number of mole of C6H12O6 =?

Number of mole = Mass /Molar Mass

Number of mole of C6H12O6 = 13.6/180

Number of mole of C6H12O6 = 0.076mole.

Therefore, 0.076mole of C6H12O6 is present in 13.6g of C6H12O6.

b. Let us calculate the molar mass of CaCl2. This is illustrated below:

Molar Mass of CaCl2 = 40 + (35.5x2) = 40 + 71 = 111g/mol

Number of mole of CaCl2 = 2.08 moles.

Mass of CaCl2 =?

Mass = number of mole x molar Mass

Mass of CaCl2 = 2.08 x 111

Mass of CaCl2 = 230.88g

Therefore, 230.88g is required.

The question is incomplete, here is the complete question:

A solution is made by mixing of 51 g of heptane and 127 g of acetyl bromide. Calculate the mole fraction of heptane in this solution. Round your answer to 3 significant digits.

Answer: The mole fraction of heptane in the solution is 0.330

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 heptane = 51 g

Molar mass of heptane = 100.2 g/mol

Putting values in equation 1, we get:

[tex]\text{Moles of heptane}=\frac{51g}{100.2g/mol}=0.509mol[/tex]

Given mass of acetyl bromide = 127 g

Molar mass of acetyl bromide = 123 g/mol

Putting values in equation 1, we get:

[tex]\text{Moles of acetyl bromide}=\frac{127g}{123g/mol}=1.032mol[/tex]

Mole fraction of a substance is given by:

[tex]\chi_A=\frac{n_A}{n_A+n_B}[/tex]

Moles of heptane = 0.509 moles

Total moles = [0.509 + 1.032] = 1.541 moles

Putting values in above equation, we get:

[tex]\chi_{(heptane)}=\frac{0.509}{1.541}=0.330[/tex]

Hence, the mole fraction of heptane in the solution is 0.330

Answer:

A. 0.064mol

B. 0.85mol

C. 1500mL

Explanation:

A. Molarity = 0.33M

Volume = 195mL = 195/1000 = 0.195L

Mole =?

Molarity = mole /Volume

Mole = Molarity x Volume

Mole = 0.33 x 0.195

Mole = 0.064mol

B. Molarity = 1.7M

Volume = 500mL = 500/1000 = 0.5L

Mole =?

Molarity = mole /Volume

Mole = Molarity x Volume

Mole = 1.7 x 0.5

Mole = 0.85mol

C. C1 = 12M

V1 = 50mL

C2 = 0.4M

V2 =?

Using the dilution formula C1V1 = C2V2, we find the volume of the diluted solution as follows:

C1V1 = C2V2

12 x 50 = 0.4 x V2

Divide both side by 0.4

V2 = (12 x 50) /0.4

V2 = 1500mL

Answer: The value of [tex]\Delta S^o[/tex] for the surrounding when given amount of hydrogen gas is reacted is -31.02 J/K

Explanation:

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

The equation used to calculate entropy change is of a reaction is:

[tex]\Delta S^o_{rxn}=\sum [n\times \Delta S^o_{(product)}]-\sum [n\times \Delta S^o_{(reactant)}][/tex]

For the given chemical reaction:

[tex]H_2(g)+F_2(g)\rightarrow 2HF(g)[/tex]

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

[tex]\Delta S^o_{rxn}=[(2\times \Delta S^o_{(HF(g))})]-[(1\times \Delta S^o_{(H_2(g))})+(1\times \Delta S^o_{(F_2(g))})][/tex]

We are given:

[tex]\Delta S^o_{(HF(g))}=173.78J/K.mol\\\Delta S^o_{(H_2)}=130.68J/K.mol\\\Delta S^o_{(F_2)}=202.78J/K.mol[/tex]

Putting values in above equation, we get:

[tex]\Delta S^o_{rxn}=[(2\times (173.78))]-[(1\times (130.68))+(1\times (202.78))]\\\\\Delta S^o_{rxn}=14.1J/K[/tex]

Entropy change of the surrounding = - (Entropy change of the system) = -(14.1) J/K = -14.1 J/K

We are given:

Moles of hydrogen gas reacted = 2.20 moles

By Stoichiometry of the reaction:

When 1 mole of hydrogen gas is reacted, the entropy change of the surrounding will be -14.1 J/K

So, when 2.20 moles of hydrogen gas is reacted, the entropy change of the surrounding will be = [tex]\frac{-14.1}{1}\times 2.20=-31.02J/K[/tex]

Hence, the value of [tex]\Delta S^o[/tex] for the surrounding when given amount of hydrogen gas is reacted is -31.02 J/K

The sum of all of the coefficients in the balanced chemical equation of AH of Fe2O3 is 9.

A chemical equation is said to be balanced when the number of atoms of each element on both sides of the equation are the same.

According to this question, the balanced equation for the enthalpy change of Fe2O3 is given as follows:

4Fe + 3O2 → 2Fe2O3

The sum of all the coefficients in this balanced equation is as follows: 4 + 3 + 2 = 9.

Therefore, the sum of all of the coefficients in the balanced chemical equation of AH of Fe2O3 is 9.

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Final answer:

The balanced chemical equation for the formation of iron(III) oxide (Fe₂O₃) from iron (Fe) and oxygen (O₂) is 4 Fe (s) + 3 O₂ (g) → 2 Fe₂O₃ (s), resulting in a sum of 9 for all coefficients in the equation.

Explanation:

The question asks for the writing and balancing of the chemical equation associated with the standard enthalpy of formation (ΔHf) of iron(III) oxide (Fe₂O₃). The standard enthalpy of formation refers to the change in enthalpy when one mole of a compound is formed from its constituent elements in their standard states. For Fe₂O₃, this reaction involves the combination of iron (Fe) and oxygen (O₂) to form iron(III) oxide.

The balanced chemical equation for this reaction is:

4 Fe (s) + 3 O₂ (g) → 2 Fe₂O₃ (s)

Here, we ensure that the atom counts for Fe and O are equal on both sides of the equation: 4 Fe atoms on the left match the 4 Fe atoms on the right (within 2 molecules of Fe₂O₃), and 6 O atoms on the left (3 O₂) match the 6 O atoms on the right (within 2 molecules of Fe₂O₃). Thus, the equation is balanced, and the sum of all coefficients is 4 + 3 + 2 = 9.

Answer :

(1) The oxidation number of P is, (+3)

The oxidation number of H is, (+1)

The oxidation number of O is, (-2)

(2) The oxidation number of Al is, (+3)

The oxidation number of O is, (-2)

(3) The oxidation number of I is, (+7)

The oxidation number of H is, (+1)

The oxidation number of O is, (-2)

Explanation :

Oxidation number : It represent the number of electrons lost or gained by the atoms of an element in a compound.

Oxidation numbers are generally written with the sign (+) and (-) first and then the magnitude.

When the atoms are present in their elemental state then the oxidation number will be zero.

Rules for Oxidation Numbers :

The oxidation number of a free element is always zero.

The oxidation number of a monatomic ion equals the charge of the ion.

The oxidation number of  Hydrogen (H)  is +1, but it is -1 in when combined with less electronegative elements.

The oxidation number of  oxygen (O)  in compounds is usually -2, but it is -1 in peroxides.

The oxidation number of a Group 1 element in a compound is +1.

The oxidation number of a Group 2 element in a compound is +2.

The oxidation number of a Group 17 element in a binary compound is -1.

The sum of the oxidation numbers of all of the atoms in a neutral compound is zero.

The sum of the oxidation numbers in a polyatomic ion is equal to the charge of the ion.

(1) The given compound is, [tex]HPO_3^{2-}[/tex]

Let the oxidation state of 'P' be, 'x'

[tex](+1)+x+3(-2)=-2\\\\x=+3[/tex]

The oxidation number of P is, (+3)

The oxidation number of H is, (+1)

The oxidation number of O is, (-2)

(2) The given compound is, [tex]Al_2O_3[/tex]

Let the oxidation state of 'Al' be, 'x'

[tex]2x+3(-2)=0\\\\x=+3[/tex]

The oxidation number of Al is, (+3)

The oxidation number of O is, (-2)

(3) The given compound is, [tex]HIO_4[/tex]

Let the oxidation state of 'I' be, 'x'

[tex](+1)+x+4(-2)=0\\\\x=+7[/tex]

The oxidation number of I is, (+7)

The oxidation number of H is, (+1)

The oxidation number of O is, (-2)

The rules of determining oxidation numbers have been applied to three compounds: hydrogen phosphate (with oxidation numbers of +1 for H, +5 for P and -2 for O), aluminum oxide (+3 for Al and -2 for O), and periodic acid (+1 for H,+7 for I, and -2 for O).

The rules to assign oxidation numbers to the elements in the given compounds may be understood as follows:

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Answer:

47.47 g of CO₂ is the amount formed.

Explanation:

The reaction is:

2C₆H₅COOH(aq) + 15O₂(g) → 14CO₂(g) + 6H₂O(l)

Let's apply the formula for the percent yield

Percent yield of reaction = (Produced yield/Theoretical yield) . 100

First of all we convert the moles of CO₂ to mass: 1.30 mol . 44 g /1 mol = 57.2 g. So now, we replace:

(Produced yield / 57.2 g ). 100 = 83

Produced yield / 57.2 g  = 83 / 100

Produced yield / 57.2 g  = 0.83

Produced yield = 0.83 . 57.2g → 47.47 g of CO₂

Answer:

The actual yield of CO2 is 1.079 moles or 47.5 grams formed

Explanation:

Step 1: Data given

Number of CO2 = 1.30 moles

Percent yield = 83.0 %

Step 2: The balanced equation

15O2(g) + 2C6H5COOH(aq) → 14CO2(g) + 6H2O(l)

Step 3: Calculate the number moles of CO2 formed

1.30 moles CO2 = 100 %

The actual amount of moles = 0.83 * 1.30 = 1.079 moles

Step 4: Calculate the percent yield of the reaction

We can control this by calculating the percent yield of the reaction

% yield = (actual yield / theoretical yield ) * 100 %

% yield = (1.079 moles / 1.30 moles ) * 100 %

% yield = 83.0 %

Step 5: Calculate thr mass of CO2 produced

Mass CO2 = moles * molar mass CO2

Mass CO2 = 1.079 moles * 44.01 g/mol

Mass CO2 = 47.5 grams

The actual yield of CO2 is 1.079 moles or 47.5 grams formed

Answer:

a. Too close to zero

b. ΔS > 0

c. Too close to zero

d. ΔS < 0

e.  ΔS > 0

Explanation:

In order to answer this question we need to compare the change in number of mol in the gas state for the products vs the reactants.

An increase in number moles gas leads to a positive change in entropy. Conversely a decrease will mean ΔS is negative.

Now we can solve the parts in this question.

a. H₂ (g) + F₂ (g)   ⇒  2 HF (g)

We have no change in the number of moles gas products minus reactants. Therefore is too close to zero to decide since there is no change in mol gas.

b. 2 SO₃ ( g) ⇒ 2 SO₂ (g) + O₂ (g)

We have three moles of gas products starting with 2 mol gas SO₂, therefore ΔS is positive.

c. CH₄ (g) + 2 O₂ (g)   ⇒ CO₂ (g) + 2 H₂O (g)

Again, we have the same number of  moles gas in the products and the reactants (3), and it is too close to zero to decide

d.  2 H₂S (g) + 3 O₂(g) ⇒ 2 H2O (g) + 2 SO₂ (g)

Here the change in number of moles gas is negative ( -1 ), so will expect ΔS < 0

e. 2H₂O₂ (l)   ⇒ 2H₂O(l) +- O2(g)

Here we have 1 mol gas and 2 mol liquid  produced from 2 mol liquid reactants, thus the change in entropy is positive.

Answer:

The answer to your question is there will be 20.5 g left of HCl

Explanation:

Data

HCl = 31.4 g

NaOH = 12 g

Excess of HCl = ?

Balanced chemical reaction

                HCl  +  NaOH  ⇒   NaCl  +  H₂O

Molar weight of HCl = 1 + 35.5 = 36.5 g

Molar weight of NaOH = 23 + 16+ 1 = 40 g

Calculate the limiting reactant

theoretical yield HCl/NaOH = 36.5/40 = 0.9125

experimental yield HCl/NaOH = 31.4/12 = 2.62

Conclusion

The limiting reactant is NaOH because the experimental yield increases.

Calculate the mass of Excess reactant

               36.5 g of HCl ---------------- 40 g of NaOH

                x                     ---------------- 12 g of NaOH

                x = (12 x 36.5) / 40

                x = 438 / 40

                x = 10.95 g of HCl

Excess HCl = 31.4 - 10.95

                   = 20.5 g

Answer:

There will remain 20.5 grams of hydrochloric acid

Explanation:

Step 1: Data given

Mass of hydrochloric acid = 31.4 grams

Mass of sodium hydroxide = 12.0 grams

Molar mass hydrochloric acid (HCl) = 36.46 g/mol

Molar mass sodium hydroxide (NaOH) = 40.0 g/mol

Step 2: The balanced equation

HCl + NaOH → NaCl + H2O

Step 3: Calculate moles

Moles = mass / molar mass

Moles HCl = 31.4 grams / 36.46 g/mol

Moles HCl = 0.861 moles

Moles NaOH = 12.0 grams / 40.0 g/mol

Moles NaOH = 0.3 moles

Step 4: Calculate the limiting reactant

The limiting reactant is NaOH. It will completely be consumed (0.3 moles). Hcl is in excess. There will react 0.3 moles. There will remain 0.861 - 0.3 = 0.561 moles

Step 5: Calculate mass HCl

Mass HCl = moles HCl * molar mass

Mass HCl = 0.561 moles * 36.46 g/mol

Mass HCl = 20.5 grams

There will remain 20.5 grams of hydrochloric acid

Answer:

The equilibrium partial pressure of NO2 is 0.152 atm

Explanation:

Step 1: Data given

Initial pressure of NO2 = 0.500 Atm

Total pressure inside the vessel at equilibrium = 0.674 atm

Step 2: The balanced equation

2 NO2(g) ⇌ 2 NO(g) + O2(g)

Step 3: The initial pressures

pNO2 = 0.500 atm

pNO = 0 atm

pO2 = 0 atm

Step 4: The pressure at the equilibrium

pNO2 = 0.500 - 2x

pNO = 2x

pO2 = x

Total pressure = 0.674 = (0.500 - 2x) + 2x + x

0.674 = 0.500 + x

x = 0.174

pNO2 = 0.500 - 2*0.174 = 0.152 atm

pNO = 2x = 0.348 atm

pO2 = x = 0.176 atm

The equilibrium partial pressure of NO2 is 0.152 atm

The equilibrium partial pressure of NO₂ in a reaction where it decomposes into NO and O₂, with a total equilibrium pressure of 0.674 atm and an initial NO₂ pressure of 0.500 atm, is calculated to be 0.152 atm using an ICE table and the stoichiometry of the reaction.

The question involves finding the equilibrium partial pressure of NO₂ in a reaction where it decomposes into NO and O₂. Given that the total pressure at equilibrium is 0.674 atm and the initial pressure of NO₂ was 0.500 atm, one can solve for the equilibrium partial pressures using the stoichiometry of the balanced equation and the properties of an equilibrium state.

To determine the equilibrium partial pressure of NO₂, one would set up an ICE (Initial, Change, Equilibrium) table. Denoting the change in NO₂ pressure as x, the change for NO would also be x, and the change for O₂ would be x/2 due to the stoichiometry of the reaction. Given the total pressure at equilibrium, we can solve for x and subsequently find the equilibrium partial pressure of each gas.

Knowing the total pressure and using the ICE table, we can calculate the equilibrium partial pressure of NO₂:
Total pressure at equilibrium = Initial pressure of NO₂ - Change in NO₂ pressure + Change in NO pressure + Change in O₂ pressure

0.674 atm = 0.500 atm - x + x + x/2
0.674 atm = 0.500 atm + x/2
x = 0.348 atm
Equilibrium partial pressure of NO₂ = Initial pressure of NO₂ - Change in NO₂ pressure

0.500 atm - x = 0.500 atm - 0.348 atm = 0.152 atm

Answer:

[tex]r=k[UO_2^+]^2[H^+][/tex]

Explanation:

The given reaction is :-

[tex]2UO_2^++4H^+\rightarrow U^{4+}+UO_2^{2+}+2H_2O[/tex]

According to the law of mass action:-

The rate of the reaction is directly proportional to the active concentration of the reactant which each are raised to the experimentally determined coefficients which are known as orders. The rate is determined by the slowest step in the reaction mechanics.

Order of in the mass action law is the coefficient which is raised to the active concentration of the reactants. It is experimentally determined and can be zero, positive negative or fractional.

The order of the whole reaction is the sum of the order of each reactant which is raised to its power in the rate law.  

m = 2 =  is order with respect to [tex]UO_2^+[/tex]

n = order with respect to H+

overall order = m+n = 3

n = 3 - m = 3 - 2 = 1

Rate law is:-

[tex]r=k[UO_2^+]^2[H^+][/tex]

The question is incomplete, here is the complete question:

At 25°C Henry's Law constant for carbon dioxide gas in water is 0.031 M/atm . Calculate the mass in grams of gas that can be dissolved in 425. mL of water at 25°C and at a partial pressure of 2.92 atm. Round your answer to 2 significant digits.

Answer: The mass of carbon dioxide that can be dissolved is 1.7 grams

Explanation:

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

[tex]C_{CO_2}=K_H\times p_{CO_2}[/tex]

where,

[tex]K_H[/tex] = Henry's constant = [tex]0.031M/atm[/tex]

[tex]C_{CO_2}[/tex] = molar solubility of carbon dioxide gas

[tex]p_{CO_2}[/tex] = partial pressure of carbon dioxide gas = 2.92 atm

Putting values in above equation, we get:

[tex]C_{CO_2}=0.031M/atm\times 2.92 atm\\\\C_{CO_2}=0.0905M[/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]

Given mass of carbon dioxide = ? g

Molar mass of carbon dioxide = 44 g/mol

Molarity of solution = [tex]0.0905mol/L[/tex]

Volume of solution = 425 mL

Putting values in above equation, we get:

[tex]0.0905mol/L=\frac{\text{Mass of carbon dioxide}\times 1000}{44g/mol\times 425}\\\\\text{Mass of solute}=\frac{44\times 425\times 0.0905}{1000}=1.7g[/tex]

Hence, the mass of carbon dioxide that can be dissolved is 1.7 grams

Answer: [tex]K_a[/tex] of PABA is 0.000022

Explanation:

[tex]NH_2C_6H_5COOH\rightarrow H^++NH_2C_6H_5COO^-[/tex]

  cM              0             0

[tex]c-c\alpha[/tex]        [tex]c\alpha[/tex]          [tex]c\alpha[/tex]

So dissociation constant will be:

[tex]K_a=\frac{(c\alpha)^{2}}{c-c\alpha}[/tex]

Give c= 0.055 M and [tex]\alpha[/tex] = ?

[tex]K_a=?[/tex]

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

[tex]2.96=-log[H^+][/tex]

[tex][H^+]=1.09\times 10^{-3}[/tex]

[tex][H^+]=c\times \alpha[/tex]

[tex]1.09\times 10^{-3}=0.055\times \alpha[/tex]

[tex]\alpha=0.02[/tex]

Putting in the values we get:

[tex]K_a=\frac{(0.055\times 0.02)^2}{(0.055-0.055\times 0.02)}[/tex]

[tex]K_a=0.000022[/tex]

Thus [tex]K_a[/tex] of PABA is 0.000022

Explanation:

        [tex]{\displaystyle {\ce {HA[/tex]  ⇄  [tex]{H^+}+{A^{-}}} }}[/tex]

Here HA is a protonic acid such as acetic acid, [tex]CH_3COOH[/tex]

        [tex]K_a[/tex] = [tex]\dfrac{[H^{+}] [A^{-}]}{[HA]}[/tex]

         [tex]\\{\displaystyle {\ce {{HA}+ H_2O}[/tex] ⇄  [tex][H_3O^+] [A^-][/tex]

        [tex]K_a[/tex] = [tex]\dfrac{[H_3O^{+}] [A^{-}]}{[HA]}[/tex]

Here, [tex]K_a[/tex] is the dissociation constant of an acid.

Answer:

Explanation:

To calculate the acid dissociation constant (Ka) of the acid, we need to know the concentration of the acid and the concentration of the conjugate base at equilibrium, as well as the concentration of H3O+ or OH- ions in the solution. The general equation for the dissociation of a weak acid, HA, is:

HA + H2O ⇌ H3O+ + A-

The acid dissociation constant can be written as:

Ka = [H3O+][A-]/[HA]

We can assume that the initial concentration of the acid is equal to the concentration of HA at equilibrium, since the acid is completely dissociated in solution. Therefore, we can write the equilibrium concentration of HA as [HA]0 - [H3O+], where [HA]0 is the initial concentration of the acid.

We are not given the values of [HA], [H3O+], or [A-], but we can use the pH of the solution to calculate the concentration of H3O+:

pH = -log[H3O+]

Solving for [H3O+], we get:

[H3O+] = 10^-pH

Substituting this expression for [H3O+] into the equation for Ka, we get:

Ka = [H3O+][A-]/([HA]0 - [H3O+])

Ka = (10^-pH)[A-]/([HA]0 - 10^-pH)

If we know the concentration of the conjugate base, [A-], we can substitute that value into the equation. If we do not know the concentration of the conjugate base, we can assume that it is small compared to [HA]0 and therefore can be neglected. In that case, the equation simplifies to:

Ka ≈ 10^-pH

Therefore, the acid dissociation constant of the acid is approximately equal to 10^-pH. We would need more information about the acid and the solution to calculate a more exact value for Ka.

Explanation:

It is known that for high concentration of [tex]M^{2+}[/tex], reduction will take place. As, cathode has a positive charge and it will be placed on left hand side.

Now, [tex]E^{o}_{cell}[/tex] = 0 and the general reaction equation is as follows.

         [tex]M^{2+} + M \rightarrow M + M^{2+}[/tex]

                3.00 M        n = 2       30 mM

         E = [tex]0 - \frac{0.0591}{2} log \frac{50 \times 10^{-3}}{1}[/tex]

            = [tex]-\frac{0.0591}{2} log (5 \times 10^{-2})[/tex]

            = 0.038 V

Therefore, we can conclude that voltage shown by the voltmeter is 0.038 V.

The covalent character of a bond usually increases with the electronegativity difference between the atoms involved in the bond. Bonds between nonmetals usually have high covalent character, whereas bonds between a metal and a nonmetal are often ionic. Therefore, the bond pairs can be ordered in increasing covalent character as: (K,Br) < (Se,S) < (S,O) < (C,H) = (H,I) < (Si,Cl) < (H,F).

The covalent character of a bond generally increases with the electronegativity difference between the two atoms involved in the bond. Using this principle, we can establish that the bonds between H and F, Si and Cl, C and H, and H and I have more covalent character compared to the rest because the pairs consist of nonmetals. The bond between K and Br is likely to have the most ionic character, as K (potassium) is a metal, and Br (bromine) is a nonmetal. This is due to the general guideline that bonds between a metal and a nonmetal are often ionic.

However, it's also important to understand that while this rule can be generally applied, there are many exceptions and other factors can influence the character of a bond.

So considering this, the bonds in increasing order of covalent character might be: (K,Br) < (Se,S) < (S,O) < (C,H) = (H,I) < (Si,Cl) < (H,F)

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In the given list of atoms, the bonds they form increase in covalent character as follows: (K,Br) < (H,F) < (Si,Cl) < (S,O) < (H,I) = (C,H) < (Se,S). This is based on factors including the atoms' electronegativity differences and bond orders.

The subject here is the covalent character of different bonds, which can be decided based on multiple factors such as electronegativity differences and the bond order.

Electronegativity is the measure of the tendency of an atom to attract shared electrons. Bonds between atoms with a large eleber of chemical bonds between a pair of atoms - greater bond orders typically result in stronger, more covalent bonds. Given these criteria, the order of increasing covalent character among the given pairs should be (K,Br) < (H,F) < (Si,Cl) < (S,O) < (H,I) = (C,H) < (Se,S). This is because K and Br have a large electronegativity difference making the bond more ionic, while Se and S are both non-metals with similar electronegativities, which makes their bond more covalent.ctronegativity difference are usually more ionic and less covalent. On the other hand, bond order refers to the num

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Answer:

Half-life of the reaction is 0.072s.

Explanation:

In a first-order reaction, half-life, t1/2, is defined as:

[tex]t_{1/2} = ln2 / k[/tex] (1)

Where k is the rate constant of the reaction.

In the problem, the thermally descomposition of cyclopropane has a rate constant of 9.6s⁻¹. Replacing in (1):

[tex]t_{1/2} = ln2 / 9.6s^{-1}[/tex]

[tex]t_{1/2} = 0.072s[/tex]

I hope it helps!

Answer:

Part A K´p = 3.4 x 10⁴

Part B K´p = 2.4 x 10⁻¹⁴

Part C K´p = 1.2 x 10⁹

Explanation:

The methodolgy to answer this question is to realize that the equilibrium in part A is the reverse of the the equilibrium given:

2PH₃(g) + As₂(g)  ⇌ 2 AsH₃(g) + P₂(g)  Kp = 2.9 x 10⁻⁵

Kp = [AsH₃]²[P₂]/[PH₃]²[As]  = 2.9 x 10⁻⁵

Likewise, part B is this equilibrium where the coefficients have been multiplied by 3,  and part C is  is the  reverse equilibrium multipled by 2.

Given the way that the equilibrium constant, Kp, is expressed mathematically, we can see that if we reserse the equilibrium its constant will be the inverse of the old Kp.

Similarly if we multiply the coefficient by a factor we have to raise the expression for Kp to that factor.

With that in mind lets answer the three parts of the questiion.

Part A

Lets call the new equilibrium K´p :

K´p = 1/ Kp = 1/2.9 x 10⁻⁵ = 3.4 x 10⁴

Part B

K´p = ( Kp )³ = ( 2.9 x 10⁻⁵ )³ = 2.4 x 10⁻¹⁴

Part C

K´p = ( 1/Kp)² = ( 1/ 2.9 x 10⁻⁵ )² = 1.2 x 10⁹

If you can not see why this is so, lets show it for part C

K´p = [As₂]²[PH₃]⁴ / [P₂]²[AsH₃]⁴

rearranging this equation:

K´p = (1 / {  [AsH₃]⁴[P]² / [As₂]²[PH₃]⁴ } =) 1 / Kp²

Final answer:

The Kp for Part A (the reverse reaction) is the reciprocal of the original Kp, for Part B (the original times 3) is the original Kp to the third power, and for Part C (exchanged reactants/products with doubled coefficients) is the reciprocal of original Kp squared.

Explanation:

The original reaction is 2PH₃(g) + As₂(g) ⇌ 2AsH3(g) + P₂(g) and has Kp=2.9×10⁻⁵ at 873 K.

Part A is the reverse of the original reaction. For the reverse reaction, the equilibrium constant Kp is the reciprocal of the original reaction. Thus, Kp for Part A is 1/(2.9×10⁻⁵).

Part B has the original reaction multiplied by 3. To find the new Kp, we raise the original Kp to the power of 3. So, Kp for Part B is (2.9×10⁻⁵)³.

Part C is the original reaction with products and reactants exchanged and coefficients doubled. This combines the effects of reversing the reaction and squaring its coefficients. Therefore, Kp for Part C is (1/(2.9×10⁻⁵))².

Answer : The molecular of the compound is, [tex]C_8H_8[/tex]

Solution :

If percentage are given then we are taking total mass is 100 grams.

So, the mass of each element is equal to the percentage given.

Mass of C = 92.26 g

Mass of H = 100 - 92.26 = 7.74 g

Molar mass of C = 12 g/mole

Molar mass of H = 1 g/mole

Step 1 : convert given masses into moles.

Moles of C = [tex]\frac{\text{ given mass of C}}{\text{ molar mass of C}}= \frac{92.26g}{12g/mole}=7.688moles[/tex]

Moles of H = [tex]\frac{\text{ given mass of H}}{\text{ molar mass of H}}= \frac{7.74g}{1g/mole}=7.74moles[/tex]

Step 2 : For the mole ratio, divide each value of moles by the smallest number of moles calculated.

For C = [tex]\frac{7.688}{7.74}=0.99\approx 1[/tex]

For H = [tex]\frac{7.74}{7.74}=1[/tex]

The ratio of C : H = 1 : 1

The mole ratio of the element is represented by subscripts in empirical formula.

The Empirical formula = [tex]C_1H_1[/tex]

The empirical formula weight = 1(12) + 1(1) = 13 gram/eq

Now we have to calculate the molecular mass of polymer.

As, the mass of polymer of [tex]7.02\times 10^{18}[/tex] molecules = 1.22 mg = 0.00122 g

So, the mass of polymer of [tex]6.022\times 10^{23}[/tex] molecules = [tex]\frac{6.022\times 10^{23}}{7.02\times 10^{18}}\times 0.00122g=104.6g/mol[/tex]

Now we have to calculate the molecular formula of the compound.

Formula used :

[tex]n=\frac{\text{Molecular formula}}{\text{Empirical formula weight}}[/tex]

[tex]n=\frac{104.6}{13}=8[/tex]

Molecular formula = [tex](C_1H_1)_n=(C_1H_1)_8=C_8H_8[/tex]

Therefore, the molecular of the compound is, [tex]C_8H_8[/tex]

The molecular formula for styrene is C8H8.

The molecular formula for styrene can be determined using the given information. Firstly, we need to find the mass of 7.02 * 10^18 molecules of styrene. Given that 7.02 * 10^18 molecules weigh 1.22 mg, we can calculate the molecular weight of the styrene as follows:

Therefore, the molecular formula of styrene is C8H8.

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Answer:

Volume at half equivalence point, that is 19.19 mL/2 = 9.595 mL.

Explanation:

From the question we are asked to find the volume that would one can use the pH to determine the pKa of the acid and the answer will be the volume at half equivalence point.

At half equivalence point, the weak acid and the conjugate base will have the same number of moles because the number of moles of sodium Hydroxide, NaOH will neutralize half of the number of moles of the weak acid which in turn will produce more conjugate base. And this concept is what is known as the buffer solution.

HA <-----------------------------> H^+ + A^-

(Note: the reaction above is a reversible reaction. Also, the concentration of HA is equal to the concentration of A^-).

Therefore, we can calculate our pka from the equation below(assuming the pH is given).

pH= pka + log ( [A^-] / log [HA].

===> At half equivalence point pH= pKa.

Recall that, pka = - log ka.

Then, ka = 10^-pka.

Where pH= pKa.

Therefore, ka = 10^-pH.

Volume at half equivalence point = 9.595 mL.

At half equivalence point, the weak acid and the conjugate base will have the same number of moles because the number of moles of sodium Hydroxide, NaOH will neutralize half of the number of moles of the weak acid which in turn will produce more conjugate base. It is also known as buffer solution.

Chemical reaction:

[tex]HA--->H^++A^-[/tex]

Calculation of pKa:

[tex]pH= pka + log\frac{[A^-]}{[HA]}[/tex]

At half equivalence point, pH= pKa.

We know, [tex]ka = 10^{-pka}[/tex]

Thus, [tex]ka = 10^{-pH}[/tex] (since, pH= pKa )

Volume is given which is 19.19mL. So, the volume that is used to determine pKa of the acid will be:

19.19/2 = 9.959mL

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Answer:

pOH = 4.8

pH = 9.2

Explanation:

Given data:

Hydrogen ion concentration = 6.3×10⁻¹⁰M

pH of solution = ?

pOH of solution = ?

Solution:

Formula:

pH = -log [H⁺]

[H⁺] = Hydrogen ion concentration

We will put the values in formula to calculate the pH.

pH = -log [6.3×10⁻¹⁰]

pH = 9.2

To calculate the pOH:

pH + pOH = 14

We will rearrange this equation.

pOH = 14 - pH

now we will put the values of pH.

pOH = 14 - 9.2

pOH = 4.8

Answer:

ΔrxnHº  =-1,124.3 kJ

Explanation:

To solve this question we first need to search  for the enthalpies of formation of reactants and products and calculate the change in enthalpy of reaction utilizing the equation

ΔrxnHº = ∑ νΔfHº reactants - Σ νΔfHº products

where ν is the stoichiometric coefficient of the compound  in the balanced equation .

ΔfHº H₂S(g) = -20.50 kJmol⁻¹

ΔfHº 0₂(g)  = 0  ( O₂ is in standard state )

ΔfHº H₂O(l) = -285.8 kJmol⁻¹

ΔfHº SO₂(g) = -296.84 kJmol⁻¹

ΔrxnHº =  [2 x  -285.8  + 2 x  -296.84] - [2 x -20.50]

           =    - 1,124.3 kJ

Answer:

A. Theoretical yield is 36800g

B. 49.73%

Explanation:

First let us generate a balanced equation for the reaction. This is illustrated below:

CO + 2H2 —> CH3OH

From the question given, we obtained the following:

Mass of CO = 37.5 kg = 37.5 x 1000 = 37500g

Mass of H2 = 4.60kg = 4.60 x 1000 = 4600g

Let us convert these Masses to mol

For CO:

Molar Mass of CO = 12 + 16 = 28g/mol

Mass of CO = 37500g

Number of mole of CO = 37500/28 = 1339.3moles

For H2:

Molar Mass of H2 = 2x1 =2g/mol

Mass of H2 = 4600g

Number of mole of H2 = 4600/2 =

2300moles

Now we can see from the equation above that for every 2moles of H2, 1mole of CO is required. Therefore, 2300moles of H2 will require = 2300/2 = 1150moles of CO. This amount(ie 1150moles) is little compared to 1339.3moles of CO calculated from the question. Therefore, H2 is the limiting reactant.

Now we can calculate the theoretical yield as follows:

CO + 2H2 —> CH3OH

Molar Mass of H2 = 2g/mol

Mass of H2 from the balanced equation = 2 x 2 = 4g

Molar Mass of CH3OH = 12 + 3 + 16 + 1 = 32g/mol

From the equation,

4g of H2 produced 32g of CH3OH

Therefore, 4600g of H2 will produce = (4600 x 32)/4 = 36800g of CH3OH

Therefore, the theoretical yield is 36800g

B. Actual yield = 1.83 x 10^4g

theoretical yield = 36800g

%yield =?

%yield = Actual yield /Theoretical yield x 100

%yield = 1.83 x 10^4/36800 x 100

%yield = 49.73%

Answer:

There is no short answer.

Explanation:

In the given example Bobby is creating a solution for his bacteria count which consists of 1% bacterial sample.

Considering that the solution was mixed homogeneously, he can apply the procedure to the remaining sample and get the results he wants.

Or if the average number of bacteria in a 1 mL sample is known, he can apply that information proportionally to the 100 mL mixture and find the original cell concentration.

I hope this answer helps.

Sulfonation of benzene involves the replacement of a hydrogen atom by a sulfonyl group using sulfuric acid. The reaction involves several steps, each with specific reactants and products. Key substances involved include sulfuric acid, sulfurous acid, benzene, and sulfur dioxide.

The question is asking for a detailed understanding of the sulfonation of benzene, a process where a hydrogen atom on benzene is replaced by a sulfonyl group. This reaction is a type of electrophilic aromatic substitution. Sulfonation involves the use of sulfuric acid, which provides H3O+ and SO3 in the first reversible step. In the second step, SO3 reacts slowly with benzene to form H(C6H5+)SO3−, and in the subsequent fast step, HSO4− reacts with H(C6H5+)SO3− to yield C6H5SO3− and H2SO4. Finally, C6H5SO3− and H3O+ react to produce the sulfonated benzene product, C6H5SO3H.

The reaction mechanism involves various substances, including sulfurous acid and sulfur dioxide. Sulfurous acid is unstable and can decompose into sulfur dioxide and water. Sulfur dioxide can further react with oxygen to form sulfur trioxide, which can participate in the sulfonation of benzene. Note that sulfur forms several compounds exhibiting positive oxidation states, unlike oxygen.

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Answer: Friction (Fk) = 46.2N.

Explanation: In the attachment, there is the drawing of the forces acting on the skiing person. As the person is in movement, the formula to calculate the friction is Fk = μk.N, where μk is the coefficient of kinetic friction and N is normal force. Normal force is the force necessary to keep a person "above" the ground. To find the normal force as suggested, we use the representation of the forces in the attachment. With it, it's shown that to calculate N:

cos 15° = [tex]\frac{N}{Fg}[/tex]

N = Fg·cos 15°

N = 608·(-0.76)

N = - 461.9N

Now, with N and knowing the coefficient, which can be found in Physics Text Books and it is 0.1, we have:

Fk = 0.1·(-461.9)

Fk = - 46.2N

The friction is negative because it is pointing to the opposite side of the reference.

Answer:

There will be no reaction between the product of hydrolysis and I2-KI (-ve). When the product of hydrolysis is tested with Benedict reagent, a brick-red precipitate is observed.

Explanation:

Benedict's reagent as a chemical reagent is a mixture of sodium carbonate, sodium citrate and copper(II) sulfate pentahydrate. It is often used in place of Fehling's solution to detect the presence of reducing sugars. A positive test with Benedict's reagent is shown by a color change from clear blue to a brick-red precipitate.

Glycogen and starch are both complex structures containing repeating units of glucose(a reducing sugar). The polysaccharides have non redusing ends and so cannot react with benedict reagent.

When they are hydrolysed, glucose which is a reducing sugar can then test positive with Benedict reagent.

Final answer:

After hydrolysis, glycogen and starch would no longer give a positive iodine test but would give a positive Benedict's test, as they break down into glucose monomers, which are reducing sugars capable of reacting with Benedict's reagent.

Explanation:

If you hydrolyzed glycogen and starch and then tested the solutions with I₂-KI (iodine test) and Benedict's solution, you would get different results compared to the tests performed on the non-hydrolyzed polysaccharides. Glycogen and starch are both polysaccharides that give a positive reaction with the iodine test due to their coiled structures, which trap iodine molecules and create a bluish-black color. However, they do not give a positive Benedict's test result in their non-hydrolyzed form because they are not free-reducing sugars and therefore do not react with the Benedict's reagent.

Once hydrolyzed, glycogen and starch break down into glucose monomers. Glucose is a reducing sugar, which will react with Benedict's reagent. Upon heating with Benedict's solution, the glucose will reduce the copper(II) ions in the reagent to copper(I) oxide, resulting in a color change from blue to greenish or orangish precipitate, depending on the amount of glucose present. Therefore, after hydrolysis, the solutions of glycogen and starch would give a negative result with the iodine test (no blue-black color) and a positive result with Benedict's test (color change indicating the presence of reducing sugar).

Answer: Concentration of the dye in solution is 1.159 M

Explanation:

According to Beer-Lambert law, the absorbance is directly proportional to the concentration of an absorbing species.

Formula used :

[tex]A=\epsilon \times C\times l[/tex]

where,

A = absorbance of solution = 1.657

C = concentration of solution = ?

l = length of the cell = 0.701 cm

[tex]\epsilon[/tex] = molar absorptivity of this solution =

[tex]1.657=2.038Lmol^{-1}cm^{-1}\times C\times 0.701cm[/tex]

[tex]C=1.159mol/L[/tex]

Thus the concentration of the dye in solution is 1.159 M

Concentration of the dye in solution is 1.159 M

According to Beer-Lambert law, the absorbance is directly proportional to the concentration of an absorbing species. The concentration can be calculated by using this law which is given as:

[tex]A=E*l*c[/tex]

where,

A = absorbance of solution = 1.657

c = concentration of solution = ?

l = length of the cell = 0.701 cm

E= molar absorptivity of this solution = [tex]2.038 L mol^{-1} cm^{-1}[/tex]

On substituting the values:

[tex]A=E*l*c\\\\1.657=2.038*0.701*c\\\\c=\frac{1.657}{2.038*0.701}\\\\ c=1.159mol/L[/tex]

Thus, the concentration of the dye in solution is 1.159 M.

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