Consider a reaction involving two reactants (A and B) in which the reaction is first-order in reactant A and second-order in reactant B.


a. write the rate law for this equation.

b. what is the overall order of the reaction?

c. identify how the reaction rate would change if...


i. [A] is doubled and [B] held constant.

ii. [A] is held constant and [B] is doubled.

iii. [A] is tripled and [B] is doubled

iv. [A] is doubled and [B] is halved.

The radioactive uranium decays to produce thorium and it emits an alpha particle or helium atom. Thus, option A is correct.

Unstable heavy isotopes of elements undergo nuclear decay to produce stable atoms by the emission of charged particle such as alpha or beta particles.

Based on the emitted particle, there are two types of decay process namely alpha decay and beta decay. In alpha decay atoms emits alpha particles which are helium nuclei and the atom losses its mass number by 4 units and atomic number by two units,

In beta decay, electrons are emitted by the atom, where no change occurs in mass number and atomic number increases by one unit. Uranium undergo alpha decay by emitting alpha particle or helium nuclei.

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Answer:

[tex]Fe\rightarrow Fe^{2+}+2e^-[/tex]

Explanation:

The given cell notation is:-

[tex]Fe_{(s)}|Fe^{3+}_{(aq)}||Cl_2_{(g)}|Cl^-_{(aq)}|Pt[/tex]

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

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

The oxidation half reaction of the cell is:-

[tex]Fe\rightarrow Fe^{2+}+2e^-[/tex]

Answer:

The particles in a gas are comparatively very far from each other. Further an ideal gas has not interactions between the particles i.e repulsion. This explains why a gas that is compressed into a liquid collapses into a volume that is small by a factor of 1000. The particles get much closer to each other to form a liquid and they do not repel each other

Answer:

The following answer are based only on emf values

a).Most anodic element ,Titanium - Ti

b).Metal That will corrode easily is : Titanium - Ti

c). Most reactive metal is Titanium-Ti

d).Al can't be used for making implants because it undergo corrosion easily and do not form strong passive oxide layer.

Ti forms very strong passive layer which prevents further corrosion.

Au is least reactive and does not corrode.

Note : You have to give answer according to E values

Explanation:

Here the oxidation potential data is given (since elements are getting oxidised)

[tex]\Delta^{0} E[/tex]  : Standard Reduction Potential : It is tendency of the element to gain electron and get reduced.

Oxidation Potential is opposite to reduction potential

Standard Oxidation Potential :It is tendency of the element to loose electron and get oxidised.

It is measured in Volts/V

The main application [tex]\Delta^{0} E[/tex] values is in electrochemical cell to predict the products of the reaction. or  predict whether the reaction is spontaneous or not.

Since [tex]Ti\rightarrow Ti^{+}[/tex] has maximum value of [tex]\Delta^{0}E[/tex]  = 2.00V , so it get oxidised as compared to other elements.

From the , E values Ti show maximum tendency of oxidation hence it also get corroded easily.

Final answer:

Titanium is the most anodic and most reactive metal given its high standard electrode potential of +2.00V, which also makes it the most prone to corrosion. Titanium and Gold are suitable for biomedical implants due to their corrosion resistance and biocompatibility, while Aluminum is not used due to potential toxicity.

Explanation:

To address the questions related to electrochemical potentials and the reactivity of metals, we must consider both the given standard electrode potentials and practical implications such as corrosion resistance in the case of biomedically important metals.

Most Anodic and Most Corrosive

The most anodic metal would be the one with the lowest reduction potential, which correlates with the highest oxidation potential. Since oxidation is the loss of electrons, the metal with the highest potential to lose electrons (most positive ΔEº) will be the most anodic and the most susceptible to corrosion. In this list, Titanium (Ti) has the highest standard electrode potential at +2.00 V, indicating it would be the most anodic and would corrode the most in an environment where it can react.

Most Reactive Metal

The reactivity of a metal in chemical terms is often reflected by how easily it can lose electrons (oxidize). By the given data, Titanium (Ti) would be the most reactive given its high standard electrode potential for oxidation.

Biomedical Implant Materials

Titanium (Ti) and Gold (Au) are used for biomedical implants because they are biocompatible and highly resistant to corrosion in the body. Titanium forms a stable, protective oxide layer when exposed to air or bodily fluids, preventing further oxidation. While Aluminum (Al) has a high standard reduction potential, which may suggest it is resistive to corrosion, it is not used for implants due to potential toxicity and less stable oxide formation than Ti, which could lead to corrosion in the body.

The balanced net ionic equations for the reactions are:

a. Cr2(SO4)3(aq) + 3(NH4)2CO3(aq) → 2Cr(CO3)3(s) + 6NH4NO3(aq),

b. Ba(NO3)2(aq) + K2SO4(aq) → BaSO4(s) + 2KNO3(aq),

c. Fe(NO3)2(aq) + 2KOH(aq) → Fe(OH)2(s) + 2KNO3(aq)

Net ionic equations represent the actual chemical change occurring in a reaction, excluding spectator ions. Examples include the exchange of ions to form water in a neutralization reaction and the formation of precipitates such as calcium phosphate or silver chloride.

Thus, the net balanced ionic equations of given reactions are:

a. Cr2(SO4)3(aq) + 3(NH4)2CO3(aq) → 2Cr(CO3)3(s) + 6NH4NO3(aq)

b. Ba(NO3)2(aq) + K2SO4(aq) → BaSO4(s) + 2KNO3(aq)

c. Fe(NO3)2(aq) + 2KOH(aq) → Fe(OH)2(s) + 2KNO3(aq)

Answer:

c. (CH3CH2CH2)2CuLi

Explanation:

(CH3CH2CH2)2CuLi

This reagent is called Lithium di(n propyl)cuprate used to covert benzoyl chloride to phenyl propyl ketone. this reagent is called Gilman Reagent and The reaction occur with nucleophilic substitution of an alkyl group for the leaving group (chloride), forming one new carbon-carbon bond.

Hence the correct answer is C that is   c. (CH3CH2CH2)2CuLi

Answer:

Ka1 = 4.27x10⁻⁷, Ka2 = 2.78x10⁻¹⁶

Explanation:

The number of moles that had reacted when the pH was 6.37 was:

nH₂A = 0.046 L * 0.15 mol/L = 6.9x10⁻³ mol

nKOH = 0.023 L * 0.15 mol/L = 3.45x10⁻³ mol

Thus, H₂A is in excess, and the base will not react with HA⁻ to form A⁻², so the first reaction is absolute, and after the reaction:

nH₂A = 6.9x10⁻³ - 3.45x10⁻³ = 3.45x10⁻³

nHA⁻ = 3.45x10⁻³

Because the volume will be the same, we can use the number of moles instead of the concentrations. So:

6.37 = pKa1 + log(3.45x10⁻³/3.45x10⁻³)

pKa1 = 6.37

Ka1 = [tex] 10^{-6.37}[/tex]

Ka1 = 4.27x10⁻⁷

After 46 mL of KOH was added:

nKOH = 0.046 * 0.15 = 6.9x10⁻³

So, all the H₂A reacts to form HA⁻, and the absolute equilibrium is:

HA⁻ ⇄ H⁺ + A⁻²

6.9x10⁻³ 0 0 Initial

-x +x x Reacts

6.9x10⁻³-x x x Equilibrium

pH = -log[H⁺]

8.34 = -log[H⁺]

[H⁺] = [tex]10^{-8.34}[/tex]

[H⁺] = 4.57x10⁻⁹ M

The total volume is 46 mL + 46 mL = 92 mL = 0.092 L

x = 0.092 L * 4.57x10⁻⁹ = 4.20x10⁻¹⁰ mol

Thus, nHA⁻ = 6.9x10⁻³ - 4.20x10⁻¹⁰ ≅ 6.9x10⁻³

Then,

8.34 = pKa2 + log(4.20x10⁻¹⁰/6.9x10⁻³)

8.34 = pKa2 - 7.216

pKa2 = 15.556

Ka2 = [tex] 10^{-15.556}[/tex]

Ka2 = 2.78x10⁻¹⁶

Final answer:

The values of Ka1 and Ka2 for the diprotic acid can be determined using the titration data provided. At the halfway point of the titration, the pH of the solution is equal to the pKa1 value. Using the given pH value, we can calculate the pKa1 and Ka1 values. At the endpoint of the titration, the pH of the solution is equal to the pKa2 value, allowing us to calculate the pKa2 and Ka2 values.

Explanation:

To determine the values of Ka1 and Ka2 for the diprotic acid, we can use the titration data provided. When 23.0 mL of 0.15 M KOH was added, the pH of the solution was 6.37. At this point, half of the acid has reacted with the base, indicating that we are at the halfway point between Ka1 and Ka2. Therefore, the pH is equal to the pKa at this point.

Using the pH value of 6.37, we can calculate the pKa by taking the negative logarithm of the [H+] concentration. Since pH = -log[H+], we can rearrange the equation to find [H+]:
[H+] = 10^(-pH).
Therefore, [H+] = 10^(-6.37) = 2.17 x 10^(-7) M.

Since this is the concentration at the halfway point, we can assume that [H2A] = [HA-]. Using the Henderson-Hasselbalch equation, we can write:
pKa1 = pH + log([HA-]/[H2A]) = 6.37 + log(2.17 x 10^(-7)/0.15).
pKa1 = 6.37 + (-3.66) = 2.71.

Therefore, the value of Ka1 for the diprotic acid is 10^(-pKa1) = 10^(-2.71).

Using similar calculations, we can determine the value of Ka2. When 46 mL of 0.15 M KOH was added, the pH of the solution was 8.34. At this point, all of the acid has reacted with the base, indicating that we are at the endpoint of the titration. Therefore, the pH is equal to the pKa2 at this point.

Using the pH value of 8.34, we can calculate the pKa2 by taking the negative logarithm of the [H+] concentration. Since pH = -log[H+], we can rearrange the equation to find [H+]:
[H+] = 10^(-pH).
Therefore, [H+] = 10^(-8.34) = 4.28 x 10^(-9) M.

Since all of the acid has reacted, we can assume that [A-] = 0.15 M. Using the Henderson-Hasselbalch equation, we can write:
pKa2 = pH + log([A-]/[HA]).
pKa2 = 8.34 + log(0.15/0.15).
pKa2 = 8.34 + 0 = 8.34.

Therefore, the value of Ka2 for the diprotic acid is 10^(-pKa2) = 10^(-8.34).

Answer:

b) C = 0.50 J/(g°C)

Explanation:

∴ Q = 50 J

∴ m = 10.0 g

∴ ΔT = 35 - 25 = 10 °C

specific heat (C) :

⇒ C = Q / mΔT

⇒ C = 50 J / (10.0 g)(10 °C)

⇒ C = 0.50 J/(g°C)

Answer:

a) 39/19 K : stable nuclide, 40/19 K  : radioactive nuclide.

b) 209B: stable nuclide, 208Bi : radioactive nuclide

c) nickel-58 : stable nuclide, nickel-65 : radioactive nuclide.

Explanation:

As per the rule, nuclides having odd number of neutrons are generally not stable and therefore, are radioactive.

Mass number (A) = Atomic number (Z) + No. of neutrons (N)

Or, N = A - Z

a)

39/19 K and 40/19 K

Calculate no. of neutrons in 39/19 K as follows:

atomic no. = 19, mass no. 39

N = 39 - 19

   = 20 (even no.)

Calculate no. of neutrons in 40/19 K as follows:

atomic no. = 19, mass no. 40

N = 40 - 19

   = 21 (odd no.)

Therefore, 39/19 K is a stable nuclide and 40/19 K is a radioactive

nuclide.

b)

209Bi and 208Bi

Calculate no. of neutrons in 209Bi as follows:

atomic no. = 83, mass no. 209

N = 209 - 83

   = 126 (even no.)

Calculate no. of neutrons in 208Bi as follows:

atomic no. = 83, mass no. 208

N = 208 - 83

   = 125 (odd no.)

Therefore, 209Bi is a stable nuclide and 208Bi is a radioactive nuclide.

c)

nickel-58 and nickel-65

Calculate no. of neutrons in nickel-58 as follows:

atomic no. = 28, mass no. 58

N = 58 - 28

   = 30 (even no.)

Calculate no. of neutrons in nickel as follows:

atomic no. = 28, mass no. 65

N = 65 - 28

   = 37 (odd no.)

Therefore,nickel-58 is a stable nuclide and nickel-65 is a radioactive nuclide.

To determine which nuclide is radioactive and which is stable, we need to analyze the composition of their nuclei. The number of neutrons in a nuclide plays a crucial role in determining its stability. Based on this, we can predict whether a nuclide is radioactive or stable.

In order to predict which nuclide is radioactive and which is stable, we need to determine the composition of their nuclei. Radioactive nuclides have an unstable nucleus that undergoes radioactive decay, while stable nuclides have a more balanced composition.

Therefore, the pairs are:
a. 39/19K (stable) and 40/19K (radioactive)
b. 209Bi (stable) and 208Bi (radioactive)
c. 58Ni (stable) and 65Ni (radioactive)

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Answer:

15,4g of ice

Explanation:

The dissolution of HNO₃ increases the temperature of the solution thus:

Q = -C×m×ΔT (1)

Where Q is heat produced:

Q = -33 kJ/mol×(0,0130L×14,0M) = -6,00kJ

C is molar heat capacity of solution: 80,8 J/mol°C

m are moles of solution ≈ moles of water = 100mL≡100g×(1mol/18g) = 5,56 moles

And ΔT is final temperature - Initial temperature (X-25°C)

Replacing in (1):

-6000J = -80,8J/mol°C×5,56mol×(X-25°C)

13,4 = X-25°C

X = 38,4°C

Knowing that you want to return the temperature of the system to 25°C, the ice must to absorb 6000J of energy (In the fussion process and increasing each temperature until equilibrium) produced in the dissolution of HNO₃:

6000J = ΔH°fus×X + C×X×ΔT

-Where X are moles of ice-

Replacing:

6000J = 6010J/mol×X + 75,3J/mol°C×X×(38,4°C-25°C)

6000J = 6010J×X + 1006J×X

0,855 = moles of X

In grams:

0,855 moles×(18g/1mol)= 15,4g of ice

I hope it helps!

Answer:

a) Nor Mg, neither I2 is the limiting reactant.

b) I2 is the limiting reactant

c) Mg is the limiting reactant

d) Mg is the limiting reactant

e) Nor Mg, neither I2 is the limiting reactant.

f) I2 is the limiting reactant

g) Nor Mg, neither I2 is the limiting reactant.

h) I2 is the limiting reactant

i) Mg is the limiting reactant

Explanation:

Step 1: The balanced equation:

Mg(s) + I2(s) → MgI2(s)

For 1 mol of Mg we need 1 mol of I2 to produce 1 mol of MgI2

a. 100 atoms of Mg and 100 molecules of I2

We'll have the following equation:

100 Mg(s) + 100 I2(s) → 100MgI2(s)

This is a stoichiometric mixture. Nor Mg, neither I2 is the limiting reactant.

b. 150 atoms of Mg and 100 molecules of I2

We'll have the following equation:

150 Mg(s) + 100 I2(s) → 100 MgI2(s)

I2 is the limiting reactant, and will be completely consumed. There will be consumed 100 Mg atoms. There will remain 50 Mg atoms.

There will be produced 100 MgI2 molecules.

c. 200 atoms of Mg and 300 molecules of I2

We'll have the following equation:

200 Mg(s) + 300 I2(s) →200 MgI2(s)

Mg is the limiting reactant, and will be completely consumed. There will be consumed 200 I2 molecules. There will remain 100 I2 molecules.

There will be produced 200 MgI2 molecules.

d. 0.16 mol Mg and 0.25 mol I2

We'll have the following equation:

Mg(s) + I2(s) → MgI2(s)

Mg is the limiting reactant, and will be completely consumed. There will be consumed 0.16 mol of I2. There will remain 0.09 mol of I2.

There will be produced 0.16 mol of MgI2.

e. 0.14 mol Mg and 0.14 mol I2

We'll have the following equation:

Mg(s) + I2(s) → MgI2(s)

This is a stoichiometric mixture. Nor Mg, neither I2 is the limiting reactant.

There will be consumed 0.14 mol of Mg and 0.14 mol of I2. there will be produced 0.14 mol of MgI2

f. 0.12 mol Mg and 0.08 mol I2

We'll have the following equation:

Mg(s) + I2(s) → MgI2(s)

I2 is the limiting reactant, and will be completely consumed. There will be consumed 0.08 moles of Mg. There will remain 0.04 moles of Mg.

There will be produced 0.08 moles of MgI2.

g. 6.078 g Mg and 63.455 g I2

We'll have the following equation:

Mg(s) + I2(s) → MgI2(s)

Number of moles of Mg = 6.078 grams / 24.31 g/mol = 0.250 moles

Number of moles I2 = 63.455 grams/ 253.8 g/mol = 0.250 moles

This is a stoichiometric mixture. Nor Mg, neither I2 is the limiting reactant.

There will be consumed 0.250 mol of Mg and 0.250 mol of I2. there will be produced 0.250 mol of MgI2

h. 1.00 g Mg and 2.00 g I2

We'll have the following equation:

Mg(s) + I2(s) → MgI2(s)

Number of moles of Mg = 1.00 grams / 24.31 g/mol = 0.0411 moles

Number of moles I2 = 2.00 grams/ 253.8 g/mol = 0.00788 moles

I2 is the limiting reactant, and will be completely consumed. There will be consumed 0.00788 moles of Mg. There will remain 0.03322 moles of Mg.

There will be produced 0.00788 moles of MgI2.

i. 1.00 g Mg and 2.00 g I2

We'll have the following equation:

Mg(s) + I2(s) → MgI2(s)

Number of moles of Mg = 1.00 grams / 24.31 g/mol = 0.0411 moles

Number of moles I2 = 20.00 grams/ 253.8 g/mol = 0.0788 moles

Mg is the limiting reactant, and will be completely consumed. There will be consumed 0.0411 moles of Mg. There will remain 0.0377 moles of I2.

There will be produced 0.0411 moles of MgI2.

The limiting reagent in a chemical reaction is the reactant that is completely consumed first. To identify the limiting reagent, we need to compare the mole ratios of the reactants to the mole ratio of the products in the balanced chemical equation.

* Reaction mixture a: neither reactant is limiting.

* Reaction mixture b: Mg is the limiting reagent.

* Reaction mixture c: I2 is the limiting reagent.

* Reaction mixture d: Mg is the limiting reagent.

* Reaction mixture e: neither reactant is limiting.

* Reaction mixture f: I2 is the limiting reagent.

* Reaction mixture g: neither reactant is limiting.

* Reaction mixture h: I2 is the limiting reagent.

* Reaction mixture i: Mg is the limiting reagent.

To identify the limiting reagent in a reaction mixture, we need to compare the mole ratios of the reactants to the mole ratio of the products in the balanced chemical equation. The limiting reagent is the reactant that will be completely consumed before the other reactants are consumed.

**Balanced chemical equation:**

Mg(s) + I2 (s) → MgI2 (s)

**Mole ratios:**

Mg : I2 : MgI2 = 1 : 1 : 1

**a. 100 atoms of Mg and 100 molecules of I2**

Since we have the same number of atoms of Mg and molecules of I2, neither reactant is limiting. The reaction will proceed until both reactants are completely consumed.

**b. 150 atoms of Mg and 100 molecules of I2**

Since we have more atoms of Mg than molecules of I2, Mg is the limiting reagent. I2 will be completely consumed before Mg is consumed.

**c. 200 atoms of Mg and 300 molecules of I2**

Since we have more molecules of I2 than atoms of Mg, I2 is the limiting reagent. Mg will be completely consumed before I2 is consumed.

**d. 0.16 mol Mg and 0.25 mol I2**

Since we have less moles of Mg than moles of I2, Mg is the limiting reagent. I2 will not be completely consumed.

**e. 0.14 mol Mg and 0.14 mol I2**

Since we have the same number of moles of Mg and moles of I2, neither reactant is limiting. The reaction will proceed until both reactants are completely consumed.

**f. 0.12 mol Mg and 0.08 mol I2**

Since we have less moles of I2 than moles of Mg, I2 is the limiting reagent. Mg will not be completely consumed.

**g. 6.078 g Mg and 63.455 g I2**

First, we need to convert the masses of Mg and I2 to moles:

Moles of Mg = 6.078 g / 24.31 g/mol = 0.250 mol

Moles of I2 = 63.455 g / 253.808 g/mol = 0.250 mol

Since we have the same number of moles of Mg and moles of I2, neither reactant is limiting. The reaction will proceed until both reactants are completely consumed.

**h. 1.00 g Mg and 2.00 g I2**

First, we need to convert the masses of Mg and I2 to moles:

Moles of Mg = 1.00 g / 24.31 g/mol = 0.0411 mol

Moles of I2 = 2.00 g / 253.808 g/mol = 0.00791 mol

Since we have less moles of I2 than moles of Mg, I2 is the limiting reagent. Mg will not be completely consumed.

**i. 1.00 g Mg and 20.00 g I2**

First, we need to convert the masses of Mg and I2 to moles:

Moles of Mg = 1.00 g / 24.31 g/mol = 0.0411 mol

Moles of I2 = 20.00 g / 253.808 g/mol = 0.0788 mol

Since we have more moles of I2 than moles of Mg, Mg is the limiting reagent. I2 will not be completely consumed.

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Answer:

K remains the same;

Q < K;

The reaction must run in the forward direction to reestablish the equilibrium;

The concentration of [tex]PCl_3[/tex] will decrease.

Explanation:

In this problem, we're adding an excess of a reactant, chlorine gas, to a system that is already at equilibrium. According to the principle of Le Chatelier, when a system at equilibrium is disturbed, the equilibrium shifts toward the side of the equilibrium that minimizes the disturbance.

Since we'll have an excess of chlorine, the system will try to reduce that excess by shifting the equilibrium to the right. Therefore, the reaction must run in the forward direction to reestablish the equilibrium.

The value of K remains the same, as it's only temperature-dependent, while the value of Q will be lower than K, that is, Q < K, as Q < K is the case when reaction proceeds to the right.

As a result, since [tex]PCl_3[/tex] is also a reactant, its concentration will decrease.

Answer:

The heat of combustion of liquid butane is: -2636 kJ/mol

Explanation:

This problem is solved by using Hess law of heats summation:

We have the heat of combustion for gaseous butane:

C₄H₁₀ (g) + 13/2 O₂ (g) ⇒ 4 CO₂ (g) + H₂O (l)     ΔHºcomb = -2658 kJ/mol

and we want the heat of combustion for liquid butane.

What we need to do is add the heat of vaporization for liquid butane to this equation:

C₄H₁₀ (l)   ⇒  C₄H₁₀ (g)   ΔHºvap =  22.44 kJ/mol

C₄H₁₀ (g) + 13/2 O₂ (g) ⇒ 4 CO₂ (g) + H₂O (l)  ΔHºcomb = -2658 kJ/mol

______________________________________________

C₄H₁₀ (l)   + 13/2 O₂ (g) ⇒ 4 CO₂ (g) + H₂O (l)  

So,

ΔH comb liq butane = -2658 kJ/mol + 22.44 kJ/mol = -2636 kJ/mol

Answer:

1. The theoretical yield is 0.2mols

2. No, aldol condensation reaction does not proceeds via an E1 mechanism.

Answer:

2 mole H2O = 18.02 g will not work.

Explanation:

Let us check the options one by one:

1 mole of O2 weighs 32 g.

⇒Hence option A is correct.

1 mole of H2O weighs 18 g.

⇒Hence option B is wrong.

From the balanced equation we can say :

1 mole of CH4 reacts with 2 moles of O2 to give 1 mole of CO2 and 2 moles of H2O.

⇒Hence option C and D are correct.

Answer:

Option B: 2 mole H2O = 18.02 g is not correct.

Explanation:

Step 1: The balanced equation

CH4 + 2O2 → CO2 + 2H2O

Step 2: Data given

Molar mass of O2 = 32 g/mol

Molar mass of H2O = 18.02 g/mol

Molar mass of CH4 = 16.04 g/mol

Step 3: Calculate number of mol

For 1 mol of CH4 we need 2 moles of O2 to produce 1 mol of CO2 and 2 moles of H2O

This means option C is correct: For 2 moles O2 consumed, we'll get 1 mol of CO2 produced.

Also Option D is corect: For 1 mole of CH4 we get 2 moles of H2O

Mass = moles * molar mass

Mass of 1 mole O2 = 1 mol * 32.00 g/mol = 32.00 grams

Mass of 2 moles H2O = 2 mol * 18.02 g/mol = 36.04 grams  (This means option B is not correct.)

Option B: 2 mole H2O = 18.02 g is not correct.

Answer:

B) ) –1615.1 kJ mol^–1

Explanation:

since

SiO2(s) + 4 HF(aq) → SiF4(g) + 2 H2O(l) ∆Hºrxn = 4.6 kJ mol–1

the enhalpy of reaction will be

∆Hºrxn = ∑νp*∆Hºfp - ∑νr*∆Hºfr

where ∆Hºrxn= enthalpy of reaction , ∆Hºfp= standard enthalpy of formation of products , ∆Hºfr = standard enthalpy of formation of reactants , νp=stoichiometric coffficient of products, νr=stoichiometric coffficient of reactants

therefore

∆Hºrxn = ∑νp*∆Hºfp - ∑νr*∆Hºfr

4.6 kJ/mol = [1*∆HºfX + 2*(–285.8 kJ/mol)] - [1*(–910.9kJ/mol) + 4*(–320.1 kJ/mol)]

4.6 kJ/mol =∆HºfX -571.6 kJ/mol + 2191.3 kJ/mol

∆HºfX = 4.6 kJ/mol + 571.6 kJ/mol - 2191.3 kJ/mol = -1615.1 kJ/mol

therefore ∆HºfX (unknown standard enthalpy of formation = standard enthalpy of formation of SiF4(g) ) = -1615.1 kJ/mol

Answer:

The ice will not melt, and the temperature will remain at 0°C.

Explanation:

The reaction of the sodium in water is exothermic because heat is being released. In an isolated system, the change in heat must be 0, so the released heat must be absorbed by the ice.

The molar mass of Na is 23 g/mol, so the number of moles that reacted was:

n = 0.250 g/ 23g/mol

n = 0.011 mol

By the reaction:

2 moles ------- -368 kJ

0.011 mol ----- x

By a simple direct three rule:

2x = -4.048

x = -2.024 kJ/mol

So the ice will absorbs 2.024 kJ/mol, which is less than the necessary to melt it (6.02 kJ/mol). Then, the ice will not melt.

The temperature of a pure substance didn't change until all of it has changed of phase, so the temperature must remain at 0°C.

Answer:

[tex]Br_2[/tex]

Explanation:

The compound [tex]X_2[/tex] is [tex]Br_2[/tex].

The reaction included are:

[tex]Br_2+NaCl--> No\ reaction[/tex]  (because  bromine is less reactive than chlorine)

But , because of hexane the solution get dilute and its color changes to orange.

Now, NaI is added and we know Br is more reactive than I.

Therefore it replace it.

Reaction: [tex]Br_2+NaI-->NaBr+I_2[/tex]

Purple color develop due to formation of Iodine.

Answer:

10 g %  w/v

Explanation:

Weight/volume percent concentration means the grams of solute, in 100 mL of solution.

If we have 1 g of B3 in 10mL of solution, the rule of three for the weight volume percent will be:

10mL ___ 1 g

100 mL ____ 10 g

Answer:

1. The difference between a chemical change and a physical change is as follows;

In a physical change no new substance is formed while in a chemical change new substance are formed

Chemical change involves great amount of heat changes while physical change does not involves great amount of heat change except the latent heat changes which occur during changes of state

Physical change is easily reversed while a chemical change is not easily reversed

There is no change in the mass of a substance involve in physical change while in chemical change there is a change in the mass of a substance that undergoes such a change

2 . examples of physical and chemical changes

Physical change.

1. The magnetization and demagnetization of iron rods( the iron can be removed and no new substance will be formed)

11. The separation of mixtures by evaporation, distillation, fractional distillation, sublimation.( the separation techniques mentioned are physical change because they can be easily revised to get the original quantity back, like when water evaporates to vapour it can be retrieved back by cooling)

111. Changes in the states of matter such as melting of solids to liquids the freezing of liquids to solids, the vaporization of liquids to gases.( also changes of states are physical change because no new substance will be formed and it is easily reversible)

Chemical change

1. Fermentation and decay of substances .( new substance is formed which is the decayed substance)

11. The rusting of iron( rusting of iron is a chemical change because a new substance is formed. Which is the rusted iron)

111. The addition of water to quick lime I.e the slacking of lime. ( is a chemical change because a new substance is formed)

3. Boiling of water in a pot is a physical change because the water will evaporate into vapour at 100° Celsius and can be regained by cooling

Food digesting is a chemical change because a new substance will be formed , which is faeces.

Cooking an egg for breakfast is a chemical change because a new substance will be formed which is boiled egg and it can not be reversed.

Souring of milk is a chemical change because a new substance will be formed which is the soured taste

1. Chemical changes alter a substance's composition (e.g., burning wood), while physical changes affect its physical properties (e.g., melting ice).

2. Examples: Physical (melting ice), Chemical (burning wood).

3. Boiling water is a physical change; food digestion, cooking an egg, and milk souring are chemical changes due to altered composition.

1. **Difference between Chemical and Physical Changes**:

  - **Chemical Change**: A chemical change involves the alteration of a substance's chemical composition. New substances with different properties are formed. This change is typically irreversible and accompanied by a release or absorption of energy.  

  - **Physical Change**: A physical change, on the other hand, does not alter the substance's chemical identity. It involves changes in physical properties, such as size, shape, phase (solid, liquid, gas), or state (e.g., melting, freezing), but the chemical composition remains the same. Physical changes are generally reversible, and no new substances are created.

2. **Examples of Physical and Chemical Changes**:

  **Physical Changes**:

  - **Ice Melting**: When ice (solid water) melts to form liquid water, it undergoes a physical change. The chemical composition of H₂O remains the same, but the phase changes from solid to liquid.

  - **Cutting Paper**: When you cut a piece of paper into smaller pieces, it's a physical change. The chemical composition of cellulose in the paper remains unchanged.

  - **Boiling Water**: Boiling water represents a physical change. Water changes from a liquid to a gas (steam) without any alteration in its chemical composition.

  **Chemical Changes**:

  - **Burning Wood**: When wood is burned, it undergoes a chemical change. The cellulose and other compounds in wood react with oxygen to produce carbon dioxide and water vapor. New substances are formed, and energy is released.

  - **Digesting Food**: The process of digesting food in your body involves chemical changes. Enzymes break down complex molecules into simpler ones, allowing your body to absorb nutrients.

  - **Milk Souring**: When milk sours, it undergoes a chemical change due to the action of bacteria. Lactose in milk is converted into lactic acid, changing the taste and composition of the milk.

3. **Boiling Water - Physical Change**:

  Boiling water is a physical change because it involves a change in the physical state of water from liquid to gas (steam), but the chemical composition remains H₂O.

  **Chemical Changes**:

  1. **Your Food Digesting**: Digestion involves chemical changes as enzymes break down complex food molecules into simpler ones, which can then be absorbed by your body.

  2. **Cooking an Egg for Breakfast**: Cooking an egg is a chemical change because heat causes proteins in the egg white to denature and coagulate, forming a solid structure with different properties than the raw egg white.

  3. **The Milk Souring in Your Fridge**: Milk souring is a chemical change. Bacteria in the milk metabolize lactose into lactic acid, changing the taste and composition of the milk.

In summary, chemical changes involve a change in chemical composition, while physical changes involve changes in physical properties without altering the chemical identity of a substance. Boiling water is a physical change, while food digestion, cooking an egg, and milk souring are chemical changes due to alterations in chemical composition and properties.

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Answer: 0.100 m [tex]K_2SO_4[/tex]

Explanation:

Elevation in boiling point is given by:

[tex]\Delta T_b=i\times K_b\times m[/tex]

[tex]\Delta T_b=T_b-T_b^0[/tex] = Elevation in boiling point

i= vant hoff factor

[tex]K_b[/tex] = boiling point constant

m= molality

1. For 0.100 m [tex]K_2SO_{4}[/tex]

[tex]K_2SO_4\rightarrow 2K^{+}+SO_4^{2-}[/tex]  

, i= 3 as it is a electrolyte and dissociate to give 3 ions. and concentration of ions will be [tex]3\times 0.100=0.300[/tex]

2. For 0.100 m [tex]LiNO_3[/tex]

[tex]LiNO_3\rightarrow Li^{+}+NO_3^{-}[/tex]  

, i= 2 as it is a electrolyte and dissociate to give 2 ions, concentration of ions will be [tex]2\times 0.100=0.200[/tex]

3.  For 0.200 m [tex]C_2H_8O_3[/tex]

, i= 1 as it is a non electrolyte and does not dissociate, concentration of ions will be [tex]1\times 0.200=0.200[/tex]

4. For 0.060 m [tex]Na_3PO_4[/tex]

[tex]Na_3PO_4\rightarrow 3Na^{+}+PO_4^{3-}[/tex]  

, i= 4 as it is a electrolyte and dissociate to give 4 ions. and concentration of ions will be [tex]4\times 0.060=0.24m[/tex]

Thus as concentration of solute is highest for [tex]K_2SO_4[/tex] , the elevation in boiling point is highest and thus has the highest boiling point.

The aqueous solution of K2SO4 has the highest boiling point among the given options.

The boiling point of a solution depends on the concentration and nature of the solute. In this case, we need to compare the boiling points of different aqueous solutions. Higher concentration and the presence of ionic solutes will generally result in a higher boiling point.

Comparing the given options, K2SO4 and Na3PO4 are both ionic compounds, and their solutions will have higher boiling points due to their strong ionic interactions. LiNO3 is also an ionic compound, but its concentration is lower than K2SO4 and Na3PO4. C3H8O3 is a molecular compound and does not dissociate into ions in solution, so its solution will have a lower boiling point than the ionic compounds. Therefore, the aqueous solution of K2SO4 (option A) will have the highest boiling point among the given options.

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Answer:

-1.37 kJ/mol

Explanation:

The expression for the calculation of the enthalpy of dissolution of [tex[NH_4NO_3[/tex] is shown below as:-

[tex]\Delta H=m\times C\times \Delta T[/tex]

Where,  

[tex]\Delta H[/tex]  is the enthalpy of dissolution of [tex[NH_4NO_3[/tex]

m is the mass

C is the specific heat capacity

[tex]\Delta T[/tex]  is the temperature change

Thus, given that:-

Mass of ammonium nitrate = 5.60 g

Specific heat = 4.18 J/g°C

[tex]\Delta T=17.9-22.0\ ^0C=-4.1\ ^0C[/tex]

So,  

[tex]\Delta H=-1.25\times 4.18\times 3.9\ J=-95.9728\ J[/tex]

Negative sign signifies loss of heat.  

Also, 1 J = 0.001 kJ

So,  

[tex]\Delta H=-0.096\ kJ[/tex]

Also,

Molar mass of [tex[NH_4NO_3[/tex] = 80.043 g/mol

The formula for the calculation of moles is shown below:

[tex]moles = \frac{Mass\ taken}{Molar\ mass}[/tex]

Thus,

[tex]Moles= \frac{5.60\ g}{80.043 \ g/mol}[/tex]

[tex]Moles= 0.06996\ mol[/tex]

Thus, [tex]\Delta H=-\frac{0.096}{0.06996}\ kJ/mol=-1.37\ kJ/mol[/tex]

The final answer for the solution is Q= -8.8171*[tex]10^{-4}[/tex] kW

Explanation:

[tex]CO +H_2O  ⇄  CO_2+H_2[/tex]

basic:

100 moles of product gas

[tex]CO_2[/tex] = 40 moles

[tex]H_2[/tex] = 40 moles

[tex]H_20[/tex] = 20 moles

for 40 moles of[tex]CO_2[/tex] we need 40 moles of CO and 40 moles of [tex]H_2O[/tex]

therefore,

Inlet :

No of moles of CO = 40

No of moles of [tex]H_20[/tex] = 40

but [tex]H_20[/tex] is in the outlet stream = 20 modes

Excess = [tex]\frac{20}{40}\ \times 100[/tex]%

= 50%

50% excess stream is fed to the reactor

a) Rate of conversion:

ρ_water (density ) =[tex]\frac{1g}{cm^{3}}[/tex]

Given volume rate = 3.5 [tex]\frac{cm^{3} }{hr}[/tex]

3.5 * density = 3.5 * [tex]10^{-3}[/tex] [tex]\frac{kg}{hr}[/tex]

b) Rate (kW)

ΔH = ΔH_1 + ΔH_2 +ΔH_3

Rate of condensation Q° =n° ΔH

Q = n° (ΔH_1 + ΔH_2 +ΔH_3)

ΔH_1 is sensible heat

V =[tex]\int\limits^T_S {C_p(T) } \, dT[/tex]

C_p(T) = a + bT + CT^2 +dT^3

H_2O

a=32.218

b=o.192 * 10^-2

c= 1.055 * 10^-5

d= -3.593 * 10^-9

ΔH_2 = ∫(a+bT +a^2 +dT^3)dT (of limits 100 to 500 )

= -13497.6

ΔH_3 =∫(a +bT +T^2 +dT^3)dT (of limit 95 to 100)

= -2751.331

phase change

ΔH_2 =  - ΔH_v

ΔH_v = 7.3  @ 100° C

liquid → vapor ⇒ ΔH_v

vapor → liquid ⇒ -ΔH_v

Q= n°(-16324.231)

=350 *[tex]10^{-5}[/tex] (-16324.231)

=[tex]\frac{350* 10^{-5} }{18}[/tex] (-16324.231) *[tex]\frac{kg}{hr}[/tex][tex]\frac{J}{mol}[/tex]

=[tex]\frac{350* 10^{-5} }{18}[/tex] (-16324.231) * \frac{J}{mol}[/tex]

=[tex]\frac{350* 10^{-5} }{18}[/tex]  (-16324.231)  *kW

Q= -8.8171*[tex]10^{-4}[/tex] kW

The solution involves various calculations related to a chemical reactor, including determining the excess steam fed into the reactor, condensation rate, heat removal rate, and heat transfer rate.

This is a complex chemical engineering problem involving numerous calculations, including finding the excess steam fed into the reactor, condensation rate, heat removal rate, and heat transfer rate. Firstly, the percentage excess steam is found by comparing the stochiometric requirement of the reaction with actual steam flow. Next, we use the molar mass of water and the gas constant to find the condensation rate of water. For the heat removal rate, we need to consider the latent heat of condensation along with the energy balance of the condenser.

Lastly, for the heat transfer rate in the reactor, we need to take into account the enthalpy of reaction, heat capacities of reactants and products, and the reactor temperature profile to decide whether heat is added or removed from the reactor.

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The concentration of free Ag+ ions at equilibrium is calculated by assuming full complexation of Ag+ with CN- due to a large formation constant, and then considering any small dissociation of the complex back to Ag+ and CN-. The exact concentration can be determined by applying the formation constant for the complex, which is not provided here.

To calculate the concentration of free Ag+ ions at equilibrium in a silver plating operation, we need to consider the following reaction:

Ag+(aq) + 2 CN-(aq) \<--> Ag(CN)2-(aq)

First, we determine the initial concentrations of Ag+ and CN-. AgNO3 completely dissociates in water, so the initial concentration of Ag+ is 0.21 M in 90.0 L. NaCN also dissociates completely, providing 5.0 M of CN- ion in 9.0 L.

Mixing the solutions, we have a total volume of 99.0 L (90.0 L + 9.0 L). The concentration of CN- is calculated as follows:

(5.0 M)(9.0 L) / 99.0 L = 0.4545 M

Since the formation constant (Kf) of the Ag(CN)2- complex is large, we can assume that all the Ag+ will react to form the complex, leaving a negligible concentration of free Ag+ ions. Any remaining CN- ions will affect the equilibrium concentration of free Ag+ ions due to the shift in reaction to the left, as described by Le Chatelier's principle.

We assume that essentially all 0.21 M of the Ag+ ion forms the complex. To find the equilibrium concentration of Ag+, we use the dissociation of the complex, simplified by assuming that the dissociation is so small that it's negligible compared to the remaining CN- concentration.

Since we do not have the exact Kf value provided here, we would normally express the equilibrium concentration of Ag+ in terms of Kf and the concentration of remaining CN-, but to give an exact numerical answer, the specific Kf value for the Ag(CN)2- complex is required from a reliable source like a textbook or scientific database.

Answer:

i=1.62 .

Explanation:

Let, i be the Van't Hoff Factor.

Moles of benzamide,=[tex]\dfrac{Mass}{molar\ mass}=\dfrac{70.4}{121.14}=0.58\ mol.[/tex]

Molality of solution, m=[tex]\dfrac{moles  }{mass\ of\ solvent}=\dfrac{0.58}{0.85}=0.68\ molal.[/tex]

Now, we know

Depression in freezing point, [tex]\Delta T=i\times K_f\times m[/tex]  .....1

It is given that,

[tex]\Delta T=2.7^o C\\i=1 ( since\ it\ is\ non\ dissociable\ solutes)\\K_f  ( freezing\ constant)\\[/tex]

Putting all these values we get,

[tex]K_f=3.949\ C/m.[/tex]

Now, moles of ammonium chloride=[tex]\dfrac{70.4}{53.49}=1.316\ mol.[/tex]

molality =[tex]\dfrac{1.316}{0.85}=1.54 molal.\\\Delta T=9.9 .[/tex]

Putting all these values in eqn 1.

We get,

i=1.62 .

Hence, this is the required solution.

Answer:

(a) [tex]2HgO_{(s)}\rightarrow 2Hg_{(l)}+O_2_{(g)}[/tex]

(b) 0.726 g

Explanation:

(a)

The balanced chemical equation is shown below as:-

[tex]2HgO_{(s)}\rightarrow 2Hg_{(l)}+O_2_{(g)}[/tex]

(b)

Given that:

Temperature = 90 °C

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

T(K) = T( °C) + 273.15  

So,  

T₁ = (90 + 273.15) K = 363.15 K

V = 50.0 mL = 0.05 L ( 1 mL = 0.001 L )

Pressure = 1 atm

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:

1 atm × 0.05 L = n ×0.0821 L atm/ K mol  × 363.15 K

⇒n = 0.001677 mol

Thus, moles of [tex]O_2[/tex] = 0.001677 mol

According to the reaction shown above,

1 mole of [tex]O_2[/tex] is produced when 2 moles of mercury(II) oxide are reacted

0.001677 mole of [tex]O_2[/tex] is produced when [tex]2\times 0.001677[/tex] moles of mercury(II) oxide are reacted

Moles of mercury(II) oxide = 0.003354 moles

Molar mass of mercury(II) oxide = 216.59 g/mol

Mass of mercury(II) oxide = [tex]Moles\times Molar\ mass[/tex] =  [tex]0.003354\times 216.59\ g=0.726\ g[/tex]

0.726 g is the mass of mercury(II) oxide that must have reacted.

The balanced chemical equation for the decomposition of mercury(II) oxide into mercury and dioxygen is: 2HgO(s) → 2Hg(l) + O2(g). The mass of mercury(II) oxide that reacted to produce 50.0 ml of dioxygen gas at 90°C and 1 atm pressure is approximately 0.97 g.

1. The balanced chemical equation for the reaction is: 2HgO(s) → 2Hg(l) + O2(g).

2. First, we need to use the ideal gas law to calculate the number of moles of dioxygen gas produced. At standard temperature and pressure (STP), one mole of any gas occupies a volume of 22.4 L. Therefore, the number of moles of dioxygen is 50.0 ml / 22.4L/mol = 0.00223 moles. Next, we use the balanced chemical equation, which tells us that one mole of HgO produces a half mole of dioxygen gas. So, the amount of HgO reacted is 2 x 0.00223 moles = 0.00446 moles. Finally, we use the molar mass of HgO (216.6 g/mol) to calculate the mass of HgO reacted: 0.00446 moles x 216.6 g/mol = 0.97 g (to three significant digits).

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The question is incomplete, the complete question is attached below.

Answer : The volumes of stock solution and distilled water will be, 20 mL and 80 mL respectively.

Explanation : Given,

Mass of NaOH = 60 g

Volume of stock solution = 300 mL

Molar mass of NaOH = 40 g/mol

First we have to calculate the molarity of stock solution.

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

Now put all the given values in this formula, we get:

[tex]\text{Molarity}=\frac{60g\times 1000}{40g/mole\times 300mL}=5mole/L=5M[/tex]

Now we have to determine the volume of stock solution and distilled water mixed.

Formula used :

[tex]M_1V_1=M_2V_2[/tex]

where,

[tex]M_1\text{ and }V_1[/tex] are the molarity and volume of stock solution.

[tex]M_2\text{ and }V_2[/tex] are the molarity and volume of diluted solution.

From data (A) :

[tex]M_1=5M\\V_1=20mL\\M_2=1M\\V_2=?[/tex]

Putting values in above equation, we get:

[tex]5M\times 20mL=1M\times V_2\\\\V_2=100mL[/tex]

Volume of stock solution = 20 mL

Volume of distilled water = 100 mL - 20 mL = 80 mL

From data (B) :

[tex]M_1=5M\\V_1=20mL\\M_2=1M\\V_2=?[/tex]

Putting values in above equation, we get:

[tex]5M\times 20mL=1M\times V_2\\\\V_2=100mL[/tex]

Volume of stock solution = 20 mL

Volume of distilled water = 100 mL - 20 mL = 80 mL

From data (C) :

[tex]M_1=5M\\V_1=60mL\\M_2=1M\\V_2=?[/tex]

Putting values in above equation, we get:

[tex]5M\times 60mL=1M\times V_2\\\\V_2=300mL[/tex]

Volume of stock solution = 60 mL

Volume of distilled water = 300 mL - 60 mL = 240 mL

From data (D) :

[tex]M_1=5M\\V_1=20mL\\M_2=1M\\V_2=?[/tex]

Putting values in above equation, we get:

[tex]5M\times 60mL=1M\times V_2\\\\V_2=300mL[/tex]

Volume of stock solution = 60 mL

Volume of distilled water = 300 mL - 60 mL = 240 mL

From this we conclude that, when 20 mL stock solution and 80 mL distilled water mixed then it will result in a solution that is approximately 1 M NaOH.

Hence, the volumes of stock solution and distilled water will be, 20 mL and 80 mL respectively.

To make a 1 M NaOH solution, you would mix 2 L of the stock solution with 2 L of distilled water.

To prepare a 1 M NaOH solution using a stock solution of 60 g NaOH in 300 mL, we need to calculate the amount of stock solution and distilled water needed.

We can start by calculating the number of moles of NaOH in the stock solution:

moles = mass / molar mass => moles = 60 g / (22.99 g/mol + 16.00 g/mol + 1.01 g/mol) = 2 mol NaOH

Since we want a 1 M NaOH solution, we can use the formula: moles = concentration x volume

2 mol = 1 M x volume => volume = 2 mol / 1 M = 2 L

Therefore, we need to mix 2 L of the stock solution with 2 L of distilled water to obtain a 1 M NaOH solution.

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Answer:

Mole fraction: A =  8.70%, B = 37.00%, C =  19.60%, D = 34.80%

K = 6.86

Standard reaction free energy change: -4.77 kJ/mol

Explanation:

Let's do an equilibrium chart for the reaction:

2A + B ⇄ 3C + 2D

1.00  2.00  0     1.00    Initial

-2x     -x     +3x   +2x    Reacts (stoichiometry is 2:1:3:2)

1-2x    2-x    3x    1+2x  Equilibrium

3x = 0.9

x = 0.3 mol

Thus, the number of moles of each one at the equilibrium is:

A = 1 - 2*0.3 = 0.4 mol

B = 2 - 0.3 = 1.7 mol

C = 0.9 mol

D = 1 + 2*0.3 = 1.6 mol

The molar fraction is the mol of the component divided by the total number of moles (0.4 + 1.7 + 0.9 + 1.6 = 4.6 mol):

A = 0.4/4.6 = 0.087 = 8.70%

B = 1.7/4.6 = 0.37 = 37.00%

C = 0.9/4.6 = 0.196 = 19.60%

D = 1.6/4.6 = 0.348 = 34.80%

The equilibrium constant is the multiplication of the concentration of the products elevated by their coefficients, divided by the multiplication of the concentration of the reactants elevated by their coefficients. Because the volume remains constant, we can use the number of moles:

K = (nC³*nD²)/(nA²*nB)

K = (0.9³ * 1.6²)/(0.4² * 1.7)

K = 6.86

The standard reaction free energy change can be calculated by:

ΔG° = -RTlnK

Where R is the gas constant (8.314 J/mol.K), and T is the temperature (25°C = 298 K)

ΔG° = -8.314*298*ln6.86

ΔG° = -4772.8 J/mol

ΔG° = -4.77 kJ/mol

The Mole fraction: A is = 8.70%, B = 37.00%, C = 19.60%, D = 34.80%

K is = 6.86

Then, The Standard reaction free energy change: -4.77 kJ/mol

Now, Let's do an equilibrium chart for the reaction:

Then, 2A + B ⇄ 3C + 2D

1.00  2.00  0     1.00    Initial

After that, -2x     -x     +3x   +2x    Reacts (stoichiometry is 2:1:3:2)

1-2x    2-x    3x    1+2x  Equilibrium

Then, 3x = 0.9

x = 0.3 mol

Thus, When the number of moles of each one at the equilibrium is:

A is = 1 - 2*0.3 = 0.4 mol

B is = 2 - 0.3 = 1.7 mol

C is = 0.9 mol

D is = 1 + 2*0.3 = 1.6 mol

When The molar fraction is the mol of the component divided by the total number of moles (0.4 + 1.7 + 0.9 + 1.6 = 4.6 mol):

A is = 0.4/4.6 = 0.087 = 8.70%

B is = 1.7/4.6 = 0.37 = 37.00%

C is = 0.9/4.6 = 0.196 = 19.60%

D is = 1.6/4.6 = 0.348 = 34.80%

When The equilibrium constant is the multiplication of the concentration of the products elevated by their coefficients, Also divided by the multiplication of the concentration of the reactants elevated by their coefficients.

Because When the volume remains constant, we can use the number of moles:

K is = (nC³*nD²)/(nA²*nB)

K is = (0.9³ * 1.6²)/(0.4² * 1.7)

K is = 6.86

When The standard reaction free energy change can be calculated by:

Then, ΔG° = -RTlnK

Where R is the gas constant (8.314 J/mol.K), and T is the temperature (25°C = 298 K)

After that, ΔG° = -8.314*298*ln6.86

Then, ΔG° = -4772.8 J/mol

Therefore, ΔG° = -4.77 kJ/mol

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Answer:

Kp = 3.62 x 10 ^ -2  

Explanation:

The relationship between Kc and Kp is given as:

Kp =  Kc (RT) ^ Δn -------- (1)

Where:

Kp = Equilibrium Constant in terms of Molar concentrations = 61

Kc = Equilibrium Constant in terms of Partial Pressures

R = Gas Constant = 0.0821 litre.atm/mole.K

Δn = (Total No. of moles of products) - (Total No. of moles of reactants)

Δn = (2) - (3+1) = -2

T = Temperature (K) = 500 K

Substituting all values in equation (1), we get,

Kp = 61 x [ 0.0821 x 500 ] ^ (-2)

Kp = 3.62 x 10 ^ -2  

The study of chemicals and bonds is called chemistry. There are two types of elements are there are these are metals and nonmetals.

The correct answer is Kp = [tex]3.62 * 10^{-2[/tex]

The relationship between Kc and Kp is given as:

Kp =  Kc (RT)^Δn -------- (1)

Where:

Δn = (2) - (3+1) = -2

Substituting all values in equation (1), we get,

[tex]Kp = 61 * [ 0.0821 * 500 ]^{-2Kp = 3.62*10^{-2[/tex]

Hence, the correct answer is [tex]Kp = 3.62*10^{-2[/tex].

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Part A:

Therefore the correct statements are as follows.

With a few exceptions, the solubility of most solid solutes in water decreases as the solution temperature increases.(True)

The solubility of gases in water increLases with increasing temperature. (False).

The solubility of a gas in any solvent is increased as the partial pressure of the gas above the solvent increases. (True)

Substances with similar intermolecular attractive forces tend to be insoluble in one another. (False)

solubility.

Part B:

Therefore, hexane (C6H14) is the best solvent for nonpolar solutes.

Part A:

The first statement is true. Generally, the solubility of solid solutes in water tends to decrease with increasing temperature, although there are exceptions.

The second statement is false. The solubility of gases in water typically decreases with increasing temperature. This is described by Henry's law.

The third statement is true. This statement is consistent with Henry's law, which states that the solubility of a gas in a liquid is directly proportional to the partial pressure of the gas above the liquid

The fourth statement is false. Substances with similar intermolecular attractive forces tend to be more soluble in one another. Like dissolves like is a general principle in solubility.

Part B:

The best solvent for nonpolar solutes is the one with similar characteristics, meaning it should be nonpolar as well

Acetone is a polar solvent. It is not the best solvent for nonpolar solutes.

Ethanol is a polar solvent. It is not the best solvent for nonpolar solutes.

Hexane is a nonpolar solvent and is the best choice for nonpolar solutes.

Water is a polar solvent. It is not the best solvent for nonpolar solutes.

Therefore, hexane (C6H14) is the best solvent for nonpolar solutes.