an acid and base react to form a salt and water in a(n) reaction. group of answer choices oxidation ionization neutralization reduction dissociation

The Cannizzaro reaction is a redox reaction in which an aldehyde (or a ketone) is simultaneously oxidized and reduced. The reaction requires a strong base such as a hydroxide ion (OH-) to deprotonate one aldehyde molecule and a second aldehyde molecule is reduced by the resulting hydride ion (H-).

The given aldehyde with a trichloromethyl group attached to the carbonyl group will undergo the Cannizzaro reaction with OH- as the strong base. After the reaction, the mixture is acidified to protonate the products.

The trichloromethyl group (-CCl3) is a strongly electron-withdrawing group that deactivates the carbonyl group toward nucleophilic attack. Therefore, the expected products from the Cannizzaro reaction of the given aldehyde are:

A carboxylic acid, is the oxidized form of the aldehyde.

Alcohol, is the reduced form of the aldehyde.

The final products after acidification will be:

A carboxylic acid with a trichloromethyl group attached to the carbon.

Alcohol with a two-carbon chain attached to a hydroxyl group.

Therefore, the expected product(s) from the reaction, after acidifying the reaction mixture are:

A carboxylic acid with a trichloromethyl group attached to the carbon, and

Alcohol with a two-carbon chain attached to a hydroxyl group.

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the pH of a saturated solution of Mg(OH)₂ is approximately 9.6.

The solubility product constant (Ksp) expression for Mg(OH)₂ is:

Ksp = [Mg²⁺][OH⁻]²

At equilibrium, the concentrations of Mg²⁺ and OH⁻ in a saturated solution of Mg(OH)₂ are equal to the solubility of the salt, which we can assume to be "s" (in mol/L):

[Mg²⁺] = s

[OH⁻] = 2s

Substituting these expressions into the Ksp expression, we get:

Ksp = s * (2s)²

Ksp = 4s³

We can rearrange this expression to solve for the solubility of Mg(OH)₂:

s = [tex](Ksp/4)^(1/3)[/tex]

s =[tex][(6.3 × 10⁻¹⁰)/4]^(1/3)[/tex]

s = 5.94 × 10⁻⁵ M

Now we can use the concentration of OH⁻ in the saturated solution to calculate the pH of the solution:

pOH = -log[OH⁻]

pOH = -log(2s)

pOH = -log(2(5.94 × 10⁻⁵))

pOH = 4.4

pH + pOH = 14

pH = 14 - pOH

pH = 14 - 4.4

pH = 9.6

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The substance that is a strong electrolyte in an aqueous solution is KI.

Strong electrolytes are substances that completely dissociate into ions when dissolved in water, resulting in a solution that can conduct electricity well.

KI is an ionic compound, meaning it is made up of a metal (potassium) and a nonmetal (iodide) that are held together by ionic bonds. When KI is dissolved in water, it dissociates into its ions (K+ and I-), which allows the solution to conduct electricity effectively.

The other options are, b. C2H6 (ethane), c. C6H6 (benzene), d. C8H18 (octane), and e. CCl4 (carbon tetrachloride), are all non-electrolytes or weak electrolytes, as they do not dissociate into ions when dissolved in water, and their solutions cannot conduct electricity well.

These substances are composed of nonmetal atoms and are held together by covalent bonds.

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The functional groups that can be oxidized to give a carboxylic acid product are:

Carboxylic acids can be prepared by oxidizing a variety of functional groups. However, not all functional groups can be oxidized to give a carboxylic acid product. In general, primary alcohols and alkynes are two types of functional groups that can be oxidized to give a carboxylic acid.

Primary alcohols are oxidized to carboxylic acids via an intermediate aldehyde using a strong oxidizing agent such as potassium permanganate (KMnO₄) or chromium trioxide (CrO₃). The aldehyde is then further oxidized to the corresponding carboxylic acid.

Alkynes, on the other hand, can be oxidized directly to carboxylic acids using a strong oxidizing agent such as ozone (O₃) or potassium permanganate (KMnO₄). Phenols and secondary alcohols, however, cannot be directly oxidized to carboxylic acids. Phenols can be oxidized to quinones, while secondary alcohols can be oxidized to ketones, but these products are different from carboxylic acids.

The complete question is

Carboxylic acids are typically prepared using oxidation reactions. Which of the functional groups below can be oxidized to give a carboxylic acid product? Select all that apply.

A) Primary alcohols

B) Phenols

C) Secondary alcohols

D) Alkynes

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The reaction of cyclohexanol with phosphoric acid catalyst typically results in the formation of cyclohexene and water as the two products.

This is a dehydration reaction in which a molecule of water is eliminated from cyclohexanol to form cyclohexene. The chemical equation for this reaction can be written as:

C6H11OH + H3PO4 → C6H10 + H2O

where C6H11OH is cyclohexanol, H3PO4 is phosphoric acid, C6H10 is cyclohexene, and H2O is water.

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a) G = 1/(s+1) and H = (s-1)/(s+2): The system is stable for all values of K.
b) G(s) x H(s) = (s^2+2s+2)/(s^2(s-1)): The system is unstable for K > 2.

To determine the values of K for which the system is unstable, we need to use the characteristic equation and analyze the closed-loop transfer function.

a) For G = 1/(s+1) and H = (s-1)/(s+2), the closed-loop transfer function is given by:
T(s) = G(s) * H(s) / (1 + G(s) * H(s))
T(s) = (1/(s+1)) * ((s-1)/(s+2)) / (1 + (1/(s+1)) * ((s-1)/(s+2)))

To find the stability, we analyze the characteristic equation:
1 + G(s) * H(s) = 0
1 + (1/(s+1)) * ((s-1)/(s+2)) = 0

Solving this equation for s, we find that the system is stable for all values of K since there are no positive real parts of s.

b) For G(s) x H(s) = (s^2+2s+2)/(s^2(s-1)), the closed-loop transfer function is given by:
T(s) = G(s) * H(s) / (1 + G(s) * H(s))
T(s) = (s^2+2s+2)/(s^2(s-1)) / (1 + (s^2+2s+2)/(s^2(s-1)))

To find the stability, we analyze the characteristic equation:
1 + G(s) * H(s) = 0
1 + (s^2+2s+2)/(s^2(s-1)) = 0

Solving this equation for s, we find that the system is unstable when the poles of the denominator have a positive real part. In this case, the denominator is:
s^3 - s^2 + s^2 - s - 2 = s^3 - s - 2

Using the Routh-Hurwitz criterion, we find that the system is unstable when there is a change in the sign of the coefficients in the first column of the Routh array. For this cubic polynomial, the system is unstable when K > 2.

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The components of the ionic formula for nickel(III) sulfide are as follows: 1. The symbol for the element that forms the positive cation is Ni (nickel). 2. The symbol for the element that forms the negative anion is S (sulfur). 3. The subscript on the cation in the neutral formula is 2, which represents the number of nickel atoms in the formula. 4. The subscript on the anion in the neutral formula is 3, which represents the number of sulfur atoms in the formula.

To find the correct formula, we need to balance the charges of the ions. Nickel(III) indicates that the nickel cation has a charge of +3 (Ni3+), and sulfide has a charge of -2 (S2-). To create a neutral compound, we need two nickel ions (2 x +3 = +6) and three sulfide ions (3 x -2 = -6). Therefore, the balanced ionic formula for nickel(III) sulfide is Ni2S3.

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Answer: group 1A (or IA)

Explanation:

The percent yield of the reaction is 52.6%. The balanced chemical equation for the reaction between NaCl and NaHCO3 is:

2 NaCl + NaHCO3 → Na3CO3 + H2O + CO2

The molar mass of NaCl is 58.44 g/mol, and 9.97 g of NaCl corresponds to 9.97 g / 58.44 g/mol = 0.1705 mol of NaCl.

According to the equation, 1 mole of NaCl reacts with 1/2 mole of NaHCO3. So, the amount of NaHCO3 needed to react completely with 0.1705 mol of NaCl is:

0.1705 mol NaCl x (1/2) mol NaHCO3 = 0.08525 mol NaHCO3

The actual amount of NaHCO3 isolated is 3.77 g / 84.01 g/mol = 0.0449 mol NaHCO3.

The percent yield of the reaction is:

(actual yield / theoretical yield) x 100%

= (0.0449 mol / 0.08525 mol) x 100%

= 52.6%

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The molarity of acetic acid in the vinegar solution is 0.666 M.

to calculate the molarity of acetic acid ch3cooh in vinegar we need to understand the molecular weight of acetic acid which is 6005 gmol first we want to determine the mass of acetic acid present in 1000 grams 1 liter of vinegar we recognize that the density of vinegar is 10 gml so one thousand ml of vinegar would weigh a thousand g since the vinegar contains 40 acetic acid by using mass we can calculate the mass of acetic acid in one thousand g of vinegar as follows

Mass of acetic acid = 4.0% × 1000 g = 40 g

Now we can calculate the number of moles of acetic acid in 1000 g of vinegar:

Number of moles = mass / molecular weight

Number of moles = 40 g / 60.05 g/mol = 0.666 mol

Finally, we can calculate the molarity of acetic acid in the vinegar solution:

Molarity = number of moles / volume in liters

Molarity = 0.666 mol / 1 L = 0.666 M

Therefore, the molarity of acetic acid in the vinegar solution is 0.666 M.

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The frequency of electromagnetic radiation with wavelength 532 nm is [tex]5.64 x 10^14 s^-1[/tex]. The correct option is (a).

According to the equation c = λν,

where c is the speed of light,

λ is the wavelength,

and ν is the frequency of the wave.

We have been given,

λ = 532 nm = [tex]532 x 10^-9 m[/tex] (since, 1 nm = 10^-9 m)

[tex]c = 3.00 x 10^8 m/s[/tex]

By substituting the values in the equation,

[tex]c = λν3.00 x 10^17 s^-1 = (532 x 10^-9 m)[/tex]

[tex]νν = (3.00 x 10^17 s^-1)/(532 x 10^-9 m)ν = 5.64 x 10^14 s^-1[/tex]

Therefore, the frequency of electromagnetic radiation with wavelength 532 nm is [tex]5.64 x 10^14 s^-1[/tex]. The correct option is (a).

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Q is the heat transferred, m is the mass, C is the specific

PV = nRT

where P is the pressure, V is the volume, n is the number of moles, R is the gas constant, and T is the temperature. Since we are given the pressure and temperature of the system, we need to find the volume and number of moles of gas.

First, we can find the number of moles of hydrogen that dissociates from the H₂ gas. If the mole fraction of H is 5%, then the mole fraction of H₂ is 95%. Let's assume we start with 1 mole of H₂ . Then, we have:

0.05 moles of H

0.95 moles of H2

Some of the H₂ gas dissociates to H, so let's say x moles of H₂ dissociate. Then, we have:

(0.05 + x) moles of H

(0.95 - x) moles of H₂

Since we are assuming that the only final product is H₂, we know that all of the H atoms will recombine to form H₂:

(0.05 + x - 2x) moles of H₂

(0.95 - x) moles of H2

Simplifying this, we get:

(0.05 - x) moles of H₂

(0.95 - x) moles of H₂

Since we started with 1 mole of H₂, we know that the total number of moles is:

(0.05 - x) + (0.95 - x) = 1 - 2x

Now, we can use the ideal gas law to find the volume of gas at the final pressure and temperature. Since we know the number of moles of gas and the temperature, we just need to find the volume. We can rearrange the ideal gas law to get:

V = (nRT)/P

At 298 K and 1 atm, the gas constant R is 0.08206 L·atm/(mol·K). So, for the initial conditions, we have:

V_initial = (1 mol * 0.08206 L·atm/(mol·K) * 298 K) / 1 atm = 24.4658 L

Now, we can use the mole fractions and the number of moles to find the final volume of gas. We have:

(0.05 - x) moles of H₂

(0.95 - x) moles of H₂

The total number of moles is 1 - 2x, so we have:

(0.05 - x)/(1 - 2x) moles of H₂

(0.95 - x)/(1 - 2x) moles of H₂

Using the ideal gas law again, we can find the final volume:

V final = [(0.05 - x)/(1 - 2x) + (0.95 - x)/(1 - 2x)] * (0.08206 L·atm/(mol·K) * T) / 1.1 atm

Simplifying this, we get:

V final = [1 - 2x] * (0.08206 L·atm/(mol·K) * T) / (1.1 atm)

Now, we need to use the fact that the specific heat is constant to find the final temperature. We can use the formula:

Q = mCΔT

where Q is the heat transferred, m is the mass, C is the specific

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Answer:

To find the empirical formula of a compound, we need to determine the simplest whole number ratio of the atoms present in the compound. Let's assume we have 100 g of this compound. Then we have 79.3 g of tungsten and 20.7 g of oxygen. We can convert these masses to moles by dividing by their respective atomic masses: Number of moles of W = 79.3 g / 183.84 g/mol = 0.431 mol Number of moles of O = 20.7 g / 16.00 g/mol = 1.294 mol To get the simplest whole number ratio of the atoms, we need to divide both of these values by the smallest value, which is 0.431 mol: 0.431 mol W / 0.431 mol = 1 1.

The dissociation equation for weak acid can be written as HB + H2O ⇌ H3O+ + B- and there will be no change in Poh at all.

The correct answer is option D.

When a stress is given to an equilibrium system, the system will move to  offset the stress and reestablish equilibrium, according to Le Chatelier's principle.    In this situation, adding a  bitsy  volume of the weak base B- to the  result will raise the  attention of B- in the  result.

This indicates that the equilibrium will move to the left to  neutralize the rise in B-  attention, performing in a drop in H3O ion  attention.   Because pH measures the  attention of H3O ions in a  result, a reduction in H3O  attention will affect in a rise in pH. As a result, the answer is pH slightly rises; pOH slightly falls.

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HF, or hydrogen fluoride, is likely to dissolve in water (H2O), a polar solvent.

This is due to the principle "like dissolves like," which means that polar solvents tend to dissolve polar solutes, and nonpolar solvents tend to dissolve nonpolar solutes.  HF is a polar molecule because of the difference in electronegativity between hydrogen (H) and fluorine (F) atoms, resulting in a polar covalent bond. This bond creates a dipole moment, causing the HF molecule to have a partial positive charge on the hydrogen end and a partial negative charge on the fluorine end.

Water, as a polar solvent, has a similar dipole moment due to the difference in electronegativity between oxygen (O) and hydrogen (H) atoms. The partial positive charges on the hydrogen atoms in water molecules can interact with the partial negative charges on the fluorine atoms in HF molecules, forming hydrogen bonds. These interactions between polar molecules make it possible for HF to dissolve in water.


In conclusion, HF is likely to dissolve in water (H2O) because both are polar substances that can form hydrogen bonds, following the "like dissolves like" principle.

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The answer is the athletic trainer will need 15 quarts of the 35% carbohydrate solution to achieve a final 27% carbohydrate-electrolyte solution when mixed with 10 quarts of the 15% solution.

To find out how many quarts of the 35% solution the athletic trainer needs, we can set up a weighted average equation using the given information:
27% = (15% * 10 quarts + 35% * x quarts) / (10 quarts + x quarts)
Step 1: Convert the percentages to decimals by dividing by 100.
0.27 = (0.15 * 10 + 0.35 * x) / (10 + x)
Step 2: Distribute the denominator to both sides of the equation.
0.27(10 + x) = 0.15 * 10 + 0.35 * x
Step 3: Expand the equation.
2.7 + 0.27x = 1.5 + 0.35x
Step 4: Subtract 0.27x from both sides.
2.7 - 1.5 = 0.35x - 0.27x
Step 5: Simplify the equation.
1.2 = 0.08x
Step 6: Divide by 0.08 to find x.
x = 1.2 / 0.08
x = 15
The athletic trainer will need 15 quarts of the 35% carbohydrate solution to achieve a final 27% carbohydrate-electrolyte solution when mixed with 10 quarts of the 15% solution.

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There are around 0.118 grammes of hydrogen in the gas mixture. We can use the ideal gas law to solve this problem.

The ideal gas law relates the pressure, volume, number of moles, and temperature of a gas:

PV = nRT

where P is the pressure, V is the volume, n is the number of moles, R is the gas constant, and T is the temperature.

First, we need to calculate the partial pressure of hydrogen in the gas mixture. The total pressure of the mixture is given as 660 mmHg, and the partial pressure of oxygen is given as 425 mmHg. Therefore, the partial pressure of hydrogen is:

P(H2) = P(total) - P(O2) = 660 mmHg - 425 mmHg = 235 mmHg

Next, we need to calculate the number of moles of oxygen in the mixture. We are given that the mixture contains 3.75 grams of oxygen. The molar mass of oxygen is 16.00 g/mol. Therefore, the number of moles of oxygen is:

n(O2) = mass/molar mass = 3.75 g/16.00 g/mol = 0.2344 mol

Since hydrogen and oxygen react in a 2:1 mole ratio to form water, the number of moles of hydrogen in the mixture is:

n(H2) = 0.5 x n(O2) = 0.5 x 0.2344 mol = 0.1172 mol

Finally, we can calculate the mass of hydrogen in the mixture using the number of moles and the molar mass of hydrogen, which is 1.008 g/mol:

mass(H2) = n(H2) x molar mass(H2) = 0.1172 mol x 1.008 g/mol = 0.118 g

Therefore, the gas mixture contains approximately 0.118 grams of hydrogen.

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Answer:

The balanced chemical equation for the formation of ammonia gas by the reaction between nitrogen gas an hydrogen gas is given. N2+3H→2NH3.

Explanation:

I hope it's correct

Answer:

The balanced equation for the reaction between nitrogen gas and hydrogen gas to produce ammonia gas is: N2(g) + 3H2(g) -> 2NH3(g) This equation shows that one molecule of nitrogen gas reacts with three molecules of hydrogen gas to produce two molecules of ammonia gas. The coefficients in the balanced equation indicate that the number of atoms of each element is conserved in the reaction.

B. N2 is the limiting reactant. The reactant that has the smaller number of moles in the balanced equation is the limiting reagent.

What is Limiting Reagent?

The limiting reagent is the reactant that is entirely consumed when the reaction goes to completion, and thus limits the amount of product that can be formed. The other reactants that are present in excess are not entirely consumed and do not limit the amount of product that can be formed.

To determine the limiting reactant, we need to calculate the amount of product that can be produced from each reactant and then choose the reactant that produces the smallest amount of product.

The balanced equation for the reaction is:

N2 + 3H2 → 2NH3

The molar mass of N2 is 28 g/mol, and the molar mass of H2 is 2 g/mol. Using these values, we can calculate the number of moles of each reactant:

Number of moles of N2 = 3.5 g / 28 g/mol = 0.125 mol

Number of moles of H2 = 3.5 g / 2 g/mol = 1.75 mol

According to the balanced equation, 1 mole of N2 reacts with 3 moles of H2 to produce 2 moles of NH3. So, the number of moles of NH3 that can be produced from each reactant is:

Moles of NH3 produced from N2 = 0.125 mol × (2 mol NH3 / 1 mol N2) = 0.25 mol NH3

Moles of NH3 produced from H2 = 1.75 mol × (2 mol NH3 / 3 mol H2) = 1.17 mol NH3

Therefore, the limiting reactant is N2, since it produces a smaller amount of NH3 compared to H2.

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Clearing forests.

According to the text, humans impact the population growth of other ecosystems by clearing forests.

This is because forests are important habitats for many different species, and when they are destroyed or degraded, those species are forced to relocate or die off.

Additionally, forests are also important carbon sinks, which means they help to absorb and store carbon from the atmosphere. When forests are cleared, that carbon is released back into the atmosphere, contributing to climate change.

Therefore, forest clearing by humans can have significant impacts on the population growth of other ecosystems

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The percent yield of HF is 15.2%.

We need to first determine which is the limiting reagent in the reaction. Let's assume that H2O is the limiting reagent. The balanced chemical equation tells us that 1 mole of H2O reacts with 1 mole of HF. The molar mass of H2O is 18 g/mol and the molar mass of HF is 20 g/mol.

4 grams of H2O is equal to 4/18 = 0.22 moles of H2O.

From the balanced equation, we know that 1 mole of H2O reacts with 1 mole of HF. Therefore, 0.22 moles of H2O should produce 0.22 moles of HF.

The theoretical yield of HF is therefore:

Theoretical yield = 0.22 moles x 20 g/mol = 4.4 grams

The actual yield of HF is given as 0.67 grams.

The percent yield is then calculated as:

Percent yield = (actual yield / theoretical yield) x 100%

Percent yield = (0.67 / 4.4) x 100% = 15.2%

The percent yield of HF is 15.2%.

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Adding nitric acid to sulfuric acid could cause the concentrated acid to boil, potentially leading to safety hazards.

The result of adding nitric corrosive to sulfuric corrosive rather than the opposite way around is that the exothermic response could make the concentrated corrosive bubble, possibly prompting security risks. Nitration responses regularly require a combination of sulfuric corrosive and nitric corrosive as the nitrating specialist, as the sulfuric corrosive goes about as an impetus to produce the electrophilic nitrating species. Assuming that nitric corrosive were added first, the concentrated sulfuric corrosive might actually bubble, which could bring about the deficiency of reactants, decline in yield, and posture wellbeing perils because of the potential for arrival of harmful exhaust and additionally blast.

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H2]eq = 1.1 × 10^−4 M.

The given equilibrium reaction is:

2H₂S(g) ⇌ 2H₂(g) + S₂(g)

The equilibrium constant, Kc, for the reaction at 700 °C is 9.0 × 10^−8.

We are given the initial number of moles of H2S as 0.700 moles in a 0.500 L flask. Let's assume that x moles of H2S is consumed at equilibrium, and therefore the equilibrium concentrations of H2 and S2 are both 2x, based on the stoichiometry of the reaction.

The expression for the equilibrium constant, Kc, is:

Kc = [H2]^2[S2]/[H2S]^2

At equilibrium, the concentration of H₂S will be (0.700 - x)/0.500 M, and the concentration of H₂ and S₂ will be 2x/0.500 M. Substituting these values in the expression for Kc and solving for x gives:

9.0 × 10^−8 = (2x/0.500 M)^2 / ((0.700 - x)/0.500 M)^2

Solving for x gives x = 5.6 × 10^−5 M.

Therefore, the equilibrium concentrations of H₂ and S₂ are both 2x = 1.1 × 10^−4 M, and the equilibrium concentration of H₂S is (0.700 - x)/0.500 M = 0.699 M.

Thus, the concentration of H2 at equilibrium is 1.1 × 10^−4 M, rounded to the correct number of significant figures.

Therefore, [H2]eq = 1.1 × 10^−4 M.

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The n-factor of [tex]Cl_2[/tex] is 2.5 in this reaction.

To calculate the n-factor of [tex]Cl_2[/tex] in the given reaction:

[tex]Cl_2[/tex] → [tex]ClO^{3-} + Cl^-[/tex]

We need to identify the oxidation states of Cl in both reactants and products.

In [tex]Cl_2[/tex], since it is a diatomic molecule, both Cl atoms have the same oxidation state, which we can represent as 0.

In [tex]ClO^{3-}[/tex], the oxidation state of Cl is +5, since the oxidation state of O is -2 and there are three O atoms bonded to Cl.

In [tex]Cl^-[/tex], the oxidation state of Cl is -1.

Now, we can calculate the n-factor of [tex]Cl_2[/tex] using the following formula:

n-factor = change in oxidation state / number of electrons transferred

In the given reaction, [tex]Cl_2[/tex] is oxidized to [tex]ClO^{3-}[/tex] and reduced to [tex]Cl^-[/tex]. The change in oxidation state of Cl is:

+5 (in [tex]ClO^{3-}[/tex]) - 0 (in [tex]Cl_2[/tex]) = +5

-1 (in [tex]Cl^-[/tex]) - 0 (in [tex]Cl_2[/tex]) = -1

The maximum of these values is +5, so we will use this value for the n-factor.

The number of electrons transferred is equal to the number of Cl atoms in the balanced equation, which is 2.

The n-factor of [tex]Cl_2[/tex] in the given reaction is:

n-factor = 5/2 = 2.5

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The pressure exerted by 0.750 mol of gas at a temperature of 0.00ºC and a volume of 5.00 L is 101325 Pa.

To calculate the pressure exerted by a gas, we can use the ideal gas law:

PV = nRT

where P is the pressure, V is the volume, n is the number of moles of gas, R is the gas constant, and T is the temperature in kelvin.

First, we need to convert the temperature from Celsius to Kelvin:

T = 0.00ºC + 273.15 = 273.15 K

Next, we can plug in the given values and solve for the pressure:

P = nRT/V

P = (0.750 mol)(8.314 J/mol·K)(273.15 K)/(5.00 L)

P = 101325 Pa

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Answer:

true.

Explanation:

Earth's lithosphere is broken into plates

Plate tectonics explains how plates move and interact with each other

Plate tectonics theory helps to understand earthquakes, volcanic eruptions, and the formation of mountain ranges.

Plate tectonics explains the process of subduction,

Subduction is where one plate is forced under another

One of the plate eventually melts into the mantle

Plate melting recycles Earth's crustal material

When Q < K (the reaction quotient is less than the equilibrium constant), the reaction will proceed in the forward direction to reach equilibrium. The correct response is option C

When Q = K (the reaction quotient is equal to the equilibrium constant), the reaction is at equilibrium and the concentrations of the reactants and products will remain constant over time.

At equilibrium, the rates of the forward and reverse reactions are equal and there is no net change in the concentrations of the reactants and products.

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Answer: d

Explanation:

The equilibrium constant, Kc, is 2.93 x 10^-4.

The balanced equation for the reaction is;

NH₃ (g) ⇌ ½ N₂ (g) + 3/2 H₂ (g)

The equilibrium expression for the reaction is;

Kc = ([N₂[tex]]^{(1/2)}[/tex] [H₂[tex]]^{(3/2)}[/tex]) / [NH₃]

We are given the initial and equilibrium concentrations of NH₃. The change in NH₃ concentration is;

Δ[NH₃] = [NH₃]equilibrium - [NH₃]initial = 7.91 g/L - 15 g/L = -7.09 g/L

Since NH₃ is a reactant, its concentration decreases at equilibrium. Therefore, we can assume that the reaction proceeds in the forward direction, and that the concentration of N₂ and H₂ at equilibrium are both equal to x.

Using the ideal gas law, we can find the concentration of NH3 at equilibrium; PV = nRT

n/V = P/RT

[n(NH₃)]equilibrium = (P/RT) x V

[n(NH₃)]initial = (P/RT) x V

Substituting the values, we get;

[n(NH₃)]equilibrium = (1 atm / (0.0821 L·atm/mol·K x 298 K)) x 1.4 L = 0.0652 mol/L

[n(NH₃)]initial = (1 atm / (0.0821 L·atm/mol·K x 298 K)) x 1.4 L = 0.0652 mol/L

Using the balanced equation, we can relate the concentration of NH3 to the concentrations of N₂ and H₂.

[NH₃]initial = 0.0652 mol/L

[N₂]equilibrium = [H₂]equilibrium = x

[N₂]initial = [H₂]initial = 0 mol/L

Substituting into the equilibrium expression, we get;

Kc = ([N₂[tex]]^{(1/2)}[/tex] [H₂[tex]]^{(3/2)}[/tex]) / [NH₃]

Kc = ([tex]X^{(1/2)}[/tex] [tex]X^{(3/2)}[/tex]) / (0.0652 - 7.09x10⁻³)

Kc = 2.93 x 10⁻⁴

To know more about equilibrium constant here

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