identify bronsted-lowry conjugate acid-base pair
A. NH3 , NH4+
B. H30+ , OH-
C. HCl , HBr
D. ClO4- , ClO3-

a. The balanced equation from the line notation Ag(s) Ag+(aq) || Cd2+(aq) Cd(s) is given below;Ag(s) + Cd2+(aq) → Ag+(aq) + Cd(s)The given line notation represents an electrochemical cell where two half-cells are separated by a salt bridge.

The anode half-cell is on the left side of the double vertical line notation while the cathode half-cell is on the right side of the double vertical line notation. In the anode, oxidation takes place, and the electrode is considered negative, whereas in the cathode, reduction takes place, and the electrode is considered positive.b. The balanced equation from the line notation Pb(s) Pb2+(aq) || MnO2(aq) | Mn2+(aq) | Pt(s) is given below;Pb(s) + MnO2(s) + 4 H+(aq) → Pb2+(aq) + Mn2+(aq) + 2 H2O(l)The given line notation represents an electrochemical cell where two half-cells are separated by a salt bridge. The anode half-cell is on the left side of the double vertical line notation while the cathode half-cell is on the right side of the double vertical line notation. In the anode, oxidation takes place, and the electrode is considered negative, whereas in the cathode, reduction takes place, and the electrode is considered positive.

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a) 2/1L t will be the volume of the balloon if 0.20 moles of gas are released

b) 1.02L is the volume (in L) at which the balloon bursts.

What is ideal gas law ?

The macroscopic characteristics of ideal gases are related by the ideal gas law (PV = nRT). A gas is considered to be ideal if its particles (a) do not interact with one another and (b) occupy no space (have no volume).

There are four guiding principles that determine if a gas is "ideal": The volume of the gas particles is quite small. The gas particles are of similar size and do not interact with one another through intermolecular forces (attraction or repulsion). The random motion of the gas particles is consistent with Newton's Laws of Motion.

a) PV = nRT

R = 0.082 atm.L/K.mol

V1 = 1.50 L

n1 = 3.00 mol

T1 = 25°C ≅ 298 K

P1 = (RT1n1)/(V1) = (0.082 *298 *4.00 )/(2)

P1 = 48.8 atm

If pressure and temperature remain constant:

T2 = T1 = 298 K

P2 = P1 = 48.8 atm

n2 = 0.20 mol + 4.00 mol = 4.20 mol

V2 = (RT2n2)/P2

V2 = (0.082 * 298 *4.20)/(48.8)

V2 = 2.1 L

b) V1 = 35.0 L

T1 = 20.0 °C = 293K.

P1 = 1.00 atm

P2 = 26.0 torr

T2 = -50.0 °C = 223K

P1V1/T1 = P2V2/T2

1*35/293 = 26*V2/223

V2 = 1.02L

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In a balanced exothermic reaction, the heat term appears on the product side of the chemical equation.

In a balanced chemical equation, the reactants and products are represented using chemical formulas and coefficients. The heat term, which represents the heat released or absorbed during the reaction, is often included as a separate term in the equation.

For an exothermic reaction, which releases heat to the surroundings, the heat term appears on the product side of the equation. It is typically denoted as a positive value since it represents the heat being released. The heat term is often written as "ΔH" or "heat" and may be accompanied by the corresponding value indicating the heat change.

The inclusion of the heat term allows us to account for the energy changes that occur during a chemical reaction. It provides information about the heat flow associated with the reaction and helps in understanding the thermodynamics of the process.

Therefore, in a balanced exothermic reaction, the heat term appears on the product side to indicate the heat being released.

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The molar solubility of strontium fluoride is 1.11 × 10⁻³ M.

The Ksp value for strontium fluoride, SrF2, is 2.6 × 10⁻⁹. The molar solubility of strontium fluoride, we use the equation for the solubility product constant (Ksp). Ksp = [Sr²⁺] [F⁻]² Ksp is the product of the concentrations of the dissolved ions, each raised to the power of its stoichiometric coefficient. The balanced equation for the dissociation of strontium fluoride is as follows:SrF₂(s) ⇌ Sr²⁺(aq) + 2 F⁻(aq)The molar solubility of strontium fluoride is represented by "s," so we will substitute "s" into the concentrations of the dissolved ions as shown below:Ksp = [Sr²⁺] [F⁻]²2.6 × 10⁻⁹ = s × (2s)²= 4s³Solving for "s" gives us the molar solubility of strontium fluoride:s = ∛(2.6 × 10⁻⁹ / 4)= 1.11 × 10⁻³ MTherefore, the molar solubility of strontium fluoride is 1.11 × 10⁻³ M.

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1) The percentage ionization is 3.2%

2) The percentage ionization is 21.9%

The percent ionization is primarily used for weak acids and bases, where only a fraction of the compound dissociates. Strong acids and bases, on the other hand, completely dissociate in solution, so the percent ionization is 100%.

We have to use the formula;

α = √ka/C * 100

Where;

α = percent ionization

Ka = The dissociation constant

C = concentration

Then;

α = √1.4×10^−4/0.135 * 100

= 3.2 %

2) α = √6.5×10^−3/0.135 * 100

= 21.9%

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pH of the solution using the Henderson-Hasselbalch equationpH = pKa + log([base]/[acid + base])pKa = 10.33 + log([CO2]/[HCO3-])For NaHCO3, [CO2] = 0.0004 M and [HCO3-] = 0.2025 MFor Na2CO3, [CO2] = 0.0004 M and [HCO3-] = 0.0 pH = 10.33 + log((0.2025)/(0.0004 + 0.2025))pH = 9.25 (Answer)

Percent ionization of 0.120 MM lactic acid(Ka=1.4×10−4): Percent ionization refers to the degree of ionization of a weak electrolyte in solution. It is calculated by taking the ratio of the concentration of ionized species to the initial concentration of the compound multiplied by 100.The formula for percent ionization is:% Ionization = (concentration of H+ ions / initial concentration of lactic acid) × 100Given that, concentration of lactic acid = 0.120 MMInitial concentration of lactic acid = 0.120 MMConcentration of H+ ions = x (as lactic acid is weak electrolyte and it is assumed that x moles of lactic acid ionizes into x moles of H+ and x moles of lactate)Ka of lactic acid = 1.4×10−4Kw/Ka = 1.0×10-14/1.4×10-4 = 7.14×10^-11Now, write the expression for Ka of lactic acid and solve for x Ka = [H+][Lactate]/[Lactic acid]1.4×10^-4 = x² / (0.120 – x)x = 3.87 × 10^-3[Moles of H+]/[Initial moles of lactic acid] x 100 = Percent ionization = 3.87×10^-3/0.120 × 100 = 3.22% (Answer)B) Calculation of percent ionization of 0.120 M lactic acid in a solution containing 8.0×10−3 M M sodium lactate:Given that, concentration of lactic acid = 0.120 MConcentration of sodium lactate = 8.0×10−3 MInitial concentration of lactic acid = 0.120 – 8.0×10−3 = 0.112 MConcentration of H+ ions = x (as lactic acid is weak electrolyte and it is assumed that x moles of lactic acid ionizes into x moles of H+ and x moles of lactate)Ka of lactic acid = 1.4×10−4Kw/Ka = 1.0×10-14/1.4×10-4 = 7.14×10^-11Now, write the expression for Ka of lactic acid and solve for xKa = [H+][Lactate]/[Lactic acid]1.4×10^-4 = x² / (0.112 – x)x = 2.73 × 10^-3[Moles of H+]/[Initial moles of lactic acid] x 100 = Percent ionization = 2.73×10^-3/0.120 × 100 = 2.28% (Answer)C) Calculation of pH of a buffer that is 0.120 MM in NaHCO3 and 0.280 M in Na2CO3:Given that, concentration of NaHCO3 = 0.120 MMConcentration of Na2CO3 = 0.280 MNow, calculate the pKa of the H2CO3/HCO3- buffer system:pKa = pH + log([HCO3-]/[H2CO3])pKa = 10.33 + log(0.280/0.120) = 10.72[HCO3-]/[H2CO3] = antilog (pKa - pH) = antilog (10.72 - 10.33) = 3.65Buffer capacity (ß) = (Change in base/Change in pH) ß = (0.120 × (3.65 + 1))/(1.5 × (10^-5)) = 11680pH = pKa + log([Salt]/[Acid])pH = 10.72 + log(0.280/0.120) = 10.97Answer: 10.97D) Calculation of pH of a solution formed by mixing 65 mL of a solution that is 0.24 M in NaHCO3 with 75 mL of a solution that is 0.17 M in Na2CO3:Step 1: Calculation of moles of NaHCO3 in the first solutionMoles of NaHCO3 = Molarity × Volume (L)Moles of NaHCO3 = 0.24 × (65/1000) = 0.0156Step 2: Calculation of moles of Na2CO3 in the second solutionMoles of Na2CO3 = Molarity × Volume (L)Moles of Na2CO3 = 0.17 × (75/1000) = 0.01275Step 3: Calculation of total moles of HCO3- and CO32-Total moles of HCO3- and CO32- = Moles of NaHCO3 + Moles of Na2CO3Total moles of HCO3- and CO32- = 0.0156 + 0.01275 = 0.02835Step 4: Calculation of new concentration of HCO3- and CO32- after the two solutions are mixedNew concentration of HCO3- and CO32- = Total moles of HCO3- and CO32- / Total volume (L)Total volume = 65/1000 + 75/1000 = 0.14 LNew concentration of HCO3- and CO32- = 0.02835 / 0.14 = 0.2025 MStep 5: Calculation of pH of the solution using the Henderson-Hasselbalch equationpH = pKa + log([base]/[acid + base])pKa = 10.33 + log([CO2]/[HCO3-])For NaHCO3, [CO2] = 0.0004 M and [HCO3-] = 0.2025 MFor Na2CO3, [CO2] = 0.0004 M and [HCO3-] = 0.0 pH = 10.33 + log((0.2025)/(0.0004 + 0.2025))pH = 9.25 (Answer)

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Answer:

Yes I agree

Explanation:

Answer:

hmm that is a very tricky question. probably from over flooding with water

Answer:

is this a poem or something?

During an elimination reaction in organic chemistry, a new double bond (π bond) is formed, resulting in the creation of an alkene functional group. This process involves the removal of a leaving group and the adjacent hydrogen atom from a molecule, resulting in the formation of a double bond between the two adjacent carbon atoms.

In elimination reactions, a strong base or acid is often used to abstract the proton from the adjacent carbon atom, generating a carbanion intermediate. The leaving group is then expelled from the molecule, and the carbanion intermediate undergoes a rearrangement to form a more stable carbocation. Finally, the base or another molecule acts as a nucleophile, capturing a proton from the carbocation to form the double bond. This newly formed double bond represents the alkene functional group and is characteristic of elimination reactions in organic chemistry.

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a) Balanced equation:

2Na2CO3 + 2HNO3 -> 2NaNO3 + H2O + CO2

b) Pathway:

To determine the mass of Na2CO3 required to react with 150 mL of 0.15 M HNO3, we need to follow these steps:

Write the balanced equation to determine the stoichiometry between Na2CO3 and HNO3.

Convert the volume of HNO3 to moles using its molarity.

Use the stoichiometry from the balanced equation to determine the moles of Na2CO3 required.

Convert the moles of Na2CO3 to grams using its molar mass.

c) Calculation:

Given:

Volume of HNO3 = 150 mL = 0.150 L

Molarity of HNO3 = 0.15 M

Step 1: Balanced equation:

2Na2CO3 + 2HNO3 -> 2NaNO3 + H2O + CO2

Step 2: Convert volume of HNO3 to moles:

Moles of HNO3 = Volume (L) × Molarity

= 0.150 L × 0.15 M

= 0.0225 moles of HNO3

Step 3: Use stoichiometry to find moles of Na2CO3:

From the balanced equation, we can see that 2 moles of Na2CO3 react with 2 moles of HNO3.

Therefore, the moles of Na2CO3 required = 0.0225 moles of HNO3

Step 4: Convert moles of Na2CO3 to grams:

Molar mass of Na2CO3 = (2 × atomic mass of Na) + atomic mass of C + (3 × atomic mass of O)

= (2 × 22.99 g/mol) + 12.01 g/mol + (3 × 16.00 g/mol)

= 105.99 g/mol

Mass of Na2CO3 = Moles × Molar mass

= 0.0225 moles × 105.99 g/mol

= 2.384 g

d) Calculation:

The mass of Na2CO3 required to completely react with 150 mL of 0.15 M HNO3 is 2.384 grams.

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Answer:

Rice is grown in rainy regions.

The river flooded the whole region.

He explored the region around the South Pole.

Explanation:

Hope this helped!

The mole fraction of hydrochloric acid (HCl) in the solution is approximately 0.460.

Molality (m) is defined as the number of moles of solute per kilogram of solvent. In this case, the molality of HCl is given as 8.56 mol/kg. Mole fraction (X) is defined as the ratio of the moles of a component to the total moles of all components in the solution.

To calculate the mole fraction of HCl, we need to know the total number of moles in the solution. However, the information provided only gives the molality of HCl, which provides the moles of HCl per kilogram of solvent, but not the total moles of the solution. Without the total moles of the solution, it is not possible to directly calculate the mole fraction of HCl. Therefore, based on the given information, it is not possible to determine the mole fraction of HCl in the solution accurately.

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Answer:

15-21 is 6

Explanation:

Answer:

[OH⁻] = 0.0627M

pOH = 1.20

pH = 12.8

[H⁺] = 1.59x10⁻¹³M

Explanation:

To solve this question we must, as first, find the molarity of Al(OH)₃ in the solution -Molar mass Al(OH)₃: 78.00g/mol-:

0.300g * (1mol/ 78.00g) = 3.846x10⁻³ moles

In 184mL = 0.184L:

3.846x10⁻³ moles / 0.184L = 0.0209M Al(OH)₃. Three times this molarity = [OH⁻]:

[OH⁻] = 0.0209M * 3

pOH = -log [OH⁻] =

pH = 14 - pOH

And [H⁺] = 10^-pH

The pressure of the gas is 5.52 atm

Mass of CO2 (dry ice) = 9.70 gVolume of the balloon after the solid CO2 sublimes = 1.00 LTemperature of the balloon = 298 KWe need to find out the pressure of the gas. The molar mass of CO2 is:Molecular mass of C = 12.01 g/molMolecular mass of O = 15.99 g/molMolecular mass of CO2 = 12.01 + (2 × 15.99) = 44.01 g/molNow, the number of moles of CO2 = mass/molar mass= 9.70/44.01 = 0.220 molThe Ideal Gas Law is represented by the formula PV = nRT,where P = pressureV = volume of the gasn = number of moles of the gasR = gas constant = 0.0821 L atm/(mol K)T = temperature of the gasNow substituting the values in the Ideal Gas Law,we getP = nRT / V= (0.220 mol × 0.0821 L atm/(mol K) × 298 K) / 1.00 LP = 5.52 atmTherefore, the pressure of the gas is 5.52 atm.

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Answer:

There are three main types of recycling: primary, secondary, and tertiary.Single-stream recycling is a system in which all recyclables, including newspaper, cardboard, plastic, aluminum, junk mail, etc., are placed in a single bin or cart for recycling. ... While collections costs are lower with a single stream system, processing costs are much higher.

Explanation:

Hope it helps u

FOLLOW MY ACCOUNT PLS PLS

A. Sucrose (aq) + H₂O(l) ⇌ Glucose (aq) + Fructose (aq) (k = 1.4 × 10⁵)

B. NH₃ (aq) + H₂O(l) ⇌ NH⁴⁺(aq) + OH⁻(aq) (k = 1.6 × 10⁻¹⁵)

C. Fe₂O₃(s) + 3 CO(g) ⇌ 2 Fe(s) + 3 CO₂(g) (k at 727°C = 24)

A. Sucrose (aq) + H₂O(l) ⇌ Glucose (aq) + Fructose (aq) (k = 1.4 × 10⁵)

The high value of the equilibrium constant (k = 1.4 × 10⁵) indicates that the reaction strongly favors the products (glucose and fructose) at equilibrium. This means that at equilibrium, there will be a high concentration of glucose and fructose compared to sucrose and water.

B. NH₃ (aq) + H₂O(l) ⇌ NH₄⁺(aq) + OH⁻(aq) (k = 1.6 × 10⁻⁵)

The low value of the equilibrium constant (k = 1.6 × 10⁻⁵) indicates that the reaction favors the reactants (NH₃ and H₂O) at equilibrium. This means that at equilibrium, there will be a higher concentration of NH₃ and H₂O compared to NH₄⁺ and OH⁻.

C. Fe2O₃(s) + 3 CO(g) ⇌ 2 Fe(s) + 3 CO₂(g) (k at 727°C = 24)

The value of the equilibrium constant (k = 24) does not provide information about whether the reaction favors the reactants or products. To determine which side is favored, one would need to compare the initial concentrations or partial pressures of the reactants and products. However, the presence of the solid Fe2O₃ indicates that it is likely the reactant side (Fe2O₃ and CO) that is favored at equilibrium, as the solid does not contribute to the equilibrium expression.

Overall,

A. The reaction strongly favors the products.

B. The reaction favors the reactants.

C. The information provided is insufficient to determine which side is favored at equilibrium.

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Factors that may influence the solubility of BaF₂ in an aqueous solution include the following substances: LiF, H₂SO₄, NaOH, BaCl₂, and KNO₃. (A,B,C,D)

Solubility is the ability of a solid to dissolve in a liquid to form a homogeneous mixture.

In an aqueous solution, the ability of a substance to dissolve is determined by various factors, including temperature, pressure, and the nature of the solvent and the solute. The concentration of the solute, pH, and the presence of other solutes or substances in the solution can all influence solubility.  (A,B,C,D)

The solubility of BaF₂, a sparingly soluble salt, is influenced by the presence of other substances. Lithium fluoride (LiF) and barium chloride (BaCl₂) both contain ions that could affect the solubility of BaF₂. Li⁺ and Ba²⁺, respectively, are cations, while F⁻ and Cl⁻ are anions.

When LiF or BaCl₂ is dissolved in water, their respective ions will react with the F⁻ and Ba²⁺ ions present in the BaF₂, respectively. These reactions result in the formation of LiBaF₃ and BaClF, respectively, and the BaF₂ becomes more soluble in the solution.

Similarly, NaOH and H₂SO₄ are strong electrolytes that dissociate in water to produce OH⁻ and H⁺ ions, respectively. These ions can react with the F⁻ ions in BaF₂, resulting in the formation of water and a soluble salt.

KNO₃, on the other hand, is a soluble salt that dissociates in water to produce K⁺ and NO₃⁻ ions. The presence of these ions can increase the solubility of BaF₂ in solution.

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The new molarity of the solution is 0.5 M after diluting 2.0 L of a 1.5 M solution with water until the final volume is 6.0 L.

To find the new molarity of the solution, we can use the formula:

M1V1 = M2V2

Where:

M1 = initial molarity of the solution

V1 = initial volume of the solution

M2 = final molarity of the solution

V2 = final volume of the solution

Given:

M1 = 1.5 M

V1 = 2.0 L

V2 = 6.0 L

Let's substitute the values into the formula and solve for M2:

M1V1 = M2V2

(1.5 M)(2.0 L) = M2(6.0 L)

3.0 mol = M2(6.0 L)

Now, let's isolate M2 by dividing both sides of the equation by 6.0 L:

M2 = 3.0 mol / 6.0 L

M2 = 0.5 M

Therefore, the new molarity of the solution is 0.5 M after diluting 2.0 L of a 1.5 M solution with water until the final volume is 6.0 L.

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Answer:

Water applied to the surface of a relatively dry soil infiltrates quickly due to the affinity of the soil particles for water. As time passes and the soil becomes wet, the force of gravity becomes the dominant force causing water to move.

Explanation:

Answer:

Gravity

Explanation:

Water applied to the surface of a relatively dry soil infiltrates quickly due to the affinity of the soil particles for water. As time passes and the soil becomes wet, the force of gravity becomes the dominant force causing water to move.

The complete sentences are:

Page 3:

The effect of energy in phase transitions of matter is that it is required to break the intermolecular forces that hold the particles of a substance together. When energy is added to a substance, the particles move faster and the intermolecular forces are broken. This can cause the substance to change phase.

The interactive demonstration on the sample of water shows that energy is required to melt ice and boil water. When the ice is heated, the particles start to move faster and the ice melts. The temperature of the water stays constant at 0°C until all of the ice has melted. This is because the energy is being used to break the intermolecular forces in the ice. Once all of the ice has melted, the temperature of the water starts to rise again. When the water is boiled, the particles move so fast that they escape from the liquid state and become a gas. The temperature of the water stays constant at 100°C until all of the water has boiled. This is because the energy is being used to break the intermolecular forces in the water. Once all of the water has boiled, the temperature of the steam starts to rise again.

The complete sentences:

Page 4:

Heating curves show the temperature of a substance as it is heated. The curve has a horizontal line at the melting and boiling points, which indicates that the temperature does not change during these phase changes.

Cooling curves show the temperature of a substance as it is cooled. The curve has a horizontal line at the melting and boiling points, which indicates that the temperature does not change during these phase changes.

Both curves show that the temperature of a substance increases as it is heated and decreases as it is cooled.

A heating curve is more choppy than a cooling curve because there are more phase changes during heating than during cooling.

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The pH of the solution after the addition of .0500 M NaOH is  12.70.

Buffer solution volume of the buffer solution = 1.50LConcentration of HF = 0.250 M

Concentration of NaF = 0.250 M

After the addition of NaOH

The concentration of NaOH = 0.0500 M

Ka of HF = 3.5 × 10⁻⁴

First of all, we have to determine the moles of HF and NaF initially present in the solution.

Initial moles of HF = Molarity × Volume of solution = 0.250

M × 1.50 L = 0.375 moles

Initial moles of NaF = Molarity × Volume of solution = 0.250 M × 1.50 L = 0.375 moles

After the addition of NaOH, HF, and NaF react with NaOH to form NaF and water as shown below.

HF + NaOH → NaF + H₂O(0.250 M) (excess)(0.0500 M) -x x

molarity of NaOH = 0.0500 M - x

[NaOH] = [NaF] = (initial moles of NaF - x)/Volume of solution

= (0.375 - x)/1.50

Initial moles of NaOH = Molarity × Volume of solution = (0.0500 M - x) × 1.50 L = 0.075 - 1.50 x

Initial moles of NaF = Initial moles of NaOH(As they react in 1:1 ratio)= 0.075 - 1.50 x

Initial moles of HF = 0.375 - x

Initial concentration of HF = (0.375 - x)/1.50= (0.250 - x/6)

After the reaction, the concentration of HF and F⁻ will change by the same amount that is - x and + x respectively.

[H⁺] [F⁻]/[HF] = Ka[H⁺] [0.375 + x]/[0.250 - x/6]

= 3.5 × 10⁻⁴[H⁺]

= 3.5 × 10⁻⁴ × (0.375 + x)/(0.250 - x/6)

As the solution is a buffer solution, pH = pKa + log [F⁻]/[HF]

pKa = -log Ka

= -log 3.5 × 10⁻⁴= 3.455

pH = 3.455 + log [0.375/(0.250 - x/6)]

The value of x can be calculated by using the ICE table. The values of [H⁺] and x will be very small as the concentration of NaOH added is very less.

So, x can be neglected.

x = [OH⁻] = 0.0500 M[H⁺] = Kw/[OH⁻] = 1.0 × 10⁻¹⁴/0.0500 M

= 2.0 × 10⁻¹³pH

= -log [H⁺] = -log (2.0 × 10⁻¹³)

= 12.70 (approx)

Therefore, the pH of the buffer solution after the addition of 0.0500 M NaOH is approximately 12.70.

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Answer: The cleaning spray might have containing alcohol.

Explanation:

Some alcohol have the tendency to give an explosive effect on heating. According to the given situation, if the cleaning fluid is containing alcohol and when it had been exposed to air, oxygen and sun heat it will become unstable by producing peroxides which can explode the can in which they have been stored. So, in the given situation, the cleaning spray containing alcohol will explode due to intense heat.

Answer:

no it's is false because an electron have negative charges and it is not inside the atom and it is found out side the nucleus

In the Newman projection, the adenine group is in the axial position, while the CHOH group is in the equatorial position. This arrangement suggests that the adenine group is pointing upward and away from the ring, while the CHOH group is pointing outward and slightly downward.

Converting the Newman projection into a bond-line chair form involves visualizing the cyclohexane ring in a chair conformation. In this conformation, the six carbon atoms form a hexagonal shape, resembling a chair, with alternating axial and equatorial positions. The adenine group, initially in the axial position, is represented as a substituent extending upward from one of the carbons in the ring, while the CHOH group, initially in the equatorial position, is depicted as a substituent extending outward and slightly downward from the ring.

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The stoichiometric concept can be employed for the 3A(g) → 3B(g) transformation.

The enthalpy change for the reaction leading from A(g) to B(g) with a value of ΔH = 70 kJ implies that the corresponding enthalpy change for the conversion of 3A(g) to 3B(g) will be threefold higher, with ΔH = 3 * 70 kJ = 210 kJ.

To determine the enthalpy change for the inverse reaction of A(g) → B(g), we can utilize the knowledge that the enthalpy change has the inverse polarity in the reverse reaction.

The enthalpy change for the conversion of gas B to gas A will result in a decrease of 70 kJ.

We can determine the enthalpy shift for the A(g) → C(g) reaction by merging the provided equations.

By combining the equations A(g) → B(g) with a heat of reaction of 70 kJ and B(g) → C(g) with a heat of reaction of -110 kJ, we can obtain a new equation.

Through this, we are presented with the generalized reaction of converting A into B, which subsequently forms C, accompanied by a change in enthalpy of -40 kJ within the range of 70 kJ-110 kJ.

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Answer:

The reaction is exothermic

Explanation: