what is occurring in the following reaction: nabr cl2→nacl br2 select the correct answer below: sodium is oxidized chlorine is oxidized bromine is reduced chlorine is reduced

Based on the chemical reaction between NaOH and HCl, we can expect the pH to increase as NaOH is added to the HCl solution.

This is because NaOH is a strong base and HCl is a strong acid, so when they react, they neutralize each other and form water and a salt. The salt formed in this reaction is typically neutral and does not contribute to the acidity or basicity of the solution. As more NaOH is added, it will continue to react with the remaining HCl until all of the HCl has been neutralized, which will result in a pH close to 7, indicating a neutral solution.

It is important to note that the rate at which the pH changes will depend on the concentration of both the NaOH and HCl solutions.

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The number of moles of gas contained in the cylinder is approximately 0.381 moles.

To calculate the number of moles of gas contained in the cylinder, we can use the ideal gas law equation:

PV = nRT

Where:

P = Pressure (in atm)

V = Volume (in L)

n = Number of moles

R = Ideal gas constant (0.0821 L·atm/(mol·K))

T = Temperature (in Kelvin)

We need to convert the temperature from Celsius to Kelvin by adding 273.15:

T = 23.8 °C + 273.15 = 296.95 K

Now, we can rearrange the ideal gas law equation to solve for the number of moles (n):

n = PV / RT

Plugging in the values:

P = 4.26 atm

V = 15.0 L

R = 0.0821 L·atm/(mol·K)

T = 296.95 K

n = (4.26 atm)(15.0 L) / (0.0821 L·atm/(mol·K))(296.95 K)

n ≈ 0.381 moles

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In the external circuit, electrons migrate from the Mn|Mn2+ electrode to the Cu+|Cu2+ electrode. In the salt bridge, anions migrate from the Mn|Mn2+ compartment to the Cu+|Cu2+ compartment.

A voltaic cell is a type of electrochemical cell that converts chemical energy into electrical energy. In this specific voltaic cell, the anode is a Mn|Mn2+ half cell, and the cathode is a Cu+|Cu2+ half cell. The two half-cell compartments are connected by a salt bridge, which allows ions to move between the compartments and maintain charge balance.
The anode reaction is: Mn(s) → Mn2+(aq) + 2e-
The cathode reaction is: Cu2+(aq) + e- → Cu+(aq)
The net cell reaction is: Mn(s) + Cu2+(aq) → Mn2+(aq) + Cu+(aq)
This is because the Mn electrode is the anode, where oxidation occurs, and the Cu electrode is the cathode, where reduction occurs. Electrons always flow from the anode to the cathode in a voltaic cell.
This is because the Mn2+ ions are positively charged and attract negatively charged anions from the salt bridge, while the Cu+ ions are positively charged and repel positively charged cations from the salt bridge. The movement of ions in the salt bridge helps to maintain charge balance in the two half-cell compartments.

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Answer:

Oil-in-water emulsion are droplets of oil dispersed in water. Water-in-oil emulsion are droplets of water dispersed in oil.

Explanation:

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Weighing by difference eliminates systematic balance errors because the mass is determined by the difference between two readings, a systematic error in the absolute mass on the balance will be removed by subtracting the final weight from the initial weight.

Option (d) is correct.

Weighing by difference involves weighing the sample container before and after adding or removing the sample. By subtracting the final weight from the initial weight, the mass of the sample can be determined. This method helps eliminate systematic balance errors because any systematic error in the absolute mass reading of the balance will affect both the initial and final weights equally.

Weighing by difference takes into account any potential errors in the balance's absolute mass reading, such as calibration or zero errors, and focuses on the change in mass instead. This approach improves the accuracy of the measurement and reduces the impact of systematic errors associated with the balance.

Therefore, the correct option is (d).

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Nitrogen makes up approximately 78% of the Earth's atmosphere by volume and more than 10% of its mass. The correct option is E.

Water vapor, argon, helium, and hydrogen each make up less than 1% of the atmosphere by volume and therefore contribute less than 10% to its mass. Nitrogen makes up approximately 78% of the Earth's atmosphere by volume and more than 10% of its mass.

Nitrogen makes up more than 10% of the mass of the atmosphere. In fact, nitrogen is the most abundant gas in the atmosphere, constituting approximately 78% of its total mass.

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The average bond order for a Br-O bond in the bromate ion, BrO3-, is 1.33. The bond order is a measure of the number of chemical bonds between a pair of atoms in a molecule.

To calculate the average bond order for a Br-O bond in the bromate ion, we need to first determine the number of bonds between the bromine and oxygen atoms. In the bromate ion, there are three oxygen atoms bonded to a central bromine atom. Each Br-O bond is a single bond, meaning that there is one bond between the bromine and each oxygen atom.

Draw the Lewis structure of the bromate ion (BrO3-). You will notice that it has resonance structures, which means that the electrons are distributed over multiple locations, and the bond order is an average value. In the resonance structures, there are 4 total bonds between Br and O atoms. Each of these bonds is a single bond, so the total bond order across all structures is 4. There are 3 oxygen atoms bonded to the bromine atom. To find the average bond order for one Br-O bond, divide the total bond order by the number of oxygen atoms.

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Dimethyl ether is a polar molecule with an oxygen atom bonded to two methyl groups. Water is also a polar molecule with two hydrogen atoms bonded to an oxygen atom.

Hydrogen bonding occurs between these molecules due to the attractive forces between the partially positive hydrogen atom in water and the partially negative oxygen atom in dimethyl ether.

In the illustration, you would expect to see hydrogen bonds forming between the oxygen atom of dimethyl ether and the hydrogen atoms of water.

The oxygen atom in dimethyl ether acts as the hydrogen bond acceptor, while the hydrogen atoms in water act as the hydrogen bond donors.

These hydrogen bonds play a crucial role in the interactions between dimethyl ether and water molecules, affecting their solubility and behavior in mixtures.

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The pOH of water at 0°C is 7.0. The answer is (a). To find the pOH of water at 0°C, we need to use the relationship between the pH and pOH of a solution, which is pH + pOH = 14.

Since the pH of pure water at 0°C is 7.0 (neutral), we can calculate the pOH as follows:

pH + pOH = 14
7.0 + pOH = 14
pOH = 14 - 7.0
pOH = 7.0

Therefore, the pOH of water at 0°C is 7.0. The answer is (a).

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The molar solubility of BaF₂ is 0.0021 M. The molar solubility of a compound can be determined using its Ksp value.

Ksp is the solubility product constant, which represents the equilibrium constant for the dissolution of a sparingly soluble compound. In the case of BaF₂, the Ksp value is 1.7*10⁻⁶.

To determine the molar solubility of BaF₂, we first write out the balanced equation for its dissolution:
BaF₂(s) ⇌ Ba²⁺(aq) + 2F⁻(aq)

The Ksp expression for this equation is:
Ksp = [Ba²⁺][F⁻]²

We can use this equation to solve for the molar solubility of BaF₂. Let x be the molar solubility of BaF₂. Then, we have:
Ksp = x(2x)² = 4x³
1.7*10⁻⁶ = 4x³
Therefore,
x = 0.0021 M

Therefore, the molar solubility of BaF₂ is 0.0021 M. This means that at equilibrium, 0.0021 moles of BaF₂ will dissolve in one liter of water.

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In the MPV (Mukaiyama-Prins-Vogel) reaction, aluminum serves as a Lewis acid catalyst that coordinates with the carbonyl group of the substrate and activates it towards nucleophilic attack by the alcohol or other nucleophile.

This coordination of the carbonyl group to aluminum enhances the electrophilicity of the carbonyl carbon and promotes the formation of the C-O bond in the product. Aluminum also helps to stabilize the intermediate species formed during the reaction and facilitates the final product formation.

As for the formal charges on the central atoms in the reducing agents, it depends on the specific reducing agent being used. For example, in the presence of a boron-based reducing agent such as borane, the central boron atom has a formal charge of +1, while the hydrogen atoms attached to it have a formal charge of 0. Similarly, in the presence of a metal hydride reducing agent such as lithium aluminum hydride, the central metal atom has a formal charge of 0, while the hydrogen atoms attached to it have a formal charge of -1.

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Three of the four salts you mentioned are expected to produce acidic solutions: NaHSO4, NaHC2O4, and NaH2PO4. These salts are formed by weak acids and strong bases, so when they dissolve in water, they will increase the concentration of H+ ions, making the solution acidic. NaHCO3, on the other hand, will produce a slightly basic solution because it is formed by a weak acid and a weak base.

Only one of the mentioned salts, NaHCO3, is expected to produce an acidic solution. This is because it is a weak acid salt, meaning it will partially dissociate in water to release H+ ions, which will make the solution acidic. On the other hand, NaHSO4, NaHC2O4, and NaH2PO4 are all acidic salts, which means they will fully dissociate in water to release H+ ions, making the solution highly acidic. However, NaHCO3 is a basic salt, which means it will partially dissociate to release OH- ions, making the solution basic. Overall, only NaHCO3 is expected to produce an acidic solution out of the four mentioned salts.
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We need 20 grams of powdered drink mix to make a 1.0 m solution of 100 ml using stoichiometry. To use stoichiometry to determine the mass of powdered drink mix needed to make a 1.0 m solution of 100 ml, we first need to know the molar mass of the drink mix. Let's say the molar mass is 200 g/mol.

We can use the formula for molarity:
Molarity = moles of solute / liters of solution
To find the moles of solute needed for a 1.0 m solution of 100 ml, we need to rearrange the formula:
moles of solute = Molarity x liters of solution
Since we have 100 ml of solution, we need to convert this to liters:
100 ml / 1000 ml/L = 0.1 L
Now we can plug in the values we know:
moles of solute = 1.0 mol/L x 0.1 L = 0.1 mol
To find the mass of solute needed, we can use the formula:
mass = moles x molar mass
mass = 0.1 mol x 200 g/mol = 20 g
Therefore, we need 20 grams of powdered drink mix to make a 1.0 m solution of 100 ml using stoichiometry.

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The correct answer is option A. 484 mmHg.

To convert the pressure from mbar (millibar) to mmHg (millimeters of mercury), we can use the conversion factor: 1 mbar = 0.7501 mmHg.

Given: Pressure in mbar = 645 mbar.

Using the conversion factor:

Pressure in mmHg = 645 mbar * 0.7501 mmHg/mbar.

Calculating the value:

Pressure in mmHg = 483.985 mmHg.

Rounding to the nearest whole number, the corresponding pressure in mmHg is approximately 484 mmHg.

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To find the volume of the "dry" oxygen, we can use the combined gas law, which relates the initial and final conditions of pressure, volume, and temperature.Therefore, the volume of the "dry" oxygen would be approximately 11.36 mL.

The combined gas law formula is as follows:

(P1 * V1) / (T1 + 273.15) = (P2 * V2) / (T2 + 273.15)

Let's plug in the given values into the formula:

(P1 * V1) / (T1 + 273.15) = (P2 * V2) / (T2 + 273.15)

P1 = 54.5 kPa

V1 = 8.44 ml

T1 = 20°C = 20 + 273.15 = 293.15 K

P2 = 116.94 kPa

T2 = -27.2°C = -27.2 + 273.15 = 245.95 K

Substituting the values:

(54.5 * 8.44) / 293.15 = (116.94 * V2) / 245.95

Now, solve for V2 (the volume of the "dry" oxygen):

V2 = (54.5 * 8.44 * 245.95) / (116.94 * 293.15)

V2 ≈ 11.36 ml

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The phases in the equation are: aq (aqueous) - for Hg2(NO3)2 and NaCl solutions, s (solid) - for Hg2Cl2 precipitate, aq (aqueous) - for NaNO3 solution

The complete ionic equation for the reaction of aqueous Hg2(NO3)2 with aqueous sodium chloride to form solid Hg2Cl2 and aqueous sodium nitrate is:
Hg2^2+ (aq) + 2Cl^- (aq) → Hg2Cl2 (s)
The reaction involves the precipitation of solid Hg2Cl2 due to the reaction of mercury(I) ions (Hg2^2+) with chloride ions (Cl^-) from the sodium chloride solution. The nitrate ions (NO3^-) from the Hg2(NO3)2 solution combine with the sodium ions (Na+) from the sodium chloride solution to form aqueous sodium nitrate (NaNO3).
The overall chemical equation for this reaction is:
Hg2(NO3)2 (aq) + 2NaCl (aq) → Hg2Cl2 (s) + 2NaNO3 (aq)
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The estimated crystal field splitting energy for the [Ti(H2O)6]³⁺ complex ion is approximately 7.18 KJ/mol, expressed to three significant figures.

To estimate the crystal field splitting energy (Δ₀) for the complex ion [Ti(H2O)6]³⁺, we can make use of the spectrochemical series.

Based on the spectrochemical series, the ligand water (H2O) is considered a weak field ligand. This means that the crystal field splitting energy for this complex ion will be relatively small.

The crystal field splitting energy (Δ₀) is typically measured in terms of the Dq parameter, which represents the energy difference between the d orbital energy levels in the presence of the ligand field.

For a complex with weak field ligands, such as [Ti(H2O)6]³⁺, the typical range for Δ₀ is around 600 to 1000 cm⁻¹.

To convert this value to kilojoules per mole, we can use the conversion factor 1 cm⁻¹ = 1.1963 x 10⁻² KJ/mol.

Therefore, the estimated crystal field splitting energy (Δ₀) for [Ti(H2O)6]³⁺ would be:

Δ₀ = 600 cm⁻¹ × 1.1963 x 10⁻² KJ/mol/cm⁻¹

Δ₀ ≈ 7.18 KJ/mol

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If your total federal tax deposit is $100,000.00 or greater, you must make the deposit no later than the next business day after the tax liability was incurred. This is also known as the semi-weekly deposit schedule.

The IRS has different deposit schedules based on the amount of tax liability. For businesses with a tax liability of less than $2,500.00, the deposit can be made quarterly. For those with a liability between $2,500.00 and $100,000.00, the deposit must be made using the monthly deposit schedule. However, for businesses with a liability of $100,000.00 or more, the semi-weekly deposit schedule applies.

Under the semi-weekly deposit schedule, deposits must be made no later than the next business day after the tax liability was incurred. The deposit due dates are based on when the payroll is paid. For example, if payroll is paid on Wednesday, Thursday, or Friday, the deposit is due the following Wednesday. If payroll is paid on Saturday, Sunday, Monday, or Tuesday, the deposit is due the following Friday.

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pH of a solution is the degree of acidity or alkalinity of a solution. It can be measured using colorimeteric method or pH meter.

pH of a solution is a measure of the relative amount of free hydrogen and hydroxyl ions in the aqueous solution of a substance.

The range of pH goes from 0 - 14, with 7 being neutral. pHs of less than 7 indicate acidity, whereas a pH of greater than 7 indicates a base.

There are two methods for measuring pH as follows;

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The estimated enthalpy change (ΔH) for the reaction of methane with fluorine is approximately -842 kJ/mol. The negative sign indicates that the reaction is exothermic, meaning it releases energy.

To estimate the enthalpy change (ΔH) for the reaction of methane (CH4) with fluorine (F2) according to the equation: CH4(g) + 2 F2(g) → CF4(g) + 2 H2(g), we need to use the average bond dissociation energies (D) of the bonds involved.The bonds that are broken in the reaction are the C-H bonds in methane and the F-F bonds in fluorine. The bonds that are formed are the C-F bonds in carbon tetrafluoride (CF4) and the H-H bonds in hydrogen gas (H2).The bond dissociation energies (in kJ/mol) are as follows:

D(C-H) = 413 kJ/mol

D(F-F) = 159 kJ/mol

D(C-F) = 485 kJ/mol

D(H-H) = 436 kJ/mol

To estimate the enthalpy change (ΔH) for the reaction, we sum up the energy required to break the bonds and subtract the energy released when new bonds are formed.

ΔH = Σ(bonds broken) - Σ(bonds formed)

ΔH = (4 * D(C-H) + 2 * D(F-F)) - (4 * D(C-F) + 2 * D(H-H))

ΔH = (4 * 413 kJ/mol + 2 * 159 kJ/mol) - (4 * 485 kJ/mol + 2 * 436 kJ/mol)

ΔH = (1652 kJ/mol + 318 kJ/mol) - (1940 kJ/mol + 872 kJ/mol)

ΔH = -842 kJ/mol

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The final answer, rounded to two significant digits, is 0.98.

When rounding a number to two significant digits, we need to look at the first two digits of the number and determine whether to round up or down based on the third digit.In the case of 0.98029, the first two significant digits are 9 and 8. The third digit, 0, is less than 5, so we round down.
It's important to note that when we round to two significant digits, we're essentially saying that we're confident in the first two digits of the number and the third digit is uncertain. So, we want to keep the first two digits as accurate as possible while also rounding to a reasonable degree of precision.
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In the log concentration versus pH diagram, the B(OH)4- species will dominate at high pH values and the B(OH)3 species will dominate at low pH values.

At the pH of seawater (pH 8.1), the concentration of B(OH)3 will be 0.04 mM and the concentration of B(OH)4- will be 0.36 mM. This is because at pH 8.1, the majority of the borate anions will have deprotonated to form B(OH)4-, leaving only a small amount of B(OH)3 present. It's worth noting that at this pH, the total concentration of boron species will equal the initial concentration of 0.4 mM. The equilibrium between the two species will shift towards B(OH)3 domination as the pH decreases below 9.2 and towards B(OH)4- domination as the pH increases above 9.2. In an aqueous solution, the borate anion (B(OH)4-) and boric acid (B(OH)3) exist in equilibrium, represented by the equation B(OH)3 + H2O ⇌ B(OH)4- + H+. The pKa value of this equilibrium is 9.2. A log concentration vs. pH diagram can be used to visualize the relative concentrations of these species as the pH changes. At the pH of seawater (8.1), the Henderson-Hasselbalch equation can be applied:
pH = pKa + log ([B(OH)4-]/[B(OH)3])
Rearranging the equation and using the given values, we find the ratio of [B(OH)4-] to [B(OH)3]. Since the total boron concentration is 0.4 mM, we can calculate the individual concentrations of B(OH)4- and B(OH)3 at pH 8.1.

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The new freezing point for a 0.73 m solution of CCl4 in benzene is 1.94 °C.

To calculate the new freezing point for a 0.73 m solution of CCl4 in benzene, we first need to find the change in freezing point caused by adding CCl4 to benzene. This can be calculated using the formula:

ΔTf = Kf × molality

where ΔTf is the change in freezing point, Kf is the freezing point depression constant for benzene (4.90 °C/m), and molality is the molality of the solution (0.73 m).

ΔTf = 4.90 °C/m × 0.73 m = 3.57 °C

This means that the freezing point of the solution will be lowered by 3.57 °C compared to pure benzene. To find the new freezing point, we simply subtract this value from the freezing point of pure benzene:

New freezing point = 5.51 °C - 3.57 °C = 1.94 °C

Therefore, the new freezing point for a 0.73 m solution of CCl4 in benzene is 1.94 °C.

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Assumes the shape of its container: Liquid (Explanation: Liquids take the shape of the container they are in and can flow freely.)

Definite volume: Solid (Explanation: Solids have a fixed volume and maintain their shape regardless of the container they are in.) Expands to fill its container: Gas (Explanation: Gases do not have a fixed shape or volume and expand to fill the available space in their container.) Strong intermolecular forces: Solid (Explanation: Solids have strong intermolecular forces that hold their particles together in a fixed arrangement.) Particles are close together but not rigidly arranged: Liquid (Explanation: In liquids, particles are close together but can move past each other, allowing the substance to flow.) Assumes the shape and volume of its container: Gas (Explanation: Gases have no fixed shape or volume and completely fill the container they are in.) Definite shape: Solid (Explanation: Solids have a definite shape and retain it regardless of the container they are placed in.) Weak intermolecular forces: Gas (Explanation: Gases have weak intermolecular forces as their particles are far apart and have high kinetic energy, allowing them to move freely.) Particles are far apart and move freely: Gas (Explanation: Gas particles are widely spaced and move rapidly in all directions, resulting in their ability to fill a large space.) Flows with viscosity: Liquid (Explanation: Liquids have viscosity, which is a measure of their resistance to flow. They flow but at a slower rate compared to gases.)

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(a) The average velocity during the period is 2.5 m/s. (b) The distance travelled by the particle during the period is 12.5 m.

It is given that a particle starting from rest moves with constant acceleration and it takes 5.0s to reach the speed 18.0 km/h.

(a) To find the average velocity, we first need to convert the final speed to m/s:

18.0 km/h * (1000 m/km) / (3600 s/h) = 5.0 m/s.

The average velocity is the average of the initial and final velocities:

(0 m/s + 5.0 m/s) / 2 = 2.5 m/s.

(b) To find the distance traveled, we can use the formula

d = (v_initial + v_final) * t / 2.

Plugging in our values, we get:

d = (0 m/s + 5.0 m/s) * 5.0 s / 2 = 2.5 m/s * 5.0 s = 12.5 m.

So, during this period, the average velocity is 2.5 m/s, and the distance traveled is 12.5 m.

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K values less than 1 (K < 1) would indicate that there is more B than A at equilibrium.

The equilibrium constant (K) represents the ratio of the concentrations of products to reactants at equilibrium. In the given reaction, A(g) ⇌ B(g), the equilibrium constant expression is K = [B(g)]/[A(g)].

When K < 1, it implies that the concentration of B (denoted as [B(g)]) is smaller compared to the concentration of A ([A(g)]) at equilibrium. Since the equilibrium constant is smaller than 1, it indicates that the reaction favors the formation of reactant A over product B.

Therefore, there is more B than A at equilibrium when the equilibrium constant (K) is less than 1 (K < 1).

Hence, among the given K values, K = 0.8 and K = 7×10^(-9) would indicate that there is more B than A at equilibrium.

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d. HF.Among the options provided, the acid commonly used for etching and frosting glass is **HF** (hydrofluoric acid).

HF is known for its ability to react with glass and create a frosted or etched appearance. This acid is particularly effective in etching glass due to its unique property of being able to dissolve silicates, which are the primary components of glass. The reaction between HF and glass involves the formation of silicon tetrafluoride, which is volatile and can be easily removed, leaving behind the etched or frosted surface.

It is important to note that handling HF requires extreme caution due to its corrosive and toxic nature. Proper safety measures, including the use of protective equipment and working in well-ventilated areas, should be followed when working with HF or any other hazardous chemicals.

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The rate of the reaction when [A] is 0.035 M and [B] is 0.070 M is approximately 0.0591 M/s.

To determine the rate of the reaction when [A] is 0.035 M and [B] is 0.070 M, we can use the initial rate data from the given experiments to calculate the rate constant.The rate of a chemical reaction is typically expressed as the change in concentration of a reactant or product per unit time. In this case, we will focus on the initial rate of the reaction, as indicated in the table.From the given data, we can observe that Experiment 1 has [A] = 0.060 M, [B] = 0.030 M, and an initial rate of 0.0744 M/s.According to the rate law for the given reaction, which can be expressed as rate = k[A]^x[B]^y, the rate is directly proportional to the concentrations of A and B.By comparing Experiments 1 and 2, we can determine the order of the reaction with respect to A:

(0.0744 M/s) / (0.00828 M/s) = (0.060 M)^x / (0.020 M)^x

9 = 3^x

x = 2

By comparing Experiments 1 and 3, we can determine the order of the reaction with respect to B:

(0.0744 M/s) / (0.0248 M/s) = (0.030 M)^y / (0.090 M)^y

3 = 1/3^y

y = 1

The overall order of the reaction is the sum of the individual orders:

Overall order = x + y = 2 + 1 = 3

Now, we can calculate the rate constant (k) using the data from Experiment 1:

0.0744 M/s = k * (0.060 M)^2 * (0.030 M)^1

k = 0.0744 M/s / (0.060 M)^2 * (0.030 M)^1

k ≈ 0.1236 M^-2s^-1

Finally, we can calculate the rate of the reaction when [A] is 0.035 M and [B] is 0.070 M:

rate = k * [A]^x * [B]^y

rate = 0.1236 M^-2s^-1 * (0.035 M)^2 * (0.070 M)^1

rate ≈ 0.0591 M/s

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Based on the information above, option (c) CH3N=NCH3, which contains a conjugated system, is more likely to have the largest molar absorptivity (€) among the given options.

The molar absorptivity (€) is a measure of how strongly a compound absorbs light at a particular wavelength. It is influenced by various factors, including the molecular structure and functional groups present in the compound. In general, compounds with extended conjugation and aromatic systems tend to have higher molar absorptivities. Based on this information, let's evaluate the given options:

(a) C6H13CH=CH2:

This compound is a linear alkene and does not possess an extended conjugated system or aromaticity. Therefore, it is unlikely to have a large molar absorptivity.

(b) CH3(C=O)H:

This compound is an aldehyde. While it contains a carbonyl group, it does not have an extended conjugated system or aromaticity. Hence, its molar absorptivity is expected to be relatively low.

(c) CH3N=NCH3:

This compound is dimethyl diazene. It contains a diazenyl functional group, which exhibits conjugation between the nitrogen atoms. The presence of this conjugated system suggests that it may have a higher molar absorptivity compared to the previous options.

(d) C2H5ONO2:

This compound is ethyl nitrite. It does not possess an extended conjugated system or aromaticity, similar to option (a) and (b). Therefore, its molar absorptivity is expected to be relatively low.

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The systematic names for the isomeric acid chlorides with the chemical formula C6H9ClO are as follows:

cis-3-chlorocyclopent-1-ene-1-carbonyl chloride

trans-3-chlorocyclopent-1-ene-1-carbonyl chloride

2-chloro-3-methylbut-2-ene-1-carbonyl chloride

The first compound is a cis isomer with a chlorine atom and a carbonyl group on the same side of the cyclopentene ring. The prefix "cis-" is used to indicate this stereochemistry.

The second compound is a trans isomer with a chlorine atom and a carbonyl group on opposite sides of the cyclopentene ring. The prefix "trans-" is used to indicate this stereochemistry.

The third compound is a 2-chloro-3-methylbut-2-ene-1-carbonyl chloride. The chlorine atom is attached to the second carbon atom of the butene chain, and the carbonyl group is attached to the first carbon atom.

The prefix "2-chloro" indicates the position of the chlorine atom, and "3-methyl" indicates the presence of a methyl group on the third carbon atom. The term "but-2-ene-1-carbonyl" describes the butene chain with a carbonyl group attached to the first carbon atom.

Overall, these names provide a clear and systematic description of the isomeric acid chlorides based on their molecular structures and stereochemistry.

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