PLEASE HELP
How do scientists determine a nucleus' location within an atom?
A. by measuring half the distance between the nuclei of two bonded atoms
B. by splitting the atom into an ion and measuring the nucleus' size
C. by dividing the atomic radius by a factor of 2 D. by sending an electric current through the atom and measuring the amount of nucleus resistance​

The total number of oxygen atoms in the mixture is 3.49 x \(10^{23\) atoms.

To determine the number of oxygen atoms in the given mixture, we need to first calculate the number of oxygen atoms in each compound and then add them up.

So, 3.49x10^23 molecules of H2O will contain:

3.49x10^23 x 1 oxygen atom = 3.49x10^23 oxygen atoms

Number of oxygen atoms in 78.1g of CH3OH:

To find the number of moles of CH3OH, we can use its molar mass:

Therefore, 78.1 g of CH3OH will contain:

2.44 mol x 1 oxygen atom = 2.44 oxygen atoms

The molecular formula of CO2 is CO2, which contains 2 oxygen atoms.

To find the number of moles of CO2, we can use the ideal gas law:

Assuming standard temperature and pressure (STP), which is 1 atm and 0°C (273 K), we have:

n = (1 atm x 14.2 L) / (0.0821 L·atm/mol·K x 273 K) = 0.58 mol

Therefore, 14.2 L of CO2 will contain:

0.58 mol x 2 oxygen atoms = 1.16 oxygen atoms

Now we can add up the number of oxygen atoms in each compound to find the total number of oxygen atoms in the mixture:

3.49x10^23 + 2.44 + 1.16 = 3.49x10^23 + 3.6 = 3.49 x \(10^{23\)

Therefore, the mixture contains 3.49 x \(10^{23\) oxygen atoms.

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Answer:

Hi there

Thermal Energy and Chemical Reactions

The device controlled the activation as shown by the water holding room temperature for 15 seconds until activation by crushing the pills. The temperature rose 30° over two minutes,showing a measurable exothermic reaction from the Calcium Chloride dissolving in the water.

Pour 5mL water in a quart bag

Place Thermometer in the bag to make contact

Fill two gelatin 000 capsules with 1 gram of Calcium Chloride

Place capsules into the bag

Wait 15 seconds to crush the capsules while they are submerged in the water

-Temperature

0 seconds 75 degrees

15 seconds - Activation 75 degrees

30 seconds 85 degrees

60 seconds 95 degrees

90 seconds 104 degrees

120 seconds 105 degrees

Conclusion: During a reaction, thermal energy from the environment is converted into chemical energy stored in the products.

Brainliest plz

Answer:

hypothesis testing is usually used to assess the plausibility of a hypothesis by using sample data

Explanation:

A 0.425g sample of propane was mixed with excess oxygen in a calorimeter containing 98.72 g of water. The system was ignited and the temperature rose from 18.29ºC to 26.85ºC.  the molar enthalpy of propane is 393851.1 J/mol

The mass of propane = 0.425 g

molar mass of propane = 44 g/mol

moles = mass / molar mass

         = 0.425 / 44

         = 0.009 mol

ΔH = mc ΔT

ΔH  = n ΔHsol

ΔHsol = mc ΔT / n

          = 98.72 × 4.18 ( 26.85 - 18.29 ) / 0.009

molar enthalpy of solution = 393851.1 J/mol

Thus, A 0.425g sample of propane was mixed with excess oxygen in a calorimeter containing 98.72 g of water. The system was ignited and the temperature rose from 18.29ºC to 26.85ºC.  the molar enthalpy of propane is 393851.1 J/mol

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To determine the grams of water produced, we need to first balance the chemical equation for the reaction between hydrogen gas (H2) and oxygen (O2) to form water (H2O). The balanced equation is:

2H2 + O2 → 2H2O. From the balanced equation, we can see that 2 moles of H2 react with 1 mole of O2 to produce 2 moles of H2O. Given that the reaction is at STP (standard temperature and pressure), we can use the molar volume of gases at STP to calculate the number of moles of hydrogen gas. The molar volume of a gas at STP is 22.4 L/mol. Number of moles of H2 = (volume of H2 gas) / (molar volume of H2 at STP) = 20 L / 22.4 L/mol = 0.8928 mol. From the balanced equation, we know that the ratio of H2 to H2O is 2:2 (1:1). Therefore, the number of moles of water produced is also 0.8928 mol. To calculate the mass of water produced, we need to use the molar mass of water (H2O), which is approximately 18.015 g/mol. Mass of water produced = (number of moles of water) * (molar mass of water) = 0.8928 mol * 18.015 g/mol = 16.075 g. Therefore, approximately 16.075 grams of water can be produced from the reaction of 20 liters of hydrogen gas with 20 grams of oxygen at STP.

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The time of burial, the organic material will be about 34,880 years old.

Half-life (t½) is the time which is required for a quantity of the substance to reduce to the half of its initial value. The term is commonly used in the nuclear physics to describe how quickly the unstable atoms or chemical elements undergo the radioactive decay or how long the stable atoms survive.

The amount of carbon 14 (14C) which can be found in the organic matter decreases due to the radiocarbon process. This process is also called as the radioactive decay. The half-life of carbon-14 (14C) is 5730 years. An organic material which was buried in the sedimentary rocks is examined, and it is the parent-daughter ratio is equal to about 1:15, indicating that there will be 1/16 of the parent element and 15/16 of the daughter element.

The organic material is supposed to have no daughter element at the time of burial. The age of this organic material is to be calculated. As given, the ratio of parent-daughter elements is 1:15 (1/16 parent, 15/16 daughter). After one half-life (i.e., 5730 years), half of the parent atoms will have decayed to the daughter atoms. Therefore, the parent-to-daughter ratio would be 1/32 parent, 31/32 daughter.

After the two half-lives (5730 + 5730 = 11460 years), 1/4 of the original parent atoms will remain, and the ratio will be 1/4 parent, 3/4 daughter. 1/4 is equal to 4/16. 4/16 + 12/16 = 16/16 = 1. This implies that the original amount of carbon 14 (14C) was about 4/16 of what it would have been if there were no daughter material present. To determine the age of the organic material, we may set up the following equation:

Parent to daughter ratio = 1:15 after 2 half-lives,

which is 5730 × 2 = 11,460.15/16 = (1/2)² × (1/16) = 1/64 (15 daughter atoms)

Therefore, there were originally 4 × 15 = 60 carbon 14 (14C) atoms.

1/64 = 1/60 × (1/2)n where n is the number of half-lives which have occurred.

Multiplying both sides by 60 × 64 gives: 1 = 64 × (1/2)n

Subtracting 64 from both sides gives: 63 = (1/2)n

Taking the natural logarithm of both sides gives: ln(-63) = n ln(1/2)

The value of ln(1/2) is -0.69315, so:

n = ln(-63)/ln(1/2)n = 6.0 half-lives have passed (rounded up).

Therefore, the organic material is 6 × 5730 = 34,380 years old.

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In metallic bonding, the valence electrons of metal atoms are free to move throughout the entire crystal structure of the metal.

This mobility of electrons in the metal crystal structure is the key feature that contributes to metallic bonding. The delocalized electrons are attracted to the positively charged metal ions, which hold the electrons together in the crystal lattice through electrostatic forces.

The unique behavior of valence electrons in metal elements makes metallic bonding one of the strongest types of chemical bonds, with high melting and boiling points and high tensile strength.

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The statement is true. A molecule with polar bonds can either be polar or nonpolar depending on the arrangement of the polar bonds. Polar bonds are formed when two atoms with different electronegativities bond, resulting in an unequal sharing of electrons.

The more electronegative atom attracts the electrons towards itself, creating a partial negative charge while the other atom has a partial positive charge.

If the polar bonds in a molecule are arranged symmetrically, such that the partial charges on each atom cancel out, then the molecule is nonpolar.

However, if the polar bonds are arranged asymmetrically, such that the partial charges do not cancel out, then the molecule is polar.

For example, carbon dioxide (CO2) has two polar bonds between the carbon and oxygen atoms, but the molecule itself is nonpolar because the bonds are arranged symmetrically around the carbon atom, canceling out the partial charges.

On the other hand, water (H2O) also has polar bonds between the hydrogen and oxygen atoms, but the molecule is polar because the bonds are arranged asymmetrically, resulting in a partial negative charge on the oxygen atom and a partial positive charge on the hydrogen atoms.

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Equilibrium. Option C is correct

This law states that a new equilibrium state is achieved if the changes in temperature, pressure, concentration and volume will cause a predictable and opposing changes in the system.

This shows that Le Chatelier's principle can be used to predict the properties above to determine the effect equilibrium have on a system.

Based on the above explanations, we can conclude that Le Chatelier's Principle governs the property of Equilibrium.

Kylie drank 35% of a 400-mL container of water. Eugenia drank 42% of a 350-mL container of water.  Eugenia drank more water.

To determine who drank more water, we need to calculate the amount of water consumed by each person.

Kylie drank 35% of a 400-mL container of water, which is equal to:

0.35 x 400 mL = 140 mL of water.

Eugenia drank 42% of a 350-mL container of water, which is equal to:

0.42 x 350 mL = 147 mL of water.

Comparing the two, we see that Eugenia drank more water than Kylie, with a total of 147 mL compared to Kylie's 140 mL. It's important to note that percentage alone is not enough to determine the amount of water consumed - the volume of the container is also a crucial factor.

In this case, although Eugenia consumed a smaller percentage of water, the volume of her container was also smaller, resulting in her consuming a larger amount of water than Kylie.

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The addition of a solution that contains a high concentration of calcium ions (Ca2+) would cause precipitation to occur more easily in sea water than in tap water.

This is because sea water already contains a significant amount of calcium ions, and the addition of more calcium ions would cause them to react with the carbonate ions (CO32-) present in sea water to form calcium carbonate (CaCO3), which is a common type of precipitation in sea water. In tap water, the concentration of calcium ions is much lower, so the addition of a calcium solution would not cause precipitation to occur as easily as in sea water. The main salt components of seawater include Na, K, Mg2+, Ca2+, Sr2+, Cl-, SO42-, HCO3-, Br-, and F-. Precipitation occurs more easily in seawater than in tap water when a solution is added that reacts with these ions, forming insoluble compounds. One such solution is one containing carbonate ions (CO32-), which can react with Ca2+ and Mg2+ ions in seawater to form insoluble calcium carbonate (CaCO3) and magnesium carbonate (MgCO3). This reaction leads to the precipitation of these salts, which is more likely to happen in seawater due to its higher concentration of Ca2+ and Mg2+ ions compared to tap water.

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An equilibrium between AgCl(s) and its dissolved ions occurs when the aqueous solution is saturated.

In a saturated solution, the maximum amount of solute, in this case AgCl, has dissolved in the solvent at a given temperature, and no more solute can dissolve. This results in a dynamic equilibrium where the rate of dissolution of AgCl(s) equals the rate of precipitation of AgCl(s). The concentrations of the dissolved ions (Ag+ and Cl-) remain constant, and the solution is neither supersaturated nor unsaturated.
In an unsaturated solution, the amount of dissolved solute is less than the maximum that the solvent can dissolve. In this case, the AgCl(s) will continue to dissolve until the solution becomes saturated. On the other hand, a supersaturated solution contains more solute than the solvent can typically dissolve at a given temperature. This occurs when the solution is prepared at a higher temperature and then cooled. Supersaturated solutions are unstable and can lead to the precipitation of excess solute, eventually reaching the saturated state and establishing equilibrium between the solid AgCl and its dissolved ions.

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Answer:

107.9

Explanation:

To calculate the amount of silver that will precipitate, we need to use Faraday's laws of electrolysis, which relate the substance produced or consumed in a cell to the amount of electric charge passed through it.

The first law of electrolysis states that the amount of substance produced or consumed in an electrolytic cell is proportional to the amount of electric charge passed through the cell. The proportionality constant is known as the Faraday constant, which is equal to 96,485 coulombs per mole of electrons.

The second law of electrolysis states that the amount of different substances produced or consumed in an electrolytic cell is proportional to their respective chemical equivalents.

In this problem, the substance being produced is silver, which has a chemical equivalent of 1 (i.e., one mole of electrons produces one mole of silver). The electric charge passed through the cell can be calculated as follows:

electric charge = current x time

electric charge = 5 A x 2.5 hours x 3600 s/hour

electric charge = 45,000 C

The amount of silver that will precipitate can be calculated using the first law of electrolysis and the Faraday constant:

moles of silver = electric charge / Faraday constant

moles of silver = 45,000 C / 96,485 C/mol

moles of silver = 0.467 mol

To convert moles of silver to mass, we need to know the molar mass of silver. The molar mass of silver is 107.87 g/mol. One teaspoon is approximately 5 mL.

mass of silver = moles of silver x molar mass

mass of silver = 0.467 mol x 107.87 g/mol

mass of silver = 50.4 g

number of teaspoons = mass of silver / (density of silver x volume of teaspoon)

Assuming the density of silver is 10.5 g/mL, we get:

number of teaspoons = 50.4 g / (10.5 g/mL x 5 mL)

number of teaspoons = 0.96 teaspoon

Therefore, approximately 0.96 teaspoons of silver will precipitate.

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a) Melting: An example of melting that occurs in our everyday life is when we heat butter on a stovetop.

b) Freezing: Freezing is the process in which a liquid transforms into a solid upon cooling.

c) Condensation: One example of condensation that we encounter regularly is when water droplets form on the surface of a cold drink on a hot day.

d) Evaporation: Evaporation is the process by which a liquid transforms into a gas or vapor.

e) Sublimation: Sublimation refers to the transformation of a substance directly from a solid to a gas without passing through the liquid state.

a) Melting:  Butter is a solid at room temperature, but when heat is applied, it melts into a liquid. This change is a result of the increase in temperature, which provides enough energy to overcome the intermolecular forces holding the butter molecules together.

b) Freezing:Eventually, the temperature reaches the freezing point of water (0°C or 32°F), at which the water molecules slow down and arrange themselves into a regular, crystalline structure. This transformation from a liquid to a solid state is accompanied by the release of heat energy.

c) Condensation: As the temperature decreases, the air's capacity to hold moisture decreases, causing the water vapor in the air to condense into liquid water droplets. This process occurs due to the transfer of heat energy from the warm air to the cold surface, leading to the saturation of the air and the conversion of water vapor into liquid form.

d) Evaporation:  As the sun's heat energy is absorbed by the water molecules on the clothes' surface, their kinetic energy increases, causing them to break free from the liquid phase and escape into the surrounding air as water vapor. This process occurs because the molecules at the liquid surface with sufficient energy can overcome the attractive forces within the liquid and enter the gas phase.

e) Sublimation: Sublimation refers to the transformation of a substance directly from a solid to a gas without passing through the liquid state. An example of sublimation is the process of dry ice (solid carbon dioxide) converting into carbon dioxide gas.

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The statement  "Believing that, with training, you will be able to run two miles without stopping by the end of the semester demonstrates self-efficacy." is True.

Self-efficacy is an individual's belief in their ability to successfully perform a specific task or complete a particular goal. Believing that, with training, you will be able to run two miles without stopping by the end of the semester is a demonstration of self-efficacy. This belief in one's abilities can have a significant impact on motivation, effort, and performance, as individuals who have high self-efficacy are more likely to set challenging goals and persist in the face of obstacles and setbacks.

Self-efficacy can be developed and strengthened through various experiences, including mastery experiences (successes), vicarious experiences (observing others), emotional and physiological states, and social persuasion. An individual's self-efficacy can also be affected by situational and dispositional factors, such as their confidence, past experiences, and the difficulty of the task at hand.

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Answer:

See explanation

Explanation:

The number of electrons in each principal energy shell increases as the number of shells increases because more electronic orbitals become available to accommodate the electrons.

For instance, the n= 2 level only accommodates eight electrons in the s and p orbitals whereas the n=3 level accommodates 18 electrons in s, p and d orbitals respectively.

Each principal level accommodates 2n^2 electrons where n= the principal energy shell.

The increase in the number of shell leads to increase in the number of electrons.

The number of electrons in each principal energy shell increase as the number of the shell increases because each cell can accommodate 8 electrons except the first shell which has only 2 electrons. The number of electrons increases when we go down the group due to addition of new shells.

The atom added new shells in the atom if the existing shells fill with electrons and no place is available for the remaining electrons so we can conclude that the increase in the number of shell leads to increase in the number of electrons.

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Answer:

The answer is C.

Explanation:

Answer:

C is right i yeah C is right

Answer:

Ammonia

Explanation:

I believe that ammonia is the correct answer to this question.

Considering the definition of STP conditions, a mass of 20.10 grams of chlorine gas is found in a 12.7 L sample at STP.

STP conditions refer to standard temperature and pressure, using 1 atmosphere and 0 °C as reference values for gases. Under these conditions, 1 mole of any gas occupies an approximate volume of 22.4 liters.

The molar mass of substance is the amount of mass that a substance contains in one mole.

In this case, you can apply the following rule of three: if by definition of STP conditions 22.4 liters of chlorine is occupied by 1 mole of chlorine, 12.7 liters is occupied by how many moles of chlorine?

moles= (12.7 liters× 1 mole)÷ 22.4 liters

moles= 0.567 moles

The molar mass of chlorine is 35.45 g/mole.

Now,  you can apply the following rule of three: If by definition of molar mass 1 mole of chlorine has a mass of 35.45 grams, 0.567 moles of chlorine has how much mass?

mass= (0.567 moles× 35.45 grams)÷ 1 mole

mass= 20.10 grams

Finally, a mass of 20.10 grams of chlorine gas is present.

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The volume of the gas is approximately 625.8 mL.

To solve this problem, we can use the combined gas law:

\((P1V1)/T1 = (P2V2)/T2\)

where P is pressure, V is volume, and T is temperature in Kelvin.

We are given the initial volume (V1) as 300.0 mL at 17.0C, which is 290.15 K (since we need to convert Celsius to Kelvin by adding 273.15). We want to find the final volume (V2) at 333.0C, which is 606.15 K.

We can assume that the pressure (P) remains constant, so we can simplify the formula to:

\(V1/T1 = V2/T2\)

Solving for V2, we get:

\(V2 = (V1 * T2)/T1\)

Plugging in the values, we get:

\(V2 = (300.0 * 606.15)/290.15\\V2 = 625.8 mL\)

Therefore, the volume of the gas at 333.0C is approximately 625.8 mL.

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iven data:Aqueous solution contains NaOH and ethyl acetate,Initial concentration of NaOH and ethyl acetate=0.1 MConversion of ethyl acetate=18%Operating temperature of reactor (T1)=300 KDesired product=C2H5OHProduction rate=10 mol/minVolumetric flow rate of ethyl acetate (V1)= 5 dm³/minVolumetric flow rate of NaOH (V2)= 5 dm³/minOperating temperature of CSTR (T2)= 310 KActivation energy(Ea)= 82,000 cal/molTo find:

A chemical reactor is a vessel where a chemical reaction takes place. The design of this reactor depends on many variables that can be studied in chemical engineering.

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Taking into account the reaction stoichiometry, 11.2 grams of BaSO₄ are produced from 10 g of barium chloride.

In first place, the balanced reaction is:

BaCl₂ + H₂SO₄ → BaSO₄ + 2 HCl

By reaction stoichiometry, the following amounts of moles of each compound participate in the reaction:

The molar mass of the compounds is:

By reaction stoichiometry, the following mass quantities of each compound participate in the reaction:

The following rule of three can be applied: if by reaction stoichiometry 208.24 grams of BaCl₂ form 233.34 grams of BaSO₄, 10 grams of BaCl₂ form how much mass of BaSO₄?

mass of BaSO₄= (10 grams of BaCl₂× 233.34 grams of BaSO₄) ÷208.24 grams of BaCl₂

mass of BaSO₄= 11.2 grams

Finally, 11.2 grams of BaSO₄ are formed.

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The molar concentration of chloride ions in a 2.6 M solution of MgCl₂ can be calculated using the molar ratio of chloride ions to MgCl₂.

To find the molar concentration of chloride ions, we need to consider the molar ratio of chloride ions to MgCl₂ in the solution. MgCl₂ dissociates in water to form one Mg²⁺ ion and two Cl⁻ ions. This means that for every one mole of MgCl₂, we have two moles of chloride ions.

Given that the solution is 2.6 M, it means that there are 2.6 moles of MgCl₂ per liter of solution. Since the molar ratio of chloride ions to MgCl₂ is 2:1, we can multiply the molar concentration of MgCl₂ by the ratio to find the molar concentration of chloride ions.

Molar concentration of chloride ions = 2.6 M MgCl₂ × 2 = 5.2 M Cl⁻

Therefore, the molar concentration of chloride ions in the solution is 5.2 M.

It's important to note that the density of the solution is not directly used in this calculation, as the molar concentration is determined by the number of moles of solute (MgCl₂) in a given volume of solution, rather than its mass.

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To determine the energy of a photon required for an electronic transition from the n=2 state to the n=5 state in a hydrogen atom, we can use the formula for the energy of a photon:

E = ΔE = hc/λ

Where:

- E is the energy of the photon

- ΔE is the change in energy between the initial and final states

- h is Planck's constant (approximately 6.626 x 10^-34 joule-seconds)

- c is the speed of light (approximately 3 x 10^8 meters per second)

- λ is the wavelength of the photon

The energy difference between two energy levels in a hydrogen atom is given by the Rydberg formula:

ΔE = Rh * (1/n_f^2 - 1/n_i^2)

Where:

- ΔE is the change in energy

- Rh is the Rydberg constant (approximately 2.18 x 10^-18 joules)

- n_f is the final energy level (n=5 in this case)

- n_i is the initial energy level (n=2 in this case)

Substituting the values into the Rydberg formula:

ΔE = Rh * (1/5^2 - 1/2^2)

= Rh * (1/25 - 1/4)

= Rh * (4/100 - 25/100)

= Rh * (-21/100)

≈ -0.0218 * Rh

Now, we can substitute this change in energy value into the energy formula for the photon:

E = hc/λ = -0.0218 * Rh

Rearranging the equation to solve for λ:

λ = hc / E

Substituting the values for h, c, and E:

λ = (6.626 x 10^-34 joule-seconds * 3 x 10^8 meters per second) / (-0.0218 * Rh)

Calculating this expression will give us the wavelength of the photon required for the electronic transition.

In the balanced chemical reaction of MnO₂, the coefficient of H⁺ is 4.

Balanced chemical reaction

The chemical equations which have the same number and same type of each atom on both sides of the equation is known as balanced chemical equation. The coefficients in a balanced equation must be always the simplest whole number ratio. Mass is always conserved in chemical reactions i.e., neither destroyed nor created.

Balanced chemical reaction of MnO₂ is given as:

5MnO₂ + 4H⁺ → 3Mn₂ + 4H₂O + 2MnO₄⁻

Hence, the coefficient of H⁺ in the balanced chemical equation of reduction MnO₂ is 4.

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Answer:

Rate = k [HO2]

Rate constant = 0.8456us-1

Explanation:

The rate law is given as;

Rate = k [HO2]^x

Where x signify the order of reaction.

For an order of reaction, the rate constant is constant for all concentrations. We are going to use this to obtain the order of reaction.

Zero Order:

[A] = [A]o -kt

5.1 = 8.5 - k(0.6)

-k (0.6) = 5.1 - 8.5

k = 5.67

3.6 = 5.1 - k(0.4)

-k (0.4) = 3.6 - 5.1

k = 3.75

The fact that the rate constant was not constant means the reaction is not a zero order reaction.

First Order:

ln[A] = ln[A]o -kt

(5.1) = ln(8.5) - k(0.6)

-k (0.6) = ln(5.1) - ln(8.5)

k = 0.8524

ln(3.6) = ln(5.1) - k(0.4)

-k (0.4) = ln(3.6) - ln(5.1)

k = 0.8708

ln(2.6) = ln(3.6) - k (0.4-k (0.4) = ln(2.6) - ln(3.6)

k = 0.8136

From the three calculations we see that the value of the rate constant is fairly constant in the range of 0.8 This means our reaction is a first order reaction.

The rate law is given as;

Rate = k [HO2]

We can represent the rate constant as the average of the three rate constants calculated above;Rate constant = (0.8136 + 0.8708 + 0.8524 / 3)

Rate constant = 0.8456us-1

Answer:

Explanation:

The cell potential under these nonstandard concentrations is 0.22 V.

To calculate the cell potential under nonstandard conditions, we need to use the Nernst equation:

Ecell = E°cell - (RT/nF)ln(Q)

where E°cell is the standard cell potential, R is the gas constant (8.314 J/mol*K), T is the temperature in Kelvin (25°C = 298 K), n is the number of electrons transferred in the balanced equation (2 for both half-reactions), F is Faraday's constant (96,485 C/mol), and Q is the reaction quotient.

The reaction quotient can be expressed as:

Q = [Ni2+] / ([VO2+] [H+] / [VO2+] [H+]°)

where [H+]° is the standard concentration of hydrogen ions, which is 1 M. Plugging in the given concentrations and standard reduction potentials, we get:

Q = (2.0 M) / [(0.012 M)(1.1 M)/(2.0 M)(1.0 M)] = 298.18

Now we can calculate the cell potential:

Ecell = 0.23 V - (8.314 J/molK / (296,485 C/mol) * ln(298.18))

Ecell = 0.23 V - (0.00573 V)

Ecell = 0.22427 V

Therefore, the cell potential under these nonstandard concentrations is 0.22 V.

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