when the equation is nonstandard (non 1 M) use equation for nonstandard cell potential

Answer:

7.1 g

Explanation:

A way to figure this out is by using the molecular formula.

\(Na + Cl\) ⇒ \(NaCl\)

(Sodium + Chloride ⇒ Sodium Chloride)

From this formula it's one mol sodium + one mol chloride = one mol of sodium chloride.

Then you can substitute the values:

4.6 grams sodium + x grams chlorine = 11.7 g of sodium chloride

By using algebra you get:

x = 11.7 - 4.6 = 7.1 g

You can check this by using the molar masses of sodium and chlorine which can take a more time and is kind of unnecessary.

Hope that helps!

The major factors influencing fusion consist of the required high temperature and high pressure which is not shown in the model and hence becomes one of its limitations.

Two major requirements for fusion are:

1) The energy from the high temperature allows the hydrogen atoms to overcome the electrical attraction between the protons. Temperatures of roughly 100 million Kelvin are necessary for fusion. Hydrogen is not a gas at these temperatures; it is a plasma. The high-energy state of matter known as plasma is one in which all atoms have had their electrons removed and are now free to move about. The sun's massive mass and the compression of that mass in the core caused by gravity allow it to reach these temperatures.

2) The hydrogen atoms are compressed together by high pressure. For Fusion, they need to be within 1x10-15 meters of one another. The sun compresses hydrogen atoms together in its core using gravity and its bulk. We must use strong magnetic fields, potent lasers, and ion beams to force hydrogen atoms together.

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There are numerous methods for remembering amino acid abbreviations. Here are some pointers:

Employ flashcards, Make a mnemonic, Employ visual aids, Regular practise.

Employ flashcards: Put the amino acid's abbreviation and full name on one side of a flashcard and its chemical structure on the other. This will assist you in associating the abbreviation with the full name as well as the chemical structure of the amino acid.

Make a mnemonic: Come up with a word or phrase that will help you remember the first letter of each amino acid abbreviation. "Goodness gracious, always have fast cars, get speed tickets," for example, can aid in remembering the initial letters of the amino acids Glycine, Alanine, Histidine, Phenylalanine, Cysteine, Glutamic acid, Aspartic acid, Glutamine, Serine, and Threonine.

Employ visual aids: Match each amino acid abbreviation with an image that illustrates the full name or the amino acid's qualities. For example, "C" (Cysteine) corresponds to a cyst, "G" (Glycine) corresponds to a jelly bean, and "W" (Tryptophan) corresponds to a waffle.

Regular practise: The more you do it, the easier it will be to memorise amino acid abbreviations. While learning amino acids, test yourself on a regular basis and utilise acronyms as much as feasible.

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Answer:

I DONT KNOW SORRY

Explanation:

Answer:

Iodine-131 is more likely to undergo beta decay than positron decay.

answer

Gives brown colour gas

The chemical process known as dehydration synthesis, also known as a condensation reaction, involves the covalent bonding of two molecules to create a new molecule while simultaneously releasing a water molecule.

Complex carbohydrates, proteins, DNA, and RNA are just a few of the molecules that are formed via dehydration synthesis processes and are crucial to human health.

Dehydration synthesis forms covalent bonds between sugar molecules. The hydroxyl (-OH) group from one reactant joins with the hydrogen atom from the other to generate water throughout the reaction, while the leftover oxygen forms a glycosidic bond that binds the two molecules together. Repeated dehydration synthesis of individual glucose molecules (monomers) can result in the formation of a polymer, which is a long chain or branched chain structure.

Dehydration synthesis processes, in which one monomer establishes a covalent connection to another monomer (or expanding chain of monomers), release a water molecule in the process, are frequently used to create large biological molecules. Dehydration stands for the loss of the water molecule and synthesis for the development of a new bond, making it easy to recall what occurs.

The methods through which monomers can combine to produce polymers are quite diverse. For instance, starch, cellulose, and glycogen are all made up of glucose monomers. These three polysaccharides, which are categorized as carbohydrates, were created as a result of several glucose monomer-to-glucose dehydration synthesis processes. These three distinct polysaccharides, however, have varied characteristics and functions because of the way that glucose monomers combine, particularly the positions of the covalent connections between linked monomers and the orientation (stereochemistry) of the covalent bonds.

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Answer:

The correct option is;

D. Manganese (Mn)

Explanation:

Manganese is very brittle, hard, iron like silvery-gray metal, that is difficult to melt. In air, Manganese slowly disintegrate in a similar manner to iron rusting in water

Manganese and iron have similar chemical and physical properties however manganese is more harder and more brittle than iron

A brittle material is one that easily breaks without deforming elastically

Therefore, manganese, due to its very iron like appearance and brittle nature will be suitable to replace the metal wall and crumble easily when the actor hits it.

Carboxylic acids are organic acids that contain the carboxyl functional group (–COOH) as their structural feature.  the structures of the carboxylic acids with the formula C6H12O2 that contain an ethyl group branching off the main chain are pentanoic acid and 3-methylbutanoic acid.

They can be found in various organic materials such as fruits, fats, and oils. The structure(s) of carboxylic acids with the formula C6H12O2 that contain an ethyl group branching off the main chain can be represented as follows:Two isomers can be possible for the given formula C6H12O2. They are pentanoic acid and 3-methylbutanoic acid.Pentanoic acid has a straight-chain of five carbon atoms (pentane) with a carboxyl group at one end and an ethyl group branching off from the fourth carbon atom. The structure of pentanoic acid is as follows:3-Methylbutanoic acid is a branched-chain carboxylic acid in which the carboxyl group is attached to the third carbon atom of a four-carbon chain, with an ethyl group attached to the second carbon atom. The structure of 3-methylbutanoic acid is as follows:Therefore, the structures of the carboxylic acids with the formula C6H12O2 that contain an ethyl group branching off the main chain are pentanoic acid and 3-methylbutanoic acid.

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Compound is a group of same molecules, Element is a group of same atoms and mixture is combination of two or more pure substances.

What are Molecules ?

Molecules are the combination of two or more different atoms combine in a fixed ratio.

Molecules can be independently founded.

Atoms are always founded in combined state and does not occurs independently except rare earth elements which belongs to 18th group of periodic table.

Chemicals are taken in compound state But they undergo chemical reaction individually in molecular level.

Hence, Compound is a group of same molecules, Element is a group of same atoms and mixture is combination of two or more pure substances.

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Answer:

(limonene) particles go from liquid to vapour

diffusion

random movement of particles / particles move anywhere / particles move in all directions

spreading out of particles / intermingling of particles / mixing of particles / particles collide / particles bounce off each

other / particles go all over

(bulk) movement of particles from higher to lower concentration / movement of particles down concentration gradient

Ques.1: Why do you think that you had to use a thin layer of onion skin, rather than a thick layer for the microscope?

Ans:- It is because, I the onion peel is thick layered, then it will be hard or impossible to see through the microscope.

Ques.2: When iodine reacts with starch, it produces a blue-black color. Starch is a white substance which plants use to store food. What structure did you see better because of the iodine? Why did you see this structure better?

Ans: Amylose in starch is responsible for the formation of a deep blue color in the presence of iodine. The iodine molecule slips inside of the amylose coil.

This makes a linear triiodide ion complex with is soluble that slips into the coil of the starch causing an intense blue-black color.

Due to this, we can see the structure better.

Ques.3: From low power to high power, what structures became clearer in the stained onion tissue?

Ans: From lower to higher power, we can see:-

-- Large, rectangular interlocking cells,

-- Clearly visible distinct cell walls surrounding the cells,

-- Dark stained nucleus,

-- Large vacuoles at the center,

-- Small granules may be observed inside the cells (within the cytoplasm)

Ques.4: What is the shape of an onion cell?

Ans: The shape of an onion cell is rectangular or square in shape.

Yes, More heat speeds up the dissolving reaction by supplying energy to break bonds in the solid. This is the most prevalent case in which an increase in temperature causes an increase in solid solubility.

As the temperature of the solution rises, so does the average kinetic energy of the solute molecules. As a result, the molecules are less able to stick together and disintegrate more easily. As a result, increasing the temperature increases the solubility of solid states.

Solute and solvent crystal structures must be identical. When the solvent and solute have the same valency, complete solubility occurs. A metal with a higher valency is more likely to dissolve a metal with a lower valency. The electronegativity of the solute and solvent should be comparable.

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1. The volume of water in the tank at time t is given by the equation

Volume(t) = 30 + t.

2.The appropriate initial condition for the ODE is S(0) = 0, as there is no salt initially in the tank.

3. It is not separable because the variables S(t) and t are not separable on opposite sides of the equation.

4. The solution can be expressed in terms of the integral as:
\(S(t) = (70 * \int e^{(t^{2/2} + 30t)} dt) / e^{(t^{2/2} + 30t)})\)

5.  the salt concentration in the tank as t→infinity is zero.

1. To determine the volume of water in the tank at time t, we need to consider the rate at which water is being added and drained. The tank is being filled at a rate of 2 liters per minute and drained at a rate of 1 liter per minute.

Since the tank starts with an initial volume of 30 liters, the volume of water in the tank at time t can be calculated using the equation:
Volume(t) = Initial volume + (Rate of filling - Rate of draining) * t

Volume(t) = 30 + (2 - 1) * t

So, the volume of water in the tank at time t is given by the equation

Volume(t) = 30 + t.

2. Let S(t) denote the amount of salt in the fish tank at time t in grams.

To show that S(t) satisfies the ODE S'(t) = 70 - (t+30)S(t),

we need to take the derivative of S(t) with respect to t and substitute it into the given ODE.

Taking the derivative of S(t), we have:

S'(t) = 0 - (1+0)S(t) + 0

S'(t) = -S(t)

Substituting this into the given ODE, we get:

-S(t) = 70 - (t+30)S(t)

Simplifying the equation, we have:

S'(t) = 70 - (t+30)S(t)

Therefore, S(t) satisfies the ODE S'(t) = 70 - (t+30)S(t).

The appropriate initial condition for the ODE is S(0) = 0,

as there is no salt initially in the tank.

3. This ODE is a first-order linear ordinary differential equation. It is not separable because the variables S(t) and t are not separable on opposite sides of the equation.

4. To solve the initial value problem for S(t) using the method of integrating factors, we first rewrite the ODE in standard form:

S'(t) + (t+30)S(t) = 70

The integrating factor is given by:
\(\mu(t) = e^{(\int (t+30) dt)} = e^{(t^2/2 + 30t)\)

Multiplying both sides of the equation by μ(t), we have:
\(e^{(t^2/2 + 30t)} * S'(t) + e^{(t^2/2 + 30t)} * (t+30)S(t) = 70 * e^{(t^2/2 + 30t)\)

Applying the product rule to the left side of the equation, we get:
\((e^{(t^{2/2} + 30t) * S(t))' = 70 * e^{(t^{2/2} + 30t)})\)

Integrating both sides of the equation with respect to t, we have:
\(\int (e^{(t^2/2 + 30t)} * S(t))' dt = \int (70 * e^{(t^2/2 + 30t))} dt\)

Using the fundamental theorem of calculus, the left side becomes:
\(e^{(t^2/2 + 30t)} * S(t) = \int (70 * e^{(t^2/2 + 30t))} dt\)

Simplifying the right side by integrating, we get:
\(e^{(t^2/2 + 30t)} * S(t) = 70 * \int e^{(t^2/2 + 30t)} dt\)

At this point, the integration of \(e^{(t^2/2 + 30t)\) becomes difficult to express in terms of elementary functions.

Hence, the solution can be expressed in terms of the integral as:
\(S(t) = (70 * \int e^{(t^2/2 + 30t)} dt) / e^{(t^2/2 + 30t)\)

5. As t approaches infinity, the exponential term \(e^{(t^2/2 + 30t)\) becomes very large, causing the salt concentration S(t) to approach zero. Therefore, the salt concentration in the tank as t→infinity is zero.

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The salt concentration in the tank as t approaches infinity is 70/3.

1. To determine the volume of water in the tank at time t, we need to consider the rate at which water is being poured into the tank and the rate at which water is being drained from the bottom.

At a rate of 2 litres per minute, water is being poured into the tank. So after t minutes, the amount of water poured into the tank is 2t litres.

At a rate of 1 litre per minute, water is being drained from the tank. So after t minutes, the amount of water drained from the tank is t litres.

Since the tank was initially filled with 30 litres of fresh water, the volume of water in the tank at time t is given by:
Volume(t) = 30 + 2t - t
Volume(t) = 30 + t

2. Let S(t) denote the amount of salt in the fish tank at time t. To determine the ODE for S(t), we need to consider the salt being poured into the tank and the salt being drained from the tank.

The salt concentration in the water being poured into the tank is 35 grams per litre. So the amount of salt being poured into the tank per minute is 35 * 2 = 70 grams.

The amount of salt being drained from the tank per minute is S(t)/Volume(t) * 1.

Therefore, the ODE for S(t) is:
S'(t) = 70 - S(t)/Volume(t)

The initial condition for this ODE is S(0) = 0, since there was no salt in the tank initially.

3. The ODE S'(t) = 70 - S(t)/Volume(t) is a first-order linear ODE. It is not separable since the variables S(t) and Volume(t) are mixed together.

4. To solve the initial value problem for S(t), we can rewrite the ODE as:
Volume(t) * S'(t) + S(t) = 70 * Volume(t)

This is a linear ODE of the form y'(t) + p(t)y(t) = g(t), where p(t) = 1/Volume(t) and g(t) = 70 * Volume(t).

To solve this type of ODE, we can multiply both sides by an integrating factor, which is the exponential of the integral of p(t).

The integrating factor is exp(integral of 1/Volume(t) dt) = exp(ln(Volume(t))) = Volume(t).

Multiplying both sides of the ODE by the integrating factor, we get:
Volume(t) * S'(t) + S(t) = 70 * Volume(t)
Volume(t) * S'(t) + Volume(t) * S(t) = 70 * Volume(t)^2
( Volume(t) * S(t) )' = 70 * Volume(t)^2

Integrating both sides with respect to t, we get:
Volume(t) * S(t) = 70/3 * Volume(t)^3 + C

S(t) = 70/3 * Volume(t)^2 + C/Volume(t)

Using the initial condition S(0) = 0, we can solve for C:
0 = 70/3 * 30^2 + C/30
C = -70000

Therefore, the solution for S(t) is:
S(t) = 70/3 * Volume(t)^2 - 70000/Volume(t)

5. As t approaches infinity, the volume of water in the tank becomes very large. In this case, we can approximate the volume of the tank as t, since the rate at which water is being poured in is 2 litres per minute. So the salt concentration in the tank as t approaches infinity is given by:
S(t)/Volume(t) = (70/3 * t^2 - 70000/t) / t
As t approaches infinity, the second term (-70000/t) approaches 0, so the salt concentration in the tank as t approaches infinity is:
S(t)/Volume(t) = 70/3 * t^2 / t = 70/3 * t

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If the ratio of acid to base in a buffer increases by a factor of 100, the pH of the buffer remains unchanged (c).

The pH of a buffer solution is determined by the ratio of acid to base components and their respective dissociation constants. In a buffer, the acid and its conjugate base exist in equilibrium, and this equilibrium determines the pH of the solution.

When the ratio of acid to base in a buffer increases by a factor of 100, it means that the concentration of the acid component becomes 100 times greater compared to the base component. This change in concentration, however, does not affect the dissociation constants of the acid and its conjugate base.

The pH of a solution is a measure of its acidity or alkalinity and is defined as the negative logarithm of the hydrogen ion concentration ([H+]). In a buffer solution, the pH is determined by the equilibrium between the acid and base components. Increasing the concentration of the acid while keeping the base concentration constant does not alter the equilibrium and, therefore, does not change the pH of the buffer solution.

Hence, when the ratio of acid to base in a buffer increases by a factor of 100, the pH of the buffer remains unchanged.

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When alkaline hydrolysis was first invented, people were hired for various roles related to the process and implementation of this technology. Some of the jobs that emerged include Chemical engineers, Technicians and operators, Waste management specialists, Scientists and researchers.

Chemical engineers: These professionals played a crucial role in developing and optimizing the alkaline hydrolysis process. They were responsible for designing the equipment, developing the necessary chemical reactions, and ensuring the efficient operation of the system.

Technicians and operators: Skilled technicians and operators were hired to operate and maintain the alkaline hydrolysis equipment. They were trained to monitor the process parameters, handle the chemicals involved, and ensure the proper functioning of the system.

Waste management specialists: With the introduction of alkaline hydrolysis as a method for disposal of organic waste, specialized professionals in waste management were employed to oversee the proper handling and treatment of the waste materials. They were responsible for implementing safety protocols, managing waste streams, and complying with environmental regulations.

Scientists and researchers: Alkaline hydrolysis required scientific expertise for continuous improvement and innovation. Scientists and researchers were hired to study the process, analyze the results, and explore potential applications in various fields such as biofuel production and chemical synthesis.

Overall, the introduction of alkaline hydrolysis created employment opportunities for professionals in engineering, chemistry, waste management, and research, among others, as this technology gained recognition and adoption.

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Answer:

Water in a freezer at -2 degrees Celsius

Answer:  Chemistry, like physics and biology, is a natural science. In fact, there is considerable overlap between chemistry and these other disciplines. Chemistry is a science that studies matter. This includes atoms, compounds, chemical reactions, and chemical bonds.

Explanation:  Chemistry is the scientific study of the properties and behavior of matter. It is a natural science that covers the elements that make up matter to the compounds made of atoms, molecules, and ions: their composition, structure, properties, behavior, and the changes they undergo during a reaction with other substances.

Answer:

Never pour water into acid but acid into water

Explanation:

If water is poured into extremely concentrated acid/bases, the rate of volatility and exothermic reaction is too rapid and might cause a chemical eruption, leading to acid burns.

Safety precautions hence dictate the reverse is practiced.

I believe this is a clear answer.

The original concentration of the protein before dilution is 0.16 mM.

Explanation:

Molar extinction coefficient is defined as the amount of light absorbed by the substance per unit concentration. If it has higher molar extinction coefficient, it indicates that the protein solution is highly concentrated. Thus, it will absorb more light.

As per the question,

Given, The molar extinction coefficient of the protein sample at 280 nm is ε = 0.25 L μmol⁻¹ cm⁻¹.

Dilution factor = 20 (50 μL of the protein sample is diluted to a final volume of 1 mL).

Measured absorbance, A = 0.40.

Path length of the cuvette, b = 0.5 cm.

Now, we need to calculate the original concentration of the protein before dilution.

So, the formula is, A = εbc

Where, ε = molar extinction coefficient of the protein.

b = path length of the cuvette.

c = original concentration of the protein.

Substituting the given values,

0.4 = 0.25 × 0.5 × c (As, we need to find the original concentration of the protein)

c = 0.4/0.125

c = 3.2 mM

This is the concentration of the protein after dilution.

But, we need to find the original concentration of the protein before dilution.

Therefore, the original concentration = diluted concentration / dilution factor

c = 3.2 mM / 20

c = 0.16 mM

Therefore, the original concentration of the protein before dilution is 0.16 mM.

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The chemical equations of the following compounds shows their respective pH. Chemical equations are given below.

Chemical Equations

A equation that consists of reactants, products and an arrow showing the direction of reaction is known as a chemical equation. The equation is said to be balanced chemical equation when the number of atoms of all the molecules is equal on both sides of the equation.

For Al(NO₃)₃

Al(NO₃)₃ + H₂O → Al³⁺ + 3NO₃⁻

Al³⁺ + 6H₂O ⇄ Al(H₂O)₆³⁺

Al(H₂O)₆³⁺ + H₂O ⇄ Al(OH)(H₂O)₅²⁺ + H₃O⁺

pH > 7

For Zn(NO₃)₂

Zn(NO₃)₂ → Zn²⁺ + 2NO₃⁻

Zn(H₂O)₆³⁺ + H₂O ⇄ Zn(OH)(H₂O)₅²⁺ + H₃O⁺

pH < 7

For NH₄NO₃

NH₄NO₃ → NH₄⁺ + NO₃⁻

NH₄⁺ + H₂O ⇄ NH₃ + H₃O⁺

pH < 7

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Answer:

check my explanation

Explanation:

it is the secondary consumer because it doesn't eat grass

Please provide the calorimeter data, and we will be able to complete the calculations and determine the latent heat of fusion of mercury.

To determine the latent heat of fusion of mercury with a specific heat of 1.38 J/kg°C, we will use the calorimeter data provided in the question.

Step 1: Identify the data given.

Unfortunately, the calorimeter data is not provided in the question. Please provide the data for us to proceed with the calculations.

The required data includes mass, initial temperature, and final temperature of mercury and the calorimeter.

Step 2: Calculate heat absorbed or released by mercury.

Once the calorimeter data is provided, we can use the formula Q = mcΔT, where Q is the heat absorbed or released, m is the mass of mercury, c is the specific heat (1.38 J/kg°C),

and ΔT is the change in temperature (final temperature - initial temperature).

Step 3: Calculate heat absorbed or released by the calorimeter.

Using the calorimeter data, we can calculate the heat absorbed or released by the calorimeter as well. The formula is the same: Q = mcΔT.

However, the specific heat and mass will be different as they correspond to the calorimeter.

Step 4: Calculate the total heat absorbed or released.

Since the heat absorbed by one substance is equal to the heat released by the other, we can add the two heats calculated in steps 2 and 3 to get the total heat absorbed or released (Q_total).

Step 5: Determine the latent heat of fusion.

Finally, we can determine the latent heat of fusion (L) using the formula L = Q_total / m, where m is the mass of mercury. This will give us the latent heat of fusion in J/kg.

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Glycinate donates an electron pair so it is a bidentate ligand.

The molecular formula is C₂H₄NO₂⁻. The octahedral complex is formed between glycinate molecules and cobalt(III) and the stoichiometry of the complex is [Co(gly)₃]. The reaction is as follows;

Co₃⁺(aq) + 3C₂H₄NO₂⁻ ⇒ [Co(C₂H₄NO₂⁻](aq)

A cobalt complex is formed when 3 glycinate ions equivalents react with one Co₃⁺ ion equivalent so, it is necessary to keep the glycinate ions concentration greater than the cobalt(III) ions at least three times more.

So, taking the concentration 4 times greater can facilitate the reaction.

For a complex whose concentration is 0.015M, 0.06M glycinate ions are required to obtain the desired cobalt(III) glycinate complex.

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Answer:

15900

Explanation:

I think hope this helps!! :D

Answer:

15900

Explanation:

In a 1.00E2 gram sample of the unknown substance, there are 252.50 grams of hydrogen

Step 1: Find the number of moles of hydrogen using the given mass of the unknown.

Moles of hydrogen} = mass of hydrogen/molar mass of hydrogen

\($$ \text{mass of hydrogen} = \text{mass of the unknown} \times \frac{\text{number of moles of hydrogen}}{\text{number of moles of unknown}} $$\)

The number of moles of hydrogen in 1 mole of the unknown is 2.50, as given in the question.

\($$ \text{mass of hydrogen} = 1.00e2 \text{ g} \times \frac{2.50 \text{ moles of hydrogen}}{1 \text{ mole of unknown}} \times \frac{1.01 \text{ g}}{1 \text{ mole}} = \boxed{252.50 \text{ g}} $$\)

Therefore, there are 252.50 grams of hydrogen in a 1.00E2 g sample of the unknown.

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The minimum frequency of light required to ionize boron is approximately 2.01 x 10^15 Hz.

To find the minimum frequency of light required to ionize boron, we can use the equation E = hf, where E is the energy required to ionize boron and f is the frequency of light.

Given that the energy required to ionize boron is 801 kJ/mol, we can convert this to joules by multiplying by the conversion factor 1000 J/1 kJ and dividing by Avogadro's number (6.022 x 10^23 mol^-1) to get the energy per atom.

801 kJ/mol * (1000 J/1 kJ) / (6.022 x 10^23 mol^-1) = 1.33 x 10^-18 J

Now, we can rearrange the equation to solve for the frequency of light:

f = E / h

Substituting the values:

f = (1.33 x 10^-18 J) / (6.626 x 10^-34 J·s) = 2.01 x 10^15 Hz

Therefore, the minimum frequency of light required to ionize boron is approximately 2.01 x 10^15 Hz.

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Ionization energy is the energy required to remove an electron from an atom or ion in its ground state. It is often expressed in units of joules (J) or kilojoules per mole (kJ/mol).  2.01 x 10¹⁵ Hz is the minimum frequency of light is required to ionize boron

Ionization energy per atom

= ionization energy per mole/Avogadro's number

= 801/6.022 x 10²³

= 1.330 x 10⁻²¹ kJ = 1.330 x 10⁻¹⁸ J

Photon energy E = hf

where h is the Planck constant and f is the frequency

Photon energy = ionization energy per atom

E=hf = 1.330 x 10⁻¹⁸

Frequency f = 1.330 x 10⁻¹⁸/6.626 x 10⁻³⁴

= 2.01 x 10¹⁵ Hz

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