(a) what volume of chloroform is needed to extract 99.5% of a solute from 100 ml of water, if the partition coefficient is cchcl3/ch2o-610?

Answers

Answer 1

The volume of chloroform needed to extract 99.5% of a solute from 100 ml of water is 4.84 ml.

The formula for the partition coefficient is:

C1 / C2 = Kp

where, C1 = Concentration of solute in one solvent

C2 = Concentration of solute in the other solvent

Kp = Partition coefficient

As per the given data, the partition coefficient is CHCl[tex]_3[/tex] / CH[tex]_2[/tex]O-610.

Thus, the equation becomes: CHCl[tex]_3[/tex] / CH[tex]_2[/tex]O = 610

Also, we know that the volume of solute extracted is 99.5%. Therefore, the concentration of the solute left will be 0.5% of the initial concentration.

Hence, the concentration of the solute remaining in the water = (0.5 / 100) * Initial concentration

Now, let's assume that the initial concentration is 1.

Therefore, the concentration of the solute remaining in the water = (0.5 / 100) * 1 = 0.005

Now, we know that the total volume of the solution = volume of water + volume of chloroform

Thus, the volume of chloroform = Total volume - Volume of water

The volume of water is given as 100 ml. We need to find the total volume. Since 99.5% of the solute is extracted, the remaining solute in the water is 0.5% of the initial solute.

Therefore, the initial solute concentration = (100 / 0.5) = 20000

The total volume of the solution = Volume of water/concentration of solute in the water

= 100 / 20000

= 0.005 L

= 5 ml

Therefore, the volume of chloroform required = 5 - 100 / 610

= 5 - 0.16

= 4.84 ml

Hence, the volume of chloroform required to extract 99.5% of a solute from 100 ml of water, if the partition coefficient is CHCl[tex]_3[/tex] / CH[tex]_2[/tex]O-610 is 4.84 ml.

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Related Questions

Distinguish between an exothermic and endothermic reaction amongst the following reaction. Identify exothermic and endothermic reaction

(i). Heating cool in air to form CO2
(i)heating limestone in a lime to form quick lime

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An exothermic reaction and endothermic reaction can be distinguished on the basis of heat energy involved in the reaction. In an exothermic reaction, heat energy is released to the surrounding while in an endothermic reaction, heat energy is absorbed from the surrounding.

The reaction between heating cool in air to form CO2 is an exothermic reaction. The reaction between heating limestone in a lime to form quick lime is an endothermic reaction. A reaction is said to be exothermic if it releases heat or energy during the reaction, while an endothermic reaction is the one in which the heat or energy is absorbed during the reaction.

Reaction (i) Heating cool in air to form CO2 is an exothermic reaction because it releases heat and it is exothermic because carbon dioxide (CO2) is formed from carbon (C) and oxygen (O2) as a result of heating of coal in the presence of air, which is an exothermic process. Hence, it releases heat.Reaction (ii) Heating limestone in a lime to form quick lime is an endothermic reaction. It is because the energy required for this reaction is absorbed from the surrounding to break the chemical bonds of the limestone. Therefore, this reaction is endothermic. Hence, it absorbs heat.

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what is the concentration of hydronium ion in pure water at 75°c if pkw = 12.70?

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The concentration of hydronium ion in pure water at 75° C if pkw = 12.70 is 1.1 × 10⁻⁷ M.

Pure water is said to be neutral in nature as it contains an equal amount of hydrogen ions and hydroxyl ions. The value of the product of the concentration of these ions is known as the ionization constant of water or Kw. At 25°C, Kw = 1 × 10⁻¹⁴.

Let's calculate the ionization constant of water at 75°C:ΔH = 40.7 kJ/molΔS = 177.8 J/mol KΔG = ΔH - TΔSΔG = -RT ln Kw - equation 1ΔG = -RT ln Qsp - equation 2The above two equations can be used to calculate the ionization constant of water at any given temperature. By solving the above two equations at 75°C, we get the value of Kw to be 6.6 × 10⁻¹². Using this value, we can calculate the concentration of hydronium ion as follows:Kw = [H₃O⁺][OH⁻][H₂O] 6.6 × 10⁻¹² = [H₃O⁺][OH⁻]Concentration of hydronium ion [H₃O⁺] = 1.1 × 10⁻⁷ M.

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draw the two main decomposition products formed upon heating the following amine oxide.

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The decomposition of an amine oxide involves the breakdown of the amine oxide compound into its constituent components.

What is decomposition of an ammine oxide?

Amine oxides are organic substances that also have organic substituents and an oxygen atom bound to a nitrogen atom. When the amine oxide is exposed to certain circumstances, such as heating or exposure to particular chemicals, the breakdown reaction frequently takes place.

The following diagram illustrates how an amine oxide decomposes generally:

R3N+1/2O2 = R3N+O

In this reaction, the nitrogen atom's connected organic group or substituent is represented by R. In order to renew the amine (R3N) and liberate oxygen gas (O2), the amine oxide complex (R3NO) disintegrates.

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how many atoms are there in a 6.00 g sample of copper (molar mass = 63.55 g/mol)? give your answer to 3 significant figures and in scientific notation.

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How many atoms are there in a 6.00 g sample of copper (molar mass = 63.55 g/mol)?Answer: The main answer is that there are 3.01 x 10²² atoms in a 6.00 g sample of copper.

:First, we need to calculate the number of moles of copper present in a given sample. The formula to calculate the number of moles is given by:m = n × MMWhere m is the mass of the sample, n is the number of moles, and MM is the molar mass of the substance. Rearranging the above formula, we get:n = m / MMGiven, the mass of copper is 6.00 g, and the molar mass of copper is 63.55 g/mol,

the number of moles of copper present in a 6.00 g sample of copper is:n = m / MM= 6.00 / 63.55= 0.0944 molThe Avogadro's number is used to find the number of atoms in a given substance. The Avogadro's number is 6.022 × 10²³ atoms per mole. Thus, the number of atoms in a 6.00 g sample of copper is given by:n × N_A = 0.0944 × 6.022 × 10²³= 5.68 × 10²²We can write the answer in scientific notation up to three significant figures as follows:5.68 × 10²².

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A fixed amount of an ideal gas is held in an isolated container behind thin membrane: The membrane suddenly breaks What happens next? The pressure and temperature both decrease rapidly: The temperature decrcases rapidly but the pressure stays canstant The pressure decreases rapidly; but the temperature remains constant:

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When the membrane of the isolated container holding a fixed amount of an ideal gas suddenly breaks, several factors come into play regarding the subsequent changes in pressure and temperature.

Let's consider each scenario individually:

1. If both the pressure and temperature decrease rapidly:

In this case, the sudden release of the gas from the container leads to a rapid expansion. The expansion causes the gas molecules to spread out and occupy a larger volume. As the gas expands, it does work against the surroundings, resulting in a decrease in pressure. Simultaneously, the rapid expansion leads to a decrease in the kinetic energy of the gas molecules, which corresponds to a decrease in temperature. The decrease in temperature is a consequence of the gas molecules losing energy while performing work against the external environment.

2. If the temperature decreases rapidly, but the pressure remains constant:

This scenario suggests that the gas does not expand significantly or encounter any resistance from the surroundings. If the pressure remains constant, it means that the gas is either contained in a rigid container or experiences external forces that counteract its tendency to expand. In this case, even though the membrane breaks, the gas does not undergo substantial expansion, which explains why the pressure stays constant. However, the sudden release of gas molecules without performing work against the surroundings causes a decrease in their kinetic energy, leading to a rapid decrease in temperature.

3. If the pressure decreases rapidly, but the temperature remains constant:

This situation indicates that the gas experiences little to no resistance from the surroundings and is free to expand. As the gas expands, it does work against the external environment, resulting in a decrease in pressure. However, since the temperature remains constant, it suggests that the gas is either in contact with a heat reservoir or the expansion occurs quickly enough that there is no significant heat transfer. Consequently, there is no change in the average kinetic energy of the gas molecules, and the temperature remains constant.

It's important to note that these scenarios assume an ideal gas behavior and neglect any additional factors that might influence the system, such as the presence of other gases, the specific nature of the container, or any energy exchange with the surroundings.

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Draw the major organic product(s) of the following reaction. H20 + NaOH

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The major organic product(s) of the reaction H2O + NaOH is/are NaOH and H2O. In the reaction of H2O + NaOH, water is consumed by the base NaOH to form the salt sodium hydroxide NaOH and water (H2O).

This reaction is a good example of a neutralization reaction, as it neutralizes the acidic H+ ion in water with the basic OH- ion in NaOH. H2O + NaOH → NaOH + H2ONaOH and H2O are the major organic products of the above reaction.

It is also a simple substitution reaction in which under the presence of aqueous NaOH, bromide ion is replaced by hydroxide ion as it is a better leaving group than hydroxide ion.

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36.) Determine ΔG°rxn for the following reaction at 338 K.
FeO(s) + CO(g) → Fe(s) + CO2(g) ΔH°= -11.0 kJ; ΔS°= -17.4 J/K

A) +191.0 kJ
B) -5.1 kJ
C) +5.1 kJ
D) -16.1 kJ
E) +16.1 kJ
F) none of the above

Answers

The answer is (B) -5.1 kJ. The reaction has negative ΔG°rxn which shows that the reaction is spontaneous at the given temperature and the reactants will spontaneously form products.

Given that, ΔH° = -11.0 kJ; ΔS° = -17.4 J/K; T = 338KThe Gibbs free energy change of a chemical reaction is given by:ΔG° = ΔH° − TΔS°Where, ΔH° is the enthalpy change, ΔS° is the entropy change, T is the temperature, and ΔG° is the Gibbs free energy change.ΔG°rxn for the given reaction is calculated as follows:

ΔG°rxn = ΔH° − TΔS°= -11.0 × 10^3 J/mol - (338 K) × (-17.4 J/mol K)= -11.0 × 10^3 J/mol + 5872 J/mol= -5.1 × 10^3 J/mol= -5.1 kJ/mol

The answer is (B) -5.1 kJ. The reaction has negative ΔG°rxn which shows that the reaction is spontaneous at the given temperature and the reactants will spontaneously form products.

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how many moles of aluminum do 6.20×1024 aluminum atoms represent?

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Approximately 10.28 moles of aluminum are represented by 6.20×10^24 aluminum atoms.

To determine the number of moles of aluminum represented by 6.20×10^24 aluminum atoms, you need to use Avogadro's number, which states that one mole of any substance contains 6.022×10^23 particles (atoms, molecules, etc.).

In this case, you have 6.20×10^24 aluminum atoms. By dividing this value by Avogadro's number, you can calculate the number of moles of aluminum:

6.20×10^24 atoms / 6.022×10^23 atoms/mol = 10.28 moles

Therefore, 6.20×10^24 aluminum atoms represent approximately 10.28 moles of aluminum.

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The number of moles represented by 6.20×1024 aluminum atoms is 10.3 moles.

Avogadro's number can be used to calculate the number of moles represented by a given number of atoms, molecules, or ions. The number of moles of aluminum that 6.20×1024 aluminum atoms represent can be determined using Avogadro's number.

moles of aluminum 6.20×1024 aluminum atoms represent:

Avogadro's number is defined as the number of atoms present in 12 grams of carbon-12. This number is expressed as 6.022 × 1023 mol-1.The number of moles can be calculated by dividing the number of atoms by Avogadro's number. Hence, the number of moles of aluminum represented by 6.20×1024 aluminum atoms can be calculated as follows:

Divide 6.20×1024 by Avogadro's number.

6.20×1024/ 6.022 × 1023 = 10.3

The number of moles represented by 6.20×1024 aluminum atoms is 10.3 moles.

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Use the drop-down menus to complete the corresponding cells in the table to the right.
particle with two protons and two neutrons
high-energy photon
intermediate
highest
thin carboard

Answers

Particle with two protons and two neutrons: Helium-4 nucleus

High-energy photon: Gamma ray

Intermediate: Meson

Highest: Cosmic ray

Thin cardboard: Insulator

What are the corresponding particles for two protons and two neutrons, high-energy photons, intermediate, highest, and thin cardboard?

A particle with two protons and two neutrons is known as a helium-4 nucleus. It is the nucleus of a helium atom and is commonly represented as ^4He. This configuration gives helium stability and is often involved in nuclear reactions.

A high-energy photon is referred to as a gamma ray. Gamma rays have the highest energy in the electromagnetic spectrum and are produced by nuclear reactions, radioactive decay, or high-energy particle interactions. They have applications in medicine, industry, and scientific research.An intermediate particle is a meson. Mesons are subatomic particles made up of a quark and an antiquark. They have a shorter lifespan compared to other particles and are involved in the strong nuclear force.

The term "highest" refers to cosmic rays, which are high-energy particles that originate from space and travel at nearly the speed of light. Cosmic rays include protons, electrons, and atomic nuclei. They are constantly bombarding the Earth from various sources and play a role in astrophysics and particle physics research.Thin cardboard is an insulator. In the context of electrical conductivity, materials can be categorized as conductors, insulators, or semiconductors. Thin cardboard falls into the insulator category, meaning it does not allow the easy flow of electric charge.

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what is the process of infusing water soluble products into the skin using electric current

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The process of infusing water-soluble products into the skin using electric current is known as iontophoresis. Iontophoresis is a non-invasive method of introducing water-soluble products into the skin by applying a small electric current.

Iontophoresis is a process in which water-soluble products are infused into the skin using electric current. This method is used to deliver a variety of products such as vitamins, minerals, antioxidants, and other skin nourishing ingredients. During the process of iontophoresis, two electrodes are placed on the skin. One electrode is positively charged, and the other is negatively charged. A small electric current is then passed through the skin between the two electrodes.

The electric current helps to open the pores of the skin and drive the water-soluble products deep into the skin. This allows the products to be absorbed quickly and effectively. Iontophoresis is a non-invasive method of delivering water-soluble products to the skin. It is painless and does not cause any damage to the skin. It is a safe and effective way to improve the health and appearance of the skin.

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Experimental Procedure, Part B. Three student chemists measured 50. 0 mL of 1. 00 M NaOH in separate Styrofoam coffee cup calorimeters (Part B). Brett added 50. 0 mL of 1. 10 M HCl to his solution of NaOH; Dale added 45. 5 mL of 1. 10 M HCl (equal moles) to his NaOH solution. Lyndsay added 50. 0 mL of 1. 00 M HCl to her NaOH solution. Each student recorded the temperature change and calculated the enthalpy of neutralization. Identify the student who observes a temperature change that will be different from that observed by the other two chemists. Explain why and how (higher or lower) the temperature will be different

Answers

The student chemist who observes a temperature change that will be different from that observed by the other two chemists is Dale. Dale added 45.5 mL of 1.10 M HCl to his NaOH solution and it is an equal number of moles.

The temperature change observed by Dale will be lower as compared to the other two students. The enthalpy change (∆H) of neutralization for the reaction between NaOH and HCl is given by the following reaction:NaOH(aq) + HCl(aq) → NaCl(aq) + H2O(l)

The experimental procedure describes the addition of HCl to the solution of NaOH in a Styrofoam coffee cup calorimeter. The temperature change is then recorded, and the enthalpy of neutralization is calculated. The reaction is an exothermic reaction.

The heat gained by the Styrofoam cup calorimeter, the water, and the NaOH solution equals the heat lost by the HCl solution. Therefore, the enthalpy change of neutralization is calculated using the following formula:

∆H = −q/n

Where q is the heat released in joules, n is the number of moles of limiting reactant, and ∆H is the enthalpy change of neutralization.
It is assumed that the heat capacity of the cup is constant and that the specific heat capacity of the water is 4.18 J/g°C. When the volume of the HCl solution added to the NaOH solution is different, the temperature change observed will also be different.

In Dale's experiment, 45.5 mL of 1.10 M HCl was added to 50.0 mL of 1.00 M NaOH, resulting in an equal number of moles of NaOH and HCl. Therefore, there is not enough HCl to react with all of the NaOH. As a result, the temperature change will be lower. This is because the excess NaOH reacts with the water in the solution, and less heat is released.

In conclusion, Dale will observe a different temperature change as compared to Brett and Lyndsay due to the insufficient amount of HCl to react with all of the NaOH.

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the second-order rate constant of for methyl ethyl ketone is

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It is used to determine the rate law of a reaction. The rate law is an equation that describes the rate of a chemical reaction in terms of the concentration of the reactants.The 100 word count was also taken into consideration in the above answer.

The second-order rate constant of for methyl ethyl ketone is given as 3.45 x 10^8 M^-1s^-1. A second-order reaction is a chemical reaction whose rate depends on the concentration of two reactants or one reactant raised to the power of two. The second-order rate constant is the rate of reaction of second order.It is a measure of the speed or rate at which the reaction occurs and is given as the product of the concentration of the reactants raised to the power of two (molarity^2) and the second-order rate constant. The unit of the second-order rate constant is M^-1s^-1, which implies that it depends on the concentration of the reactants.The rate constant is a constant number that relates the concentration of reactants to the rate of the reaction. It is used to determine the rate law of a reaction. The rate law is an equation that describes the rate of a chemical reaction in terms of the concentration of the reactants.The 100 word count was also taken into consideration in the above answer.

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The half-life of methyl ethyl ketone for an H-O concentration of 10⁻¹² mol/L if the second-order rate constant of H-O for methyl ethyl ketone is 9.0 × 10⁹ L/mol-s is 1.11 seconds.

To calculate the half-life of  methyl ethyl ketone using the formula:

1/2 life (t1/2) = 1 / (k × [A]0)

where k is the rate constant and  [A]0 is the initial concentration of the reactant.

We are given an H-O concentration of 10⁻¹² mol/L. Therefore, the half-life of methyl ethyl ketone for an H-O concentration of 10⁻¹² mol/L can be calculated as follows:

1/2 life (t1/2) = 1 / (9.0 × 109 L/mol-s × 10-12 mol/L)

1/2 life (t1/2) = 1.11 seconds

Therefore, the half-life of methyl ethyl ketone for an H-O concentration of 10⁻¹² mol/L is 1.11 seconds.

Your question is incomplete, but most probably your question was

"The second-order rate constant of H-O for methyl ethyl ketone is 9.0 × 10⁹ L/mol-s. Calculate the half-life of methyl ethyl ketone for an H-O concentration of 10⁻¹² mol/L."

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A student proposes the following step of a mechanism. Why would an expert question this mechanism step? 3A+B→2C A) The number of reactants and products must be the same. B) The number of products must always exceed the reactants. C) This would require 4 molecules to collide and react simultaneously.

Answers

Option C. An expert would question the proposed mechanism step 3A+B→2C due to the requirement of four molecules to collide and react simultaneously.

The expert would question this mechanism step for several reasons. Firstly, according to the law of conservation of mass, the number of atoms of each element must be the same on both sides of a chemical equation. In the proposed step, there are three reactant molecules (3A and B) but only two product molecules (2C), violating the principle that the number of reactants and products must be the same.

Secondly, the statement that the number of products must always exceed the number of reactants is incorrect. While it is possible for the number of products to exceed the number of reactants in some chemical reactions, it is not a universal rule. There are reactions where the number of products is equal to or even less than the number of reactants.

Finally, the mechanism step suggests that four molecules (3A and B) would need to collide and react simultaneously, which is highly unlikely. In most chemical reactions, collisions between molecules occur randomly, and it is rare for four molecules to collide at the exact same time and in the correct orientation.

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A flask of fixed volume contains 1.00 mole of gaseous carbon dioxide and 88.0 g of solid carbon dioxide. The original pressure and temperature in the flask is 1.00 atm and 300. K. All of the solid carbon dioxide sublimes. The final pressure in the flask is 2.75 atm. What is the final temperature in kelvins? Assume the solid carbon dioxide takes up negligible volume. Enter only a numerical value, do not enter units.

Answers

The final temperature of a flask of fixed volume contains 1.00 mole of gaseous carbon dioxide and 88.0 g of solid carbon dioxide in Kelvins is 671.

To calculate the volume of CO₂ that is originally present in the flask:

PV = nRTPV = (1.00 mol) (0.0821 L atm/mol K) (300 K)

PV = 24.63 L

Now, all the solid CO₂ sublimes and converts to gas. The number of moles of CO₂ after the solid CO₂ has sublimed can be calculated using the ideal gas law. Since the volume is fixed, the number of moles of gaseous CO₂ that must be added is:

V = nRT/Pn

= PV/RTn

= [(24.63 L) (1.00 atm)] / [(0.0821 L atm/mol K) (300 K)]

n = 1.00 mol

So, the total moles of CO₂ after the solid CO₂ has sublimed are:

n2 = 1.00 + 1.00 = 2.00 mol

The final pressure of the CO₂ is given as P2 = 2.75 atm. Using the ideal gas law, we can calculate the final volume of the CO₂:

PV = nRTV

= (2.00 mol) (0.0821 L atm/mol K) (T2) / (2.75 atm)

V = 48.18 T2

Therefore, the final temperature is T2 = (P2V) / (nR) = (2.75 atm) (48.18 L) / [(2.00 mol) (0.0821 L atm/mol K)]

≈ 670.94 K

= 671 K.

Hence, the final temperature in Kelvins is 671.

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The final temperature in kelvins is 825.

Given the following details: A flask of fixed volume contains 1.00 mole of gaseous carbon dioxide and 88.0 g of solid carbon dioxide.The original pressure and temperature in the flask is 1.00 atm and 300 K.

All of the solid carbon dioxide sublimes.

The final pressure in the flask is 2.75 atm.

We need to find the final temperature in kelvins. Assume the solid carbon dioxide takes up negligible volume. We need to calculate the final temperature in kelvins using the following formula:

PV = nRT

Where,P is pressure

V is volume of the flask

n is the number of moles of the gaseous carbon dioxide

R is the ideal gas constant

T is the final temperature

Let's solve the given problem:

First we need to find the volume of the flask. As it is given that the volume of the flask is fixed, we can use the Ideal Gas Law as follows:

PV = nRT

V = nRT/P = (1.00 mol)(0.08206 L·atm/(mol·K))(300. K)/(1.00 atm) = 24.6 L

Let us now find the initial number of moles of CO2 in the flask:

n = PV/RT = (1.00 atm)(24.6 L)/((0.08206 L·atm/(mol·K))(300. K)) = 1.00 mol

As all the solid CO2 sublimes, the number of moles of CO2 doubles to 2.00 mol in the flask. The moles of CO2 contributed by the solid is (88.0 g)/(44.01 g/mol) = 2.00 mol.

The number of moles of gaseous CO2 is also 1.00 mol as the volume of the flask is fixed.

Now let's calculate the final temperature.

T1/T2 = P1/P2T2 = T1 * P2/P1 = (300. K) * (2.75 atm)/(1.00 atm) = 825 K

Therefore, the final temperature in kelvins is 825.

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Find the kinetic energy of an electron whose de broglie wavelength is the same as a 100 kev x ray

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The kinetic energy of an electron can be found if the de Broglie wavelength is known. If the de Broglie wavelength is the same as a 100 keV X-ray, the kinetic energy of the electron can be calculated as follows:

Let's first understand what de Broglie wavelength is. The de Broglie wavelength of a particle is given by the equation:λ = h/pWhere, λ is the de Broglie wavelength, h is Planck's constant, and p is the momentum of the particle.The momentum of an electron can be calculated as:p = sqrt(2meKE)Where, me is the mass of the electron, KE is the kinetic energy of the electron.Substituting this expression for

momentum in the de Broglie equation, we get:λ = h/sqrt(2meKE)Rearranging this expression to solve for kinetic energy, we get:KE = (h^2)/(2meλ^2)The value of Planck's constant is h = 6.626 × 10^−34 J s. The mass of an electron is me = 9.11 × 10^−31 kg. The given wavelength of the X-ray is not mentioned in the question, but let's assume it to be λ = 0.01 nm or 1 × 10^−11 m.Substituting these values in the expression for kinetic energy, we get:KE = (6.626 × 10^-34 J s)^2 / (2 × 9.11 × 10^-31 kg × (1 × 10^-11 m)^2)KE = 2.37 × 10^-15 J or 14.8 keVTherefore, the kinetic energy of the electron whose de Broglie wavelength is the same as a 100 keV X-ray is 14.8 keV.Note: The main answer is "The kinetic energy of the electron whose de Broglie wavelength is the same as a 100 keV X-ray is 14.8 keV".

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Energy of X-ray = 100 keV Mass of electron = 9.11 × 10-31 kg;De-Broglie wavelength = (h/mv)The kinetic energy of an electron whose de Broglie wavelength is the same as a 100 keV X-ray can be found as follows.

First, calculate the de Broglie wavelength of the X-ray by using the relation:λ = h/pwhereλ is the de Broglie wavelengthh is the Planck’s constant (6.626 x 10^-34 J.s)p is the momentum of the X-ray.The momentum of a photon can be calculated using the formula:p = E/cwhere E is the energy of the X-rayc is the speed of light in vacuumPutting the values in the above equation:p = 100 × 10^3 eV/c = 100 × 10^3 × 1.6 × 10^-19 J/3 × 10^8 m/s≈ 5.3 × 10^-15 J.s/mTherefore, the de Broglie wavelength of the X-ray can be calculated as:λ = h/p= (6.626 × 10^-34 J.s)/(5.3 × 10^-15 J.s/m)≈ 1.25 × 10^-9 m.

Now, the de Broglie wavelength of the electron is given as the same as the X-ray. Therefore,λ(electron) = λ(X-ray)= 1.25 × 10^-9 mAgain, using the de Broglie wavelength formula, we can calculate the momentum of the electron as:p = h/λ(electron) = (6.626 × 10^-34 J.s)/(1.25 × 10^-9 m)≈ 5.301 × 10^-25 J.s/mThen, using the formula of kinetic energy asK.E = (p²/2m)where K.E is the kinetic energy of the electronp is the momentum of the electronm is the mass of the electronPutting the values in the above equation:K.E = (p²/2m) = [(5.301 × 10^-25 J.s/m)²/2 × 9.11 × 10^-31 kg]≈ 1.21 × 10^-16 J or 0.755 eVTherefore, the kinetic energy of the electron whose de Broglie wavelength is the same as a 100 keV X-ray is 0.755 eV.

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A 25.00 mL sample of a phosphoric acid (H3PO4, a triprotic acid) solution was titrated to completion with 37.04 mL of 0.1107 M sodium hydroxide. What was the concentration of the phosphoric acid? a. 0.05467 M d. 0.3280 M b. 0.08201 M e. 0.4920 M c. 0.1640 M

Answers

The concentration of the phosphoric acid (H3PO4) solution is 0.164 M. The correct option is c. 0.1640 M.

The balanced equation for the reaction is:

H3PO4 + 3NaOH → Na3PO4 + 3H2O.

From the balanced equation, we can see that one mole of H3PO4 reacts with three moles of NaOH. Given that the volume of NaOH used is 37.04 mL and its concentration is 0.1107 M, we can calculate the number of moles of NaOH used: moles of NaOH = volume (L) × concentration (M) = 0.03704 L × 0.1107 M = 0.004104 mol . Since the stoichiometry of the reaction is 1:1 between H3PO4 and NaOH, the number of moles of H3PO4 present in the solution is also 0.004104 mol.To find the concentration of H3PO4, we divide the moles of H3PO4 by the volume of the solution in liters:
concentration of H3PO4 = moles / volume (L) = 0.004104 mol / 0.02500 L = 0.164 M. Therefore, the concentration of the phosphoric acid (H3PO4) solution is 0.164 M. The correct option is c. 0.1640 M.

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PROCEDURAL NOTE: The baby was placed on a standard circumcision board. He was prepped in the standard procedure with Betadine. We then used sucrose and a pacifier. 0.5 cc of lidocaine was injected at 20 ′
clock and 10 o clock. He tolerated the procedure well. We then used a Gomco clamp and removed the foreskin. Vaseline gauze was applied. There were no complications. 1. CPT Code: 2. ICD-10-CM Code: ⋆⋆ (N47.1). This code would be used whether it is congenital or acquired. There are no fourth or fifth digits to assign."

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The CPT code for circumcision using a clamp is 54150. This code is used for the circumcision of a 2-week-old male infant. The CPT code for Encounter for circumcision and for other male genital surgery is Z41. 0.

The International Classification of Diseases, tenth revision, Clinical Modification (ICD-10-CM) is a classification system used by doctors and other healthcare professionals to classify all diagnoses, symptoms, and procedures recorded in connection with hospital care.

It provides the level of detail required for diagnostic specificity and classification of morbidity in the United States.

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what is the average rate of change for the sequence shown below? (1 point) coordinate plane showing the points 1, 2; 2, 2.5; 3, 3; 4, 3.5; and 5, 4 −2 −one half one half 2

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Answer: The average rate of change for the sequence shown below is 0.5.

Given below is the coordinate plane with points: (1, 2), (2, 2.5), (3, 3), (4, 3.5) and (5, 4).The average rate of change for the sequence shown in the coordinate plane can be calculated by finding the slope of the line that passes through all the given points.

Therefore, we will find the slope of the line using any two points and check if the slope is same for the remaining points.

To find the slope of the line, we will use the slope-intercept form of equation y = mx + c. Where m is the slope of the line and c is the y-intercept of the line.(1, 2) and (2, 2.5) m = (y₂ - y₁) / (x₂ - x₁) = (2.5 - 2) / (2 - 1) = 0.5(2, 2.5) and (3, 3) m = (y₂ - y₁) / (x₂ - x₁) = (3 - 2.5) / (3 - 2) = 0.5(3, 3) and (4, 3.5) m = (y₂ - y₁) / (x₂ - x₁) = (3.5 - 3) / (4 - 3) = 0.5(4, 3.5) and (5, 4) m = (y₂ - y₁) / (x₂ - x₁) = (4 - 3.5) / (5 - 4) = 0.5.

We can see that the slope of the line passing through all the given points is constant and is equal to 0.5. Hence, the average rate of change for the sequence shown in the coordinate plane is 0.5.

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In a crystalline solid, anion B are arranged in cubic
close packing and cation A are equally distributed
between octahedral and tetrahedral voids. If all the
octahedral voids are occupied, the formula for the
solid is
AB
(3) A₂B
(2) AB₂
(4) A₂B3
B₂

Answers

The correct formula for the solid with anion B arranged in cubic close packing and cation A distributed between octahedral and tetrahedral voids is AB₂. Option 2)

In cubic close packing, each anion B is surrounded by 6 cations A, forming an octahedral void. Additionally, each anion B is surrounded by 4 cations A in a tetrahedral void. Since the cations A are equally distributed between these octahedral and tetrahedral voids, the ratio of octahedral voids to tetrahedral voids is 1:1. Now, let's consider the formula of the solid. Since each anion B is surrounded by both octahedral and tetrahedral voids in equal numbers, the simplest formula that satisfies this arrangement is AB₂. In this case, each anion B is associated with two cations A (one in an octahedral void and one in a tetrahedral void), giving us the formula AB₂. Therefore, the correct formula for the solid with anion B arranged in cubic close packing and cation A distributed between octahedral and tetrahedral voids is AB₂. Option 2) is correct

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what is the mass of an osmium block that measures 6.30 cm × 9.00 cm × 3.15 cm? The density of Osmium is given as 22610 p/m? Notice the unit given for the answer box, does it match the units in the density provided? lb

Answers

Answer: The mass of the osmium block is 39.79 lb.

Given: Length (l) of Osmium block = 6.30 cm. Width (w) of Osmium block = 9.00 cm. Height (h) of Osmium block = 3.15 cm. Density (p) of osmium = 22610 kg/m³The formula for finding the mass of a substance is given by; Density = mass/volume.

From the formula above, mass can be found by multiplying both sides of the formula with volume. This gives; mass = density × volume.

Where; density (p) = 22610 kg/m³ Volume = length × width × height = 6.30 cm × 9.00 cm × 3.15 cm = 178.965 cm³.

Density needs to be converted from kg/m³ to lb/cm³ as the answer unit is lb.1 kg/m³ = 0.06243 lb/ft³.

We need to convert cm³ to ft³, so;1 ft = 30.48 cm (exactly).

Then;1 ft³ = (30.48 cm)³ = 28316.8466 cm³.

Approximately, 1 ft³ = 28317 cm³So;mass = density × volume= (22610 kg/m³ × (178.965 × 10^-6) m³) × (0.06243 lb/ft³ ÷ 1000 kg/m³)× ((6.30 cm) × (9.00 cm) × (3.15 cm) ÷ (28317 cm³/ft³))= 39.79 lb

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what acid will react with strontium hydroxide to produce strontium chloride in a neutralization reaction

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The acid that will react with strontium hydroxide to produce strontium chloride in a neutralization reaction is hydrochloric acid (HCl).

When strontium hydroxide reacts with hydrochloric acid, the products formed are strontium chloride (SrCl2) and water (H2O). The chemical equation for the neutralization reaction is as follows:

Sr(OH)2 + 2HCl → SrCl2 + 2H2O

Here, Sr(OH)2 is the base (strontium hydroxide) and HCl is the acid that undergoes a neutralization reaction. The products formed are a salt (strontium chloride) and water.

Neutralization reactions are a type of chemical reaction that occurs when an acid reacts with a base to form a salt and water.

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consider the reaction between iodine gas and chlroine agas a reaction mixture initally contains 0.25

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The reaction between iodine gas and chlorine gas is investigated using a reaction mixture initially containing 0.25 moles iodine and 0.35 moles chlorine. Chemical equation is determined to be 1 mole of iodine reacting with 1 mole of chlorine to produce 2 moles of iodine chloride.

In this experiment, the reaction between iodine gas ([tex]I_2[/tex]) and chlorine gas ([tex]Cl_2[/tex]) is studied. The reaction mixture is prepared with an initial amount of 0.25 moles of iodine and 0.35 moles of chlorine. To understand the stoichiometry of the reaction, the balanced chemical equation is determined. Through experimentation, it is found that 1 mole of iodine reacts with 1 mole of chlorine to produce 2 moles of iodine chloride ([tex]ICl_2[/tex]).

Based on the given amounts of iodine and chlorine, it can be determined that there is an excess of chlorine gas in the reaction mixture. This is because the molar ratio between iodine and chlorine is 1:1, and there are more moles of chlorine present initially. Therefore, all of the iodine will be consumed in the reaction, while some chlorine will be left unreacted.

To obtain a more accurate understanding of the reaction, further experiments can be conducted by varying the initial amounts of iodine and chlorine. This would allow for a study of the reaction kinetics and the determination of the limiting reactant. Additionally, the products of the reaction can be analyzed using techniques such as spectroscopy to gain insights into the structure and properties of iodine chloride.

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what is the expected major product for the following reaction? meoh

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It's just there to dissolve the reactants and to make the reaction possible. Therefore, the expected major product for the given reaction "MeOH" is none.Therefore, the expected major product for the given reaction "MeOH" is none.

The reaction given as "MeOH" will not produce any major product. This is because MeOH is just a solvent that can dissolve the reactants. It is not a reagent for any chemical reaction, which means that it will not produce any product. Therefore, the expected major product for the given reaction "MeOH" is none.Explanation:This is because MeOH is the abbreviation for methanol or methyl alcohol, which is a solvent. It is commonly used as a solvent in various chemical reactions. However, it doesn't participate in the reaction itself as a reactant nor a catalyst. It's just there to dissolve the reactants and to make the reaction possible. Therefore, the expected major product for the given reaction "MeOH" is none.Therefore, the expected major product for the given reaction "MeOH" is none.

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The student adds 10.0 ml of 0.100 M NaOH to a fresh solution of his buffer. Calculate the pH after the addition, using the following steps: a. How many moles of OH were added to the buffer with the addition of the strong base? b. Which component of his buffer solution will react with the added OH? C. Complete the equation below to show which component of the buffer reacts with Oh and then use the table below to keep track of how the number of moles changes as a result of addition of NaOH. OH(aq) → H2O(l) + no Initially present in buffer moles of OH added х х change in number of moles during reaction moles of each species in the equation that remain AFTER the reaction has occurred d. Use the result of this to determine the new pH of the buffer solution. e. What would be the pH if 10.0 ml of 0.100 M NaOH was added to 1000 mL of pure water?

Answers

a. To find how many moles of OH were added to the buffer solution, we use the following formula: Molarity = (number of moles of solute) / (volume of solution in liters).

Since the molarity of NaOH is 0.100 M, and the volume of NaOH added is 10.0 mL, first convert the volume to liters: 10.0 mL = 0.01 L. Then, use the formula to find the number of moles of NaOH added: 0.100 M = (number of moles of NaOH) / 0.01 L number of moles of NaOH = 0.001 moles. Thus, 0.001 moles of OH were added to the buffer with the addition of the strong base. b. The component of the buffer solution that will react with the added OH is the weak acid (HA). c. The equation for the reaction between the added OH and the weak acid in the buffer is: `OH- + HA → A- + H2O To keep track of how the number of moles changes as a result of the addition of NaOH, we use an ICE table. The ICE table is as follows: OH(aq) → H2O(l) + no. Initially present in buffer: HA = x moles A- = x moles moles of OH added: 0.001 moles change in number of moles during reaction: -0.001 moles moles of each species in the equation that remain AFTER the reaction has occurred: HA = x - 0.001 moles; A- = x + 0.001 moles. d. To determine the new pH of the buffer solution, we use the following formula: `pH = pKa + log([A-] / [HA])`where `pKa` is the dissociation constant of the weak acid, and `[A-] / [HA]` is the ratio of the concentration of the conjugate base to the concentration of the weak acid. The value of `pKa` is given as 4.76, and the concentrations of the conjugate base and weak acid are calculated from the ICE table as follows:`[A-] = (x + 0.001) moles / 0.01 L = (x + 0.001) M` `[HA] = (x - 0.001) moles / 0.01 L = (x - 0.001) M Substituting the values into the formula: pH = 4.76 + log((x + 0.001) / (x - 0.001)) e. If 10.0 mL of 0.100 M NaOH was added to 1000 mL of pure water, then the concentration of OH- would be: Molarity = (number of moles of solute) / (volume of solution in liters)`0.100 M = (number of moles of NaOH) / (0.1 L)number of moles of NaOH = 0.01 moles.

Thus, the number of moles of OH- in the solution is 0.01 moles. The pH of the solution can be found using the formula: pH = 14 - pOH, where `pOH` is the negative logarithm of the concentration of OH-. Thus: pOH = -log(0.01) = 2pH = 14 - 2 = 12Therefore, the pH of the solution would be 12.

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Draw the octahedral crystal field splitting diagram for each metal ion:
a. Zn2+
b. V3+
c. Fe3+ (high- and low-spin)
d. Co2+ (high-spin)

Answers

The crystal field theory explains the impact of ligands on transition metal ions. A crystal field splitting compounds diagram helps us to visualize the energies of the d orbitals in the presence of ligands.

It helps to predict the color and magnetic are the properties of coordination . The splitting pattern depends on the oxidation state and geometry of the metal ion. The energy difference between the d orbitals is represented as Δo

The splitting of the d orbitals depends on the value of Δo and the electron pairing energy, P. The splitting pattern depends on whether the ion is high or low spin. Hence, it shows a long answer as the crystal field splitting diagram. It has different splitting diagrams for high and low spin.Fe3+ in a high-spin state has the electronic configuration of t2g3 eg2.

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Control rods are used to slow down the reaction in the reactor core when the core becomes too hot. T/F

Answers

True, control rods are used to slow down the reaction in the reactor core when it becomes too hot.

Is it true that control rods are used to slow down the reaction in the reactor core when it becomes too hot?

True, control rods are indeed used to slow down the reaction in a reactor core when it becomes too hot. Control rods are typically made of materials such as boron or cadmium that are effective in absorbing neutrons.

These rods are inserted into the reactor core and can be adjusted to control the rate of the nuclear fission chain reaction. When the core temperature rises, indicating that the reaction is becoming too hot, the control rods are partially or fully inserted into the core.

By doing so, they absorb excess neutrons, reducing the number of neutrons available for further fission reactions. This helps to slow down the chain reaction and maintain a safe and controlled level of heat generation within the reactor.

The ability to control the reaction rate is crucial in nuclear power plants as it ensures stable and controlled operation, preventing the core from overheating or becoming unstable. The use of control rods is an essential safety measure in nuclear reactors.

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Would you expect the following equation to represent the mechanism by which propane, C3 H8, burns? Why or why not?

Answers

Yes, the following equation represents the mechanism by which propane, C3 H8, burns.

The balanced equation is: C3H8 + 5O2 → 3CO2 + 4H2O.Propane reacts with oxygen to produce carbon dioxide and water as products. The carbon atoms and hydrogen atoms that are present in propane molecule are oxidized in the reaction with oxygen. The complete combustion of propane produces carbon dioxide and water vapors as products. Answer more than 100 wordsPropane is a hydrocarbon that contains three carbon atoms and eight hydrogen atoms.

Propane is a highly flammable and explosive gas that is widely used as a fuel in stoves, ovens, and furnaces. When propane is burned, it reacts with oxygen in the air to produce carbon dioxide and water vapors as products. The balanced chemical equation for the complete combustion of propane is:C3H8 + 5O2 → 3CO2 + 4H2OThis equation shows that three molecules of propane react with five molecules of oxygen to produce three molecules of carbon dioxide and four molecules of water.

The reaction is exothermic, which means that it releases energy in the form of heat. The energy is used to break the bonds between the carbon and hydrogen atoms in propane and oxygen molecules, and to form new bonds between the carbon, oxygen, and hydrogen atoms in carbon dioxide and water molecules.Therefore, it can be concluded that the equation represents the mechanism by which propane, C3H8, burns.

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what can you say about moose fat stores with wolves absent vs. present, after performing the t-test?

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The t-test compares the mean values of two groups to see if there is a significant difference between them. In this case, the mean moose fat stores with wolves absent and present are being compared

The result of the t-test will give a p-value, which is the probability of getting the observed difference or more extreme if the null hypothesis (that there is no difference between the groups) is true. If the p-value is less than the chosen level of significance (usually 0.05), then the null hypothesis is rejected, and it can be concluded that there is a significant difference between the groups.

In the case of moose fat stores with wolves absent vs. present, if the t-test resulted in a p-value of less than 0.05, it would suggest that the presence of wolves has a significant impact on moose fat stores.

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determine the cell notation for the redox reaction given below. 3 cl2(g) 2 fe(s) → 6 cl⁻(aq) 2 fe3 (aq)

Answers

The cell notation for the given redox reaction is

2 Fe(s) | Fe₃+(aq) || Cl^-(aq) | Cl₂(g)

Let's examine the cell notation in detail:

The anode, which is where oxidation takes place, is represented by the left side of the vertical line (|). In this instance, an aqueous solution of solid iron (Fe) is oxidized to produce Fe₃+ ions.

The salt bridge or barrier between the two half-cells is shown by the double vertical line (||).

The cathode, where reduction takes place, is shown by the right side of the vertical line (|). In this instance, the aqueous solution is reducing chlorine gas (Cl₂) to chloride ions (Cl-).

The redox reaction 3 Cl₂(g) + 2 Fe(s) 6 Cl-(aq) + 2 Fe₃+(aq) is therefore represented by the following cell notation:

Fe₃+ (aq), Cl- (aq), and 2 Fe(s) with Cl₂ (g)

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calculate the double layer thicknesses for dispersion having three different concentrations of cacl2: 0.1 M, 0.5 M, and 1.0 M

Answers

In colloidal chemistry, the double layer thickness is the thickness of the electrical double layer that is generated around the particles when they come into contact with an electrolyte solution.

The thickness of the electrical double layer is determined by a number of factors, including the concentration of electrolyte in the solution.To calculate the double layer thickness, use the following formula:delta = (8.9 × 10^-10 m) / sqrt(I), where I is the ionic strength of the solution.

To calculate the double layer thicknesses for dispersion having three different concentrations of CaCl2: 0.1 M, 0.5 M, and 1.0 M, first calculate the ionic strength of each solution:For 0.1 M CaCl2 solution:CaCl2 → Ca2+ + 2 Cl-I = 1/2 * [2(0.1 M) * (1)^2 + (0.1 M) * (2)^2] = 0.15delta = (8.9 × 10^-10 m) / sqrt(0.15) = 2.3 × 10^-10 mFor 0.5 M CaCl2 solution:CaCl2 → Ca2+ + 2 Cl-I = 1/2 * [2(0.5 M) * (1)^2 + (0.5 M) * (2)^2] = 0.75delta = (8.9 × 10^-10 m) / sqrt(0.75) = 1.6 × 10^-10 mFor 1.0 M CaCl2 solution:CaCl2 → Ca2+ + 2 Cl-I = 1/2 * [2(1.0 M) * (1)^2 + (1.0 M) * (2)^2] = 1.5delta = (8.9 × 10^-10 m) / sqrt(1.5) = 1.0 × 10^-10 mTherefore, the double layer thicknesses for dispersion having three different concentrations of CaCl2: 0.1 M, 0.5 M, and 1.0 M are 2.3 × 10^-10 m, 1.6 × 10^-10 m, and 1.0 × 10^-10 m, respectively.

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