F₂ and Cl₂ are produced by electrolytic oxidation, whereas Br₂ and I₂ are produced by chemical oxidation of the halide ions in a concentrated aqueous solution (brine) by a more electronegative halogen. Give two reasons why Cl₂ isn’t prepared this way.

Answer:

It would tend to lose two electrons and form a +2 ion. Potassium is in the first column and therefore has 1 electron in its outermost shell. It would tend to lose one electron and form a +1 ion.

Answer: It has to be stratosphere because This region of the atmosphere is called the stratosphere. It is present in stratosphere According to NASA earth have 6 atmospheric layers..the layer just above the earth is troposphere and above this layer is the stratosphere.

Explanation:

The third quantum number (m_l) of a 3s² electron in phosphorus is 0.

The third quantum number, denoted as m_l, represents the magnetic quantum number and describes the orientation of an orbital within a subshell. It can have integer values ranging from -l to +l, where l is the azimuthal quantum number.

In the electron configuration of phosphorus, we see that the 3s subshell is being filled. The azimuthal quantum number (l) for the 3s subshell is 0. Since the electron is in the 3s² subshell, there are two electrons present in the 3s orbital.

For the two electrons in the 3s orbital, they will have opposite spins due to the Pauli exclusion principle. However, the magnetic quantum number (m_l) for both electrons in the 3s orbital will be the same, which is 0.

Therefore, the third quantum number (m_l) of a 3s² electron in phosphorus is 0. This means that both electrons in the 3s orbital have the same orientation within the subshell.

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Here are the given properties. The task is to decide whether each property is a physical property or a chemical property:

1. Solubility: Physical property

2. Flammability: Chemical property

3. Reactivity: Chemical property

4. Color: Physical property

5. Density: Physical property

6. Melting point: Physical property

There are two types of properties: Physical property and Chemical property. In this task, we have to identify which property among Solubility, Flammability, Reactivity, Color, Density, Melting point is Physical property or Chemical property.

A Physical property is a characteristic that defines the behavior or appearance of a substance without any changes in its composition. Physical properties can be identified by measuring, observing, or using physical methods.

A Chemical property is a characteristic that defines the behavior or the ability of a substance to undergo a chemical change or to transform into a new substance. The correct answer is given above.

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Answer:

This is an abbreviation for "parts per million" and it also can be expressed as milligrams per liter (mg/L). This measurement is the mass of a chemical or contaminate per unit volume of water. Seeing ppm or mg/L on a lab report means the same thing.

The regiochemistry of hydroboration/oxidation of alkenes is (a) Markovnikov.

The hydroboration/oxidation reaction follows Markovnikov's rule, which states that the electrophile (in this case, the boron atom) adds to the carbon atom of the double bond that has the greater number of hydrogen atoms attached to it. The regioselectivity of the reaction is determined by the relative stability of the carbocation intermediates formed during the reaction.

In hydroboration, the boron atom adds to the less substituted carbon atom of the double bond, leading to the formation of a boron-alkyl bond and a boron-hydrogen bond. Subsequently, in the oxidation step, the boron-alkyl bond is replaced with an alcohol group (-OH) while maintaining the regiochemistry established during hydroboration.

Therefore, the regiochemistry of hydroboration/oxidation of alkenes is Markovnikov, where the electrophilic addition occurs preferentially at the carbon atom of the double bond that has the greater number of hydrogen atoms attached to it.

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Answer:

pints

Explanation:

Answer:

B: pints

fl.oz.

Explanation:

The pH of the base before the titration begins is 7.

To calculate the pH, we need to consider the concentration of the conjugate acid. Since propylamine is initially present in excess, the concentration of the conjugate acid will be negligible. Thus, the pH of the base before the titration begins will be determined by the dissociation of water.

Water undergoes autoionization to form hydronium (H3O+) and hydroxide (OH-) ions.

At 25°C, the concentration of both ions is 1.0 x 10^-7 M.

To find the pH, we can calculate the concentration of hydronium ions and take the negative logarithm

(pH = -log[H3O+]).

Therefore, the pH of the base before the titration begins is 7.

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Answer:

The information that they will learn from the graph is how fast the waves will move and which one has the fastest velocity. What they will notice ia that the P wave will be moving faster and arrive first.

i think A but if I'm wrong sorry

then if increasing then becoming spread out so letter A

Answer:

Energy can exist in many forms within a system and may be converted from one form to another within the constraint of the conservation law. These different forms include gravitational, kinetic, thermal, elastic, electrical, chemical, radiant, nuclear, and mass energy.

Explanation:

the answer is 6.0°C that is the solution

To neutralize the mixture of 0.160 M HNO3 and 0.100 M KOH, 67.0 mL of 2.00 M HCl must be added.

The balanced chemical equation for the neutralization reaction between HNO3 and KOH is:

HNO3 + KOH → KNO3 + H2O

From the balanced equation, we can see that one mole of HNO3 reacts with one mole of KOH to form one mole of water. Therefore, the number of moles of HNO3 present in the solution is given by:

moles of HNO3 = 0.160 M × 0.100 L = 0.016 mol

Similarly, the number of moles of KOH present in the solution is given by:

moles of KOH = 0.100 M × 0.400 L = 0.040 mol

Since the reaction is a 1:1 stoichiometric ratio, the number of moles of HCl required to neutralize the mixture is also 0.016 mol.

To calculate the volume of 2.00 M HCl required, we can use the following formula:

moles of HCl = Molarity of HCl × volume of HCl

0.016 mol = 2.00 M × volume of HCl

volume of HCl = 0.008 L = 8.0 mL

Therefore, 67.0 mL of 2.00 M HCl must be added to neutralize the mixture of 0.160 M HNO3 and 0.100 M KOH.

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Answer:

as molecular mass increases, more molecules have speeds close to their average speed.

Explanation:

A plot of the Maxwell distribution of velocities for several different gases measured at the same temperature shows that "as molecular mass increases, more molecules have speeds close to their average speed."

This is evident in the fact that the Maxwell distribution of speeds for a given selection of gas at a given temperature shows certain traits if one of the main variables is changed, such as

1. Mass: The proportion of speeds for a given selection of gas is narrower and the average speed is lower for gas of higher molecular weight.

2. Temperature: The proportion of speeds for a given selection of gas is wider and the average speed is higher as the temperature of a gas increases.

Answer:

d. 1 mole Al₄C₃ and 12 moles H₂O

Explanation:

The coefficients in a balanced chemical equation indicate the number of moles of reactants and products. If a substance does not have a coefficient written in front of it, it is assumed to be 1.

The change in internal energy of the gas is 123.7 J.

The change in internal energy of a gas can be calculated using the equation: ΔU = Q - W, where ΔU is the change in internal energy, Q is the heat added to or removed from the gas, and W is the work done on or by the gas.

In this case, the work done on the gas can be calculated as W = P * ΔV, where P is the pressure and ΔV is the change in volume. Since the pressure is 1.00 bar and the change in volume is 5.50 L - 1.20 L = 4.30 L, the work done on the gas is W = 1.00 bar * 4.30 L = 4.30 J.

Therefore, the change in internal energy of the gas is ΔU = Q - W = 128 J - 4.30 J = 123.7 J.

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Because of their differing nuclear charges, and as a result of shielding by inner electron shells, the different atoms of the periodic table have different affinities for nearby electrons.

The electron affinity is the potential energy change of the atom when an electron is added to a neutral gaseous atom to form a negative ion. So the more negative the electron affinity the more favourable the electron addition process is.

Not all elements form stable negative ions in which case the electron affinity is zero or even positive.

Electron affinity depends on the nuclear charge of an atom.

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After the third dilution, the fraction of the original concentration is the concentration of the final solution is A. 1/8.

To solve this problem, we need to understand that each dilution reduces the concentration by half. So after the first dilution, the concentration is 1/2 of the original. After the second dilution, it is 1/4 of the original. And after the third dilution, it is 1/8 of the original.

To see why, consider that removing half the solution and adding solvent effectively cuts the concentration in half. For example, if we start with a solution that is 10% solute and 90% solvent, removing half the solution will give us 5% solute and 95% solvent. Adding back the same amount of solvent will give us 10% solute and 90% solvent again, but with half the amount of solute that we started with.

Therefore, the answer is A. 1/8. This means that the concentration of the final solution is only 1/8th of the original concentration. It's important to note that diluting a solution multiple times will continue to decrease the concentration by half each time.

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In terms of increasing strength of intermolecular forces, with the weakest at the top of the list, the order for the given compounds would be:

A. CH3CH2CH2CH3 (n-butane)

B. CH3COON (ethyl ethanoate)

C. CH3CH2OCH3 (dimethyl ether)

D. CH3CH2CH2OH (1-butanol)

The intermolecular forces in these compounds can be ranked as follows:

1. A. CH3CH2CH2CH3 (n-butane): This compound consists of nonpolar covalent bonds and experiences only weak London dispersion forces between its molecules.

2. B. CH3COON (ethyl ethanoate): Ethyl ethanoate contains polar covalent bonds, and the intermolecular forces include London dispersion forces as well as dipole-dipole interactions. However, the dipole-dipole interactions are relatively weaker than the next two compounds.

3. C. CH3CH2OCH3 (dimethyl ether): Dimethyl ether has polar covalent bonds, leading to London dispersion forces and dipole-dipole interactions. The oxygen atom creates a stronger dipole moment than in ethyl ethanoate, resulting in slightly stronger intermolecular forces.

4. D. CH3CH2CH2OH (1-butanol): 1-butanol contains polar covalent bonds and experiences London dispersion forces, dipole-dipole interactions, and hydrogen bonding. Hydrogen bonding is the strongest intermolecular force among the four compounds, as the hydrogen atoms are directly bonded to the highly electronegative oxygen atom.

Thus, the correct order from weakest to strongest intermolecular forces is A < B < C < D.

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Answer:

\(P_{N_2}=285 torr\)

Explanation:

Given that

Volume of cylinder = 3 L

Temperature= 25 degrees

Partial pressure of the gases

\(P_{He}=115 torr\)

\(P_{N_2}=285 torr\)

\(P_{Ar}=325 torr\)

By using Dalton's law , the total pressure of the non reacting gas is the sum of the partial pressure of all gases.

\(P_{total}=P_{He}+P_{N_2}+P_{Ar}\)

\(P_{total}=115+285+325=725 torr\)

When all the He gas will removed then partial pressure of He will be zero.

\(P_{total}=P_P_{N_2}+P_{Ar}\)

\(P_{total}=285+325=610 torr\)

Partial pressure of N₂

\(P_{N_2}=610-325=285 torr\)

\(P_{N_2}=285 torr\)

The Ksp value of Mg(OH)2 is A. 2.4 X 10^11.

To find the Ksp of Mg(OH)2, we need to use the equation:

Ksp = [Mg2+][OH-]^2

Since Mg(OH)2 is a strong electrolyte, it will dissociate completely in water, meaning that the concentration of Mg2+ will be equal to the concentration of Mg(OH)2, which is 3.63 times 10-4 M.

We can then use the concentration of Mg2+ to find the concentration of OH- using the equation:

Mg(OH)2 ⇌ Mg2+ + 2OH-

Since Mg(OH)2 is a 1:2 electrolyte, the concentration of OH- will be twice the concentration of Mg2+, or 2(3.63 times 10-4 M) = 7.26 times 10-4 M.

Plugging these values into the Ksp equation, we get:

Ksp = (3.63 times 10-4 M)(7.26 times 10-4 M)^2 = 2.4 times 10^11

Therefore, the answer is A. 2.4 X 10^11.

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Answer:

\(tiempo=0.72s\)

Explanation:

¡Hola!

En este caso, como podemos observar, la relación entre la rapidez de reaction, las moles y el tiempo de reacción es:

\(rapidez=\frac{mol}{tiempo}\)

De este modo, despejamos el tiempo en esta ecuación para obtener:

\(tiempo=\frac{mol}{rapidez}\)

Así, reemplazamos la rapidez y moles dadas para obtener:

\(tiempo=\frac{0.323mol}{0.450mol/s}\\\\tiempo=0.72s\)

¡Saludos!

Answer: 1)  Caboxylic acid

2)  Ether

3)  Ester

4)  Aldehyde

5)  Amine

6)  Alcohol

Explanation:  The type of a molecule is based it's primary functional group.  

https://en.wikipedia.org/wiki/Functional_group

The identities of the given molecules shown in the drawing are:

1) Carboxylic acid.

2)  Ether

3)  Ester

4)  Ethane

5) Aldehyde

6)  Amine

7)  Alcohol

Organic compounds are those compounds, which contain carbon-carbon and carbon-hydrogen bonds.

The five organic compounds are lipids, carbohydrates, protein, nucleotide, etc.

Thus, the correct option is 1) Carboxylic acid.

2)  Ether

3)  Ester

4)  Ethane

5) Aldehyde

6)  Amine

7)  Alcohol

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Taking into account the direction the meter points will be from left to right, the copper (||) ions are being reduced and Zn is being oxidized in the zinc and copper galvanic cell.

The direction of the flow of elections will be indicated by the direction of the needle, which will move from left to right. At this point, oxidation takes place at the Zn electrode, resulting in the release of e-, whilst the reduction of Cu+2 ion takes place at the Cu electrode, leading to the movement of electrons from left to right.

The correct answer is that the copper (||) ions are being reduced while the zinc ions are being oxidized.

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A system's equilibrium will move to the right, or toward the side of the products, in accordance with Le Chatelier's principle, if more reactants are added. ... The equilibrium will move to the left if we add more product to a system, producing more reactants.

Solution: By increasing the number of reactants, the equilibrium moves to the right and in the direction of the products.

Thus, if a reactant is added, equilibrium shifts to the right, away from the reactant. Equilibrium shifts to the left, away from the product, when a product is added. If we take away the product, equilibrium returns and produces the product. Reactant is created if reactant is removed, breaking the equilibrium.

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When a strong acid and a strong base are titration together, the acid and base will mix to form a neutral solution. The hydronium (H+) and hydroxide (OH-) ions will unite to create water.

Titration, also known as titrimetry, is a technique for analyzing a mixture's composition chemically to ascertain the concentration of a certain analyte. Volumetric analysis is another name for titration, a key analytical chemistry technique.

A standard solution with a set volume and concentration is created as part of the titration process and is known as a titrant or titrator. When an endpoint or equivalence point is attained, the titrant is made to continue reacting with the analyte, and at that point, the concentration of the analyte can be determined from the amount of titrant consumed. Titration is another method for using the stoichiometry principle to figure out the concentration of an unknown solution.

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The boiling point is the characteristic on which a by-products of the refining of petroleum are separated.

For separating different types of petroleum from one another, we use the boiling method in which those substance having lower boiling point can be removed from the combine solution. In this way, all of the materials are separated from one another.

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Answer:

It is a good conductor of heat and electricity

Its melting point is 1453°C and its boiling point is 2913°C

It's bivalent and it has a valency of two

Step 1

The reaction is written and balanced:

4 Rb + O2 =>2 Rb2O

-----------

Step 2

Define % yield of product (Rb2O) = (Actual yield/Theoretical yield) x 100

The actual yield is provided by the exercise = 39.7 g

----------

Step 3

Determine the limiting reactant. The molar masses are needed to solve this:

For Rb) 85.4 g/mol

For O2) 32 g/mol

Procedure:

4 Rb + O2 =>2 Rb2O

4 x 85.4 g Rb ----- 32 g O2

82.4 g Rb ----- X = 7.72 g O2 are needed

For 82.4 g Rb, 7.72 g O2 is needed, but there is 11.6 g O2. Therefore, O2 is the excess agent. Rb is the limiting reactant.

--------

Step 4

Determine the theoretical yield from the limiting reactant:

The molar mass Rb2O) 187 g/mol

Procedure:

4 x 85.4 g Rb ------ 2 x 187 g Rb2O

82.4 g Rb ------ X = 90.2 g Rb2O = Theoretical yield

---------

Step 5

% yield = Actual y./Theoretical y. x 100 = (39.7 g/90.2 g) x 100 = 44 % approx.

Answer: % yield = 44 %

Answer:

Explanation:

Because of the decrease in the oxidation state.