A Friedel-Crafts alkylation is an electrophilic aromatic substitution in which the electrophile (E ) is a carbocation.


a. True

b. False

Explanation:

It's none of these.

Cr- 1s2 2s2 2p6 3s2 3p6 4s1 3d5

Mg- 1s2 2s2 2p6 3s2

Mn - 1s2 2s2 2p6 3s2 3p6 4s2 3d5

Fe- 1s2 2s2 2p6 3s2 3p6 4s2 3d6

I think there's a mistake in the question.

The approximately 4 moles of gas in the container at standard temperature (ST) with a volume of 99.2 L.

To determine the number of moles of gas in a container at standard temperature (ST) with a volume of 99.2 L, we need to use the ideal gas law equation:

PV = nRT

Where:

P = pressure (atmospheres)

V = volume (liters)

n = number of moles

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

T = temperature (Kelvin)

At standard temperature (ST), the temperature is 273.15 K.

Assuming the pressure is also at standard conditions (1 atm), we can rearrange the ideal gas law equation to solve for the number of moles:

n = PV / RT

Substituting the given values:

P = 1 atm

V = 99.2 L

R = 0.0821 L·atm/mol·K

T = 273.15 K

n = (1 atm) × (99.2 L) / ((0.0821 L·atm/mol·K) × (273.15 K))

Simplifying the calculation:

n ≈ 4 moles

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Answer:

C - raising the temperature of a gas

Explanation:

as you raise temperature, kinetic energy rises, and so does pressure

Answer:

C.

raising the temperature of the gas

Explanation:

Edg. 2020

Other guy is right i got 100%

Answer:

a. 0.02 moles  of NH₃, 0 moles of O₂, 0.08 moles of NO, 0.12 moles of H₂O

b. \(P_{NH_3}\) = 12,576.5 Pa, \(P_{NO}\) = 50,306.05 Pa, \(P_{H_2O}\) =  74,459.1 Pa

c. The total pressure is 138,341.64 Pa

Explanation:

a. NH₃ + O₂  → NO + H₂O

The balanced chemical equation is first found to be

4NH₃ + 5O₂  → 4NO + 6H₂O

Therefore, we have;

4 moles of NH₃ reacts with 5 moles of O₂ to form 4 moles of NO and 6 moles H₂O

Dividing by the reactant with the highest number of moles which is 5 moles of oxygen gives;

4/5 moles  of NH₃ reacts with 5/5 moles of O₂ to form 4/5 moles of NO and 6/5 moles H₂O

Which is the same as 4/5 moles  of NH₃ reacts with 1 mole of O₂ to form 4/5 moles of NO and 6/5 moles H₂O

Multiplying by 0.100 gives;

0.1×4/5 moles  of NH₃ reacts with 0.1 mole of O₂ to form 0.1×4/5 moles of NO and 0.1×6/5 moles H₂O

The quantity of each gas in the container upon completion of the reaction is therefore;

(0.1 - 0.1×4/5) = 0.02 moles  of NH₃

0 moles of O₂

0.08 moles of NO

0.12 moles H₂O

b. Given that the temperature = 105°C, we have;

PV = nRT

P = nRT/V

Where:

n = Total number of moles = 0.02 + 0.08 + 0.12 = 0.22 moles

R = Universal gas constant = 8.3145 J/(mol·K)

T = Temperature = 105°C = 378.15 K

V = Volume = 5 litre = 0.005 m³

P = 0.22×8.3145×378.15/0.005 = 138,341.64 Pa

From Dalton's law of partial pressure, we have;

Partial pressure Pₓ = Xₓ × P

Where:

Xₓ = Mole fraction

Which gives for ammonia NH₃ with 0.02 moles;

Mole fraction = 0.02/0.22 = 1/11

\(P_{NH_3}\) = 1/11 × 138,341.64  = 12,576.5 Pa

For the 0.08 moles of NO, we have

Mole fraction = 0.08/0.22 = 4/11

\(P_{NO}\) = 4/11 × 138,341.64  = 50,306.05 Pa

For the 0.12 moles H₂O

P = 0.12×8.3145×378.15/0.005 = 74,459.1 Pa

Mole fraction = 0.12/0.22 = 6/11

\(P_{H_2O}\) = 6/11 × 138,341.64  = 74,459.1 Pa

c. The total pressure = 12,576.5 Pa + 50,306.05 Pa + 74,459.1 Pa = 138,341.64 Pa.

Answer:  yes, Air is a homogeneous mixture of the gaseous substances nitrogen, oxygen, and smaller amounts of other substances. Salt, sugar, and substances dissolve in water to form homogeneous mixtures. A homogeneous mixture in which there is both a solute and solvent present is also a solution

Explanation:

Explanation:

The ionization potential of Ba(OH)2 was found to be 8±1 eV, with a value of 6±1 eV for the monohydroxide.

Answer:

chemical is fire

and ductility is physical

Explanation:

give me brainliest please

Answer:

Astatine: Halogen

Nitrogen: Non-Metal

Krypton: Non-Metal, Noble Gas

Chlorine: Non-Metal

Sulfur: Non-metal

Explanation:

Answer:

the remaining books weigh 4 kilograms or 4000 grams

The covalent radius of atom X in an XY molecule with a bond length of 212 and covalent radius of Y atom being 93 is 119.

To find the covalent radius of atom X, we need to understand that the bond length of an XY molecule is equal to the sum of the covalent radii of atoms X and Y. We can represent this relationship using the formula: bond length = covalent radius of X + covalent radius of Y.

Given that the bond length of the XY molecule is 212, and the covalent radius of Y is 93, we can use the formula to find the covalent radius of X:

212 = covalent radius of X + 93

To find the covalent radius of X, we can simply subtract the covalent radius of Y from the bond length:

covalent radius of X = 212 - 93

covalent radius of X = 119

So, the covalent radius of atom X is 119.

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Answer:

This should be correct "The biggest advantage of sexual reproduction versus asexual reproduction is genetic diversity. When a single organism reproduces asexually it produces an exact copy of the parent organism."

Explanation:

sorry that I'm late but brainliest ....... please...................., maybe............... ?????

The original volume of HBr that was titrated can be calculated as the ratio of the moles of HBr to its concentration.

To determine the original volume of HBr that was titrated, we can use the concept of stoichiometry and the equation balanced for the neutralization reaction between HBr and KOH.

The balanced equation is:

HBr + KOH → KBr + H₂O

From the balanced equation, we can see that the stoichiometric ratio between HBr and KOH is 1:1. This means that for every mole of HBr, we need an equal number of moles of KOH to complete neutralization.

First, let's determine the moles of KOH used in the titration:

Moles of KOH = 0.500 M × 0.075 L = 0.0375 mol

Since the stoichiometric ratio is 1:1, this also represents the number of moles of HBr that were neutralized.

Now, we can calculate the original volume of HBr using the concentration of the unknown solution:

Moles of HBr = 0.0375 mol

Concentration of HBr = unknown (let's assume it is C mol/L)

Volume of HBr = Moles of HBr / Concentration of HBr = 0.0375 mol / C mol/L

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the answer for this question is Mister

The breakdown of a molecule resulting from a chemical reaction can be described as chemical decomposition. In this process, the molecule undergoes a transformation that leads to its fragmentation into simpler substances.

Chemical decomposition occurs when bonds within the molecule are broken, leading to the formation of new products. This reaction can be initiated by various factors such as heat, light, or the presence of catalysts. The breakdown may involve different types of reactions, including hydrolysis, oxidation, or thermal decomposition.

During chemical decomposition, the original molecule is transformed into one or more smaller molecules, atoms, or ions. This reaction is essential in various biological, industrial, and environmental processes. Understanding the mechanisms and factors that influence chemical decomposition is crucial for studying the behavior of substances and developing practical applications in fields such as chemistry, biology, and materials science.

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Answer:

1.Set your equations Raquel to each other

2.given

3.subtract 7x from both sides

4. Add 12 to both sides

5.divide both sides by four and you get negative 5

Explanation:

this was great and fun to do

Answer: There are \(0.292 \times 10^{-4} mol\) are present in \(4.97 \times 10^{-4} g\) of ammonia.

Explanation:

Given: Mass of ammonia = \(4.97 \times 10^{-4} g\)

As number of moles is the mass of substance divided by its molar mass.

Hence, moles of ammonia (molar mass = 17 g/mol) are calculated as follows.

\(No. of moles = \frac{mass}{molar mass}\\= \frac{4.97 \times 10^{-4} g}{17 g/mol}\\= 0.292 \times 10^{-4} mol\)

Thus, we can conclude that there are \(0.292 \times 10^{-4} mol\) are present in \(4.97 \times 10^{-4} g\) of ammonia.

To determine the best buffer with a basic pH using the given pKa value for HC3H2O2, we need to find a pair of substances where one acts as a weak acid (HC3H2O2) and the other as its conjugate base (C3H2O2-).

The pKa of HC3H2O2 represents the pH at which the acid is 50% ionized. Since we want a basic pH, we need a pKa value that is slightly higher than the desired pH. Let's assume the desired pH is around 9.

A quick calculation shows that a pKa of 8.5 would be suitable for our purpose.

Now, we need to find a conjugate base with a pKa close to 8.5. One example is ammonium acetate (NH4C2H3O2) with a pKa of 9.25. When ammonium acetate is dissolved in water, it dissociates into NH4+ (conjugate acid) and C2H3O2- (conjugate base).

Therefore, the best buffer pair for a basic pH would be HC3H2O2 (acetic acid) and NH4C2H3O2 (ammonium acetate).

The pKa value of HC3H2O2 is not provided in the question. However, assuming we have the pKa value of HC3H2O2, we can use it to calculate the pH range over which the buffer will be effective.

The Henderson-Hasselbalch equation is commonly used to calculate the pH of a buffer solution:

pH = pKa + log ([A-]/[HA])

In this equation, [A-] represents the concentration of the conjugate base, and [HA] represents the concentration of the weak acid.

To create a buffer with a basic pH, we need a pKa slightly higher than the desired pH. Assuming a desired pH of 9, we can use a pKa value around 8.5.

Let's consider ammonium acetate (NH4C2H3O2) as a potential conjugate base for HC3H2O2. The pKa value of ammonium acetate is 9.25.

Using the Henderson-Hasselbalch equation, we can determine the pH range over which the buffer will be effective. For a basic pH, we want the [A-]/[HA] ratio to be high, indicating a significant concentration of the conjugate base.

With a pKa of 8.5 for HC3H2O2 and a pKa of 9.25 for NH4C2H3O2, we can calculate the pH range as follows:

pH = pKa + log ([A-]/[HA])

pH = 8.5 + log ([C2H3O2-]/[HC3H2O2])

To ensure a high [C2H3O2-]/[HC3H2O2] ratio, we can adjust the concentrations of the weak acid and its conjugate base accordingly. By choosing appropriate concentrations, we can achieve a pH in the desired range.

Based on the given pKa value for HC3H2O2, the best buffer pair for a basic pH would be HC3H2O2 (acetic acid) and NH4C2H3O2 (ammonium acetate) with a pKa of 8.5 for HC3H2O2 and a pKa of 9.25 for NH4C2H3O2. By adjusting the concentrations of the weak acid and its conjugate base

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Answer:

All unauthorized experiments are prohibited. You are allowed to enter the chemical preparation/storage area any time you need to get an item. Laboratory aprons should be worn during all lab activities. It's okay to pick up broken with our bare hand as long as the glass is placed in the trash.

Explanation:

Answer:

it have been already balanced

2H2 + O2 = 2H2O.

Answer:

the correct answer is 5

Explanation:

i just know

The minimum thickness of a quarter-wave antireflection film of an index of refraction 1.3 to minimize reflection of 560 nm light is 107.7 nm, and the smallest thicknesses for a quarter-wave film of an index of refraction 1.7 to minimize reflection are 82.4 nm (destructive interference) and 248.2 nm (constructive interference).

The formula for reflection in a thin film is given by; R = [(n2 - n1)/(n2 + n1)] 2 where R is the amount of reflected light and n1 and n2 are the indexes of refraction of the two media. The light reflected from the upper surface of the film interferes with the light reflected from the lower surface. There is constructive interference if the path difference between the two reflected rays is equal to an integral number of wavelengths (λ). If the path difference between the two reflected rays is equal to a half-integral number of wavelengths (λ/2), destructive interference occurs.

The minimum thickness of a quarter-wave film is; Tmin = (λ/4) / n film. If the wavelength is λ = 560 nm, then the minimum thickness of a quarter-wave antireflection film of an index of refraction 1.3 is; Tmin = (560 nm / 4) / 1.3 = 107.7 nm. If a film with a refractive index of 1.7 is used, then the smallest thicknesses for a quarter-wave film are; For destructive interference: Tmin = (λ/4) / film = (560/4) / 1.7 = 82.4 nm. For constructive interference: Tmin = (3λ/4) / film = (3 * 560/4) / 1.7 = 248.2.

Therefore, the minimum thickness of a quarter-wave antireflection film of an index of refraction 1.3 to minimize reflection of 560 nm light is 107.7 nm, and the smallest thicknesses for a quarter-wave film of an index of refraction 1.7 to minimize reflection are 82.4 nm (destructive interference) and 248.2 nm (constructive interference).

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There are two types of chemical compound one is covalent compound and other is ionic compound, covalent compound formed by sharing of electron and ionic compound formed by complete transfer of electron. Therefore,  the bond can be represented as M·×X.

Chemical Compound is a combination of molecule, Molecule forms by combination of element and element forms by combination of atoms in fixed proportion.

An ionic compound is a metal and nonmetal combined compound.  Ionic compound are very hard. They have high melting and boiling point because of strong ion bond.

The metal belonging to group 1 will have 1 electron in the outermost shell which can be represented as M·

Halogens have also 1 electron in its outermost shell which can be represented as X×

When these combine to form bond then the bond can be represented as M·×X.

Therefore,  the bond can be represented as M·×X.

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Yes they are

According to definition of an element

An element consists of several atoms of one type

So they are identical

Answer:

Although it may seem weird, the answer is actually yes.  An element is a pure substance consisting only of atoms that all have the same numbers of protons in their nuclei.  This means that all of the atoms are all the same because they contain the same amount of protons.

In conclusion, the atoms that make up an element are identical.

use the formula:

average atomic mass= atomic mass1 × abundance1/100 + atomic mass2 × abundance2/100

62.9296 × 69.15/100 + 64.9278 × 30.85/100=

62.9296 × 0.6915 + 64.9278 × 0.3085=

43.5158 + 20.0302=

63.546 amu

so the answer is 63.546 amu

(the element is europium)

Explanation:

a. qualitative

b. quantitative

c. quantitative

d. quantitative

Hope this helps you.

The strength of a bond can be calculated by dividing the energy required to break the bond by the number of bonds broken. In this case, if 0.5 kilocalories of energy are required to break 6 x 10^23 bonds of a particular type, the strength of the bond is approximately 8.33 x 10^-24 kilocalories per bond.

To calculate the strength of the bond, we divide the energy required to break the bond by the number of bonds broken. In this case, the energy required is 0.5 kilocalories and the number of bonds broken is 6 x 10^23. Dividing the energy by the number of bonds gives us the strength of the bond.

Strength of the bond = Energy required / Number of bonds broken

                   = 0.5 kilocalories / (6 x 10^23 bonds)

                   ≈ 8.33 x 10^-24 kilocalories per bond

Therefore, the strength of the bond is approximately 8.33 x 10^-24 kilocalories per bond. This value represents the energy required to break a single bond of the particular type.

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