The sketch below shows two marbles . The arrows show the size and the direction of the momentum of the two marbles . Draw arrows in the space below that show what will happen to these two marbles because of the law of conservation of momentum when they collide ​

The order of the reaction is 1.

The given information implies that the rate of the reaction is directly proportional to the concentration of A, i.e., Rate ∝ [A]¹. This indicates that the reaction is a first-order reaction with respect to A. In a first-order reaction, the rate of the reaction is directly proportional to the concentration of a single reactant.

When the concentration of this reactant is doubled, the rate of the reaction also doubles, as observed in the given information. Therefore, the order of the reaction with respect to A is 1.

The overall order of the reaction may be different if there are other reactants involved, but based on the given information, we can conclude that the order of the reaction with respect to A is 1.

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There are created around 125.8 grammes of magnesium metal.

By running electricity through molten magnesium chloride, magnesium metal may be produced. At the cathode, magnesium metal is produced, and at the anode, chlorine gas is developed. Therefore, the appropriate metal is formed at the cathode during the electrolytic reduction of molten salts (the negative electrode).

moles of substance = (electric charge in coulombs) / (Faraday's constant)

where Faraday's constant is the amount of electric charge carried by one mole of electrons, which is 96,485 C/mol.

electric charge = current x time = 65.7 A x 4.50 hr x 3600 s/hr = 1.00 x 10⁶ C

Next, we can calculate the moles of magnesium produced:

moles of Mg = electric charge / (Faraday's constant x 2)

(we divide by 2 because the balanced equation shows that 2 electrons are required to produce 1 mole of Mg)

moles of Mg = 1.00 x 10⁶ C / (96,485 C/mol x 2) = 5.18 mol

Finally, we can calculate the mass of magnesium produced using its molar mass, which is 24.31 g/mol:

mass of Mg = moles of Mg x molar mass of Mg

mass of Mg = 5.18 mol x 24.31 g/mol = 125.8 g

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The reason why acetyl chloride reacts with water almost violently, but you have to warm and shake the mixture of water and benzoyl chloride is due to the difference in their reactivity and molecular structure.

Acetyl chloride (CH3COCl) is a more reactive compound than benzoyl chloride (C6H5COCl). This higher reactivity is due to the presence of an electron-donating methyl group (CH3) in acetyl chloride, which increases the electrophilicity of the carbonyl carbon (C=O) and makes it more susceptible to nucleophilic attack by water.

On the other hand, benzoyl chloride has an electron-withdrawing phenyl group (C6H5), which reduces the electrophilicity of the carbonyl carbon, making it less reactive towards nucleophilic attack by water.

As a result, acetyl chloride reacts with water almost violently, forming acetic acid (CH3COOH) and hydrogen chloride (HCl) gas. In contrast, benzoyl chloride requires warming and shaking to facilitate the reaction with water, ultimately producing benzoic acid (C6H5COOH) and hydrogen chloride (HCl) gas.

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Approximately 109.7 liters of O₂ are required to completely react with 56.0 L of CH4 gas at STP to produce CO₂ and H₂O. This is calculated using the balanced chemical equation and the ideal gas law to determine the number of moles and volume of O₂ needed for the reaction.

The balanced chemical equation for the reaction is:

CH₄ + 2O₂ → CO₂ + 2H₂O

According to the equation, 1 mole of CH₄ reacts with 2 moles of O₂. At STP, 1 mole of any gas occupies 22.4 L. Therefore, 56.0 L of CH₄ at STP is equal to 56.0/22.4 = 2.50 moles of CH₄.

Since 1 mole of CH₄ requires 2 moles of O₂, then 2.50 moles of CH₄ will require 2.50 x 2 = 5.00 moles of O₂.

Using the ideal gas law, PV = nRT, we can calculate the amount of O₂ required in liters at STP:

n(O₂) = PV/RT = (1 atm)(5.00 mol)(22.4 L/mol)/(0.08206 L·atm/mol·K)(273 K) ≈ 109.7 L

Therefore, approximately 109.7 liters of O₂ gas are needed to react completely with 56.0 liters of CH₄ gas at STP to produce CO₂ and H₂O.

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The gas will exert a pressure of 0.0062 torr when transferred to a container with a volume of 18.5 L.

The pressure and volume of a gas are inversely proportional, according to Boyle's law. Therefore, we can use the formula P1V1 = P2V2 to solve this problem, where P1 and V1 are the initial pressure and volume, and P2 and V2 are the final pressure and volume.

Substituting the given values, we get:

P1 = 215 torr

V1 = 51.0 mL = 0.051 L

V2 = 18.5 L

Solving for P2, we get:

P2 = P1V1/V2 = 215 torr x 0.0510 L / 18.5 L = 0.595 torr

As a result, when transferred to an 18.5 L container, the gas will impose a pressure of 0.595 torr. It is important to note that the units of volume must be consistent (either both in mL or both in L) in order to obtain the correct answer.

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To make sweet tea, a cook dissolved 152.395 grams of sugar  in 5.19 l of water at 32.34 °c. The molality of this sugar solution 0.275 mol/kg

To find the molality of the sugar solution, we need to first calculate the moles of sugar and the mass of solvent (water) in kilograms:

We know that molar mass of sugar= 110 g/mol

Moles of sugar = Mass of sugar / Molecular weight of sugar

Moles of sugar = 152.395 g / 110 g/mol

Moles of sugar = 1.385 mol

Mass of solvent (water) = Volume of solution - Mass of solute (sugar)

Mass of solvent (water) = 5.19 L - 0.146739 kg (since density of water is approximately 1 kg/L)

Now we can calculate the molality:

Molality = Moles of solute / Mass of solvent (in kg)

Molality = 1.385 mol / 5.043261 kg

Molality = 0.275 mol/kg

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The pH after the addition of 38.0 mL of 0.50 M KOH to the 76.0 mL of 0.25 M HBr is 7.

To find the pH after the addition of 38.0 mL of 0.50 M KOH to the 76.0 mL of 0.25 M HBr:

1. Calculate the moles of HBr and KOH before the reaction.
2. Determine the limiting reactant and the remaining moles of the excess reactant.
3. Calculate the concentration of the remaining reactant.
4. Determine the pH using the concentration of the remaining reactant.

Calculate moles of HBr and KOH
Moles of HBr = volume (L) × concentration (M)
Moles of HBr = 0.076 L × 0.25 M = 0.019 moles

Moles of KOH = volume (L) × concentration (M)
Moles of KOH = 0.038 L × 0.50 M = 0.019 moles

Determine the limiting reactant
In this case, both HBr and KOH have the same moles (0.019 moles), so they will react completely with each other, leaving no excess reactant.

Calculate the concentration of the remaining reactant
Since both reactants have been completely consumed in the reaction, there are no remaining reactants. The reaction produces the salt KBr and water.

Determine the pH
As there are no remaining acidic or basic reactants in the solution, the pH of the resulting solution is neutral, with a pH of 7.

Therefore, 7 is the pH after the addition of 38.0 mL of 0.50 M KOH to the 76.0 mL of 0.25 M HBr.

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The only empirical formula among the given choices is e) HNF2.

The empirical formula is the simplest whole-number ratio of atoms in a compound. Therefore, to determine which of the given options is the empirical formula, we need to simplify each formula to its simplest ratio.

To find the empirical formula, we need to check which of these choices have the elements in their simplest ratio:

a) N2F2: This can be reduced to NF, so it's not the empirical formula.
b) N2F4: This can be reduced to NF2, so it's not the empirical formula.
c) H2C2: This can be reduced to HC, so it's not the empirical formula.
d) H2N2: This can be reduced to HN, so it's not the empirical formula.
e) HNF2: This formula is already in its simplest whole-number ratio, so it is the empirical formula.

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An ambidentate ligand is a ligand that can bind to a metal ion (M) through two different donor atoms.

Two examples of ambidentate ligands (other than NO₂) are SCN⁻ and NCS⁻. SCN⁻ can bind through the sulfur atom (S) or the nitrogen atom (N), forming [M-SCN] or [M-NCS] complexes. The IUPAC names for these ligands are thiocyanato (SCN⁻) and isothiocyanato (NCS⁻).

A polydentate ligand is a ligand that can bind to a metal ion (M) through multiple donor atoms simultaneously. Two examples are ethylenediamine (en) and ethylenediaminetetraacetic acid (EDTA).

The chelate effect occurs when a polydentate ligand forms a ring with a metal ion, which stabilizes the complex and lowers its overall energy. The IUPAC names for these ligands are ethane-1,2-diamine (en) and (ethylenedinitrilo)tetraacetate (EDTA).

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The ranking from highest to lowest pH at the equivalence point would be: a > b > c.

To rank the solutions from highest to lowest pH at the equivalence point, we need to consider the acid-base reactions involved in each titration.
a. In the first titration, NaOH (a strong base) reacts with acetic acid (a weak acid) to form sodium acetate and water. At the equivalence point, all of the acetic acid has been neutralized by NaOH, and we are left with a solution of sodium acetate. Sodium acetate is the conjugate base of acetic acid, and because acetic acid is a weak acid, its conjugate base is a relatively strong base. Therefore, the pH of the solution will be relatively high at the equivalence point.
b. In the second titration, NaOH reacts with chloroacetic acid (a stronger acid than acetic acid) to form sodium chloroacetate and water. At the equivalence point, all of the chloroacetic acid has been neutralized by NaOH, and we are left with a solution of sodium chloroacetate. Like sodium acetate, sodium chloroacetate is the conjugate base of a weak acid, so the pH of the solution will be relatively high at the equivalence point, but slightly lower than in the first titration because chloroacetic acid is a stronger acid than acetic acid.
c. In the third titration, NaOH (a strong base) reacts with HCl (a strong acid) to form sodium chloride and water. At the equivalence point, all of the HCl has been neutralized by NaOH, and we are left with a solution of sodium chloride. Because both HCl and NaCl are strong acids/bases, there will be no residual acidity/basicity in the solution, and the pH will be neutral (around 7) at the equivalence point.

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The given reaction represents a combustion reaction.

In this reaction, the hydrocarbon compound C6H14 reacts with oxygen gas (O2) to form carbon dioxide gas (CO2) and water vapor (H2O). The reactants are a hydrocarbon and oxygen, and the products are carbon dioxide and water.

Combustion reactions are exothermic reactions that involve the reaction of a fuel with oxygen, resulting in the release of heat and light energy.

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Comparing the amounts of product produced, we can see that reactant B produces the least amount of product (0.30 mol of D), while reactants A and C produce more than that. Therefore, reactant B is the limiting reactant in this scenario.

To determine the limiting reactant, we need to compare the amount of each reactant to the stoichiometric coefficients in the balanced equation. The reactant that produces the least amount of product, or runs out first, is the limiting reactant.

Let's start by calculating the amount of product that each reactant can produce:

For reactant A:
- 0.20 mol A produces 2 × 0.20 mol = 0.40 mol of D
- 0.20 mol A produces 1 × 0.20 mol = 0.20 mol of E

For reactant B:
- 0.60 mol B produces 1/2 × 0.60 mol = 0.30 mol of D
- 0.60 mol B produces 1/3 × 0.60 mol = 0.20 mol of E

For reactant C:
- 0.75 mol C produces 1/3 × 0.75 mol = 0.25 mol of D
- 0.75 mol C produces 1/3 × 0.75 mol = 0.25 mol of E

The limiting reactant is the one that produces the least amount of product. Comparing the amounts of product produced, we can see that reactant B produces the least amount of product (0.30 mol of D), while reactants A and C produce more than that. Therefore, reactant B is the limiting reactant in this scenario.

So the answer is (E) B only.

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The proper IUPAC names of tert-Butyl alcohol is 2-methylpropan-2-ol and of tert-Butyl chloride is 2-chloro-2-methylpropane.

The common names, tert-butyl alcohol and tert-butyl chloride, can be represented by their proper IUPAC names as follows:

1. tert-Butyl alcohol, also known as tert-butanol, has the IUPAC name 2-methylpropan-2-ol. This name is derived from its structure, where a methyl group (CH3) is attached to the second carbon of a three-carbon chain (propane). The alcohol functional group (-OH) is also attached to the second carbon, hence the "-2-ol" suffix.

2. tert-Butyl chloride, also known as tert-butylchloride or t-BuCl, has the IUPAC name 2-chloro-2-methylpropane. Similar to the alcohol, its structure consists of a methyl group (CH3) attached to the second carbon of a three-carbon chain (propane). Instead of an alcohol functional group, there is a chlorine atom (Cl) attached to the same second carbon, hence the "2-chloro" prefix in the name.

These IUPAC names provide a more systematic and descriptive way to identify the chemical structures of these compounds compared to their common names.

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The formal potential for the given reaction is 1.34 V.

The Nernst equation relates the standard potential, E∘′, to the actual cell potential, E, and the reaction quotient, Q:

E = E∘′ - (RT/nF)ln(Q)

where R is the gas constant, T is temperature, n is the number of moles of electrons transferred in the reaction, F is the Faraday constant, and ln is the natural logarithm.

At equilibrium, the reaction quotient Q is equal to the equilibrium constant K:

K = [HNO2][H2O]/[NO-3][H+]^3

The relationship between K and the acid dissociation constant, Ka, for the reaction HNO2 + H2O ⇌ H3O+ + NO2- is:

K = [H3O+][NO2-]/[HNO2] = Ka/[HNO2]

Substituting the expression for K into the equation for Q gives:

Q = [HNO2]/Ka[HNO2] = 1/Ka

Substituting the values given in the problem into the Nernst equation and solving for E∘′ gives:

E = 0.940 V = E∘′ - (0.0257 V/K)(298 K)/(2 mol)(96485 C/mol)ln(1/Ka)
E∘′ = 0.940 V + (0.0257 V/K)(298 K)/(2 mol)(96485 C/mol)ln(1/Ka)
E∘′ = 0.940 V + (0.0592 V)ln(1/7.1×10^-4)
E∘′ = 0.940 V + 0.0592(7.09)
E∘′ = 1.34 V

Therefore, the formal potential for the given reaction is 1.34 V.

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If all the mitochondria within intestinal cells were destroyed, glucose absorption would decrease.

If all the mitochondria within the intestinal cells were destroyed, glucose absorption would decrease due to the lack of ATP production. Active transport is responsible for the absorption of glucose in intestinal cells, and this process requires energy in the form of ATP. Mitochondria are the main organelles responsible for ATP production in cells, and without them, the energy supply for active transport would be insufficient. This would lead to a decrease in glucose absorption by the intestinal cells. Since glucose is an important source of energy for the body, reduced glucose absorption can have negative consequences on the overall health and well-being of an individual.

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The para-nitro phenyl substituent causes a higher λmax value than the meta-nitro phenyl substituent because it has a greater electron-withdrawing effect due to its position relative to the benzene ring.

The nitro group contains both electron-withdrawing (NO₂) and electron-donating (O) groups, which can affect the electron density of the benzene ring through resonance.

In the para-nitro phenyl substituent, the nitro group is positioned directly opposite to the hydrogen on the carbon that is attached to the ring. This allows for maximum overlap between the nitro group's electron-withdrawing pi-system and the pi-system of the benzene ring, resulting in a greater degree of electron withdrawal from the ring. This reduces the electron density of the ring and causes the λmax value to shift to a higher wavelength.

In the meta-nitro phenyl substituent, the nitro group is positioned one carbon away from the hydrogen on the carbon that is attached to the ring. This results in less efficient overlap between the pi-systems of the nitro group and the benzene ring, resulting in a weaker electron-withdrawing effect on the ring. This leads to a smaller shift in λmax compared to the para-nitro phenyl substituent.

Below are the structures and resonance contributors for para-nitro phenyl substituent and meta-nitro phenyl substituent.

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Explanation:

pH+pOH=14

pH=14-pOH

=14-5.039e-3

=13.9

The standard enthalpy change for the reverse reaction is the negative of the given enthalpy change. Therefore, the standard enthalpy change for the reaction  K(s) + 1/2 I2(s) → KI(s) is -656 kJ at 328K.

To find the standard enthalpy change for the given reaction at 298 K, we can manipulate the original reaction and divide its enthalpy change by 2.

The original reaction is:
2 KI(s) → 2 K(s) + I2(s) ΔH° = 656 kJ

Divide the entire reaction by 2:
KI(s) → K(s) + 1/2 I2(s)

Now, divide the enthalpy change by 2:
ΔH° = 656 kJ / 2 = 328 kJ

So, the standard enthalpy change for the reaction K(s) + 1/2 I2(s) → KI(s) at 298 K is 328 kJ.

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Sodium hydroxide (NaOH) and hydrochloric acid (HCl) cannot exist together in their pure form because they react with each other to form a highly exothermic reaction that generates a large amount of heat and releases hydrogen gas.

This reaction is known as a neutralization reaction, where the sodium ions (Na⁺) from the NaOH combine with the chloride ions (Cl⁻) from the HCl to form sodium chloride (NaCl), which is a salt.

The hydrogen ions (H⁺) from the HCl combine with the hydroxide ions (OH⁻) from the NaOH to form water (H₂O).

This reaction is so powerful that it can cause an explosion or fire if not handled carefully. Therefore, NaOH and HCl are stored separately and only mixed in controlled conditions.

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a. pOH of 0.014 M KOH at 25°c is 1.85 and PH of 0.014 M KOH is 12.15.

b. pOH of 1.65 M NaOH is 0.18 and pH of 1.65 M KOH is 13.82.

c. pOH of 0.084 M Ba(OH)₂ is 0.77 and pH of 0.084 M Ba(OH)₂ is 13.23.

a. 0.014 M KOH:
Since KOH is a strong base, its pOH can be calculated as the negative logarithm of its concentration:

pOH = -log(0.014) ≈ 1.85

To find the pH, use the formula: pH = 14 - pOH

pH = 14 - 1.85 ≈ 12.15

b. 1.65 M NaOH:
NaOH is also a strong base, so follow the same process:

pOH = -log(1.65) ≈ 0.18

pH = 14 - 0.18 ≈ 13.82

c. 0.084 M Ba(OH)₂:

Since each Ba(OH)₂ molecule releases 2 OH⁻ ions when it dissociates, the concentration of OH⁻ ions is 2 × 0.084 = 0.168 M.

pOH = -log(0.168) ≈ 0.77

pH = 14 - 0.77 ≈ 13.23

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The greener choice in a chemical process between a reaction that can be run at 300 K for 1 hour with a catalyst (Option A).

In terms of green chemistry, the reaction that can be run at 300 K for 1 hour with a catalyst is the greener choice. This is because using a catalyst can increase the reaction rate and efficiency, thus reducing the amount of energy and resources needed to run the reaction. Additionally, running the reaction at a lower temperature can reduce energy consumption and decrease the carbon footprint of the process. Overall, using a catalyst and optimizing reaction conditions for efficiency and sustainability is a key aspect of green chemistry.

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As a buffer, ammonium acetate in water can be used. Ammonium acetate is a salt solution that can function as a buffer by itself.

As we now know, p K a ​=−logK a K a =1.34 x 10 -5 (Given) pK a =log(1.34 x 10 - 5) pK a =5 x 0.127 =4.873 pH is now calculated as pH=pK a + log( [acid] [salt] )

Given:- [salt]=[acid]=0.5M ∴pH=4.87+log

0.5 0.5​=4.873

As a result, the solution's pH is 4.873.

Drive Me Tile Online has a concentration of 0.075 and Try Metal Ammonium Chloride has a concentration of 0.13 M I.

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The graph that best represents the strength of the gravitational force between the planets and the star is D. Z.

The greater the mass of an object, the greater its gravitational force on other objects. This is because mass determines the strength of an object's gravitational field.

The more massive an object is, the more space it curves around itself, and the more it attracts other objects towards it. This is why planets with larger masses have stronger gravitational forces than smaller objects like asteroids or comets. This is shown in Graph Z.

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The correct answer is A) BF₃ bond angles > SO₂ bond angle.

BF₃ has a trigonal planar molecular geometry with bond angles of approximately 120 degrees. This is due to the three electron pairs surrounding the central boron atom, which repel each other and arrange themselves as far apart as possible.

SO₂ has a bent molecular geometry with bond angles of approximately 119 degrees. This is due to the two electron pairs surrounding the central sulfur atom, which repel each other and arrange themselves as far apart as possible while also accounting for the lone pairs on the sulfur atom.

Therefore, the bond angles in BF₃ are larger than the bond angles in SO₂, making option A the correct answer.

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The concentration of Allura Red in the undiluted unknown is 0.00556 mg/mL.

To determine the concentration of Allura Red in the undiluted unknown, we can use the diluted equation, which relates the concentration of the diluted solution with the concentration of the undiluted solution, as well as the dilution factor:

C1V1 = C2V2

where:

C1 is the concentration of the undiluted solution (what we want to find)

V1 is the volume of the undiluted solution

C2 is the concentration of the diluted solution (known)

V2 is the volume of the diluted solution (known)

the dilution factor is V1/V2

Assuming we know the concentration and volume of the diluted solution, as well as the dilution factor, we can rearrange the diluted equation to solve for the concentration of the undiluted solution:

C1 = C2 x V2/V1

Let's say we have a diluted solution of Allura Red with a concentration of 0.05 mg/mL and a volume of 10 mL, and we diluted it 1:10 (meaning we added 90 mL of solvent to make a total volume of 100 mL). We can use the diluted equation to calculate the concentration of Allura Red in the undiluted solution:

C1 = 0.05 mg/mL x 10 mL / 90 mL

C1 = 0.00556 mg/mL

Therefore, the concentration of Allura Red in the undiluted unknown is 0.00556 mg/mL.

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A minimum of 25.6 mL of 3.5 M HI is needed to thoroughly dissolve 2.42 g of silver.

The less active metals, including zinc and magnesium, are easily dissolved by hydrochloric acid. Iron, copper, and other harder metals are less easily or completely disintegrated by it. While hydrochloric acid won't dissolve some metals, other chemicals, such nitric acid, will.

The chemical equation for the reaction of copper with HI is balanced as follows:

Cu(s) + 2HI(aq) → CuI2(aq) + H2(g)

The following formula can be used to determine the minimum volume of 3.5 M HI needed to completely dissolve 4.85 g Cu:

Moles of Cu = 4.85 g / 63.55 g/mol = 0.0763 mol

Moles of HI required = 2 × moles of Cu = 0.1526 mol

Volume of 3.5 M HI required = moles of HI required / 3.5 M = 0.0436 L or 43.6 mL

Therefore, 43.6 mL of 3.5 M HI is the lowest amount needed to thoroughly dissolve 4.85 g of copper. The chemical equation for the silver-HI reaction is balanced as follows:

2Ag(s) + 4HI(aq) → 2AgI(s) + 2H2(g)

The following formula can be used to determine the minimum volume of 3.5 M HI needed to completely dissolve 2.42 g of Ag:

Moles of Ag = 2.42 g / 107.87 g/mol = 0.0224 mol

Moles of HI required = 4 × moles of Ag = 0.0896 mol

Volume of 3.5 M HI required = moles of HI required / 3.5 M = 0.0256 L or 25.6 mL

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There are 0.465 moles of neon in 9.4 g of neon.

A mole is a unit of measurement used in chemistry to express amounts of a substance. One mole of a substance is defined as the amount of that substance that contains the same number of entities, such as atoms, molecules, or ions, as there are atoms in exactly 12 grams of carbon-12. This number is known as Avogadro's number and is approximately 6.022 x 10²³ entities per mole.

To calculate the number of moles of neon in 9.4 g of neon, we need to use the atomic mass of neon. The atomic mass of neon is 20.18 g/mol.
9.4 g Neon / 20.18 g/mol Neon = x moles Neon
Solving for x, we get:

x = (9.4 g Neon) / (20.18 g/mol Neon)
x = 0.465 moles Neon

Therefore, the number of moles of neon in 9.4 g of neon is 0.465 moles (rounded to three significant figures).

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To ionize an electron from the ground level, it typically takes around 10 electron volts of energy.

we first need to know the ionization energy of the specific atom or element you're referring to, as this value varies for different elements. Ionization energy is the amount of energy required to remove an electron from the ground level of an atom.

Once you have the ionization energy in joules, you can convert it to electron volts (eV) using the following conversion factor: 1 electron volt (eV) = 1.602 x 10^-19 joules (J).

So, to calculate the energy required to ionize an electron from the ground level in electron volts, follow these steps:

1. Find the ionization energy of the element in joules (J).
2. Convert the ionization energy to electron volts (eV) using the conversion factor mentioned above.

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Therefore, it takes 165.308 calories of heat to warm 14.9 g of brick from 21.4 °C to 47 °C.

The amount of heat needed to raise a substance's temperature by one degree Celsius in one gramme, also known as specific heat.

Using the following calculation, we can determine how much heat is needed to raise 14.9 g of brick from 21.4 to 47.4 degrees Celsius:

q = m * c * ΔT

where q is the required quantity of heat, m is the substance's mass, c is its specific heat, and T is the temperature change.

Brick has a specific heat of 0.20 cal/g.°C, so when the given values are substituted, we obtain:

q = 14.9 g * 0.20 cal/g.°C * (47.4°C - 21.4°C)

q = 14.9 g * 0.20 cal/g.°C * 26°C

q = 165.308 cal

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Therefore, it takes 165.308 calories of heat to warm 14.9 g of brick from 21.4 °C to 47 °C.

The amount of heat needed to raise a substance's temperature by one degree Celsius in one gramme, also known as specific heat.

Using the following calculation, we can determine how much heat is needed to raise 14.9 g of brick from 21.4 to 47.4 degrees Celsius:

q = m * c * ΔT

where q is the required quantity of heat, m is the substance's mass, c is its specific heat, and T is the temperature change.

Brick has a specific heat of 0.20 cal/g.°C, so when the given values are substituted, we obtain:

q = 14.9 g * 0.20 cal/g.°C * (47.4°C - 21.4°C)

q = 14.9 g * 0.20 cal/g.°C * 26°C

q = 165.308 cal

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The statement "Draw an arrow pushing mechanism for the formation of the acylium ion when acetic anhydride reacts with phosphoric acid" is true.

To draw an arrow pushing mechanism for the formation of the acylium ion when acetic anhydride reacts with phosphoric acid, follow these steps:

1. Draw the structures of acetic anhydride and phosphoric acid.

2. Locate the electrophilic carbonyl carbon in acetic anhydride and the nucleophilic oxygen atom in phosphoric acid.

3. Draw an arrow from the lone pair of electrons on the oxygen atom of phosphoric acid to the electrophilic carbonyl carbon in acetic anhydride, indicating nucleophilic attack.

4. Draw an arrow from the pi bond between the carbonyl carbon and oxygen in acetic anhydride to the carbonyl oxygen, representing the movement of electrons and formation of a negatively charged oxygen atom.
5. Draw the intermediate structure formed after the nucleophilic attack.

6. Draw an arrow from the negatively charged oxygen atom to reform the carbonyl double bond and simultaneously break the carbon-oxygen bond adjacent to the carbonyl carbon, pushing electrons to the neighboring oxygen atom.

7. Draw the final acylium ion and byproduct formed in the reaction.

This mechanism illustrates how the acylium ion is formed when acetic anhydride reacts with phosphoric acid.

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