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Answer:

-191 kJ

Explanation:

The given reaction is:

N₂(g) + 3H₂(g) → 2NH₃(g) ΔH = -76.4 kJ/mol

From the balanced equation, we can see that the stoichiometric ratio between hydrogen (H₂) and ammonia (NH₃) is 3:2. This means that 3 moles of hydrogen react to produce 2 moles of ammonia.

To determine the heat energy when 5.0 g of hydrogen (H₂) burns, we need to follow these steps:

Step 1: Calculate the moles of hydrogen (H₂)

Using the molar mass of hydrogen (H₂), which is 2 g/mol, we can calculate the moles of hydrogen (H₂) in 5.0 g of hydrogen:

Moles of H₂ = Mass of H₂ / Molar mass of H₂

Moles of H₂ = 5.0 g / 2 g/mol

Moles of H₂ = 2.5 mol

Step 2: Use the stoichiometry of the reaction

Based on the stoichiometry of the reaction, we know that 3 moles of hydrogen (H₂) react to produce 2 moles of ammonia (NH₃), and the enthalpy change (ΔH) is -76.4 kJ/mol.

Step 3: Calculate the heat energy

The heat energy for 2.5 moles of hydrogen (H₂) can be calculated using the given enthalpy change (ΔH) and the stoichiometry of the reaction:

Heat energy = Moles of H₂ x ΔH

Heat energy = 2.5 mol x -76.4 kJ/mol

Heat energy = -191 kJ (rounded to three significant figures)

So, the heat energy when 5.0 g of hydrogen (H₂) burns is -191 kJ (rounded to three significant figures), and the negative sign indicates that the reaction is exothermic, releasing heat.

Answer:

Explanation:

Hydrogen bonds are described as a force between molecules. However, hydrogen bonds can also exist as a force within a molecule under certain conditions. This is called intramolecular hydrogen bonding. It occurs when a hydrogen atom is sandwiched between two strongly electronegative atoms (such as F, O, N) in the same molecule1.

Answer:

6.3

Explanation:

multiply the length value by 100

Urinary tract infection is the most common nosocomial infection in patients admitted to the hospital. Surgical site wound infections, bacteremia, and gastrointestinal and skin infections are among the most common nosocomial infections.

A hospital-acquired infection, also known as a nosocomial infection, occurs in a hospital or other healthcare setting. It is sometimes referred to as a healthcare-associated infection to emphasize both hospital and nonhospital settings.

Nosocomial infections, also known as healthcare-associated infections (HAI), are infections acquired while receiving healthcare that was not present at the time of admission.

Healthcare-Acquired Infections (HAIs), also known as Healthcare-Associated Infections, are infections contracted while receiving treatment at a healthcare facility, such as a hospital, or from a healthcare professional, such as a doctor or nurse.

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Endothermic since it requires heat from the surrounding

The percent transmittance (%T) and absorbance (A) of a solution are related by an equation which can be used to solve this question.

The percent transmittance (%T) and absorbance (A) of a mixture are associated by the following equation:

%T = 100 x 10^(-A)

We are given that the %T value of the solution is 51.6% at a wavelength of 550 nm. To find the absorbance (A), we can rearrange the equation above:

A = -㏒(%T / 100)

On substituting the value in the given %T value, we get:

A = -㏒(51.6 / 100) = -㏒(0.516) = 0.286

Therefore, the absorbance of the solution at a wavelength of 550 nm is 0.286.

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The pH after 15.0 mL of 0.210 KOH is added in the titration of 55.0 mL of 0.210 M HClO is 4.56.

To solve this problem, we need to use the balanced chemical equation for the reaction between KOH and HClO:

HClO + KOH → KClO + H2O

We can see that for every mole of KOH added, one mole of HClO will react. Therefore, the number of moles of HClO in 55.0 mL of 0.210 M HClO is:

n(HClO) = M(HClO) x V(HClO) = 0.210 mol/L x 0.0550 L = 0.0116 mol

When 15.0 mL of 0.210 M KOH is added, the number of moles of KOH added is:

n(KOH) = M(KOH) x V(KOH) = 0.210 mol/L x 0.0150 L = 0.00315 mol

Since the reaction is a neutralization reaction, the moles of HClO left after the reaction will be:

n(HClO) = n(HClO)initial - n(KOH) = 0.0116 mol - 0.00315 mol = 0.00845 mol

We can now use the equilibrium expression for the ionization of HClO in water to calculate the pH of the solution:

HClO + H2O ⇌ H3O+ + ClO-

Ka = [H3O+][ClO-]/[HClO]

At equilibrium, the concentrations of H3O+ and ClO- can be assumed to be equal to the concentration of HClO that remains unreacted, since HClO is a weak acid and does not dissociate completely in water. Therefore:

[H3O+] = [ClO-] = [HClO] = 0.00845 mol / (0.0550 L + 0.0150 L) = 0.105 M

Substituting these values into the equilibrium expression for Ka:

Ka = [H3O+][ClO-]/[HClO] = (0.105 M)² / 0.00845 M = 1.31 x 10⁻⁶

pKa = -log(Ka) = -log(1.31 x 10⁻⁶) = 5.88

pH = 1/2(pKw - pKa) = 1/2(14.00 - 5.88) = 4.56

Therefore, the pH after 15.0 mL of 0.210 KOH is added in the titration of 55.0 mL of 0.210 M HClO is 4.56.

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To calculate the cell potential, Ecell, for the given reaction at 298K, we need to use the Nernst equation. The Nernst equation relates the cell potential to the standard cell potential, temperature, and the concentrations of the reactants and products. The Nernst equation is given as follows:

Ecell = E°cell - (RT/nF) ln(Q)

where,

Ecell = cell potential

E°cell = standard cell potential

R = gas constant (8.314 J/K.mol)

T = temperature (298 K)

n = number of electrons transferred in the balanced redox reaction

F = Faraday constant (96,485 C/mol)

Q = reaction quotient

The given reaction is a redox reaction, which involves the transfer of two electrons from Co to Ag+. The balanced half-reactions are as follows:

Co(s) → Co2+(aq) + 2 e-

Ag+(aq) + e- → Ag(s)

The standard reduction potentials for these half-reactions are:

Co2+(aq) + 2 e- → Co(s) E°red = -0.28 V

Ag+(aq) + e- → Ag(s) E°red = +0.80 V

The overall standard cell potential can be calculated by subtracting the standard reduction potential of the anode from that of the cathode:

E°cell = E°red,cathode - E°red,anode

= +0.80 V - (-0.28 V)

= +1.08 V

Now we need to calculate the reaction quotient Q using the concentrations of the reactants and products. According to the given information, [Ag+] = 0.010 M and [Co2+] = 0.015 M.

Q = ([Co2+][Ag+]^2)/([Ag+]^2)

= ([0.015][0.010]^2)/([0.010]^2)

= 0.015 M

Substituting the values in the Nernst equation, we get:

Ecell = E°cell - (RT/nF) ln(Q)

= 1.08 - (8.314 x 298 / (2 x 96485)) ln(0.015)

= 0.829 V

Therefore, the cell potential, Ecell, for the given reaction at 298K is 0.829 V.

The rate law equation for this reaction is:

rate = (1.99 x 10^68 L^12/mol^12 s)[F]^21[H]^-9

To find the rate law equation for this reaction, we'll use the given experimental data to determine the order of the reaction with respect to F and H. The rate law equation will be in the form:

rate = k[F]^x[H]^y

We can use the data from the first two experiments to determine the order of the reaction with respect to F:

Exp1: rate1 = k(0.000345)^x(0.000765)^y

Exp2: rate2 = k(0.000690)^x(0.000765)^y

Divide rate2 by rate1:

(rate2/rate1) = (0.000690/0.000345)^x

(3.24 x 10^8)/(3.24 x 10^2) = (2)^x

2.0 x 10^6 = 2^x

Since 2^21 = 2097152, which is approximately 2.0 x 10^6, we can conclude that x = 21. So, the reaction is 21st order with respect to F.

Now, we can use the data from experiments 1 and 3 to determine the order of the reaction with respect to H:

Exp1: rate1 = k(0.000345)^21(0.000765)^y

Exp3: rate3 = k(0.000345)^21(0.00765)^y

Divide rate3 by rate1:

(rate3/rate1) = (0.00765/0.000765)^y

(3.24 x 10^-7)/(3.24 x 10^2) = (10)^y

1.0 x 10^-9 = 10^y

From this, we can conclude that y = -9. So, the reaction is -9th order with respect to H.

Now, we can write the rate law equation:

rate = k[F]^21[H]^-9

Next, we'll calculate the rate constant k using the data from any of the experiments. Let's use the data from Experiment 1:

rate1 = 3.24 x 10^2 M/sec

[F]1 = 0.000345 mol/L

[H]1 = 0.000765 mol/L

3.24 x 10^2 = k(0.000345)^21(0.000765)^-9

After calculating, we find:

k ≈ 1.99 x 10^68 L^12/mol^12 s

So, the rate law equation for this reaction is:

rate = (1.99 x 10^68 L^12/mol^12 s)[F]^21[H]^-9

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Addition of a catalyst, at 15 °C, a certain reaction is able to produce 0.80 moles of product per minute at 25 °C it will produce at a rate of 0.40 moles per minute. The correct option to this question is C.

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The polymer that includes a monomer with a benzene ring is called a phenolic polymer.

A polymer is a large molecule made up of repeating subunits called monomers. These monomers can be identical or different, and they are linked together through chemical bonds to form a long chain. Polymers can be natural, such as proteins, cellulose, and DNA, or they can be synthetic, such as plastics, synthetic rubber, and fibers.

Polymers have a wide range of properties and can be tailored to meet specific requirements for various applications. For example, some polymers are strong and tough, making them suitable for use in structural materials, while others are flexible and elastic, making them useful in materials such as rubber and elastomers.

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Determine the amount of KOH present in the resulting solution. KOH was initially 0.00875 mol, then 0.00525 mol of it interacted with HCl. As a result, 0.00875 mole - 0.00525 mol (= 0.00350 mol of KOH is left. The resulting solution has a volume of 70.0 mL (35.0 mL plus 35.0 mL).

The titrant (NaOH), which is added gradually throughout the course of a titration, is added to the unknown substance. The equivalency point is the moment at which precisely the right quantity of titrant (NaOH) has indeed been added that react to the entire analyte (HCl).

What took place during titration: One mole of NaOH interacts with one mole of HCl inside the reaction between the two substances. NaOH with HCl equals NaCl plus H2O. (NaOH and HCl have a mole ratio of 1:1.) • The NaOH concentration is 0.1 M, or 0.1 molecules per litre.

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Answer: C)Deposition (gas to solid)

Explanation: According to the phase diagram, at room temperature (25°C) and 1 atm, the sample is in the gas phase.  As the temperature decreases to 80 K, it falls below the sublimation curve. T he sublimation curve represents the conditions at which a substance can change directly from a solid to a gas or from a gas to a solid without passing through the liquid phase.

Since the sample is in the gas phase at room temperature, cooling it to 80 K would cause it to go through the process of deposition, where the gas particles directly transform into a solid without first becoming a liquid.  This is indicated by the section of the phase diagram below the sublimation curve.

The molar mass of the unknown gas is 28 g/mol.

Graham's law of effusion states that the rate of effusion (the escape of gas molecules through a tiny hole) of a gas is inversely proportional to the square root of its molar mass, at constant temperature and pressure. This means that lighter gases effuse (escape through a hole) faster than heavier gases, all other factors being equal.

We can use Graham's law of effusion to solve this problem:

Rate of effusion is inversely proportional to the square root of molar mass.

Thus, we can write:

(rate of effusion of O₂) / (rate of effusion of the unknown gas) =

sqrt(molar mass of the unknown gas) / sqrt(molar mass of O₂)

Let's call the molar mass of the unknown gas "M". We can set up the following system of equations using the information given in the problem:

4/1 = (rate of effusion of O₂) / (rate of effusion of the unknown gas)

(rate of effusion of O₂) = 1/220

(rate of effusion of the unknown gas) = 1/245

Plugging in these values, we get:

4/1 = (1/220) / (1/245)

4/1 = 49/44

Solving for the rate of effusion of the unknown gas, we get:

(rate of effusion of the unknown gas) = (1/245) / (49/44) = 4/539

Now we can use Graham's law of effusion to find the molar mass of the unknown gas:

(rate of effusion of O₂) / (rate of effusion of the unknown gas) =

sqrt(molar mass of the unknown gas) / sqrt(molar mass of O₂)

(1/220) / (4/539) = sqrt(M) / sqrt(32)

Solving for M, we get:

M = 28 g/mol

Therefore, the molar mass of the unknown gas is 28 g/mol.

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While the length is less than a double bond, the strength exceeds that of a double bond.

Due to the presence of two bonds rather than one, triple bonds are stronger than double bonds. An sp-sp sigma bond is created when one of each carbon atom's two sp hybrid orbitals intersects with the corresponding orbital from the other carbon atom.

Six electrons are shared by a sigma bond, two bonds, and a triple bond. Double bonds are more powerful than single bonds, and triple bonds are more powerful than double bonds, according to experiments.

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Answer: third

Explanation:

a)  The conjugate base of aspirin (acetylsalicylic acid) is formed when the acidic proton (H+) is removed from the carboxylic acid group (-COOH) in the molecule.

b)  More aspirin will be available for absorption in the duodenum (97%) compared to the stomach (12%).

a. The conjugate base of aspirin (acetylsalicylic acid) is formed when the acidic proton (H+) is removed from the carboxylic acid group (-COOH) in the molecule.

b. The percentage of aspirin available for absorption depends on the degree of ionization of the molecule, which is related to the pH of the surrounding medium. At pH values below the pKa (2.97), most of the molecules exist in the protonated form (HA), while at pH values above the pKa, most of the molecules exist in the deprotonated form (A-).

Using the Henderson-Hasselbalch equation:

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

We can calculate the ratio of deprotonated (A-) to protonated (HA) forms at different pH values. At pH 2.0, the ratio is:

2.0 = 2.97 + log([A-]/[HA])

log([A-]/[HA]) = -0.97

[A-]/[HA] = 0.12

So, at pH 2.0, only 12% of the aspirin molecules are in the deprotonated form and available for absorption.

At pH 4.5, the ratio is:

4.5 = 2.97 + log([A-]/[HA])

log([A-]/[HA]) = 1.53

[A-]/[HA] = 31.6

So, at pH 4.5, 97% of the aspirin molecules are in the deprotonated form and available for absorption.

Therefore, more aspirin will be available for absorption in the duodenum (97%) compared to the stomach (12%).

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[tex]V_{tot} = \frac{587 g}{1,07 g/mL} = 549 mL[/tex]

0,022 kg = 22 g

[tex]\frac{m}{V} = \frac{22 g × 100}{549 mL} = 4,0 % [/tex]

Melting point is the property of the wax and the sulfur should be investigated to provide evidence of the relative strengths of their intermolecular forces. Therefore, the correct option is option D.

The electrical forces of attraction and repulsion that act between atoms along with other types of nearby particles, such as atoms or ions, are examples of intermolecular forces (IMFs), also known as secondary forces. In comparison to intramolecular forces, which bind a molecule together, intermolecular forces are weak.

For instance, the forces between adjacent molecules are substantially weaker than the covalent bond, which involves sharing pairs of electrons between atoms. Both types of forces are crucial components of the force fields that molecular mechanics typically use. Melting point is the property of the wax and the sulfur should be investigated to provide evidence of the relative strengths of their intermolecular forces.

Therefore, the correct option is option D.

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Light takes 20,100 seconds or 5.583 hours to travel 6.03 billion km.

To calculate the time it takes for light to travel 6.03 billion km, we can use the formula:

time = distance / speed of light

where distance is 6.03 x 10^9 km and the speed of light is approximately 299,792,458 meters per second (m/s).

First, we need to convert the distance from kilometers to meters:

distance = 6.03 x 10^9 km x 10^3 m/km = 6.03 x 10^12 m

Now we can calculate the time:

time = distance / speed of light

= 6.03 x 10^12 m / 299,792,458 m/s

= 20,107.394 seconds

To 3 significant figures, the answer is 20,100 seconds or 5.583 hours (since there are 3600 seconds in an hour).

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The percent yield of copper(I) sulfide in this experiment is 31.83%.

What is percent yield?

To calculate the percent yield, we need to compare the actual yield (the amount of product that was obtained in the experiment) with the theoretical yield (the amount of product that should have been obtained if the reaction had gone to completion).

The balanced chemical equation for the reaction between copper and sulfur to form copper(I) sulfide is:

Cu + S →  [tex]Cu_{2}S[/tex]

The molar mass of Cu is 63.55 g/mol, and the molar mass of S is 32.06 g/mol. The molar mass of  [tex]Cu_{2}S[/tex]  is 159.17 g/mol.

First, we need to calculate the theoretical yield of copper(I) sulfide using the amount of copper used in the experiment:

5 g Cu × (1 mol Cu / 63.55 g Cu) × (1 mol [tex]Cu_{2}S[/tex] / 1 mol Cu) × (159.17 g  [tex]Cu_{2}S[/tex] / 1 mol [tex]Cu_{2}S[/tex] ) = 12.57 g  [tex]Cu_{2}S[/tex]

So the theoretical yield of copper(I) sulfide is 12.57 g.

The actual yield obtained in the experiment is 4 g.

The percent yield is then:

percent yield = (actual yield / theoretical yield) × 100%

percent yield = (4 g / 12.57 g) × 100%

percent yield = 31.83%

Therefore, the percent yield of copper(I) sulfide in this experiment is 31.83%.

What is theoretical yield ?

The theoretical yield is the amount of product that would be obtained in a chemical reaction if it went to completion, meaning that all the limiting reactant was used up and no product was lost. It is calculated using stoichiometry, which involves balancing the chemical equation for the reaction and using the coefficients to determine the mole ratio between the reactants and products.

Theoretical yield is often used as a reference value to compare with the actual yield obtained in an experiment, which is the amount of product actually obtained from the reaction. The percent yield can then be calculated by dividing the actual yield by the theoretical yield and multiplying by 100%.

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Answer:

Neutrons and Protons

Explanation:

Different elements can have subatomic particles of varying sizes. The size of an atom is defined by the size of its electron cloud, which is composed of electrons, and the size of its nucleus, which is composed of protons and neutrons. The atomic number and subsequently the identity of an element are determined by the number of protons in the nucleus. The quantity of protons and neutrons in the nucleus determines its size. The quantity of electrons in the electron cloud and the energy levels they are located at define its size. The size of atoms can differ depending on the element due to differences in the amount of protons, neutrons, and electrons.

Here's a more detailed step-by-step calculation for the theoretical yield and percent yield of chromium (Cr) in the given reaction:

Given: Mass of chromium(III) oxide (Cr2O3) = 80.0 g Mass of carbon (C) = 8.00 g Actual yield of Cr = 21.7 g

Step 1: Calculate the molar mass of Cr2O3 and C. Molar mass of Cr2O3 = 2 x (51.996 g/mol) + 3 x (15.999 g/mol) = 151.996 g/mol Molar mass of C = 12.011 g/mol

Step 2: Convert the masses of Cr2O3 and C to moles. Moles of Cr2O3 = Mass of Cr2O3 / Molar mass of Cr2O3 = 80.0 g / 151.996 g/mol = 0.527 mol (rounded to three decimal places)

Moles of C = Mass of C / Molar mass of C = 8.00 g / 12.011 g/mol = 0.666 mol (rounded to three decimal places)

Step 3: Determine the limiting reactant. The limiting reactant is the one that is completely consumed and determines the maximum amount of product that can be formed. In this case, we compare the moles of Cr2O3 and C to see which one is limiting.

From the balanced equation: Cr2O3 + 3C -> 2Cr + 3CO

We can see that 1 mol of Cr2O3 requires 3 moles of C to react completely and produce 2 moles of Cr. Therefore, the limiting reactant is C, as we have 0.666 mol of C, which is less than the 0.527 mol of Cr2O3.

Step 4: Calculate the theoretical yield of Cr. The theoretical yield of Cr is the maximum amount of Cr that can be obtained based on the limiting reactant.

Moles of limiting reactant (C) = 0.666 mol Molar mass of Cr = 51.996 g/mol

Theoretical yield of Cr = Moles of limiting reactant (C) x Molar mass of Cr = 0.666 mol x 51.996 g/mol = 34.65 g (rounded to two decimal places)

Step 5: Calculate the percent yield of Cr. The percent yield is a measure of how much of the theoretical yield was actually obtained.

Actual yield of Cr = 21.7 g Theoretical yield of Cr = 34.65 g

Percent yield = (Actual yield / Theoretical yield) x 100% = (21.7 g / 34.65 g) x 100% = 62.7% (rounded to three significant figures)

Therefore, the percent yield for the reaction is approximately 62.7%.

The order for reactant A is 2 and the order for reactant B is 1. For the first reaction, the overall order of the reaction is 3 and for the second reaction, the overall order of the reaction is 5.

The order of a reaction is the sum of the exponents in the rate law expression that relates the rate of a chemical reaction to the concentrations of the reactants.

To determine the order of each reactant, we need to compare the initial rates of reaction at different concentrations while keeping the concentration of the other reactant constant.

For reactant A:

Exp#1 (0.100 M A, 0.100 M B): initial rate = k(0.100)^2(0.100) = 0.001 k

Exp#2 (0.200 M A, 0.100 M B): initial rate = k(0.200)^2(0.100) = 0.004 k

Exp#3 (0.100 M A, 0.200 M B): initial rate = k(0.100)^2(0.200) = 0.002 k

We can see that when the concentration of A doubles (Exp#1 to Exp#2), the initial rate quadruples, which indicates that A is second order. When the concentration of B doubles (Exp#1 to Exp#3), the initial rate doubles, which indicates that B is first order.

Therefore, the order for reactant A is 2 and the order for reactant B is 1.

To determine the overall order of the reaction, we add the orders of the reactants:

Overall order = 2 (order of A) + 1 (order of B) = 3

Therefore, the overall order of the reaction is 3.

For the second reaction, we can see that the rate depends on the concentration of both reactants, and we cannot determine their individual orders without further information or experiments. However, we can determine the overall order of the reaction by adding the exponents of the concentration terms in the rate law:

Overall order = 1 + 4 = 5

Therefore, the overall order of the reaction is 5.

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The addition of concentrated HCl to the equilibrium mixture will result in the precipitation of more NH₄Cl(s) as the equilibrium shifts towards the left. This can be observed as cloudiness or precipitation forming in the solution.

When concentrated hydrochloric acid (HCl) solution is added to the equilibrium mixture of NH₄Cl(s) + heat <=> NH₄+(aq) + Cl-(aq), the equilibrium will shift towards the left, meaning more solid NH₄Cl will be formed.

This is because HCl is a strong acid that will react with NH₄+ ion to form NH₄Cl(s) and H+ ion:

NH₄+(aq) + Cl-(aq) + HCl(aq) → NH₄Cl(s) + H₂O(l)

The increase in H+ ion concentration due to the addition of HCl will result in the shift of the equilibrium to the left to reduce the excess H+ ion concentration. This will favor the formation of more solid NH₄Cl.

Therefore, the addition of concentrated HCl to the equilibrium mixture will result in the precipitation of more NH₄Cl(s) as the equilibrium shifts towards the left. This can be observed as cloudiness or precipitation forming in the solution.

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Water produced by 6.32 grams of [tex]CaCO_{3}[/tex] reacts with [tex]HCl[/tex] reaction is 0.11376 L as the density of water is 1.

[tex]CaCO_{3}[/tex]'s molar mass is equal to 100 g/mol (40 + 12 + 16 3 g/mol).

HCl's molar mass is (1 + 35.5) g/mol, or 36.5 g/mol.

Water's molecular weight is (12 + 16) g/mol, or 18 g/mol.

[tex]CaCO_{3}[/tex]initial mole number = (6.32g) / (100 g/mol) = 0.00632mol

Initial number of moles of HCl = (35.5 g/mol/36.8 g) = 1.04 mol

Mole ratio:  [tex]CaCO_{3}:HCl:CaCl_{2}:H_{2}O[/tex]= 1: 2: 1: 1. [tex]CaCO_{3}:HCl:CaCl_{2}:H_{2}O[/tex]

If [tex]CaCO_{3}[/tex] fully reacts, [tex]HCl[/tex] is required at a rate of (0.00632mol) x 2 (equals 0.01264mol <1.04 mol).

[tex]HCl[/tex] is hence abundant. As a reactant, [tex]CaCO_{3}[/tex] is the limiting one.

Total number of [tex]CaCO_{3}[/tex] reactions: 0.00632moles

No. of moles of water created / No. of moles of [tex]CaCO_{3}[/tex] reacting / 0.00632mol

0.11376 g of [tex]H_{2}O[/tex]were generated from (0.00632 mol) by (18 g/mol).

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The main branches of mathematics are algebra, number theory, geometry and arithmetic.

In a reversible reaction, the relative amounts of reactant and product at the end of the reaction depend on the reaction conditions such as temperature, pressure, and concentration of the reactants and products.

At equilibrium, the forward and reverse reactions occur at equal rates, and the concentrations of the reactants and products no longer change over time. The position of equilibrium, that is, the relative concentrations of reactants and products at equilibrium, is determined by the equilibrium constant (K), which is specific to each reversible reaction.

If the equilibrium constant (K) is greater than 1, the equilibrium favors the products, and at equilibrium, the concentration of products will be greater than the concentration of reactants. Conversely, if the equilibrium constant is less than 1, the equilibrium favors the reactants, and at equilibrium, the concentration of reactants will be greater than the concentration of products.

Therefore, the relative amounts of reactant and product at the end of a reversible reaction depend on the equilibrium constant (K) and the reaction conditions.

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The eutectic point is the lowest temperature at which the liquid phase is constant at a particular pressure.

A melting composition known as a eutectic consists of at least two components that melt and freeze at the same rates. The components combine during the crystallisation phase, operating as a single component as a result.

The eutectic is the system's lowest melting point under its own pressure; it has a matching temperature called the eutectic temperature and produces the eutectic liquid as a result. In terms of composition, eutectic liquids are located between the system's solid phases.

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K and Br since an halogen and a metal make a salt