Which of the following metals would act as a sacrificial anode (cathodic protection) for iron? Check all that apply.
a. Al
b. Zn
c. Pb
d. Na
e. Ni

The chemical formula of pamitoleic acid is CH₃(CH₂)5CH=CH(CH₂)₇COOH. The structure of pamitoleic acid is given in the image attached below.

Palmitoleic acid has been known to have a lot of beneficial effects on insulin sensitivity, cholesterol metabolism, and hemostasis. It has been concluded that palmitoleic acid may prevent beta-cell apoptosis induced by glucose or saturated fatty acids.

Recently, studies have reported that palmitoleic acid is useful in treating disorders related to skin hyperpigmentation, fibrosis and as an adjuvant in formulations for the treatment of secondary infections caused by gram-positive bacteria.

Palmitoleic acid is one of the most common acids which is found in omega-7 fatty acids.

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Throughout the Korean War, South Korea received significant support from the United States.

Throughout the Korean War, the United States significantly supported South Korea militarily, economically, and politically. This involved deploying US soldiers to fight alongside South Korean forces, supplying them with equipment and supplies, and taking part in talks with North Korea. As part of its military assistance, the United States sent personnel and supplies, supported air and sea operations, and provided arms and ammunition. In the years following the war, the U.S. also gave South Korea economic and military assistance, assisting in the nation's reconstruction and fortifying its defences.

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complete question:What part did the United States play in the Korean War when it came to supporting South Korea? a.Building up a supply of hydrogen bombs. b. Supporting South Korea in the Korean War. c. Having a moon landing as a goal. d. Starting the Airlift to Berlin.

Change in energy = 207. kJ - 181. kJ = 26. kJ

What is energy?

Energy is the ability to do work. It is the capacity to do something or to cause change. It can be found in many forms such as kinetic, electrical, chemical, nuclear, gravitational, thermal, and sound energy. It is all around us and is necessary for life. It is used to power everyday activities and to provide heat, light and motion. It can be produced by natural sources such as the sun or through human efforts such as burning fossil fuels. Energy can also be stored in batteries, fuel cells and other forms. Knowing how to use energy efficiently is important in today's world.

Therefore, Change in energy = 207. kJ - 181. kJ = 26. kJ

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The result from conversion 1.077 moles of magnesium phosphate to mass in grams is 283.14 grams. The molar mass of magnesium phosphate is 262.9 g/mol.

Formula:

[tex]\boxed{\tt g=n\times Mr}[/tex]

Given:

Therefore,

First, calculate the molar mass of Mg₃(PO₄)₂.

Second, calculate mass of Mg₃(PO₄)₂.

So, 1.077 moles of magnesium phosphate correspond to 283.14 grams of magnesium phosphate.

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When 5-methyl-3-hexyne (or 5-methylhex-3-yne) reacts with Cl2, the triple bond will be broken, and two Cl atoms will be added to the two adjacent carbons, forming a dihalogenated alkene.

This will result in the formation of a trans-dihalogenated alkene, where the two Cl atoms are on opposite sides of the double bond. The major product for the first reaction will be 5-methyl-3-hexene-1,2-dichloro, which is formed by the addition of two Cl atoms to the carbons 3 and 4, and the formation of a double bond between carbons 2 and 5. The major product of this reaction will be 5-methyl-3,4-dichloro-2-hexene, which is formed by the addition of two Cl atoms to the carbons 2 and 5, and the addition of two more Cl atoms to the carbons 3 and 4.

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MgCl2: In an aqueous solution of MgCl2, the ions occur as Mg2+ and Cl–.

Na2SO4: In an aqueous solution of Na2SO4, the ions occur as Na+ and SO42-.

NH4Br: In an aqueous solution of NH4Br, the ions occur as NH4+ and Br–.

1. MgCl2: In an aqueous solution of MgCl2, the ions occur as Mg2+ and Cl–. This is represented by the scene "A" where two positive (+) ions and two negative (-) ions are shown. The equation for this is: MgCl2 → Mg2+ + 2Cl–

2. Na2SO4: In an aqueous solution of Na2SO4, the ions occur as Na+ and SO42-. This is represented by the scene "C" where two positive (+) ions and two negative (-) ions are shown. The equation for this is: Na2SO4 → 2Na+ + SO42-

3. NH4Br: In an aqueous solution of NH4Br, the ions occur as NH4+ and Br–. This is represented by the scene "B" where one positive (+) ion and one negative (-) ion are shown. The equation for this is: NH4Br → NH4+ + Br–

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Elevation of ATP (adenosine triphosphate) would be an example of a compound that will reduce activity through the TCA cycle.

The TCA cycle also known as the citric acid cycle or Krebs cycle, is a series of metabolic reactions that occur in the mitochondria of cells and play a crucial role in the production of energy. However, under certain metabolic conditions, the flux through the TCA cycle can be reduced due to an increase in TCA cycle inhibitors.

One such inhibitor is ATP, which is the primary energy currency of the cell.

When ATP levels are high, it signals to the cell that there is enough energy available and there is no need to produce more through the TCA cycle.

As a result, the activity of the TCA cycle is reduced, leading to a decrease in the production of energy.

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The correct total ionic equation for the reaction between sodium hydroxide and magnesium chloride is

option f. MgCl2 (aq) + 2 NaOH (aq) → Mg(OH)2 (s) + 2 NaCl (aq).

Here's the step-by-step explanation:

1. First, we need to write the balanced chemical equation for the reaction between sodium hydroxide and magnesium chloride. This equation is: MgCl2 (aq) + 2 NaOH (aq) → Mg(OH)2 (s) + 2 NaCl (aq)

2. Next, we need to write the total ionic equation by breaking all of the soluble compounds into their respective ions. This gives us: Mg2+ (aq) + 2 Cl- (aq) + 2 Na+ (aq) + 2 OH- (aq) → Mg(OH)2 (s) + 2 Na+ (aq) + 2 Cl- (aq)

3. Finally, we can cancel out the spectator ions, which are the ions that appear on both sides of the equation and do not participate in the reaction. In this case, the spectator ions are Na+ and Cl-. This gives us the net ionic equation: Mg2+ (aq) + 2 OH- (aq) → Mg(OH)2 (s)

So, the correct total ionic equation for the reaction between sodium hydroxide and magnesium chloride is option f. MgCl2 (aq) + 2 NaOH (aq) → Mg(OH)2 (s) + 2 NaCl (aq).

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The most volatile of the following compounds is CH4.

Volatile compounds are those that have a high vapor pressure and can easily evaporate into a gas at room temperature. This is typically due to weak intermolecular forces, such as dispersion forces, that allow the molecules to easily escape from the liquid phase.

In this case, CH4 (methane) is the most volatile because it has the weakest intermolecular forces among the compounds listed. It is a nonpolar molecule with only dispersion forces between its molecules, making it easy for it to evaporate into a gas.

The other compounds listed, CBr4, CCl4, CF4, and C6H14, all have stronger intermolecular forces, such as dipole-dipole forces or hydrogen bonding, that make them less volatile and less likely to evaporate at room temperature.

Therefore, the correct answer is A. CH4.

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C): Pectin is a heteropolysaccharide.

Pectin is a heteropolysaccharide, which is a complex carbohydrate made up of different types of monosaccharides. It is commonly found in the cell walls of plants and is used as a gelling agent in the food industry. Amylopectin and cellulose are both homopolysaccharides, meaning they are made up of only one type of monosaccharide (glucose). Chitin is a polymer of N-acetylglucosamine, which is a type of amino sugar, but it is not classified as a heteropolysaccharide.

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The 1.0 m calcium chloride solution has the lowest freezing point and the greatest boiling point among the four solutions. In solution, calcium chloride, a salt, splits into three ions (Ca2+ and two Cl-).

A colligative property of a solution is a physical property that depends on the number of solute particles present in the solution and not on their chemical identity. Examples of colligative properties include the lowering of the freezing point, the elevation of the boiling point, vapor pressure lowering, and osmotic pressure. These properties are related to the changes in the physical behavior of a solvent caused by the presence of solute particles, such as a reduction in the solvent's vapor pressure or an increase in its boiling point. Colligative properties are important in various fields, including chemistry, materials science, and biology.

Hence, the answer is, of the four solutions, 1.0 m calcium chloride has both the lowest freezing point and the highest boiling point.

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The mechanism of the reactions of both the subparts of the questions is below:

A)The formation of the acylium ion from the reaction between acetyl chloride and aluminium chloride can be described by the following mechanism:

Step 1: Formation of the complex

AcCl + AlCl3 → [AcCl-AlCl3]

Step 2: Deprotonation to form the acylium ion

[AcCl-AlCl3] → Ac+ + AlCl4-

In step 1, the acetyl chloride molecule reacts with aluminium chloride to form a complex. In step 2, the complex undergoes deprotonation to form the acylium ion, which is stabilized by the aluminium chloride.

(b) The mechanism for the Friedel-Crafts acylation of bromobenzene using the acylium ion formed in (a) can be described by the following steps:

Step 1: Formation of the complex

Ac+ + AlCl4- → [Ac-AlCl4]

Step 2: Attack of the ortho-carbon of bromobenzene

C6H5Br + [Ac-AlCl4] → C6H4Br-2-COCH3 + AlCl3

Step 3: Regeneration of the catalyst

AlCl3 + [Ac-AlCl4] → AcCl + AlCl4-

In step 1, the acylium ion formed in (a) reacts with aluminium chloride to form a complex. In step 2, the complex reacts with bromobenzene, and the ortho-carbon attacks the carbonyl carbon of the acylium ion to form a new carbon-carbon bond. In step 3, the catalyst is regenerated, and acetyl chloride is produced as a byproduct.

The arrow formalism for the three steps can be written as follows:

Step 1:

Ac+ AlCl4- [Ac-AlCl4]

↑ ↓ ↑

Step 2:

C6H5Br [Ac-AlCl4] C6H4Br-2-COCH3 AlCl3

↑ ↓ ↑ ↑

Nucleophile Electrophile Product Catalyst

Step 3:

AlCl3 [Ac-AlCl4] AcCl AlCl4-

↑ ↓ ↑ ↓

Catalyst Complex Byproduct Catalyst

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Answer:          5 × 10^3 g

Or, without significant figures, 5000 g.

To find the mass when given volume and density, multiply the volume with density.

We can solve using dimensional analysis or the 'cancelling out units' method.

[tex]m = \frac{50.0\ mL}{1}*\frac{100\ g}{1\ mL}[/tex]

[tex]m = \frac{50.0}{1}*\frac{100\ g}{1}[/tex]          Since mL is in the top and bottom, they cancel out.

[tex]m = 5000\ g[/tex]                 Multiply to find the answer.

[tex]m = 5 * 10^{3}\ g[/tex]              Write the answer with significant figures.

Therefore, the mass of 50.0 mL of water is 5 × 10^3 g.

This is the same as 5000 g if we do not consider significant figures.

If we use significant figures in this answer, we only keep 1 significant figure. When multiplying numbers, the final answer will have the same amount of significant figures as the numbers in the question with the least amount of significant figures.

If the numbers in the question have different amounts of significant figures, we will take the least amount of significant figures from the problem.

We had two numbers from the problem and they had different amounts of significant figures:

So, our final answer should only have 1 significant figure.

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In a redox reaction, there is a transfer of electrons

The question is incomplete but it seems to have something to do with redox reactions.

A balanced redox reaction is a chemical equation that represents a redox (reduction-oxidation) reaction with the same number of atoms of each element on both the reactant and product sides and the same total charge on both sides. In a redox reaction, there is a transfer of electrons from one substance (the reducing agent) to another substance (the oxidizing agent).

In order to balance a redox reaction, you need to balance the number of atoms of each element in the reaction, and also balance the charges.

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The answers to the virtual lab that specifies the use of a virtual spectrometer is given below:

2. Space consists of bodies with different types of electromagnetic spectrum.

This includes high-energy bodies emitting radiation in short wavelengths and extremely short wavelengths such as in UV spectrum, X rays, and gamma rays.

Conversely, other bodies might be emitting radiations in longer wavelengths such as Microwaves and Radio waves.

Since these radiations are at two different ends of the electromagnetic spectrum, a large number of equipment would be required for individually studying these space-based radiation sources and their characteristics.

3. The element missing from the moons and the planets would be Oxygen. It is to be remembered that Oxygen forms the base of the sustenance of life forms on Earth and forms an indispensable part of the carbon cycle. In the absence of oxygen, these planets and moons remain lifeless.

4. Stars emit heat and light. Along with the heat and light, radiations are emitted by the star. These radiations travel outward from stars and work as the signature of the stars. By analyzing the radiations from the stars, scientists back on Earth could deduce the physical conditions in the heart of a star including its constitution, temperature, and surface conditions.

E.g., If the star is emitting radiation in longer wavelengths, this is an indication that the star is cooling down and the temperature is relatively low.

5. The knowledge of the constitution of the elements making up the moon or planet is necessary to ascertain the life-sustaining capability of the same. Through adequate knowledge of the composition of the moon/planet, the possibility of the life-sustaining ability of the same could be deduced.

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Here are four possible empirical formula of binary ionic compounds that can be formed from the given ions MgI2, Al2S3, MgS, AlI3.

The empirical formula of a compound is the simplest whole-number ratio of atoms in the compound. To determine the empirical formula, we need to divide the subscripts in the molecular formula (if given) by the greatest common factor to get the simplest ratio. If we are not given a molecular formula, we can determine the empirical formula from the masses or percentages of the elements in the compound.

To write the empirical formula of a binary ionic compound, we need to combine a cation (positively charged ion) with an anion (negatively charged ion) in a way that balances the charges.

Magnesium iodide: Mg2+ + 2I- -> MgI2 (the charges balance as 2(+2) = 2(-1))

Aluminum sulfide: Al3+ + 3S2- -> Al2S3 (the charges balance as 3(+3) = 2(-2) + 3(-2))

Magnesium sulfide: Mg2+ + S2- -> MgS (the charges balance as 1(+2) = 1(-2))

Aluminum iodide: Al3+ + 3I- -> AlI3 (the charges balance as 3(+3) = 3(-1))

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The balanced form of the equations given above is as follows:

A chemical equation is a symbolic representation of a chemical reaction where reactants are represented on the left, and products on the right.

A chemical equation is said to be balanced when the number of atoms of each element on both sides of the equation are the same.

According to this question, some equations are given. The balanced form of these equations using coefficients are given in the main answer part.

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1) Element 35 is a gas

2) It would be in period 7

3) It would be a non metal

4) At would look more like iodine because both of them are closer down the group.

5) It can act both as metal and non metal.

6) Atom is neutral while ions have a charge

7) The second ionization energy is larger because it would take more energy to remove an electron from a positive specie

a) Cl has the largest atomic radius

b) Ca has the largest atomic radius

c) Cl has the largest atomic radius

8) In order of increasing atomic radius we have; Ar < Kr < Xe

The periodic table is a tabular arrangement of the chemical elements, organized on the basis of their atomic number, electron configurations, and chemical properties.

The periodic table is one of the most important tools in chemistry, as it provides a convenient way to identify and categorize the elements. The elements are arranged in rows and columns, with elements in the same row having similar chemical and physical properties. The periodic table is used by scientists and students to understand the properties of elements, predict the chemical behavior of substances, and form hypotheses about new elements that have not yet been discovered.

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Molecular equation: AgNO3(aq)+HCl(aq)→AgCl(s)+HNO3(aq), ionic equation: Ag+(aq)+ NO3-(aq)+H+(aq)+Cl-(aq) -> AgCl(s)+H+(aq)+ NO3-(aq), Net ionic equation : Ag+(aq)+Cl-(aq) -> AgCl(s).

The equation is known as a molecular solution when complex formulae are written as though they're still molecules. When dissolved ionic chemicals are represented as free ions, the equation is said to be ionic. Demonstrated were examples of both molecular and ionic relations.

Even soluble ionic chemicals and strong acids are treated as neutral molecules in the molecular formula for a reaction's reactants and products. Strong acids and soluble ionic compounds are rewritten as split ions in the full ionic equation.

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The formula given relates the frequency of the photon emitted to the difference in energy levels of the hydrogen atom. The energy of the photon is given by

E = hν

where h is Planck's constant and ν is the frequency.

Part A:

         v~ = RH(1/12 - 1/22) = 3.28984 x 1015 cm-1

         v~ = RH(1/22 - 1/32) = 1.09737 x 1015 cm-1

          v~ = RH(1/32 - 1/42) = 4.34013 x 1014 cm-1

To find the highest value of energy (E) for each series, we can use the relationship E = hν. Using Planck's constant (h) and the highest frequency (v~) for each series, we get:

Hence, the answer is,

         v~ = RH(1/12 - 1/22) = 3.28984 x 1015 cm-1

         v~ = RH(1/22 - 1/32) = 1.09737 x 1015 cm-1

          v~ = RH(1/32 - 1/42) = 4.34013 x 1014 cm-1

Write a short note on the hydrogen spectrum.

The hydrogen spectrum refers to the pattern of lines that are observed when the emission spectrum of hydrogen is analyzed. It is caused by the excitation of electrons in hydrogen atoms, which emit light of specific frequencies as they return to lower energy states. The hydrogen spectrum consists of several series of lines, including the Lyman, Balmer, and Paschen series, which correspond to different electron transitions. The study of the hydrogen spectrum played a crucial role in the development of quantum mechanics and our understanding of atomic structure.

To find the highest value of energy (E) for each series, we can use the relationship E = hν. Using Planck's constant (h) and the highest frequency (v~) for each series, we get:

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The option that would only dump the data definitions and not the data of the mydatabase to backup.sql is pg_dump -s mydatabase > backup.sql.

a. pg_dump mydatabase > backup.sql will dump the entire contents of the database, including both the data and the data definitions.

b. pg_dump -a mydatabase > backup.sql will dump only the data (i.e., it will append to the file) and not the data definitions.

c. pg_dump -s mydatabase > backup.sql will dump only the data definitions (i.e., the schema) and not the data.

d. pg_dump -d mydatabase > backup.sql will dump both the data and the data definitions.

Hence the correct option is C.

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The balanced nuclear equation for the alpha decay of polonium-84, and the nuclear equation for thorium-229 are:

Polonium-84 alpha decay:

^84Po --> ^4He + ^80Hg

In this equation, an alpha particle, which is a helium nucleus with a mass number of 4 and atomic number of 2, is emitted from the polonium-84 nucleus, resulting in the formation of a mercury-80 nucleus.

Thorium-229 nuclear equation:

^229Th

Thorium-229 is a stable isotope and does not undergo alpha decay or any other form of radioactive decay under normal circumstances.

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The option where the most likely oxidation state and electronic configuration for a ruthenium cation is 3+ oxidation state, [Kr] 5s04d5. The correct answer C.

Ruthenium is a transition metal that belongs to Group 8 of the periodic table. It has several possible oxidation states, including +2, +3, +4, +5, and +6, with the +3 oxidation state being the most common. In this state, ruthenium has lost three electrons and has a +3 charge.

The +3 oxidation state is the most stable for ruthenium because it is close to a half-filled 4d subshell, which is a particularly stable electronic configuration. In the +3 oxidation state, ruthenium has three electrons in the 4d orbital, which is one electron away from a half-filled subshell.

Complete question:

Predict which of these is the most likely oxidation state and electronic configuration for a ruthenium cation:

A)2+ oxidation state, [Kr] 5s04d5

B)1+ oxidation state, [Kr]5s24d5

C)3+ oxidation state, [Kr] 5s04d5

D)3+ oxidation state, [Kr] 5s24d6

E)8+ oxidation state, [Kr]5s04d0

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The number of CFUs in the original sample for Plate 3 is 460 CFUs.

CFU stands for "colony-forming unit". It is a unit of measurement to estimate the number of viable bacteria or fungal cells in a sample. When a small number of microorganisms are present in a sample, it is not practical to count each individual cell. Instead, a dilution series is typically performed, where the sample is diluted in a series of steps and then plated onto a culture medium. Each viable cell is capable of growing into a visible colony on the culture medium. By counting the number of colonies formed on the plates, the number of CFUs per unit volume of the original sample can be estimated. The CFU is a useful measure because it reflects the number of viable cells that are capable of growing and potentially causing an infection or disease.

To calculate the number of CFUs in the original sample, we can use the following formula:

Number of CFUs = (colony count / volume plated) x dilution factor

Where:

colony count: the number of colonies counted on the plate

volume plated: the volume of the diluted sample plated on the plate (in mL)

dilution factor: the dilution factor of the sample (i.e. the reciprocal of the dilution used to prepare the sample)

Using the given data, we can calculate the number of CFUs in the original sample as follows:

For Plate 3:

Volume plated = 0.1 mL (since only 0.1 mL was inoculated instead of 1 mL)

Dilution factor = 1/1,000,000 = 0.000001 (since the sample was diluted 1:1,000,000)

Colony count = 46

Number of CFUs = (46 / 0.1) x 0.000001 = 460 CFUs

Therefore, the number of CFUs in the original sample for Plate 3 is 460 CFUs.

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5 ppm of Cr means 5 mg is present in 1 L of water. Then, 1000 g of water is containing 5 mg or 0.005 g of chromium.

PPM or part per million is a term used to express the concentration of a solution. 1 ppm is equal to 1 mg of the solute in 1000 ml or in 1 L of the solution.

ppm  = weight of solute/mass of solution ×10⁶. = mg/L

ppm is used to express very minute quantity of ay sample. Usually used in the concentration of heavy metals in water.

It is given that the concentration of Cr in water is 5 ppm, which means 5 mg per one liter of water. 1 l of water is 1000 ml or 1000 g of water. Hence, 1000 g of water contains 5 mg or 0.005 g of Cr.

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The orbital can only take two electrons.

The maximum number of electrons that an orbital can contain is determined by its angular momentum quantum number, "l", and its spin quantum number, "m_s". For a given value of "l", an orbital can contain 2l + 1 electrons with different values of "m_s", each of which can have either spin "up" (m_s = +1/2) or spin "down" (m_s = -1/2). The maximum number of electrons in an orbital is thus 2(2l + 1), with half of the electrons having one spin orientation and the other half having the opposite spin orientation.

We know that we can be able to obtain the number of electrons by looking at the quantum numbers;

n = 6

l = 1

ml = -1

Since we know that it is a px orbital then it can take two electrons.

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The pH is determined by the relative amounts of acid and base in each circumstance.

a) Before addition of HBr;

Number of moles of C5H5N = 25/1000 L × 0.180 M = 0.0045 moles

pOH = -log[0.0045 M]

pOH = 2.35

pH = 14 - 2.35 = 11.65

b) After addition of 12.5 mL of HBr;

Since the reaction is 1:1

Number of moles of HBr = 12.5/1000 L × 0.180 M = 0.00225 moles

The HBr is the limiting reactant

Amount of excess acid = 0.0045 M - 0.00225 = 0.00225 moles

Total volume of solution = 25mL + 12.5 mL = 37.5mL or 0.0375 L

molarity of excess H+ = 0.00225 moles / 0.0375 L = 0.06 M

pH = - log[0.06 M]

pH = 1.22

c) After addition of 24.0 mL of HBr;

Number of moles of acid = 24/1000  × 0.180 M = 0.00432 moles

Number of moles of excess acid =  0.0045 moles  - 0.00432 moles = 0.00018 moles

Total volume of solution =  24.0 mL +  25.0 mL = 49mL or 0.049 L

Molarity of excess acid = 0.00018 moles/ 0.049 L = 0.00367 M

pH = -log[0.00367 M] = 2.4

d) After addition of 25.0 mL of HBr

Number of moles of acid = 25/1000  × 0.180 M = 0.0045 moles

We have an equal number of acid and base hence pH = 11.65

e) After addition of 34.0 mL of HBr;

Number of moles of acid = 34/1000  × 0.180 M =0.00612 moles

We can see that the base is the limiting reactant.

Number of moles of excess base = 0.00162 moles

Total volume of solution = 34 mL + 25 mL = 59mL or 0.059 L

Molarity of excess base = 0.00162 moles/0.059 L = 0.027 M

pOH = -log[ 0.027 M]

pOH = 1.56

pH = 14 - 1.56 = 12.44

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The coefficient of [tex]$\mathrm{HNO}_3$[/tex] exists 8 , and [tex]$\mathrm{H}_2 \mathrm{O}$[/tex] exists 4.

The ratio between a sample's mass and its substance content is known as the molar mass of a chemical compound in chemistry. A substance's molar mass is a bulk attribute rather than a molecular one.

One mole of a material is weighed to determine its molar mass. Add the masses of all the components in a molecule to determine the molecular mass of the molecule by multiplying the subscript (number of atoms) by the atomic mass of each element in the molecule.

The first step will be to convert 144.1 grams of [tex]$\mathrm{HNO}_3$[/tex] to moles. To estimate the molar mass

[tex]$$1(1.0)+1(14)+3(16)=1+14+48=63 \mathrm{~g} / \mathrm{mol}$$[/tex]

[tex]$144.1 \mathrm{gHNO}_3 \times \frac{1 \mathrm{~mol} \mathrm{HNO}_3}{63 \mathrm{gHNO}_3}=2.287 \mathrm{~mol} \mathrm{HNO}_3$$[/tex]

Moles of [tex]$\mathrm{HNO}_3$[/tex] you're starting with, utilize the ratio of moles in the equation to find the moles of [tex]$\mathrm{H}_2 \mathrm{O}$[/tex] produced. The coefficient of [tex]$\mathrm{HNO}_3$[/tex] exists 8 , and [tex]$\mathrm{H}_2 \mathrm{O}$[/tex] exists 4

[tex]$$\begin{aligned}& 2.287 \mathrm{moHHO}_3 \times \frac{4 \mathrm{~mol}}{ 8 \mathrm{moH}_3} \mathrm{H}_2 \underline{\mathrm{O}}=1.144 \mathrm{~mol} \mathrm{H}_2 \mathrm{O} \\&\end{aligned}$$[/tex]

Rounded to the nearest tenth, 1.1 moles [tex]$\mathrm{H}_2 \mathrm{O}$[/tex].

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