46. Sulfuric acid (H₂SO4) reacts with aqueous sodium cyanide, forming hydrogen cyanide gas and aqueous sodium sulfate.​

Tar is not a lucrative unconventional reserve as it is not easy to extract tar sands.

The given statement is referring to tar sands (oil sands) which is a group of sedimentary rocks containing sand, clay, water, and a thick, molasses-like substance called bitumen (tar). Tar sand is the residue of larger molecules left behind after microbes attack existing underground oil reserves. Tar sand is found in sand or sandstone that contains high concentrations of macerals (complicated carbon molecules). It is not easy to extract tar sands as it requires expensive extraction methods that are not yet viable due to low oil prices. Therefore, it is not a lucrative unconventional reserve.

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Answer:

46g

Explanation:

22.4L of NO2 at STP = 1mol of NO2 = atomic mass of NO2 = 46g

The NaCl solution has a 30 g/L concentration.

Sodium chloride, or NaCl, is a chemical compound that is frequently referred to as table salt. It is an ionic compound comprised of chloride anions and sodium cations (Na+) (Cl-). At room temperature, NaCl is a white, crystalline solid that is very soluble in water.

Let's start by converting the solution's volume from milliliters to liters:

100 ml = 100/1000 L = 0.1 L

Then, we can use the following formula to determine the solution's concentration:

concentration = amount of solute / volume of solution

replacing the specified values:

concentration = 3 g / 0.1 L = 30 g/L

As a result, the NaCl solution has a 30 g/L concentration.

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The total alkalinity [Alk] in eq/L of the original sample is 0.00211 eq/L,  the carbonate concentration (Ct) in the original sample is approximately 2.51 × 10⁻³ M, and the dissolved carbonate species in the original sample cannot be attributed solely to atmospheric CO₂, and another source of carbonate must be present as well.

To determine the total alkalinity [Alk] in eq/L of the original sample, we can use the following equation;

[Alk] = (Vb × Nb - Va × Na) / V

Where,  Vb = volume of acid added (mL), Nb = normality of acid, Va = volume of sample (mL), Na = normality of sample (in this case, assumed to be zero), V = total volume (mL)

Substituting the given values, we get;

Vb = 5.28 mL

Nb = 0.0200 N

Va = 50.00 mL

Na = 0

V = 50.00 mL

[Alk] = (5.28 mL × 0.0200 N - 50.00 mL × 0) / 50.00 mL = 0.00211 eq/L

Therefore, the total alkalinity [Alk] in eq/L of the original sample will be 0.00211 eq/L.

At pH 6.7, the dominant form of alkalinity is likely bicarbonate (HCO₃⁻).

To determine the carbonate concentration (Ct) in the original sample, we can use the following equation;

pH = pKa + log([HCO₃⁻] + 2[CO₃²⁻] / [H₂CO₃])

At pH 6.7, we can assume that [HCO₃⁻] is much greater than [CO₃²⁻] and [H₂CO₃]. Therefore, we can simplify the equation to;

pH ≈ pKa + log([HCO₃⁻])

pKa for the bicarbonate system is 6.35.

Rearranging the equation, we get;

[HCO₃⁻] = [tex]10^{(pH-pKa)}[/tex] = [tex]10^{(6.7-6.35)}[/tex] = 2.51 × 10⁻³ M

[CO₃²⁻] = (Kw / K₂) × [HCO₃⁻] = (10⁻¹⁴ / 4.45 × 10⁻⁷) × 2.51 × 10⁻³ = 5.66 × 10⁻¹² M

Ct = [HCO₃⁻] + [CO₃²⁻] = 2.51 × 10⁻³ M + 5.66 × 10⁻¹² M ≈ 2.51 × 10⁻³ M

Therefore, the carbonate concentration (Ct) in the original sample is approximately 2.51 × 10⁻³ M.

At atmospheric pressure and temperature, the partial pressure of CO₂ in air is approximately 0.0004 atm, which corresponds to a dissolved CO₂ concentration of about 10 ppm. The dissolved CO₂ concentration in the original sample is much higher than this (418 ppm), which suggests that there must be another source of carbonate in the sample besides atmospheric CO₂.

In addition, the calculated Ct in the original sample is higher than the dissolved CO₂ concentration, further supporting the conclusion that there must be another source of carbonate in the sample.

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we need to measure 83.33 ml of 18 M sulfuric acid and dilute it to 250 ml with water to prepare a 6.0 M solution of sulfuric acid.

we can use the formula:

M1V1 = M2V2

Where:

M1 is the initial concentration of the sulfuric acid solution

V1 is the volume of the sulfuric acid solution needed

M2 is the final concentration of the sulfuric acid solution

V2 is the final volume of the sulfuric acid solution

In this case, we want to prepare a 6.0 M solution of sulfuric acid with a final volume of 250 ml. We need to find out how much concentrated 18 M sulfuric acid solution is needed.

Using the formula above, we can rearrange it to solve for V1:

V1 = (M2 x V2) / M1

Substituting the values we have:

V1 = (6.0 M x 250 ml) / 18 M

V1 = 83.33 ml

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At equilibrium, reactants will be present in the larger amount compared to the products.

The equilibrium constant (Kc) is a measure of the relative concentrations of products and reactants at equilibrium. Kc is defined as the product of the concentrations of the products raised to their stoichiometric coefficients, divided by the product of the concentrations of the reactants raised to their stoichiometric coefficients.

In this case, Kc = 3.8 x 10^-4 at 500°C. This value of Kc tells us that the concentration of products at equilibrium is relatively low compared to the concentration of reactants. In other words, the reaction strongly favors the formation of reactants over products.

Therefore, at equilibrium, reactants will be present in the larger amount compared to the products.

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To solve this problem, we can use the formula for radioactive decay and after 4 half-lives, there will be 6,250 atoms of uranium remaining for Parents and  93,750 atoms of lead produced for daughters.

The formula is [tex]N = N0 * (1/2)^{(t/t1/2)} ^[/tex]

where N is the final amount, N0 is the initial amount, t is the time elapsed, and t1/2 is the half-life of the radioactive isotope.

Given that the half-life of the uranium-lead (U-Pb) system is 1.6 billion years, the time required for 4 half-lives to occur is:

t = 4 * t1/2 = 4 * 1.6 billion years = 6.4 billion years

Therefore, it would take 6.4 billion years to get through these 4 half-lives.

Using the formula for radioactive decay, we can calculate the amount of uranium and lead remaining after 4 half-lives have occurred, starting with 100,000 atoms of uranium:

For the parents (uranium):

[tex]N = N0 * (1/2)^{(t/t1/2)} ^ = 100,000 * (1/2)^4 = 6,250[/tex]atoms

Therefore, after 4 half-lives, there will be 6,250 atoms of uranium remaining.

For the daughters (lead):

[tex]N = N0 - (N0 * (1/2)^{(t/t1/2)} = 100,000 - 6,250 = 93,750[/tex] atoms

Therefore, after 4 half-lives, there will be 93,750 atoms of lead produced.

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These rules described by specific cases of an ideal gas law, which states PV = nRT. When temperature rises, molecules become more energised, and lose their attraction to one another, causing pressure to fall.

As long as the volume remains constant, the pressure of a given quantity of petrol is precisely proportional towards its absolute temperature (Amontons' law). Under constant pressure, the volume of the a given gas is proportional to its exact temperature.

Now, let's review PV = nRT, our ideal petrol law. In this equation, P denotes pressure in atmospheres, V denotes volume in litres, n denotes moles of particles, T denotes temperature in Kelvin, and R denotes the ideal gas constant.

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Answer: yes

Explanation: Petroleum solvents are hydrocarbon mixtures which can be grouped into three broad categories on the basis of their boiling ranges and solvent strengths, as follows: special boiling range solvents, boiling range, 30-160 oC; white spirits, 130-220 oC; and high-boiling aromatic solvents, 160-300 oC.

To investigate the effects of anions on pH, option d- Liam should choose salts that have acidic or basic anions.

This is because the anions in a salt can react with water to produce either an acidic or basic solution. The choice of cation is not as important, as neutral cations will not affect the pH of the solution. Therefore, option (d) is the most appropriate choice for Liam.

To elaborate further, when a salt is dissolved in water, it dissociates into its respective cation and anion. The anion, in particular, can interact with water to form either an acidic or basic solution. For example, chloride ions (Cl-) do not have any acidic or basic properties, and therefore, will not affect the pH of a solution.

However, nitrate ions (NO) are basic anions and can hydrolyze to form hydroxide ions (OH-) and increase the pH of the solution.

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There is 21.77 g of excess SO₂ left over after the reaction is complete.

Reagents are substances used in laboratory tests and can be used in a chemical process to find, quantify, or create other substances.

2SO₂ + O₂ → 2SO₃

Molar mass of SO₂ = 32.06 g/mol

Molar mass of O₂ = 31.9988 g/mol

Number of moles of SO₂ = 21.71 g / 32.06 g/mol = 0.677 mol

Number of moles of O₂ = 21.71 g / 31.9988 g/mol = 0.678 mol

2 moles of SO₂ : 1 mole of O₂

0.678 mol O₂ × (2 mol SO₂ / 1 mol O₂) = 1.356 mol SO₂

Therefore, 1.356 mol of SO₂ are required to react completely with 0.678 mol of  O₂. However, we only have 0.677 mol of SO₂, so there is an excess of:

Excess SO₂ = 1.356 mol – 0.677 mol = 0.679 mol

Molar mass of SO₂ = 32.06 g/mol

Excess SO₂ = 0.679 mol × 32.06 g/mol = 21.77 g

Therefore, there is 21.77 g of excess SO₂ left over after the reaction is complete.

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Answer:

a. FeCl3 + 3NaOH → Fe(OH)3 + 3NaCl

b. CS₂ + 3Cl₂ → CCl₄ + S₂Cl₂

c. 2KI + Pb(NO₃)₂ → PbI₂ + 2KNO₃

d. C₂H₂ + 2.5O₂ → 2CO₂ + H₂O

Note: In equation d, the coefficients have to be balanced using fractional numbers.

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2) Weakened microorganisms that promote the production of antibodies may be included in vaccines.

Microorganisms that cause disease are called pathogens. Bacteria, viruses, fungus, and parasites are examples of pathogens. By contact with an infected person or animal, as well as through contact with contaminated food, drink, or soil, they can enter the body.In order to create antibodies against particular infections, the immune system of the body is stimulated by vaccines.A vaccine typically contains a weakened or killed form of the pathogen that is responsible for the disease it is designed to protect against. When the body is exposed to this weakened or killed form of the pathogen, the immune system responds by producing antibodies which help to protect against the disease.

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One mole of a monoprotic acid would be needed to react completely with 25 ml of a 1.10 M NaOH solution.

An acid is a compound that releases a proton (H+) when dissolved in water, whereas a base is a compound that accepts a proton when dissolved in water. When an acid and a base combine, a neutralization reaction occurs. The products of this reaction are water and a salt (ionic compound).

Neutralization is a chemical reaction that occurs between an acid and a base, resulting in a salt and water. This reaction entails the transfer of electrons from the base to the acid, forming water molecules and an ionic compound known as a salt. This reaction can be categorized into five types, depending on the combination of acid and base used.

The general formula for the neutralization reaction is:

HA + BOH → BA + H2O , where HA represents an acid, BOH represents a base, BA represents a salt, and H2O represents water. To react completely with 25 mL of a 1.10 M NaOH solution, you would need 0.0275 moles of a monoprotic acid.

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Inferring from this, the sample contains roughly 1.186 moles of xenon gas.

Number of moles = Molar volume at STP litres/V o l u m e ITP litres is the formula to calculate the number of moles at STP.

We need to utilise the molar mass of xenon, which is roughly 131.3 g/mol, to calculate the amount of xenon gas in the sample. The number of moles is then determined by dividing the gas's mass, which is given, by its molar mass:

Number of moles = Mass of gas / Molar mass of gas

Number of moles = 155.7 g / 131.3 g/mol

Number of moles = 1.186 moles (rounded to three decimal places)

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Bill Nye is an American scientific educator, engineer, and television host who goes by the moniker "Bill Nye the Science Man. "Children were the target audience for the program, which used humor, experiments, and demonstrations to teach scientific principles in a fun and approachable way.

Phases of matter are the various phases that matter can take on depending on its physical characteristics as well as the ambient temperature and pressure. Matter exists in three basic states: solid, liquid, and gas.

The "Phases of Matter" video by Bill Nye the Science Man teaches you the following 10 things:

There are three basic phases of matter: solid, liquid, and gas. Matter is everything that has mass and occupies space.

Because their molecules are closely packed together and immobile, solids have a definite shape and volume.

Because their molecules can move and slide past one another, liquids have a constant volume but adopt the shape of the container.

Because their molecules are spaced far apart and are free to flow in any direction, gases don't have a set shape or volume.

By adding or subtracting heat energy, a substance can transition from one phase to another.

A substance's melting point is the temperature at which it turns from a solid to a liquid, and its boiling point is the temperature at which it turns from a liquid to a gas.

Because water has a higher melting and boiling point than other molecules of a similar size, it is a unique substance.

Moreover, a change in pressure or a chemical reaction can cause a substance to transition from one phase to another.

In extremely high temperatures and pressures, matter can exist in other states like plasma and Bose-Einstein condensate.

For the purpose of comprehending the behavior of materials and creating new technologies, the study of matter and its phases is crucial.

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1. Motion energy is properly called kinetic energy. It is proportional to the mass of the moving object and grows with the square of its speed.

2. When an object is raised, energy is released when the object is lowered or falls. This demonstrates the gravitational relationship between Earth and objects in motion.

3. The faster an object moves, the more energy it has; a ball rolling across the floor at five meters per second has less energy than a ball rolling down a steep ramp at twenty meters per second, even though they are the same

object.

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The pH of the solution prepared in trimethylamine, a weak base, with trimethylammonium chloride is 11.10 (rounded to the nearest decimal place).

The pH of the solution, the concentration of each ion and the pKa of the acid must be determined.

Trimethylamine + H₂O ⇌ Trimethylammonium ion + OH⁻ the acid-base equilibrium can be expressed as follows: Trimethylamine + H₂O ⇌ Trimethylammonium ion + OH⁻Ka = Kw / Kb = 1.00 × 10⁻¹⁴ / 6.3 × 10⁻⁴ = 1.58 × 10⁻¹¹

The initial concentration of the weak base (Trimethylamine) is determined using the molarity equation. Molarity = number of moles / liters of solution0.125 M Trimethylamine (CH₃)₃N

The concentration of the Trimethylammonium ion (conjugate acid) can be calculated using the mass balance equation. Molarity = number of moles / liters of solution0.125 M Trimethylamine (CH₃)₃N

The hydroxide ion concentration. OH⁻ concentration = Kb x [A] = 6.3 × 10⁻⁴ x 0.125 = 7.87 × 10⁻⁵ M

pH is calculated using the hydroxide ion concentration.

The pH of a solution is determined using the formula:pH = 14 - pOH= 14 - log[OH⁻]= 14 - log(7.87 × 10⁻⁵)= 11.10. Therefore, the pH of the solution is 11.10 (rounded to the nearest decimal place).

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The pH of the solution prepared by adding 20.0 mL of 0.100 M HCl to 80.0 mL of a buffer that is comprised of 0.25 M NH₃ and 0.25 M NH₄Cl is 3.79.

The balanced chemical equation for the reaction between NH₃ and HCl is:

NH₃ (aq) + HCl (aq) → NH₄⁺ (aq) + Cl⁻ (aq)

Before any HCl is added, the solution contains 80.0 mL of a buffer solution that is comprised of 0.25 M NH₃ and 0.25 M NH₄Cl. NH₃ is a weak base, and its dissociation in water can be represented as follows:

NH₃ (aq) + H₂O (l) ⇌ NH₄⁺ (aq) + OH⁻ (aq)

The base dissociation constant (Kb) for NH₃ is 1.8 x 10⁻⁵ at 25°C.

After adding 20.0 mL of 0.100 M HCl to the buffer solution, the amount of NH₃ remaining in the solution will react with the HCl to form NH₄⁺ and Cl⁻. The amount of HCl added to the solution is:

moles of HCl = M x V = 0.100 mol/L x 0.020 L

                                   = 0.002 mol

Since NH₃ is a weak base, the buffer will resist changes in pH upon addition of HCl. The added HCl will react with NH₃ in the buffer solution to form NH₄⁺ and Cl⁻ ions. The NH₄⁺ ion is the conjugate acid of NH₃ and will slightly increase the acidity of the solution.

The amount of NH₃ that reacts with the HCl is:

Moles of NH₃ = moles of HCl

                        = 0.002 mol

The remaining amount of NH₃ in the solution is:

Initial moles of NH₃ - moles of NH₃ that reacted = (0.25 mol/L x 0.080 L) - 0.002 mol = 0.018 mol

The amount of NH₄⁺ that forms is equal to the amount of NH₃ that reacted:

0.002 mol of NH₃ = 0.002 mol of NH₄⁺

The concentration of NH₃ in the final solution is:

[ NH₃ ] = moles of NH₃ / total volume of solution

           = 0.018 mol / 0.100 L = 0.18 M

The concentration of NH₄⁺ in the final solution is:

[ NH₄⁺ ] = moles of NH₄⁺ / total volume of solution

            = 0.002 mol / 0.100 L = 0.02 M

The concentration of H⁺  in the final solution can be calculated using the equilibrium constant expression for NH₃:

Kb = [ NH₃ ][ OH⁻ ] / [ NH₃ ]

[ H⁺ ] = Kb x [ NH₃ ] / [ NH₃ ] = (1.8 x 10⁻⁵) x (0.18 mol) / (0.02 mol) = 0.000162 M

The pH of the final solution can be calculated as:

pH = -㏒[H⁺] = -log(0.000162) = 3.79

Therefore, the pH of the solution obtained by adding 20.0 mL of 0.100 M HCl to 80.0 mL of a buffer containing 0.25 M NH₃ and 0.25 M NH₄Cl is 3.79.

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The concentration of the hydroxide ion in the saturated Ca(OH)₂ solution is 0.080 M.

To calculate the concentration of hydroxide ion in the saturated Ca(OH)₂ solution, we need to use the results of the titration.

Assuming that the titrant is a strong acid, the balanced chemical equation for the reaction is; H⁺ + OH⁻ → H₂O

The equivalence point of the titration is reached when all of the hydroxide ions in the saturated Ca(OH)₂ solution have reacted with the acid. At this point, the moles of H⁺ added equals the moles of OH⁻ in the original solution. Therefore, we can use the volume and concentration of the titrant, along with the volume of the saturated Ca(OH)₂ solution, to calculate the concentration of OH⁻.

Let's assume that the titrant used is 0.1 M HCl, and that it took 20.0 ml of HCl to reach the equivalence point. This means that 20.0 ml of HCl contains 0.002 moles of H⁺ ions. Since the reaction is 1:1, there are also 0.002 moles of OH⁻ ions in the original 25.0 ml of saturated Ca(OH)₂ solution.

To calculate the concentration of OH⁻ in the saturated Ca(OH)₂ solution, we can use the following equation.

OH⁻ concentration = moles of OH⁻ /volume of solution

OH⁻ concentration = 0.002 moles / 0.025 L

OH⁻ concentration = 0.080 M

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The pH of the resultant solution is 3.16.

To determine the pH of the resultant solution after mixing the two solutions, we need to calculate the concentrations of all the ions present in the final solution.

First, let's calculate the moles of HCl added;

moles of HCl = concentration x volume = 0.050 M x 0.010 L = 0.00050 moles

Next, we need to calculate the moles of NaNO₂ and HNO₂ in the 20.0 mL solution;

moles of NaNO₂ = concentration x volume = 0.10 M x 0.020 L = 0.0020 moles

moles of HNO₂ = concentration x volume = 0.10 M x 0.020 L = 0.0020 moles

Since NaNO₂ is a salt and dissociates completely in water, it will provide Na⁺ and NO₂⁻ ions in solution. HNO₂, on the other hand, is a weak acid that will partially dissociate into H⁺ and NO₂⁻ ions. The equilibrium reaction for the dissociation of HNO₂ is;

HNO₂ + H₂O ⇌ H₃O⁺ + NO₂⁻

The acid dissociation constant, Ka, for HNO₂ is 4.5 x 10⁻⁴.

Let's assume x is the concentration of H⁺ ions produced by the dissociation of HNO₂. The NO₂⁻ concentration will be equal to the initial concentration of HNO₂ (0.10 M) minus the H+ ion concentration (x). Therefore, [NO₂⁻] = [HNO₂] - [H⁺]

[NO₂⁻] = 0.10 M - x

The equilibrium expression for the dissociation of HNO₂ can be written as;

Ka = [H⁺][NO₂⁻] / [HNO₂]

Substituting the expressions we found for [NO₂⁻] and [HNO₂]:

4.5 x 10⁻⁴ = x(0.10 M - x) / 0.0020 M

Solving for x using the quadratic formula;

x = 1.4 x 10⁻⁴ M

Now we can calculate the concentration of H⁺ ions in the final solution:

[H⁺] = [HCl] + [HNO₂]

[H⁺] = 0.00050 M + 1.4 x 10⁻⁴ M

[H⁺] = 6.9 x 10⁻⁴ M

Finally, we calculate the pH of  solution by using the formula; pH = -log[H⁺]

pH = -log(6.9 x 10⁻⁴)

pH = 3.16

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False. During proper cool-down activities, the body's aerobic energy system can help remove lactic acid from the muscles, but it does not convert it into fuel.

Lactic acid is converted into lactate, which is then transported to the liver where it is converted back into glucose through a process called gluconeogenesis. This glucose can then be used as fuel by the body. During intense exercise, the body relies heavily on the aerobic energy system to produce energy quickly. This results in the production of lactic acid, which can lead to muscle fatigue and soreness.

During the cool-down period, the body's aerobic energy system takes over to restore energy stores and remove waste products, including lactic acid. However, instead of converting lactic acid to fuel, it is actually converted into pyruvate, which then enters the mitochondria of the cell and undergoes aerobic respiration to produce ATP (adenosine triphosphate) – the primary source of energy for the body. The aerobic energy system helps remove lactic acid by converting it into pyruvate, which is then used as fuel for the production of ATP through aerobic respiration.

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We can use the ideal gas law, PV = nRT, to solve for the number of moles of gas present:

P = 732.5 mmHg = 0.963 atm (converting to atm)

V = 1.28 L

T = 293.15 K (20.0°C = 293.15 K)

R = 0.08206 L·atm/(mol·K)

PV = nRT

n = PV/RT = (0.963 atm)(1.28 L)/(0.08206 L·atm/mol·K)(293.15 K) = 0.0557 mol

Next, we can use the formula for molar mass, M = m/n, where m is the mass of the gas, to solve for the molar mass of the gas:

M = m/n = 2.50 g/0.0557 mol = 44.9 g/mol

Therefore, the molar mass of the gas is 44.9 g/mol.

The molar mass of the gas is 34.75 g mol^-1. To find the molar mass of the gas, the formula for the ideal gas law can be used.

The ideal gas law is given as

PV = nRT

Where,

P = pressure of the gas

V = volume of the gas

n = number of moles of the gas

R = gas constant

T = temperature of the gas

The given values of pressure, volume, and temperature are

P = 732.5 mmHg

V = 1.28 L

R = 62.36 L mmHg mol^-1

KT = 20°C = 20 + 273.15 = 293.15 K

Substituting these values in the ideal gas law equation and solving for n will give us the number of moles of the gas in the sample.

n = (PV) / (RT)

n = (732.5 mmHg × 1.28 L) / (62.36 L mmHg mol^-1 K × 293.15 K)

n = 0.07195 mol

To find the molar mass of the gas, we can use the formula

m = (mass of gas) / (number of moles of gas)

m = 2.50 g / 0.07195 mol

m = 34.75 g mol^-1.

Therefore, the molar mass of the gas is 34.75 g mol^-1.

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You would need 2.922 grams of solid NaCl to prepare 100 mL of a 0.50 M NaCl aqueous solution.

To calculate the mass of solid NaCl needed to prepare a 0.50 M solution with a volume of 100 mL, we need to use the following formula;

moles of solute = Molarity × volume (in liters)

We can rearrange this formula to solve for the mass of solute (NaCl) needed;

mass of NaCl=moles of NaCl × molar mass of NaCl

First, let's calculate the moles of NaCl needed;

moles of NaCl = Molarity × volume (in liters) = 0.50 mol/L × 0.1 L = 0.05 moles

Next, let's calculate the mass of NaCl needed;

mass of NaCl = moles of NaCl × molar mass of NaCl = 0.05 moles × 58.44 g/mol

= 2.922 g

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Answer:

78,300 cals.

Explanation:

540 cals of heat reqd to convert1 gm of water at 100 deg C so 540 x 145 = 78,300 cals.

Answer:

The answer is 78,300 calories.

540 calories of heat are required to transform one gram of water at 100 degrees Celsius, so 540 x 145 = 78,300 cal

Explanation:

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The pH before any HCl is added will be; 12.91, and the pH after the addition of 4.00 mL HCl is 1.17.

To solve this problem, we need to use the balanced chemical equation for the reaction between M(OH)₂ and HCl;

M(OH)₂ + 2HCl → 2H₂O + MCl₂

The balanced chemical equation shows that the reaction between M(OH)₂ and HCl is a 1:2 reaction. Therefore, the number of moles of HCl needed to react with the M(OH)₂ is twice the number of moles of M(OH)₂.

Before any HCl is added, the solution contains only M(OH)₂. The concentration of M(OH)₂ is 0.0811 M. We will use the following equation to calculate the pOH:

pOH = -log[OH⁻]

Since M(OH)₂ is a strong base, it completely dissociates in water, so [OH⁻] = [M(OH)₂]. Therefore;

[OH⁻] = 0.0811 M

pOH = -log(0.0811) = 1.09

The pH can be calculated using the following equation;

pH + pOH = 14

pH = 14 - pOH = 14 - 1.09 = 12.91

After the addition of 4.00 mL of HCl, the total volume of the solution is 5.00 mL + 4.00 mL = 9.00 mL. The number of moles of M(OH)₂ in the solution is;

moles M(OH)₂ = concentration x volume = 0.0811 M x 5.00 mL = 0.0004055 moles

The number of moles of HCl added to the solution is

moles HCl = concentration x volume = 0.0512 M x 4.00 mL = 0.0002048 moles

Since the reaction is a 1:2 reaction, the number of moles of HCl required to react with all of the M(OH)₂ is:

moles HCl required = 2 x moles M(OH)₂ = 2 x 0.0004055 = 0.000811 moles

The remaining moles of HCl in the solution after the reaction is;

moles HCl remaining = moles HCl added - moles HCl required = 0.0002048 - 0.000811 = -0.0006062 moles

Since the remaining moles of HCl is negative, it means that all of the M(OH)₂ has reacted and there is an excess of HCl in the solution. Therefore, the pH of the solution is determined by the concentration of the excess H⁺ ions. The concentration of excess H⁺ ions can be calculated using the following equation;

[H⁺] = concentration of HCl remaining / total volume of solution

[H⁺] = (-0.0006062 moles / 9.00 mL) x (1000 mL / 1 L) = -0.0674 M

The pH can be calculated using the following equation:

pH = -log[H⁺] = -log(-0.0674)

= 1.17 (Note: The negative sign indicates that the solution is acidic.)

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--The given question is incorrect, the correct question is

"Using the concentration values determined in this investigation for M(OH)₂  and HCl, calculate the pH for a strong acid + strong base titration in which 5.00 mL of the M(OH)₂ was transferred via pipet to a beaker and HCl was added from the burret. Calculate the pH: a) before any HCl is added, b) after the addition of 4.00 mL HCl. Concentration of HCl = 0.0512 M, Concentration of M(OH)₂  = 0.0811 M."--

The pressure exerted by the female's stiletto heel on the hard maple floor is approximately 76.1 psi.

The force that the woman is applying to the door = 300 N/m²  * 0.0200 m²

Force is 6 N

First, we need to convert the weight of the female from pounds to pounds-force.

Since weight is a force that is due to gravity, we can use the equation:

force = mass x acceleration due to gravity

The acceleration due to gravity is approximately 32.2 feet per second squared, so we can calculate the force as:

force = mass x acceleration due to gravity

force = 120 lbs x (1 lb-force/32.2 ft/s^2)

force = 3.73 lb-force

Next, we need to calculate the area of the heel in contact with the floor. We can use the formula for the area of a circle:

area = π x (diameter/2)^2

Plugging in the values given, we get:

area = π x (0.25 in/2)^2 = 0.049 in^2

Now we can calculate the pressure as the force divided by the area:

pressure = force/area = 3.73 lb-force / 0.049 in^2

pressure  = 76.1 psi

2. Force = Pressure * area

The force that the woman is applying to the door = 300 N/m²  * 0.0200 m²

Force = 6 N

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Answer:

The chemical equation 2AgNO3 + Mg(OH)2 → 2AgOH + Mg(NO3)2 represents a double replacement reaction. In this type of reaction, the cations and anions of two different compounds exchange places with each other to form two new compounds. In the given equation, silver nitrate (AgNO3) reacts with magnesium hydroxide (Mg(OH)2) to form silver hydroxide (AgOH) and magnesium nitrate (Mg(NO3)2).

Answer:

double replacement

Explanation:

Answer:

the freezing point of the 5.6 molal solution of ethylene glycol in ethanol is -125.744 °C

Explanation:

To calculate the freezing point depression of a solution, we can use the formula:

ΔTf = Kf * molality

where:

ΔTf is the freezing point depression

Kf is the freezing point depression constant for the solvent (in this case, ethanol)

molality is the number of moles of solute per kilogram of solvent.

The freezing point depression constant for ethanol is 1.99 °C/m.

We are given that the solution is 5.6 molal, which means there are 5.6 moles of ethylene glycol per kilogram of ethanol.

So, we can calculate the freezing point depression as:

ΔTf = 1.99 °C/m * 5.6 mol/kg = 11.144 °C

The freezing point depression is 11.144 °C.

To find the freezing point of the solution, we can subtract this value from the freezing point of pure ethanol, which is -114.6 °C.

Freezing point of solution = -114.6 °C - 11.144 °C = -125.744 °C

Therefore, the freezing point of the 5.6 molal solution of ethylene glycol in ethanol is -125.744 °C.