274.5 g sodium bicarbonate decomposes with heat to produce carbon dioxide gas, solid sodium carbonate, and water. What is the percent yield if you captured and condensed 19.0 g of water as a product of the reaction?

Answer:

Balanced equations and names of salts:

a) CH3COOH + NaOH → CH3COONa + H2O (sodium ethanoate)

b) 2CH3COOH + Ca(OH)2 → Ca(CH3COO)2 + 2H2O (calcium ethanoate)

c) C2H5COOH + NH4OH → NH4C2H5COO (ammonium propanoate)

d) 2HCOOH + Na2CO3 → 2NaHCOO + CO2 + H2O (sodium formate)

e) C2H5COOH + NaHCO3 → NaC2H5COO + H2O + CO2 (sodium propanoate)

f) 2CH3COOH + CaCO3 → Ca(CH3COO)2 + CO2 + H2O (calcium ethanoate)

The displayed formula of the ethanoate anion is CH3COO-

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The volume of the container that  that contains 24. 0 grams of N2 gas at 328K at 884 atm is found to be 2.65 Liters.

The ideal gas equation is given as, PV = nRT, where, P is pressure V is volume, n is the moles, R is the gas constant and T is the temperature.

It is given to us that 24 grams of N₂ is to be stored in the container with 884 atm pressure and 328K temperature.

Now, putting all the values in the equation to find the volume of the container,

884V = 24/28(8.34)(328)

V = 2.65 Liters,

So, the volume of the container will be 2.65 Liters.

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Answer:0.40 M

Explanation:

M = moles/L

moles = 59.0 / 74( molar mass from the Periodic table for KCl) =0.797 moles

M=0.797/2 = 0.3986 = 0.40 M

To determine the gas in g/mL, we need to calculate the volume of the gas first. We can use the formula for the volume of a rectangular room to find the volume of the room:

Volume = length x width x height = 2.00 m x 2.00 m x 5.00 m = 20.00 m³

Next, we need to convert the mass of the gas from grams to kilograms, since density is usually expressed in units of kg/m³. We can do this by dividing the mass by 1000:

Mass = 3175 g ÷ 1000 = 3.175 kg

Finally, we can calculate the density of the gas using the formula:

Density = Mass ÷ Volume

Density = 3.175 kg ÷ 20.00 m³ = 0.1588 kg/m

To convert the density from kg/m³ to g/mL, we need to multiply by 1000 and then divide by 1000:

Density = 0.1588 kg/m³x 1000 g/kg ÷ 1000 mL/L = 0.1588 g/mL

Therefore, the gas has a density of 0.1588 g/mL. Without additional information, we cannot determine the identity of the gas.

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This is calculated by dividing the mass of aluminum (10.7 g) by its molar mass (26.98 g/mol), resulting in 0.395 mol.

To find the amount of pure substance (in moles) in 10.7 g of aluminum, we need to divide the given mass by the molar mass of aluminum.

The molar mass of aluminum is given as 26.98 g/mol.

Therefore, the amount of pure substance (in moles) in 10.7 g of aluminum can be calculated as:

moles = mass/molar mass

moles = 10.7 g/26.98 g/mol

moles = 0.396 moles (rounded to three significant figures)

So, there are approximately 0.395 moles of pure substance (aluminum) in 10.7 g of aluminum.

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Answer:

After neutralizing the sodium hydroxide solution with hydrochloric acid, the student should remove any excess hydrochloric acid and water from the solution to obtain a pure solution of sodium chloride. This can be done by a process called evaporation or by using a separating funnel.

To prepare a pure solution of sodium chloride, the student should follow the steps below:

Transfer the neutralized solution of sodium chloride into an evaporating dish.

Heat the evaporating dish over a low flame, stirring continuously until all the water evaporates.

Once all the water has evaporated, the student will be left with a dry residue of sodium chloride in the evaporating dish.

The sodium chloride can then be transferred to a clean, dry container and stored for use.

Alternatively, the student can use a separating funnel to remove excess hydrochloric acid and water from the solution. In this case, the neutralized solution is transferred to a separating funnel, and an equal volume of water is added. The funnel is then shaken, and the two layers are allowed to separate. The bottom layer containing the excess hydrochloric acid is drained off, leaving behind a pure solution of sodium chloride.

Explanation:

In this system, the equilibrium lies to the left , and the reaction favors the reactants

An equilibrium reaction system refers to a chemical system where the forward and reverse reactions occur at the same rate, resulting in no net change in the concentration of reactants and products over time.

In an equilibrium reaction system, the reactants and products are in a dynamic state where they are constantly interconverting. The equilibrium point is the point at which the concentrations of reactants and products remain constant, and the rate of the forward reaction is equal to the rate of the reverse reaction.

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Answer:

Explanation:

Mercury is a naturally occurring element that can be found in small quantities throughout the Earth's crust. It is typically found in cinnabar deposits, which are rocks or ores that contain high levels of mercury sulfide. These deposits can be found in regions around the world, including Spain, Italy, China, and the United States. Mercury can also be found in small amounts in other minerals and ores, including zinc, gold, and silver. However, due to its toxic nature, mercury is not typically extracted or mined for commercial purposes.

C₆H₁₂O₆ has a molar mass of 180.18 g/mol.

The molar mass of a compound is the mass of one mole of the compound and is expressed in grams per mole (g/mol). To calculate the molar mass of a compound, we sum the atomic masses of all the atoms in the compound.

For NO₂, we have:

Molar mass of N = 14.01 g/mol

Molar mass of O = 16.00 g/mol

Therefore, the molar mass of NO₂ = 14.01 g/mol + 2(16.00 g/mol) = 46.01 g/mol.

For C₆H₁₂O₆ (glucose), we have:

Molar mass of C = 12.01 g/mol

Molar mass of H = 1.01 g/mol

Molar mass of O = 16.00 g/mol

Therefore, the molar mass of C₆H₁₂O₆  = 6(12.01 g/mol) + 12(1.01 g/mol) + 6(16.00 g/mol) = 180.18 g/mol.

Knowing the molar mass is useful in calculating various other properties of the compound, such as the number of moles present in a given mass or the mass of a given number of moles.

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The molar mass of NO2 is 46 g/mol.

The molar mass of C6H12O6 is 180 g/mol.

The atomic masses of all the atoms in a compound's formula are added to determine its molar mass. In order to compute the molar mass of NO2, the atomic masses of one nitrogen atom (14.01 g/mol) and two oxygen atoms (2 x 16.00 g/mol) are added together, yielding a result of 46.01 g/mol.

Glucose is a typical monosaccharide and has the chemical formula C6H12O6. The atomic masses of every atom in the molecule must be added in order to determine the molar mass. Hydrogen, oxygen, and carbon all have atomic masses of 1.01 g/mol, 12.01 g/mol, and 16.00 g/mol, respectively. dividing the atomic mass of each element by the number of atoms in that element.

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The molar mass of A is 56.9 g/mol from the calculation below.

The Graham's law of diffusion, also known as Graham's law of effusion, is a scientific principle that describes the relationship between the rate of diffusion or effusion of gases and their molecular weights. It states that "the rate of diffusion or effusion of a gas is inversely proportional to the square root of its molecular weight."

We know that;

RA/RB = √MB/MA

0.75/1 = √32/MA

(0.75)^2 = 32/MA

MA = 32/(0.75)^2

MA = 56.9 g/mol

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The solution's pH value is 5.08.

To solve this problem, we need to use the principles of acid-base chemistry and the equilibrium constant expression for the dissociation of acetic acid (CH₃COOH) to its conjugate base, acetate (CH₃COO⁻).

First, we need to calculate the concentration of acetic acid in the solution. Since the initial concentration of acetic acid is given as 0.10 M and the volume of the solution is 92.00 mL, we can calculate the number of moles of acetic acid as follows:

moles of acetic acid = 0.10 M x 0.0920 L = 0.00920 moles

Next, we need to calculate the number of moles of sodium acetate in the solution. The molar mass of sodium acetate is 82.03 g/mol, so we can convert the mass of sodium acetate to moles as follows:

moles of sodium acetate = 1.60 g / 82.03 g/mol = 0.0195 moles

Since sodium acetate dissociates completely in water to produce acetate ions and sodium ions, the concentration of acetate ions in the solution is equal to the number of moles of sodium acetate divided by the total volume of the solution:

acetate ion concentration = moles of sodium acetate / total volume of solution

acetate ion concentration = 0.0195 moles / 0.0920 L = 0.212 M

Now we can use the equilibrium constant expression for the dissociation of acetic acid to calculate the pH of the solution:

Ka = [H⁺][CH₃COO-]/[CH₃COOH]

Since we are assuming that the concentration of acetic acid is still 0.10 M, and we now know that the concentration of acetate ions is 0.212 M, we can rearrange the equation and solve for [H⁺]:

[H⁺] = (Ka x [CH₃COOH])/[CH₃COO-]

[H⁺] = (1.75 x 10⁻⁵ x 0.10)/0.212

[H⁺] = 8.23 x 10⁻⁶ M

Finally, we can calculate the pH of the solution using the equation:

pH = -log[H⁺]

pH = -log(8.23 x 10⁻⁶)

pH = 5.08

Therefore, the pH of the solution is 5.08.

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Answer:

1. is "It produces sodium carbonate, a bitter metallic-tasting compound."
2. is 6:1

Explanation:

1. While baking powder does produce sodium bicarbonate or at least plays a part in creating sodium bicarbonate, its alkaline nature allows the dish or the reaction to avoid tasting metallic.

2. 6:1 because logically the more water there is in the flour, the less viscous it would be.

At time = 1, one of the H₂ molecules has reacted, producing one molecule of HI. This means that there are now 0.5 moles of H₂, 0.5 moles of I₂, and 1 mole of HI in the container.

At time = 3, the reaction has reached equilibrium. This means that the concentrations of all three gases have stabilized. Without additional information about the reaction conditions and equilibrium constants, we cannot determine the exact concentrations of each gas at equilibrium.

Between time = 1 and time = 2, the forwards reaction (H₂ + I₂ → 2HI) is favored because one of the H₂ molecules has reacted to produce HI. This means that the concentration of H₂ has decreased while the concentration of HI has increased, driving the reaction forwards according to Le Chatelier's principle.

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Answer: An object's speed and motion can be changed by a force acting on it. This is described by Newton's Second Law of Motion, which states that the acceleration of an object is directly proportional to the force applied to it, and inversely proportional to its mass. Therefore, the greater the force applied to an object, the greater its acceleration will be, and the greater the change in speed or motion will be. Additionally, the direction of the force also plays a role in changing an object's motion, as it can cause the object to change direction or rotate.

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Explanation:

Answer: 0.925 L

Explanation:

M=moles /L

L= moles/M

3.70/4 = 0.925 L

Answer:

see explanations

Explanation:

1. pH = -log [H+]

so [H+] = 10^(-pH) = 10^(-3.57) = 2.7 x 10^(-4) M

2. neutralize implies mol acid = mol base

so [HNO3] = (1 M) * (0.080 L) / (0.250 L) = 0.32 M HNO3

3. mol HNO3 = molarity of HNO3 * volume of HNO3

= 0.32 M HNO3 * 0.250 L HNO3 = 0.080 mol HNO3

The population density of the hawks is 6.73 hawks/sq. km.

Population density is defined as the number of individuals of a species per unit area.

Population density = Number of individuals / Area

In this problem, we are given:

Substituting these values in the formula, we get:

Population density = 148 hawks / 22 sq. km

Simplifying this expression, we get:

Population density = 6.73 hawks/sq. km

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TLE 9-Agri Crops might teach students important concepts or skills such plant propagation, soil preparation, insect control, irrigation management, and harvesting methods.

The purpose of the TLE 9-Agri Crops course is to introduce students to the basic ideas and procedures involved in crop production. Plant propagation, which includes the reproduction of plants through various ways such as seeds, cuttings, or grafting, is one of the most crucial skills covered in this course. Another crucial idea is soil preparation, which deals with managing the texture, structure, and fertility of the soil to assist plant development. Although irrigation management ensures that crops receive an appropriate quantity of water, insect control is crucial for defending crops against pests that might kill or severely harm them. Last but not least, harvesting techniques are crucial to ensuring a plentiful and high-quality crop.

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Answer:

Hydrogen is collected over water with a total pressure of 695.00 mmHg at 22.0 °C. The partial pressure of hydrogen gas is 675.173 mmHg.

Explanation:

Dalton's law states that the total pressure of a gaseous mixture is equal to the sum of the partial pressures of the gases.

The total pressure of gas collected over water is 695.00 mmHg and the temperature is 22.0 °C.

At that temperature, the vapor pressure of water is 19.827 mmHg. (sourced from a reference, look it up)

P = pGas + pH₂O

pGas = P - pH₂O = 695 mmHg - 19.827 mmHg = 675.173 mmHg

Hydrogen is collected over water with a total pressure of 695.00 mmHg at 22.0 °C. The partial pressure of hydrogen gas is 675.173 mmHg.

Sealing the flask with the stopper is an important step in conducting a controlled and accurate experiment. Without it, the reaction may be affected by outside factors and the results may be inaccurate or inconsistent.

The flask must be sealed with the stopper while the reaction takes place to prevent any outside air or substances from entering the flask and potentially interfering with the reaction. It also helps to prevent any of the reactants or products from escaping the flask, ensuring that the reaction can proceed as intended.

By sealing the flask with the stopper, you can more accurately measure and observe the reaction without any outside factors affecting it. In addition, sealing the flask with the stopper can also help to control the pressure and temperature of the reaction, which can be important for certain types of reactions.

By keeping the flask sealed, you can maintain a consistent environment within the flask and more accurately predict the outcome of the reaction.

Overall, sealing the flask with the stopper is an important step in conducting a controlled and accurate experiment. Without it, the reaction may be affected by outside factors and the results may be inaccurate or inconsistent.

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The sentence that is true about flammable materials is: option d.

have a lower flash point than combustibles.

Combustion is a chemical reaction that occurs between a fuel and an oxidizing agent, usually oxygen, in which energy in the form of heat and light is released. During combustion, the fuel undergoes a chemical reaction with oxygen, resulting in the production of new compounds such as carbon dioxide, water vapor, and other combustion by-products. This process is also commonly referred to as burning and is essential to many forms of energy production, such as in the combustion of fossil fuels in power plants and the combustion of gasoline in internal combustion engines.

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Look at your parts per million (ppm) calculations in the lab worksheet. What did you find the alkalinity to be for the 3rd sample (from Lake Ray Roberts) in parts per million? 220 ppm 540 ppm 260 ppm

The alkalinity for the 3rd sample from Lake Ray Roberts was 260 ppm.

Alkalinity is a measure of the capacity of water to neutralize acids, which is a property of water that helps to maintain a stable pH. It is a measure of the concentration of various ions in the water that can react with hydrogen ions (H+) to prevent a drop in pH.

In other words, alkalinity is the ability of water to resist changes in pH when an acid is added to it. It is usually expressed in terms of milligrams per liter (mg/L) of calcium carbonate (CaCO3) or parts per million (ppm) of CaCO3 equivalents.

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Correct answer should be

Look at your parts per million (ppm) calculations in the lab worksheet. what did you find the alkalinity to be for the 3rd sample (from lake ray roberts) in parts per million?

question 5 options:

540 ppm

220 ppm

260 ppm

Answer:

B

Explanation:

because it's an element

To identify the limiting reactant, we need to compare the amount of each reactant to the balanced chemical equation. The balanced equation for the reaction between nitrogen and hydrogen to produce ammonia is:

N2 + 3H2 → 2NH3

From the equation, we can see that one mole of nitrogen reacts with three moles of hydrogen to produce two moles of ammonia.

First, let's convert the mass of nitrogen to moles:

45.25 g N2 x (1 mol N2/28.02 g N2) = 1.612 mol N2

Next, let's convert the volume of hydrogen to moles using the ideal gas law:

PV = nRT

n = PV/RT

n = (52.5 L) (1 atm) / (0.0821 L·atm/mol·K) (273 K) = 2.19 mol H2

Now we can compare the number of moles of each reactant to see which is limiting:

N2 : H2 = 1.612 mol : 2.19 mol

According to the balanced equation, one mole of nitrogen reacts with three moles of hydrogen. So we need 4.836 moles of hydrogen to completely react with 1.612 moles of nitrogen. However, we only have 2.19 moles of hydrogen, which means it is the limiting reactant.

Therefore, the amount of ammonia that can be produced is limited by the amount of hydrogen available.

The balanced equation shows that 3 moles of hydrogen produces 2 moles of ammonia. So, with 2.19 moles of hydrogen, we can produce:

2.19 mol H2 x (2 mol NH3/3 mol H2) = 1.46 mol NH3

Now, let's convert the moles of ammonia to liters using the ideal gas law:

PV = nRT

V = nRT/P

V = (1.46 mol) (0.0821 L·atm/mol·K) (273 K) / (1 atm) = 31.2 L

Therefore, 31.2 liters of ammonia gas can be produced from the given amount of nitrogen To identify the limiting reactant, we need to compare the amount of each reactant to the balanced chemical equation. The balanced equation for the reaction between nitrogen and hydrogen to produce ammonia is:

N2 + 3H2 → 2NH3

From the equation, we can see that one mole of nitrogen reacts with three moles of hydrogen to produce two moles of ammonia.

First, let's convert the mass of nitrogen to moles:

45.25 g N2 x (1 mol N2/28.02 g N2) = 1.612 mol N2

Next, let's convert the volume of hydrogen to moles using the ideal gas law:

PV = nRT

n = PV/RT

n = (52.5 L) (1 atm) / (0.0821 L·atm/mol·K) (273 K) = 2.19 mol H2

Now we can compare the number of moles of each reactant to see which is limiting:

N2 : H2 = 1.612 mol : 2.19 mol

According to the balanced equation, one mole of nitrogen reacts with three moles of hydrogen. So we need 4.836 moles of hydrogen to completely react with 1.612 moles of nitrogen. However, we only have 2.19 moles of hydrogen, which means it is the limiting reactant.

Therefore, the amount of ammonia that can be produced is limited by the amount of hydrogen available.

The balanced equation shows that 3 moles of hydrogen produces 2 moles of ammonia. So, with 2.19 moles of hydrogen, we can produce:

2.19 mol H2 x (2 mol NH3/3 mol H2) = 1.46 mol NH3

Now, let's convert the moles of ammonia to liters using the ideal gas law:

PV = nRT

V = nRT/P

V = (1.46 mol) (0.0821 L·atm/mol·K) (273 K) / (1 atm) = 31.2 L

Therefore, 31.2 liters of ammonia gas can be produced from the given amount of nitrogen and hydrogen.

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The particles in a suspension can be separated from heterogeneous mixtures by passing the mixture through a filter.

Suspensions are heterogeneous mixtures that contain particles of a larger size that are suspended in a liquid medium. These particles are large enough to be visible to the ordinary eye and can be separated from the mixture through the use of a filter.

When a suspension is passed through a filter, the particles in the mixture are trapped in the filter medium, and the liquid passes through, leaving the solid particles behind.

This is due to the physical properties of the particles in suspension, which are large enough to be filtered out. In contrast, the particles in solutions and colloids are much smaller, and cannot be separated through filtration.

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Answer:

the right answer is first one

Molality is 10.8m, Freezing point is -09.8c ,Certain pharmaceuticals, cosmetics, or food products use it to preserve moisture by absorbing more water.

The U.S. Food and Drug Administration (FDA) has deemed propylene glycol to be "generally regarded as safe," and it believes a daily food intake of 23 mg/kg of body weight to be safe for people ages 2-65. Propylene glycol can be found in a variety of foods, cosmetics, and medication.

A humectant is a substance that is added to cosmetics to promote the retention of moisture in the skin and hair. Propylene glycol falls under this category. Propylene glycol is well accepted by skin and shouldn't irritate or produce redness.

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Answer: The pincushion moss, the echeveria, and the moon cactus in a terrarium make up a community.

:)

I think this is the answer btw Sorry if i'm wrong I have no clue.

In a terrarium, the pincushion moss, the echeveria, and the moon cactus make up the community.

Moon cactus, also known as Gymnocalycium mihanovichii or Hibotan cactus, is native to South American deserts in places like Brazil and Argentina.

Leucobryum glaucum, commonly known as leucobryum moss or pin cushion moss, is a species of haplolepideous mosses with a wide distribution in eastern North America and Europe.

Echeveria is a large genus of flowering plants in the family Crassulaceae, native to semi-desert areas of Central America, Mexico and northwestern South America.

Therefore, In a terrarium, the pincushion moss, the echeveria, and the moon cactus make up the community.

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A soil texture triangle is used to classify a soil's texture class. The percentages of sand, silt, and clay are scaled on the sides of the soil texture triangle.

Clay percentages are read across the triangle from left to right (dashed lines). The soil texture triangle identifies soil based on the amount of clay, silt, and sand. The angle of the numbers indicates the direction of the lines for each type of particle.

Clay lines, for example, go straight across, silt lines run diagonally down, and sand lines run diagonally up. M. Whitney of the USDA invented the textural triangle method in 1911 [21]. The USDA ternary diagram is divided into 12 textural classifications based on soil fraction %.

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Answer:

Here are the answers:

7. phosphorus molybdenum diboride

8. barium phosphite

9.  sulfur molybdenum

10.  dibromine heptoxide

11.   tribromine phosphide

12.  phosphorus tetrasulfide