The temperature of an ideal gas in a sealed container of volume V is increased from T to 3T. What will happen to the pressure, P of this gas?




a. It increases from P to 2P



b. There is no change



c. It increases from P to 3P.



d. It decreases from P to P/3

Answer:

b) The amount of energy required to eject an electron from an atom.

Glass Because wood and other materials can be electrified through induction, but glass can't.

Is specific heat capacity C or Q?

The equation q = mcT may be used to compute the amount of heat acquired or lost by a sample (q), where m is the sample's mass, c is the specific heat, and T is the temperature change.

The quantity of heat required to raise the temperature of a given amount of stuff by one degree Celsius is referred to as heat capacity. The heat capacity of one gram of a material is known as its specific heat capacity (or specific heat), whereas the heat capacity of one mole is known as its molar heat capacity.

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Answer: 0.925 L

Explanation:

M=moles /L

L= moles/M

3.70/4 = 0.925 L

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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Answer:

1. 79.2 g of CO2

2. the reaction is spontaneous and favors the formation of products

3. CH2O

Explanation:

To solve the problem, we first need to calculate the limiting reagent by comparing the amount of glucose and oxygen available for the reaction. We will assume that the reaction goes to completion.

The balanced equation shows that for every 1 mole of glucose, 6 moles of oxygen are required. Therefore, the moles of oxygen required for 9.0 grams of glucose is:

moles of glucose = mass/molar mass = 9.0/180.16 = 0.0499 mol

moles of oxygen = 6 x moles of glucose = 6 x 0.0499 = 0.2994 mol

Since we have 0.2994 moles of oxygen available, and only 0.2000 moles of oxygen are required to react with 0.0499 moles of glucose to produce 0.0270 moles of water (according to the balanced equation), oxygen is the limiting reagent.

Using the balanced equation, we can now calculate the moles of CO2 produced:

moles of water produced = mass/molar mass = 5.4/18.02 = 0.2997 mol

moles of CO2 produced = 6 x moles of water produced = 6 x 0.2997 = 1.7982 mol

Finally, we can calculate the mass of CO2 produced:

mass of CO2 produced = moles x molar mass = 1.7982 x 44.01 = 79.2 g

Therefore, the mass of CO2 produced when 9.0 grams of glucose completely reacts with 9.6 grams of oxygen to produce 5.4 grams of water is 79.2 g.

To compare the entropy of the reactants to the entropy of the products, we can use the equation:

ΔS = ΣS(products) - ΣS(reactants)

The entropy of a substance depends on its state and temperature, and can be looked up in tables. At standard conditions (298 K and 101.3 kPa), the molar entropy of glucose, oxygen, CO2, and liquid water are:

S(C6H12O6) = 212.8 J/(mol K)

S(O2) = 205.0 J/(mol K)

S(CO2) = 214.8 J/(mol K)

S(H2O) = 69.9 J/(mol K)

Using the above values and the balanced equation, we can calculate the entropy change:

ΔS = (6 x S(CO2) + 6 x S(H2O)) - (S(C6H12O6) + 6 x S(O2))

ΔS = (6 x 214.8 + 6 x 69.9) - (212.8 + 6 x 205.0)

ΔS = 287.4 J/(mol K)

Since ΔS is positive, the entropy of the products is greater than the entropy of the reactants. This means that the reaction is spontaneous and favors the formation of products.

The empirical formula for glucose can be determined by dividing the subscripts by their greatest common factor. In this case, the empirical formula is:

C6H12O6 ÷ 6 = CH2O

Therefore, the empirical formula for glucose is CH2O.

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

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 right answer is first one

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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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.

Your welcome (:

Explanation:

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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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

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.

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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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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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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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:

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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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 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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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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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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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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Silver sulfide (Ag2S) is the common tarnish on silver objects. The mass of silver sulfide that can be made from 1.53 x 10^-3g of hydrogen sulfide (H2S) obtained from a rotten egg is 3.67 g.

Let's understand this in detail:

The balanced chemical equation for the reaction is:

2 AgNO3 + H2S → Ag2S + 2 HNO3

The molar mass of H2S is 34.08 g/mol.

1.53 x 10^-3g of H2S = (1 mol / 34.08 g) * (1.53 x 10^-3g) = 4.49 x 10^-5 mol H2S.

From the balanced equation, it is clear that one mole of Ag2S is formed from one mole of H2S.

Therefore, the number of moles of Ag2S formed from 1.53 x 10^-3g of H2S is 4.49 x 10^-5 mol.

The molar mass of Ag2S is 247.8 g/mol. Therefore, the mass of Ag2S formed from 1.53 x 10^-3g of H2S is

(247.8 g/mol) * (4.49 x 10^-5 mol) = 0.0111 g or 11.1 mg.

To convert to grams, divide by 1000:11.1 mg ÷ 1000 = 0.0111 g or 11.1 mg = 1.11 x 10^-2

Therefore, the mass of silver sulfide that can be made from 1.53 x 10^-3g of hydrogen sulfide obtained from a rotten egg is 3.67 g (approx).

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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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a.  Total number of ions in 47. 8 g of SrF2 = 1.635 ions.

b.  Mass (kg) of 4. 90 mol of CuCl2 · 2 H2O = 0.833 kg

c.  Mass (mg) of 2. 67 1022 units of Bi(NO3)3 · 5 H2O = 21,700 mg

a. Total number of ions in 47. 8 g of SrF2

The molar mass of SrF2 per gram = 87.62 g/mol

The number of moles of SrF2 in 47.8 g is calculated by using the formula,

The number of moles = n = (m/M) = 47.8 g / 87.62 g/mol

The number of moles  = 0.545 mol

Total number of ions present in the molecule SrF2 = (1 Sr2+ ion/mol) x (0.545 mol) + (2 F- ions/mol) x (0.545 mol)

The Total number of ions = 0.545 + 2(0.545)

The total number of ions = 1.635 ions.

b. Mass (kg) of 4. 90 mol of CuCl2 · 2 H2O

Total number of ions in CuCl2 · 2 H2O = (63.55 g/mol Cu + 2 x 35.45 g/mol Cl + 2 x 18.02 g/mol H + 16.00 g/mol O) + 2 x 18.02 g/mol H + 2 x 16.00 g/mol O

Total number of ions = 170.48 g/mol

The mass of 4.90 mol of CuCl2 · 2 H2O is:

Molar mass (m) = nM = 4.90 mol x 170.48 g/mol

Molar mass (m)  = 833.8 g = 0.833 kg

c.  Mass (mg) of 2.67 10^22 units of Bi(NO3)3 · 5 H2O

Total number of ions in Bi(NO3)3 · 5 H2O = (208.98 g/mol Bi + 3 x 62.01 g/mol N + 9 x 16.00 g/mol O) + 5 x 18.02 g/mol H + 5 x 16.00 g/mol O

Total number of ions = 485.09 g/mo

Total number of moles (n) = N/NA = (2.67 × 10^22)/6.022 × 10^23

Total number of moles = 0.0444 mol

The mass of 0.0444 mol of Bi(NO3)3 · 5 H2O is:

Molar mass (m) = nM = 0.0444 mol x 485.09 g/mol

Molar mass = 21.7 g =  21,700 mg

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Answer:

B

Explanation:

because it's an element

The only one I know is NaC2HO4. 1 atom in sodium, 2 atoms in carbon, 1 atom in hydrogen and 4 atoms in oxygen completeting the total of 8 atoms in this element.

The first stage in calculating the number of atoms is to determine the number of molecules. To determine the number of moles in an element or compound, reduce the specified mass by the element or compound's molar mass. The number of atoms in 1 mole of a material is or. 023 10 23 atoms.

An element is a particle. Since the two terms are identical, the answer is always one, and only one, if you're searching for the number of atoms in an element.

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Full Question: What’s the elements found in these formulas?

NaC2HO4

H2F5BLi

2He2PSO4

3He2O4PH

Answer:

Relative atomic mass is the average mass of an atom of an element, compared to 1/12th the mass of carbon 12 atom. Relative molecular mass is the ratio of the average mass of one molecular of an element or compound to1/2 of the mass of an atom of carbon 12.