The table shows the concentration of a reactant in the reaction mixture over a period of time.

What is the average rate of the reaction over the first 450 seconds?

A. 1.7 × 10−3
B. 1.9 × 10−3
C. 2.0 × 10−3
D. 2.2 × 10−3

Lactic acid fermentation is a process by which glucose is converted into lactic acid.  This occurs when glucose is broken down without the use of oxygen, and is used by cells to produce energy.

Lactic acid fermentation produces lactic acid and energy in the form of ATP. During this process, glucose is converted into pyruvate, which is then converted into lactate in the absence of oxygen. This process is carried out by certain bacteria, fungi, and animals, including humans.Lactic acid fermentation occurs when oxygen is not present, which means that the pyruvate produced during glycolysis cannot be converted into acetyl-CoA for the Krebs cycle.

As a result, the cell converts the pyruvate into lactic acid instead, which is a type of organic acid. Lactic acid fermentation is commonly used in the production of various foods, such as cheese, yogurt, and sauerkraut. It is also used in the production of alcoholic beverages like beer and wine. In addition, lactic acid fermentation is used in some industrial processes, such as the production of bioplastics and other materials.

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To convert 25 g of liquid water to vapor at 100°C, we need 67.7 kJ of energy.

To change 25 g of liquid water to steam, we need to calculate the energy required for the following two processes:

Heating the water from 100°C to its boiling point at atmospheric pressure (100°C).

Vaporizing the water at its boiling point at atmospheric pressure (100°C) to steam at the same temperature.

Let's start with the first step. The specific heat capacity of water is 4.18 J/g°C, so we need:

q1 = m * c * ΔT

where

m = mass of water = 25 g

c = specific heat capacity of water = 4.18 J/g°C

ΔT = change in temperature = (100 - 0)°C = 100°C

q1 = 25 g * 4.18 J/g°C * 100°C

q1 = 10450 J

This means that we need 10450 J of energy to heat 25 g of water from 0°C to 100°C.

Now let's move on to the second step, which is vaporizing the water. The molar heat of vaporization of water is 40.6 kJ/mol.

Since we know the mass of water (25 g), we need to convert it to moles:

n = m / M

where

m = mass of water = 25 g

M = molar mass of water = 18.015 g/mol

n = 25 g / 18.015 g/mol

n = 1.387 mol

The energy required to vaporize the water is:

q2 = n * Δ[tex]H_v_a_p[/tex]

where

Δ[tex]H_v_a_p[/tex] = molar heat of vaporization of water = 40.6 kJ/mol

q2 = 1.387 mol * 40.6 kJ/mol

q2 = 56.3 kJ

Therefore, the total energy required to change 25 g of liquid water to steam at 100°C is the sum of q1 and q2:

q = q1 + q2

q = 10450 J + 56.3 kJ

q = 67.7 kJ

So, we need 67.7 kJ of energy to change 25 g of liquid water to steam at 100°C.

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We must take into account the following stages in order to determine the amount of energy needed to convert 25 g of liquid water into steam at 100°C: Calculate how many moles of water (H2O) are contained in 25 g.

Number of moles of H2O = mass/molar mass = 25 g / 18.015 g/mol = 1.388 mol. The molar mass of H2O is 18.015 g/mol. Determine the amount of energy needed to evaporate one mole of water. Water has a molar heat of vaporization (Hvap) of 40.6 kJ/mol. Determine the amount of energy necessary to evaporate 1.388 moles of water. 1.388 moles of water must be vaporized in order to produce 40.6 kJ/mol of energy, which equals 56.4 kJ. Hence, 56.4 kJ of energy are needed to convert 25 g of liquid water to steam at 100°C. The molar heat of vaporization for liquid water is 40.6 kj/mole.

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The number of moles of N2 that can be made when 19 moles of CuO are consumed is 6 moles.

The law of conservation of mass can alternatively be referred to as the law of conservation of moles because mass can be mathematically translated into mass (the total moles on one side of the equation must be equal to the total moles on the other side of the equation). As a result, we can balance equations thanks to this law.

The balanced chemical equation is:

2 NH3 + 3 CuO → 3 Cu + N2 + 3 H2O

From the equation, we can see that 3 moles of CuO react with 1 mole of N2, which means that:

1 mole N2 = 3 moles CuO

So, if 3 moles of CuO produce 1 mole of N2, then 19 moles of CuO would produce:

1/3 x 19 = 6.33 moles of N2

However, we cannot have a fraction of a mole of a substance, so we need to round the answer to the nearest whole number:

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The organelle heavily involved in providing runner's cells with energy during the last leg of a 5K run is the mitochondria

The organelle heavily involved in providing the runner's cells with energy during the last leg of a 5K run is the mitochondria. Mitochondria are referred to as the "powerhouses" of the cell because they are responsible for producing the energy currency of cell, ATP (adenosine triphosphate), through the process of cellular respiration. During exercise, demand for ATP increases, and mitochondria works harder to meet this demand by breaking down glucose and other fuel molecules to generate ATP. Therefore, runner's cells would require a high level of mitochondrial activity to provide them with energy needed to complete the run.

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The net chemical equation for the production of hydrogen from propane and water is:

[tex]C_{3}H_{8}+3H_{2}O -- > 4H_{2}+3CO_{2}[/tex]

The reaction between propane and water can be used to produce hydrogen gas. The production of hydrogen gas from propane and water is an important reaction that has several applications. The balanced equation for the production of hydrogen gas from propane and water is given as follows:

[tex]C_{3}H_{8} +3H_{2}O-- > 4H_{2}+3CO_{2}[/tex]

The reaction between propane and water can be catalyzed by several metals such as nickel or platinum. The process is called steam reforming, and it is used to produce large quantities of hydrogen gas. Hydrogen is used as a fuel in fuel cells and combustion engines. It is also used in the production of ammonia and methanol, and as a reducing agent in various chemical processes.

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The substance that contains iron has a yield of 30.7%.

To determine the percent yield of an Fe-containing product, we first need to determine the theoretical yield. This is the maximum amount of product that can be obtained based on the balanced chemical equation and the amount of reactants used.

The balanced chemical equation for the reaction between Ca₃(PO₄)₂ and Fe(NO₃)₂ is:

3Ca₃(PO₄)₂ + 10Fe(NO₃)₂ → 2Fe₃(PO₄)₂ + 30Ca(NO₃)₂

Using the given masses, we can determine the limiting reactant and then calculate the theoretical yield of Fe₃(PO₄)₂

First, we need to convert the masses of Ca₃(PO₄)₂ and Fe(NO₃)₂ to moles:

24.1 gCa₃(PO₄)₂ × (1 mol Ca₃(PO₄)₂/310.18 g) = 0.0778 mol Ca₃(PO₄)₂

15.2 gFe(NO₃)₂ × (1 mol Fe(NO₃)₂/179.86 g) = 0.0845 mol Fe(NO₃)₂

The stoichiometric ratio betweenCa₃(PO₄)₂ and Fe(NO₃)₂ is 3:10, so the amount of Fe(NO₃)₂used is limiting. Therefore, the theoretical yield of Fe₃(PO₄)₂ is:

0.0845 mol Fe(NO₃)₂ × (2 mol Fe₃(PO₄)₂/10 mol Fe(NO₃)₂) × (559.68 g Fe₃(PO₄)₂/1 mol Fe₃(PO₄)₂) = 9.43 g Fe₃(PO₄)₂

Now, to calculate the percent yield, we divide the actual yield (2.9 g) by the theoretical yield (9.43 g) and multiply by 100:

Percent yield = (2.9 g/9.43 g) × 100 = 30.7%

Therefore, the percent yield of the Fe-containing product is 30.7%. This indicates that the reaction did not proceed to completion, and some Fe3(PO4)2 was left unreacted or lost during the isolation process.

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The statement that the volume of space in the tube would be equal to the sum of the volumes of hydrogen and water vapor is true if the tube is exposed to the atmosphere.

The assertion might not be accurate, though, if the tube is inside a closed system.

It relies on the circumstances surrounding the tube's observation. Assuming the tube is at equilibrium and the hydrogen and water vapor are thoroughly mixed inside the tube, if the tube is open to the atmosphere, then the amount of space in the tube would indeed equal the combined volume of hydrogen and water vapor.

However, if the tube is in a closed system, like a sealed container, the volume of the space inside the tube might not be the same as the volume of the hydrogen and water vapor because the gas molecules there might be interacting with the container's walls or going through pressure or temperature changes. Consequently, the claim may be accurate or untrue depending on the specific context.

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Answer: 2.22 moles

Explanation:

Molar mass

The molar mass of water, or H2O, can be found by adding the molar mass of two hydrogen atoms and the molar mass of one oxygen atom, which comes out to be 18.0 g/mol.

To find the moles in 40.0 grams of water, we will set up a proportion.

[tex]\frac{18.0 g}{1 mol} =\frac{40.0g}{xmol} \\[/tex]

Then, we just solve for x.

[tex]18=\frac{40}{x} \\x=\frac{40}{18} \\x=2.22 mol[/tex]

40.0 g of water is 2.22 moles.

Fastest to Slowest:- Hexane 1-butanol Ethanol Methanol Water Acetone

The ranking of the compounds from fastest to slowest can be determined by their molecular structure, intermolecular forces, and boiling points. Hexane, with its non-polar structure, has only weak London dispersion forces between its molecules, making it the fastest compound. 1-butanol has a longer carbon chain than ethanol, allowing for stronger intermolecular forces, which causes it to be slower.

Ethanol, methanol, and water have polar molecules and are capable of forming hydrogen bonds. However, methanol and ethanol have smaller molecular weights and shorter carbon chains, which causes them to be faster than water. Finally, acetone has a polar carbonyl group, but its smaller molecular weight and lack of hydrogen bonding capability causes it to be the slowest compound.

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Stoichiometry and the idea of percent yield can be used to calculate the mass of K required to produce 27.6 g of H2. The reaction's balanced chemical equation is 2 K + 2 HCl 2 KCl + H2.

Response and justification Hence, 126.23 g of potassium chlorate are needed to generate 32 g of oxygen.

Divide the mass of the reactant by the molecular weight to get the mass per mole. As an alternative, we can multiply the liquid solution's density in grammes per millilitre by the amount of reactant solution in millilitres. Next, divide the outcome by the reactant's molar mass.

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The  codeine's pKa is 8.20

To calculate the expected pH of a solution of 120 mg of codeine hydrochloride in 10.0 ml of water, we first need to determine the concentration of codeine hydrochloride in the solution. We can do this by converting the mass of codeine hydrochloride to moles and dividing by the volume of the solution in liters:

moles of codhcl = 120 mg / (299.36 g/mol) = 0.000401 mol

concentration of codhcl = 0.000401 mol / 0.01 L = 0.0401 M

Next, we need to determine the pKa of codeine, which can be used to calculate the expected pH of the solution. The pKa of codeine is related to its pKb by the equation:

pKa + pKb = 14

Therefore, the pKa of codeine can be calculated as:

pKa = 14 - 5.80 = 8.20

Using this pKa value, we can calculate the expected pH of the solution using the Henderson-Hasselbalch equation:

pH = pKa + log([A⁻]/[HA])

where [A-] is the concentration of the codeine anion and [HA] is the concentration of the codeine acid (codeine hydrochloride).

The codeine anion is formed when codeine hydrochloride dissociates in water, so we need to determine the extent of dissociation. The dissociation constant (Ka) can be calculated from the pKa using the equation:

Ka = [tex]10^{-pKa}[/tex]

Ka = [tex]10^{-8.20}[/tex] = 6.31 x 10⁻⁹.

Therefore, the pKa of codeine hydrochloride is 8.20

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The temperature of the gas sample was [tex]269 Kelvin (K).[/tex]

To solve this problem, we can use the ideal gas law:

[tex]PV = nRT[/tex]

where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant, and T is the temperature in Kelvin.

First, we need to rearrange the equation to solve for T:

[tex]T = PV/nR[/tex]

Now we can plug in the given values and solve for T:

[tex]T = (0.925 atm)(3.51 L)/(1.350 mol)(0.0821 L·atm/mol·K)T = 269 K[/tex]

Therefore, the temperature of the gas sample was [tex]269 Kelvin (K).[/tex]

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None of these compounds are soluble in water. Benzene (C6H6) and ethyl methyl ketone (CH3CH2COCH3) are both non-polar and insoluble in water, while acetic acid (CH3CO2H) and pentanol (CH3CH2CH2CH2CH2OH) are both polar and soluble in water.

Non-polar molecules, such as benzene and ethyl methyl ketone, cannot interact with the polar water molecules, meaning that they do not dissolve in water. Polar molecules, like acetic acid and pentanol, can interact with water molecules due to their partial positive and negative charges.

This allows them to dissolve in water. Therefore, of the compounds given, benzene and ethyl methyl ketone are least soluble in water.

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The evidence that supports the claim that social media had more of an influence outside of the Arab world than inside it is Eighty four percent of internet users visit social networking sites for political news. Option B is correct.

This statistic suggests that social networking sites have a significant influence on the dissemination of political news and information, and it implies that individuals who use the internet are more likely to consume political news through social media than through traditional media sources.

However, option (a) is also relevant because it suggests that internet penetration is lower in the Arab world, which could limit the impact of social media in the region compared to other parts of the world. Option (c) is not directly related to the claim, while option (d) is only tangentially related to the spread of information, rather than the influence of social media on society.

Hence, B. Eighty four percent of internet users visit social networking sites for political news is the correct option.

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Explanation:

There are few simple rules to follow.

(1) The atomic number is equal to the number of protons.

Z

= number of protons

(2) In neutrally charged elements, the number of electrons is the same as the number of protons.

Z

= number of protons = number of electrons (no charge)

Otherwise, positive charge means that the element lost an electron and negative charge means it gained an electron.

(3) The atomic mass is equal to the sum of the number of protons and number of neutrons.

A

= number of protons + number neutrons

or

A

=

Z

+ number neutrons

So if you say that

Z

= 20 and

A

= 40, then

A

=

Z

+ number neutrons

40 = 20 + number of neutrons

40 - 20 = number of neutrons

Therefore,

number of neutrons = 20

Answer:

Explanation:

When Iron is dipped in Aluminium Chloride, a redox reaction occurs. Basically iron reacts with Chloride ion and forms a solid precipitate of Aluminium.

End result :Iron (II) Chloride forms and Aluminium forms the precipitate

The molarity of this solution is 1.39 M.

Molarity, on the other hand, refers to the content of a substance or ion in a solution, whereas normality only refers to the molar concentration of the solution's acid or base components.

The formula is as follows:

Molarity (M) = moles of solute/liters of solution

We must first locate the copper(II) sulphate molecules. By splitting the solute's mass by its molar mass, we can determine this.

The molar mass of copper(II) sulfate is:

63.55 g/mol (for copper) + 2(32.06 g/mol) (for sulfur) + 4(16.00 g/mol) (for oxygen) = 159.61 g/mol

So, moles of copper(II) sulfate = 55.423 g / 159.61 g/mol = 0.347 moles

The volume of solution is 250.0 mL=0.250 L

Use the formula to calculate the molarity:

Molarity=0.347 moles / 0.250 L

= 1.39 M.

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It is a non-electrolyte is not an attribute of an acid.

What are Acids?

Acids are characterized as materials that can discharge hydrogen particles when they break up in a water-based solution. An acidic solution is one that contains more hydrogen particles, H+ than hydroxide particles, OH-. Acids have a pH value of less than 7.0.

The strength of the acid is determined by the concentration of hydrogen ions, H+.

When acids are mixed with a base, they react to produce a salt and water. Acidic solutions have a sour taste and can dissolve metals. Acids turn blue litmus paper red. Acids are used in numerous industrial applications, such as the production of fertilizers, plastics, and dyes, as well as the production of fuels from oil and gas.

They are employed in numerous chemical reactions and cleaning operations. They are widely used in the food industry as a preservative and to add flavor. They are also utilized in medicine to make supplements and for the treatment of diseases.

Thus, it is a non-electrolyte is not an attribute of an acid.

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The empirical formula for magnesium oxide is MgO, indicating that there is one atom of magnesium and one atom of oxygen in the compound. We can find it in the following manner.

To find the empirical formula for magnesium oxide, we need to determine the mole ratio between magnesium and oxygen in the reactants and use that to write the simplest whole-number ratio of atoms in the compound.

First, we need to convert the masses of magnesium and oxygen to moles using their atomic masses:

Moles of magnesium = 48.62 g / 24.31 g/mol = 2.00 mol

Moles of oxygen = 32.00 g / 15.99 g/mol = 2.00 mol

Next, we need to determine the mole ratio by dividing each of the mole amounts by the smallest mole amount:

Moles of magnesium / smallest mole amount = 2.00 mol / 2.00 mol = 1

Moles of oxygen / smallest mole amount = 2.00 mol / 2.00 mol = 1

The mole ratio is 1:1, which means that the empirical formula of magnesium oxide is MgO.

Therefore, the empirical formula for magnesium oxide is MgO, indicating that there is one atom of magnesium and one atom of oxygen in the compound.

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Fractional precipitation is a method used to separate different cations from a mixture by selectively precipitating one type of cation while leaving the others in solution. This method is based on the solubility differences of the cations' salts in a solvent, typically water.

To remove one type of cation from an aqueous mixture of multiple cations, a suitable reagent is added to selectively precipitate the desired cation. The precipitate formed can be separated from the solution by filtration or centrifugation. The remaining solution still contains the other cations that did not precipitate.

For example, to remove calcium ions (Ca2+) from a mixture containing calcium, magnesium, and zinc ions, ammonium oxalate can be added. This will form a white precipitate of calcium oxalate, which can be separated from the solution. The remaining solution will still contain the magnesium and zinc ions.

Fractional precipitation is a useful technique in analytical chemistry for the separation and identification of cations in a mixture.

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In Rutherford's gold foil experiment, alpha particles were fired against a thin sheet of gold foil, and their scattering patterns were then recorded on a screen.

In Thomson's "plum custard" atom model, a positively charged "soup" was surrounded by negatively charged electrons. Rutherford's gold foil experiment proved that an atom is mostly empty space with a tiny, dense, positively-charged nucleus.

Because of Rutherford's discoveries, Thomson's plum pudding model was flawed. An atom's positive charge is not distributed uniformly. Instead, everything is gathered in the tiny nucleus. Except for the electrons that are dispersed across it, the remaining space in an atom is empty.

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The correct answer is: "Bromine adds to alkenes instead of substituting for hydrogen."

In free-radical reactions with alkenes, a halogen such as bromine can be used as a halogenating agent to add a halogen atom to the alkene, producing a dihaloalkane. However, pure bromine cannot be used as a reagent in these reactions because it adds to the alkene rather than substituting for a hydrogen atom. This is because the bond between the two bromine atoms is relatively weak and can be broken easily, allowing bromine to add to the alkene to form a vicinal dibromide.

To overcome this problem, a source of bromine atoms that is less reactive and more selective is used in these reactions, such as N-bromosuccinimide (NBS) or a similar compound. NBS is a solid reagent that slowly releases small amounts of Br· radicals in solution. These radicals can react with alkenes to form vicinal dibromides, without adding to the alkene.

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Specific heat of the metal is approximately 0.301 J/gºC (option a).

Let's discuss it further below.

The specific heat of the metal can be calculated using the formula:

q = mcΔT

where q is the heat absorbed (10000 J), m is the mass of the metal (400 g), c is the specific heat, and ΔT is the change in temperature (103.0 ºC - 20.0 ºC).

Step 1: Calculate ΔT:
ΔT = 103.0 ºC - 20.0 ºC = 83.0 ºC

Step 2: Rearrange the formula to solve for c:
c = q / (mΔT)

Step 3: Plug in the given values and solve for c:
c = 10000 J / (400 g * 83.0 ºC) = 10000 J / 33200 gºC ≈ 0.301 J/gºC

So, the specific heat of the metal is approximately 0.301 J/gºC (option a).

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-227.3 kJ/mol.

To calculate the standard enthalpy of formation (ΔHf) of nitromethane, we need to use Hess's Law, which states that the total enthalpy change during a chemical reaction is independent of the pathway between the initial and final states. In other words, if we can find a series of reactions whose enthalpy changes add up to the enthalpy change of the desired reaction, we can use them to calculate ΔHf.

First, let's write the combustion equation in terms of the standard enthalpies of formation:

ΔHrxn = ΣnΔHf(products) - ΣnΔHf(reactants)

where n is the stoichiometric coefficient of each compound in the balanced chemical equation.

Substituting the standard enthalpies of formation of CO2 and H2O into the equation, we get:

-1418 kJ/mol = 2ΔHf(CO2) + 3ΔHf(H2O) + ΔHf(N2) - 2ΔHf(CH3NO2)

(From these standard enthalpies given)

ΔHf(CO2) = -393.5 kJ/mol

ΔHf(H2O) = -285.8 kJ/mol

ΔHf(N2) = 0 kJ/mol

Now we need to solve for ΔHf(CH3NO2):

-1418 kJ/mol = 2(-393.5 kJ/mol) + 3(-285.8 kJ/mol) + 0 kJ/mol - 2ΔHf(CH3NO2)

ΔHf(CH3NO2) = [2(-393.5 kJ/mol) + 3(-285.8 kJ/mol) - 1418 kJ/mol]/2

= -227.3 kJ/mol

Therefore, the standard enthalpy of formation of nitromethane is -227.3 kJ/mol. This means that if one mole of nitromethane is formed from its elements in their standard states at a temperature of 25°C and a pressure of 1 atm, -227.3 kJ of heat will be absorbed by the system.

In simpler terms, the energy required to form one mole of nitromethane from its elements is -227.3 kJ/mol.

Nitromethane has a standard enthalpy of production of -490.2 kJ/mol.

Nitromethane fuel may be used in one run 8.6 times more frequently than gasoline for a given cylinder volume. This indicates that using nitromethane instead of gasoline and the same amount of air results in around 2.3 times higher power.

We may get the standard enthalpy of production of nitromethane by using the values for the products and reactants' standard enthalpies of formation:

ΔHf° = Σ(nΔHf° products) - Σ(nΔHf° reactants)

ΔHf° for CO2(g) = -393.5 kJ/mol

ΔHf° for H2O(l) = -285.8 kJ/mol

ΔHf° for N2(g) = 0 kJ/mol

ΔHf° for O2(g) = 0 kJ/mol

The values in the formula above can be changed as follows:

ΔHf° for CH3NO2 = [2(-393.5 kJ/mol) + 3(-285.8 kJ/mol) + 1(0 kJ/mol)] - [2ΔHf° for CH3NO2 + 1(0 kJ/mol)]

If we simplify, we get:

ΔHf° for CH3NO2 = -1418 kJ/mol - 6(-285.8 kJ/mol) + 2(-393.5 kJ/mol)

ΔHf° for CH3NO2 = -1418 kJ/mol + 1714.8 kJ/mol - 787 kJ/mol

ΔHf° for CH3NO2 = -490.2 kJ/mol

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Added sugars are various types of sugars and syrups that are added to food products during processing . Here are five common names for added sugars: Sucrose, Dextrose, Fructose, Maltose, High fructose corn syrup .

Sucrose: it is a disaccharide made up of glucose and fructose .

High fructose corn syrup: a sweetener made from cornstarch that has been processed to convert some of its glucose into fructose.

Dextrose: a simple sugar that is chemically identical to glucose and is commonly used in processed foods.

Fructose: a simple sugar found naturally in fruits, vegetables, and honey.

Maltose: a disaccharide made up of two glucose molecules and is commonly used as a sweetener in some beers and candies.

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The given statements are accordingly true and false: Part A: True, Part B: False, Part C: False, Part D: False, Part E: False.

Part A: True. To ensure that  entire inner surface of the buret is rinsed, water should be added while rotating the buret.

Part B: False. Rinsing the buret with water alone is not always enough , residual substances may adhere to the surface of the buret.

Part C: False. It is not always necessary, and may even be harmful if the soap is not thoroughly rinsed off.

Part D: False. The buret should be thoroughly rinsed with distilled water.

Part E: False. While it is important to rinse the buret with the titration solution before beginning a titration.

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Answer:

The balanced chemical equation for the combustion of ethane is

2C₂H₆(g) + 7O₂(g) → 4CO₂(g) + 6H₂O(l)

The stoichiometric ratio between C₂H₆(g) and CO₂(g)  is 1 : 2

Hence,

   moles of CO₂(g) produced = moles of reacted C₂H₆(g) x 2

                                                 = 1.00 mol x 2

                                                 = 2.00 mol

Hence, the correct answer is "C".

Using a 100 g sample of a compound with 25.9% nitrogen and 74.1% oxygen, we determine the empirical formula to be N₂O₅.

To determine the empirical formula of a compound, we need to find the ratio of the number of atoms of each element in the compound, in its simplest whole-number ratio. Here's how we can do it:

Assume we have a 100 g sample of the compound. Then, 25.9 g of it is nitrogen and 74.1 g is oxygen.

Convert the mass of each element to moles, using their respective molar masses:

Nitrogen: 25.9 g / 14.01 g/mol = 1.85 mol

Oxygen: 74.1 g / 16.00 g/mol = 4.63 mol

Find the ratio of the moles of each element by dividing both values by the smallest one:

Nitrogen: 1.85 mol / 1.85 mol = 1

Oxygen: 4.63 mol / 1.85 mol = 2.50

If necessary, adjust the ratio to the nearest whole number. Since we can't have half an atom, we need to multiply both values by 2:

Nitrogen: 1 x 2 = 2

Oxygen: 2.50 x 2 = 5

The empirical formula of the compound is therefore N₂O₅.

Therefore, the empirical formula of the compound is N₂O₅.

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