37) Determine the molecular formula of a compound that has a molar mass of 183.2 g/mol and an empirical formula of C2H5O2.A) C2H5O2B) C6H15O6C) C3H7O3D) C4H10O4E) C8H20O8

All of the following are characteristics of a Wittig  chemical reaction except it involves an SN² reaction between a carbanion and an aldehyde or ketone.

Chemical reactions are defined as changes which occur when a substance combines with another substance to form a new substance.Alternatively, when a substance breaks down or decomposes to give new substances it is also considered to be a chemical reaction.

There are several characteristics of chemical reactions like change in color, change in state , change in odor and change in composition . During chemical reaction there is also formation of precipitate an insoluble mass of substance or even evolution of gases.It involves SN¹ or SN²reactions.

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To make pure crystals of magnesium chloride from magnesium and dilute hydrochloric acid follow the procedure listed.

The apparatus needed inculde the following;

Some of the procedure include the following;

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The three individual collagen chains are themselves polypeptides that differ from each other in their amino acid sequence. Collagen is a fibrous protein that provides strength and structure to various tissues in the body, including skin, bones, and tendons.

It is made up of three individual chains of amino acids that are coiled around each other to form a triple helix structure. These three chains are known as alpha chains, and there are different types of alpha chains that can combine to form various types of collagen.

The differences in the amino acid sequences of the alpha chains lead to differences in the properties of the collagen, such as its strength and flexibility.

Therefore, the specific combination of alpha chains that make up a particular collagen molecule determines its function and location in the body.

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For the following reaction, you have 8 grams of hydrogen and 2 grams of oxygen. 2H2 + O2 → 2H2O. The limiting reagent is the oxygen is True.

1. Determine the molar mass of each reactant:
  H2 (hydrogen) = 2.02 g/mol
  O2 (oxygen) = 32.00 g/mol

2. Calculate the number of moles for each reactant:
  Moles of H2 = (8 grams) / (2.02 g/mol) = 3.96 moles
  Moles of O2 = (2 grams) / (32.00 g/mol) = 0.0625 moles

3. Determine the stoichiometric ratio of the reactants:
  For every 2 moles of H2, 1 mole of O2 is needed.

4. Calculate the amount of O2 needed for complete reaction with the given H2:
  Amount of O2 needed = (3.96 moles H2) * (1 mole O2 / 2 moles H2) = 1.98 moles

5. Compare the amount of O2 needed with the amount of O2 given:
  1.98 moles (needed) > 0.0625 moles (given)

Since there is less O2 than needed for a complete reaction, O2 is the limiting reagent.

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The pH at the equivalence point is 4.48.

In this problem, we can use the Henderson-Hasselbalch equation to find the pH at the equivalence point. The Henderson-Hasselbalch equation is;

pH = pKa + log([A⁻]/[HA])

where pKa is the acid dissociation constant of the acid, [A⁻] is the concentration of the conjugate base, and [HA] is the concentration of the acid.

Now, we can calculate the concentration of the conjugate base at the equivalence point;

moles of acetic acid = concentration x volume = 0.567 mol/L x 0.025 L = 0.01418 mol

moles of NaOH added = concentration x volume = 0.432 mol/L x volume

At the equivalence point, moles of acetic acid = moles of NaOH added, so;

0.01418 mol = 0.432 mol/L x volume

volume = 0.0328 L

The total volume of the solution at the equivalence point is the sum of the volumes of acetic acid and NaOH solutions;

[tex]V_{total}[/tex] =[tex]V_{(aceticacid)}[/tex] + [tex]V_{(NaOH)}[/tex] = 0.025 L + 0.0328 L

= 0.0578 L

The concentration of acetate ion at equivalence point is;

[acetate] = moles of acetate / [tex]V_{total}[/tex] = 0.01418 mol / 0.0578 L = 0.245 M

Now we can use the Ka expression for acetic acid to find the pKa;

Ka = [H⁺][CH₃COO⁻]/[CH₃COOH] = 1.8 x 10⁻⁵

At equilibrium, the concentration of H+ equals the concentration of acetate ion;

[H⁺] = [acetate] = 0.245 M

Substituting these values into the Ka expression and solving for [CH₃COOH], we get;

1.8 x 10⁻⁵ = (0.245)² / [CH₃COOH]

[CH₃COOH] = (0.245)² / 1.8 x 10⁻⁵

= 3.34 M

Now we can use the Henderson-Hasselbalch equation to find the pH at the equivalence point;

pH = pKa + log([A⁻]/[HA]) = pKa + log(1) = pKa

pH = -log(3.34 x 10⁻⁵)

= 4.48

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The octet rule states that second-row elements B and Be can have less than eight electrons in their valence shell because they have fewer available orbitals.

The octet rule is a guideline that suggests that atoms tend to gain, lose, or share electrons in order to have eight electrons in their outermost shell, or valence shell. This is because having eight valence electrons makes the atom more stable, due to a full outer shell.

However, second-row elements Beryllium (Be) and Boron (B) can have less than eight electrons in their valence shell because they have fewer available orbitals. Be has only four valence electrons and two available orbitals, while B has only three valence electrons and three available orbitals.

These elements may form compounds where they have fewer than eight electrons in their valence shell. For example, Be typically forms compounds where it has only four electrons in its valence shell, such as BeH₂, while B typically forms compounds where it has six electrons in its valence shell, such as BF₃.

Therefore, the octet rule is a useful guideline, but there are exceptions, such as second-row elements Be and B.

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The correct answer is E) 0.829 moles.

The given problem involves determining the number of moles of N2O4 in a given amount of the compound. To solve this problem, the molar mass of N2O4 is required, which is given as 92.02 g/mol.

To determine the number of moles of N2O4 in 76.3 g N2O4, you can use the molar mass of N2O4, which is 92.02 g/mol. Use the formula:

moles = mass (g) / molar mass (g/mol)

moles = 76.3 g / 92.02 g/mol

moles ≈ 0.829 moles

So the correct answer is E) 0.829 moles.

This means that there are approximately 0.829 moles of N2O4 in 76.3 grams of the compound.

The mole is a unit of measurement used in chemistry to express amounts of a chemical substance. It is defined as the amount of substance that contains as many elementary entities (such as atoms or molecules) as there are atoms in 12 grams of pure carbon-12.

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The gas which have the largest Van der Waals constant a is ammonia, NH₃.

The Intermolecular forces in between the molecules are forces that are the  attractive forces which make them together or will be impart physical properties. The Stronger the forces and the closer the molecules of the substance.

The molecule that have the strongest forces and that will have the largest value of the Van der Waals constant a. The gas which have the largest Van der Waals constant a is ammonia, NH₃ as the ammonia has the strongest force because of the hydrogen bonding. Therefore, the NH₃  has the largest value of the a.

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This question is incomplete, the complete question is :

Real gases and vapors deviate from ideal behavior on account of intermolecular interactions. One equation of state for a real gas is the van der Waals equation, which is expressed in terms of two parameters, a and b. Which gas would you expect to have the largest Van der Waals constant a: F₂, Ne, NH₃.

A 1 gram sample of Olestra would contribute 0 dietary calories to a human consumer because Olestra is not metabolized by the human body due to the additional fatty acid units that block the approach of digestive enzymes to the cleavage sites. The correct answer is (a) 0.

Olestra, also known as Olean, is a fat substitute that was developed to replace traditional fats in food products. Unlike traditional fats, Olestra is not metabolized by the human body because the additional fatty acid units in its molecular structure block the approach of digestive enzymes to the cleavage sites.

Therefore, it passes through the digestive system without being absorbed or broken down into calories. The correct answer is 0.

This means that Olestra is not absorbed in the small intestine and passes through the digestive system without being broken down into calories.

As a result, Olestra has a negligible caloric value and does not contribute to the overall calorie content of the food products in which it is used. This makes it an attractive alternative to traditional fats for food manufacturers who want to reduce the calorie content of their products without sacrificing taste or texture.

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The most invasive type of nuclear radiation is typically thought to be gamma radiation.

The term "radiation" describes the release or transmission of electrical energy through area or matter in the form either particles or waves. Ionising radiation and non-ionizing radiation are the two basic categories into which it can be divided.Non-ionizing radiation has a lower energy and is generally thought to be less dangerous to human health. Examples include visible sunlight, radio waves, and microwaves. However, at high enough doses, such as those associated with coming into contact with intense infrared radiation, even radiation that is not ionising can harm tissue.

A tissue is a collection of cells with a common structure and function that work together to carry out a single duty or set of duties for the organism.

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The molar mass of [tex]Al(C_2H_3O_2)_3[/tex] is 204.13 g/mol. The correct option is B.

To calculate the molar mass of [tex]Al(C_2H_3O_2)_3[/tex], we need to determine the atomic masses of each element present in the compound and add them up, taking into account the number of each element in the formula.

26.98 g/mol is the atomic mass of Aluminum (Al).

12.01 g/mol is the atomic mass of Carbon (C).

1.01 g/mol is the atomic mass of Hydrogen (H).

16.00 g/mol is the atomic mass of Oxygen (O).

The compound contains one aluminum atom, three carbon atoms, nine hydrogen atoms, and six oxygen atoms.

Molar mass of [tex]Al(C_2H_3O_2)_3[/tex] = (1 × 26.98 g/mol) + (3 × 2 × 12.01 g/mol) + (9 × 1.01 g/mol) + (6 × 16.00 g/mol) = 204.13 g/mol

Therefore, the correct answer is (B) 204.13 g/mol.

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The effect we see from changing the independent variable is called the dependent variable.

In scientific experiments, the independent variable is the variable that is deliberately changed or manipulated by the researcher, while the dependent variable is the variable that is being observed or measured to see how it responds to the changes in the independent variable.

For example, in a study to determine the effect of different doses of a medication on blood pressure, the independent variable would be the dose of the medication, while the dependent variable would be the blood pressure readings. By changing the dose of the medication (independent variable), we can observe how the blood pressure readings (dependent variable) respond.

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The Hazardous Material class that includes compressed gases, dissolved gases, and gases liquefied by compression or refrigeration is Class 2: Gases. This class is further divided into three divisions:

1. Division 2.1 - Flammable Gases: These are gases that can burn in the presence of an ignition source. Examples include propane, butane, and hydrogen.
2. Division 2.2 - Non-Flammable, Non-Toxic Gases: These are gases that do not burn and are not toxic, but may still pose risks due to their physical properties, such as high pressure or low temperature. Examples include nitrogen, helium, and carbon dioxide.
3. Division 2.3 - Toxic Gases: These are gases that are harmful or even fatal when inhaled. Examples include chlorine, ammonia, and phosgene.

The proper handling, storage, and transportation of these gases are essential to minimize the risks associated with their hazardous properties. Regulations and guidelines are in place to ensure the safety of those working with and around these materials.

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Explanation:

How many joules are required to convert 325g of water at 12 degrees Celsius to steam at 176 degrees Celsius

To calculate the energy required to convert a given mass of water from a lower temperature to steam at a higher temperature, we need to consider two processes: (1) heating the water from its initial temperature to its boiling point, and (2) vaporizing the water at its boiling point to steam at the final temperature.

The amount of heat required for each process can be calculated separately using the following formulas:

(1) Q1 = m * c * ΔT

(2) Q2 = m * L

where Q1 is the heat required to raise the temperature of the water, Q2 is the heat required for the water to vaporize, m is the mass of water, c is the specific heat of water, ΔT is the temperature change, and L is the heat of vaporization of water.

Given:

Mass of water (m) = 325 g

Initial temperature of water = 12°C

Final temperature of steam = 176°C

Specific heat of water (c) = 4.184 J/g°C

Heat of vaporization of water (L) = 2260 J/g (at standard pressure)

To find the energy required to convert 325g of water at 12°C to steam at 176°C, we need to calculate Q1 and Q2 separately and then add them together.

(1) Heating the water:

Q1 = m * c * ΔT

Q1 = 325 g * 4.184 J/g°C * (100°C) [since the boiling point of water is 100°C at standard pressure]

Q1 = 136292 J

(2) Vaporizing the water:

Q2 = m * L

Q2 = 325 g * 2260 J/g

Q2 = 735500 J

Total heat required = Q1 + Q2

Total heat required = 136292 J + 735500 J

Total heat required = 871792 J

Therefore, it would require 871792 J of energy to convert 325g of water at 12°C to steam at 176°C.

The overall structure is dictated by their interaction with the cell membrane.

Proteins can be broadly divided into two main categories based on their folding patterns:

1) Fibrous proteins: These proteins have a long, thin, and elongated shape and usually have a structural role in the body. Examples include keratin, collagen, and elastin.

2) Globular proteins: These proteins have a compact, roughly spherical shape and are typically involved in metabolic and enzymatic functions. Examples include enzymes, antibodies, and hormones.

There is also a third category of proteins, known as membrane proteins, which are embedded in cell membranes and have a different folding pattern than fibrous or globular proteins. Membrane proteins can have both globular and fibrous regions.

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The final pH of the solution is 1.75.The reaction between nitrous acid (HNO2) and sodium hydroxide (NaOH) can be written as:

HNO₂ + NaOH → NaNO₂ + H₂ O

We can use the balanced equation to determine the moles of HNO₂ that react with NaOH. Since NaOH is a strong base, it will react completely with HNO₂, so we can assume that the amount of NaOH remaining after the reaction is negligible.

Moles of HNO₂ = 0.529 mol

Moles of NaOH = 0.246 mol

Since HNO₂ is a weak acid, it will partially dissociate in water according to the equation:

HNO₂ + H₂O ⇌ H₃O+ + NO2-

The equilibrium constant (Ka) for this reaction is 4.0 x 10¯⁴.

We can use the initial moles of HNO2 and the Ka value to determine the concentration of H₃O at equilibrium.

Ka = [H₃O+][NO₂-] / [HNO₂]

Assuming x moles of HNO2 dissociate, we get:

Ka = [H₃O] * [NO2-] / [HNO₂ - x]

At equilibrium, [HNO₂] = 0.529 - x

[NO₂-] = x

[H₃O] = x

Substituting these values in the equilibrium constant expression and solving for x, we get:

4.0 x 10¯⁴ = x² / (0.529 - x)

Solving for x gives x = 0.0179 M.

Therefore, the concentration of H₃O at equilibrium is 0.0179 M, and the pH is:

pH = -log[H₃O]

pH = -log(0.0179)

pH = 1.75

Therefore, the final pH of the solution is 1.75

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The new pH of the solution after adding 6.30 mL of 0.490 M HCl to a 200.0 mL acetate/acetic acid buffer solution with a total molarity of 0.100 M and a pH of 5.000.

To solve this problem, we need to use the Henderson-Hasselbalch equation:

pH = pKa + log([A⁻]/[HA])

Total Molarity = [A⁻] + [HA]

0.100 M = [A⁻] + [HA]

We also know that the pH of the buffer solution is 5.000. The pKa of acetic acid is 4.76, so we can use this information to find the ratio of [A⁻] to [HA]:

5.000 = 4.76 + log([A⁻]/[HA])

0.24 = log([A⁻]/[HA])

[HA]/[A⁻] = 10⁰.²⁴

We can use this ratio and the total molarity equation to find the initial concentrations:

[A⁻] = (10⁰.²⁴/[HA]) * (0.100 M - [HA])

- The HCl will react with the acetate to form acetic acid and chloride ions:

HCl + A⁻ → HA + Cl⁻

The amount of HCl added is:

0.490 M * 6.30 mL = 0.00309 moles HCl

The amount of acetate consumed and the amount of acetic acid formed will be 0.00309 moles.

The new concentration of acetic acid is:

[HA] = [HA]initial + 0.00309 moles / 0.2000 L = 0.0155 M

The new concentration of acetate is:

[A⁻] = [A⁻]initial - 0.00309 moles / 0.2000 L = 0.0485 M

We can now use the Henderson-Hasselbalch equation with the new concentrations to find the new pH:

pH = 4.76 + log(0.0485/0.0155) = 4.50

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The structure of the isomerization product is similar to that of isopentenyl pyrophosphate, except that one of the double bonds has shifted to a different position in the molecule.

This is due to the acid-catalyzed process, which involves the transfer of a proton from the phosphate group to the carbon-carbon double bond, causing it to shift to a different position in the molecule. This results in the formation of dimethylallyl pyrophosphate, which is an important precursor in the biosynthesis of many essential compounds, including sterols, carotenoids, and tocopherols.

The isomerization process is a crucial step in the biosynthesis of these compounds, as it helps to ensure the correct formation of the carbon-carbon double bonds that are necessary for their biological activity.

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The crystal structure of NaCl (sodium chloride) consists of a repeating pattern of positively charged sodium ions and negatively charged chloride ions. This repeating pattern, known as a crystal lattice, forms a three-dimensional structure that extends throughout the entire sample of NaCl.

In the context of materials science, a phase is defined as a homogeneous and physically distinct portion of a material that has uniform physical and chemical properties. Therefore, the NaCl crystal structure is a single-phase material because it is a uniform and homogeneous arrangement of sodium and chloride ions throughout the crystal lattice.

However, it is important to note that NaCl can exist in different phases under different conditions of temperature and pressure. For example, at high temperatures and pressures, NaCl can exist in a cubic close-packed (ccp) structure rather than its usual face-centered cubic (fcc) structure.

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H₂ identifies the correct limiting reactant because it has the lower starting mass. The correct option is C.

In the described reaction, 2 moles of hydrogen and 1 mole of oxygen combine to form 2 moles of water. H₂ to H₂O has a stoichiometric ratio of 2:1 (or 1:1), whereas O₂ to H₂O has a stoichiometric ratio of 1:2.

We can determine the amount of moles of accessible H2 and O2 based on the facts given:

0.4g of H₂ corresponds to (0.4 g / 2.016 g/mol) = 0.198 mol of H₂.

1.8g of O₂ corresponds to (1.8 g / 32.00 g/mol) = 0.056 mol of O₂.

We determine that H₂ is the limiting reactant by comparing the moles of H₂ and O₂ that are accessible.

The amount of product (H₂O) that can be created is constrained because the reactant is fully consumed first. The stoichiometric ratio and the fact that there are less moles of H₂ than O₂ are the foundations for the reasoning.

Thus, the correct option is C.

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The average composition of the vapor condensed will be around 70-80% n-hexane and 20-30% n-octane, depending on the boiling points of the two compounds.

To separate the 40-60 mixture of n-hexane and n-octane using simple batch distillation, the liquid mixture is heated until it boils and the vapor fraction is condensed and collected. After 10% of the original liquid is distilled, the vapor fraction that is collected will have a higher concentration of n-hexane, as it has a lower boiling point than n-octane.

The remaining liquid will have a higher concentration of n-octane. The remaining liquid will have an average composition of around 20-30% n-hexane and 70-80% n-octane.

It is important to note that this separation is not perfect, and some amount of the other compound will still be present in both the vapor and the remaining liquid. The efficiency of the separation can be improved by repeating the process multiple times.

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The addition of solute molecules can lead to a decrease in the rates of both evaporation and condensation.

How does the solute affect the rate of various reactions?

1. Evaporation: The addition of solute molecules to a solvent reduces the rate of evaporation. This is because solute molecules occupy spaces at the liquid surface and decrease the surface area available for solvent molecules to escape into the vapor phase. Additionally, solute-solvent interactions can lower the kinetic energy of the solvent molecules, making it harder for them to overcome the attractive forces and evaporate.

2. Condensation: The presence of solute molecules can also affect the rate of condensation. The solute molecules in the solution occupy space, reducing the amount of space available for vapor molecules to condense back into the liquid phase. As a result, the condensation rate might be reduced.

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Answer:

False.

Explanation:

The atomic radius tends to decrease as progression is made across a period from left to right.

The more information is needed to determine the pressure of nitrogen in the larger flask after the transfer.

Unfortunately, without additional information such as the volume of each flask, the initial pressure of the nitrogen gas, and the conditions under which the transfer occurs, it is not possible to determine the pressure of nitrogen in the larger flask.

The pressure of a gas is affected by several factors, including the volume of the container, the number of gas molecules present, and the temperature of the gas. Additionally, if the transfer occurs through a valve or some other type of opening, the pressure may change due to the change in volume and the escape of some gas molecules.

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When 52 ml of nitric acid is taken, then the temperature increases by 41.34 C.

The reaction between nitric acid and magnesium is shown as,

Mg + 2HNO₃ → H₂ + Mg(NO₃)₂

Given

Volume of nitric acid = 52 ml

Molarity of nitric acid = 0.75 M

Mass of solid magnesium= 0.849 gm

Enthalpy of the reaction is = -462.0 kJ/mol

First, the moles of nitric acid are calculated as,

Moles = molarity × volume

= 0.75 M × 52 ÷ 1000

= 0.039 mol

Secondly, the moles of magnesium is calculated as,

Moles = mass ÷ molar mass

= 0.849 ÷ 24.30

= 0.0349

In the reaction, nitric acid is the limiting reagent that affects and controls the formation of magnesium ions.

2 mol HNO₃ → 1 mol Mg ions

1 mol of HNO₃ = 0.5 mol Mg ions

So, 0.039 mol of HNO₃ will result in 0.0195 moles of Mg ions.

It is known that 1 mol of magnesium ion releases 462.0 kJ/mol.

Therefore, the heat is calculated as:

= 462 × 10³ J/mol × 0.0195 mol

= 9009 J

Lastly, the increase in the temperature is given as:

q = mcΔT

9009 J = (52+0.849) × 4.184 J/ g°C × Δ T

9009 = 217.92 × ΔT

ΔT = 9009 ÷ 217.92 °C

= 41.34 °C

Therefore, the temperature increases by 41.34 °C.

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The best description of chromosomes by the end of anaphase II of meiosis is option D) The chromosomes reach opposite sides of the cell, incorporating into two haploid nuclei.

What happens to chromosomes by the end of anaphase II in meiosis?

During anaphase II of meiosis, the sister chromatids of each chromosome are separated and pulled towards opposite poles of the cell. At the end of anaphase II, the chromosomes reach opposite sides of the cell, where they become incorporated into two separate haploid nuclei. The other options mentioned describe the events that occur during earlier stages of meiosis or during mitosis.

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The molar mass of argon (Ar) is 39.95 g/mol or 0.0368 g

First, we need to determine the molar mass of argon, which is found by adding up the atomic masses of its constituent atoms. From the periodic table, we see that the atomic mass of argon is 39.95 g/mol.

Next, we can use the given number of moles of argon (9.2 × 10⁻⁴ mol) and the molar mass of argon to calculate the mass of argon present in one liter of air:

mass of Ar = number of moles of Ar × molar mass of Ar

mass of Ar = 9.2 × 10⁻⁴ mol × 39.95 g/mol

mass of Ar = 0.0368 g

Therefore, the mass of argon in one liter of air is 0.0368 g.

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TSL stands for Test and Set Lock. It is a synchronization primitive used in computer programming to prevent multiple threads from accessing a shared resource simultaneously.

The way TSL works is by first checking the value of a shared variable, and if it is not being used by another thread, the TSL operation sets the value to indicate that it is now in use by the current thread. This atomic operation ensures that no other thread can access the shared resource until the current thread has finished using it and releases the lock.

TSL is considered atomic because it happens without a switch in processes, meaning it is a single, indivisible operation that cannot be interrupted or modified by other threads. This property is essential for ensuring thread safety and preventing race conditions in concurrent programming.

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Raoult's Law is a simple yet fundamental law used in chemistry to calculate the vapor pressure of an ideal solution.

It states that the partial pressure of each component in a solution is proportional to its mole fraction in the solution, and its vapor pressure in the pure state.

This law holds for ideal solutions where the intermolecular forces between the different components of the solution are the same as those within each pure component.

The law is particularly useful in the study of colligative properties, where the change in the vapor pressure of a solution is dependent on the mole fraction of the solute.

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Answer:

394 grams initially in the rock

The activity of a radioactive isotope is measured by the isotope's half-life. Half-life is the time it takes for half of the nuclei in a radioactive sample to undergo radioactive decay. Radioisotopes can have half-lives from fractions of a second, to billions of years.

The formula for exponential decay, studied in mathematics, can be used to  describe the amount of undecayed radioisotopes present after a certain amount of time:

[tex]N_t = N_0\,e^{-\lambda t[/tex]

When one half-life (denoted by [tex]t_\frac{1}{2}[/tex] ) has elapsed, half the radioisotopes would have undergone radioactive decay. Hence after one half-life:

Nt = N₀/2, and we can write this equation as:

[tex]\frac{N_t}{2} = N_0\,e^{-\lambda t_\frac{1}{2}[/tex]. Isolating λ to one side, we are left with the formula for the decay constant:

[tex]\lambda = \frac{ln2}{t_\frac{1}{2} }[/tex]

Therefore, if potassium-40 decays with a half-life of 1.25 billion:

We can calculate decay constant:

[tex]\lambda = \frac{ln2}{t_\frac{1}{2} }\\\lambda = \frac{ln2}{1.25 }\\\\\lambda \approx 0.5545[/tex]

Hence:

[tex]98.4= N_0\,e^{-\frac{ln2}{1.25}\times2.50 }\\N_0 = 393.6\\N_0 \approx 394 \,\,grams[/tex]