the relative formula masses (Mr) are: CaCo3 = 100; CaO =56 ; Co2=44

describe how this experiment could be used to provide evidence for the law of conservation of mass.

[6 marks]
include your answer:

-method

-which measurements should eb taken

-how the student could show evidence for the conservation for mass

The crystal I chose is sodium chloride crystals and it is beneficial for man as it is used in the preservation of food as well as in seasoning of food.

A solid whose components are arranged in a highly ordered microscopic structure to form an all-pervasive crystal lattice is referred to as a crystal.

Sodium chloride also referred to as common salt is an ionic compound that has the chemical formula NaCl.

Sodium chloride is an essential nutrient employed in healthcare. It is employed as a spice to improve flavor and as a food preservative. Additionally, sodium chloride is employed in the production of plastics and other goods and is applied to de-ice sidewalks and roadways

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To decrease 0.10 L of a 0.60 M KIO₃ solution, 670 mL of 0.45 M Na₂CrO₂ solution are required.

The balanced chemical equation for the reaction is:

6 IO₃⁻ + 3 CrO₂²⁻ + 24 OH⁻ → 3 CrO₄²⁻ + 6 I⁻ + 12 H₂O

First, we need to determine the limiting reactant between KIO₃ and Na₂CrO₂. To do this, we can use the stoichiometry of the balanced equation to convert the number of moles of each reactant to the number of moles of CrO₂²⁻ required:

0.60 mol KIO₃ x (3 mol CrO₂²⁻ / 6 mol IO₃⁻) = 0.30 mol CrO₂²⁻

We can also calculate the number of moles of CrO₂²⁻ available in the Na₂CrO₂ solution using its concentration and volume:

0.45 mol/L x V(L) = 0.30 mol CrO₂²⁻

Solving for V, we get:

V = 0.30 mol / 0.45 mol/L = 0.67 L = 670 mL

Therefore, 670 mL of the 0.45 M Na₂CrO₂ solution is needed to reduce 0.10 L of a 0.60 M KIO₃ solution.

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Heat is most accurately described as "the energy transferred between objects at different temperatures" (C). Until they reach thermal equilibrium, or the same temperature, heat is a type of energy that flows freely from a hotter to a colder item.

Heat can be transferred through conduction, convection, or radiation. The temperature differential between the items and the thermal conductivity of the materials involved determine how much heat is transported.

Temperature,  a measurement of the average kinetic energy of the particles in an item, is not the same as heat. Internal energy is the entire amount of energy held by an object's particles, which includes both their kinetic and potential energies.

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The final temperature must be approximately 248 K for the pressure to remain constant. B is correct option.

We can use the combined gas law to solve this problem, which states that: (P1V1)/T1 = (P2V2)/T2

where P1 and V1 are the initial pressure and volume, respectively, T1 is the initial temperature, P2 and V2 are the final pressure and volume, respectively, and T2 is the final temperature.

We know that the initial volume V1 is 0.250 L and the final volume V2 is 0.285 L. The pressure P is constant, so we can set P1 = P2. The initial temperature T1 is 10°C, which is equivalent to 283 K (10°C + 273 = 283 K). Substituting these values into the combined gas law and solving for T2, we get: (P1V1)/T1 = (P2V2)/T2

(P1V1) = (P2V2)(T1/T2)

T2 = (P2V2)(T1)/(P1V1)

T2 = (P1V1)(T2)/(P2V2)

T2 = (283 K × 0.250 L)/(0.285 L)

T2 = 248 K.

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0.361 moles of N₂ were required to produce 12.3 g of NH₃, using the balanced chemical equation N₂ + 3H₂ → 2NH₃.

The balanced chemical equation for the reaction is N₂ + 3H₂ → 2NH₃. We can use the balanced equation and the molar mass of NH₃ to calculate the number of moles of N₂ required to produce 12.3 g of NH₃,

1 mol NH₃ = 2 mol N₂ (from the balanced equation)

molar mass of NH₃ = 17.03 g/mol

moles of NH₃ = 12.3 g / 17.03 g/mol

moles of NH₃ = 0.722 mol

moles of N₂ = (0.722 mol NH₃) / 2

moles of N₂ = 0.361 mol

Therefore, 0.361 moles of N₂ were needed to produce 12.3 grams of NH₃.

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Complete question - For the reaction, N₂ + 3H₂ → 2NH₃. Given 12.3 grams of NH3, how many moles of N₂ were needed?​

Answer: copper hydroxide

Explanation:

Copper(II) hydroxide is the hydroxide of copper with the chemical formula of Cu(OH)2. It is a pale greenish-blue or bluish-green solid. Some forms of copper(II) hydroxide are sold as "stabilized" copper(II) hydroxide, although they likely consist of a mixture of copper(II) carbonate and hydroxide.

Example:

2NaOH+CuSO4→Na2SO4+Cu(OH)2

The balanced equation for the reaction between sulfur (S₈) and oxygen (O₂) to form sulfur trioxide (SO₃) is 2S₈ + 16O₂ → 16SO₃.

Balancing chemical equations involves the addition of stoichiometric coefficients to the reactants and products.

The balanced equation for the reaction between sulfur (S₈) and oxygen (O₂) to form sulfur trioxide (SO₃) is determined as;

2S₈ + 16O₂ → 16SO₃

From the reactants side we can see that sulfur is 16 and also 16 in the product side. The number of oxygen in the reactant side is 32 and also 32 in the product side.

Thus, the balanced equation for the reaction between sulfur (S₈) and oxygen (O₂) to form sulfur trioxide (SO₃) is 2S₈ + 16O₂ → 16SO₃.

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The solubility of[tex]LaF_3[/tex] in water is 3.04 x 10^-6 mol/L.

The solubility of [tex]LaF_3[/tex] in water can be determined using the Ksp expression:

[tex]Ksp = [La^{3+}][F^-]^3[/tex]

Where [tex][La^{3+}][/tex]and [tex][F^-][/tex] are the molar concentrations of the [tex]La^{3+}[/tex] and [tex]F^-[/tex] ions in the solution.

Since each [tex]LaF_3[/tex] formula unit dissociates into one [tex]La^{3+}[/tex] ion and three [tex]F^-[/tex] ions, the molar solubility of [tex]LaF_3[/tex] can be represented as x. Thus, the molar concentrations of [tex]La^{3+}[/tex] and [tex]F^-[/tex] ions in the solution can be written as x and 3x, respectively.

Substituting these values into the Ksp expression gives:

Ksp = x*(3x)^3 = 27x^4

Now, we can solve for x:

x = (Ksp/27)^(1/4)

= (2 x 10^-19 / 27)^(1/4)

= 3.04 x 10^-6 mol/L

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The temperature of the oxygen gas sample is 609 K, which is approximately 336°C or 637°F. The answer is A.

We can use the ideal gas law equation, PV = nRT, to solve for the temperature of the oxygen gas sample.

First, we need to calculate the number of moles of oxygen gas present in the sample using its mass and molar mass:

n = m/M

where:

n = number of moles

m = mass (in grams)

M = molar mass (in g/mol)

The molar mass of oxygen gas (O2) is 32.00 g/mol.

n = 48 g / 32.00 g/mol = 1.50 mol

Next, we can rearrange the ideal gas law equation to solve for temperature (T):

T = (PV) / (nR)

where:

T = temperature (in Kelvin)

P = pressure (in atm)

V = volume (in liters)

n = number of moles

R = gas constant (0.0821 L atm/mol K)

Plugging in the given values, we get:

T = (3.0 atm x 25 L) / (1.50 mol x 0.0821 L atm/mol K)

T = 609 K

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The gas would occupy approximately 3.00 L at 24.0 °C and 670 torr.

To solve this problem, we can use the combined gas law, which relates the pressure, volume, and temperature of a gas at different conditions. The combined gas law is expressed as:

(P1 × V1) / (T1) = (P2 × V2) / (T2)

where P1, V1, and T1 are the initial pressure, volume, and temperature, respectively, and P2, V2, and T2 are the final pressure, volume, and temperature, respectively.

Using the given values, we can plug them into the equation and solve for V2:

(P1 × V1) / (T1) = (P2 × V2) / (T2)

(934 torr × 4.76 L) / (279.25 K) = (670 torr × V2) / (297.15 K)

Simplifying and solving for V2, we get:

V2 = [(934 torr × 4.76 L) / (279.25 K)] × (297.15 K / 670 torr)

V2 ≈ 3.00 L

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4) The volume of the HCl used is 9.500 mL while the volume of the NaOH used is 3.800 mL.

5) Molarity of sodium hydroxide is obtained from; Molarity of HCl * 1/2

By reacting an unknown component with a known quantity of a different chemical known as a titrant, titration is a laboratory procedure used to measure the concentration of an unknown substance, often a solute dissolved in a liquid.

The endpoint of a titration can be detected in a number of ways, depending on the specific titration being performed.

4)

Volume of the Acid used = Initial reading - Final reading = 25.00 - 15.50 = 9.500 mL

Volume of the base used = 8.80 - 5.00 = 3.800 mL

5)

We know that the mole ratio is 1:2 and the implication of this is that the set up to obtain the molarity of the sodium hydroxide solution is Molarity of HCl * 1/2

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Cuso4 + NaoH -》cu(oH)2 +Na2So4

Cuso4 + 2NaoH -》cu(oH)2 +Na2So4

Explanation:

this is balanced equation

The difference between q for the two-step process and q for the one-step process is 220.38 joules.

To solve this problem, we use the first law of thermodynamics, which states that change in internal energy (ΔU) of system will be equal to the heat (q) added or removed from the system, minus the work (w) done by or on the system;

[tex]Δ_{U}[/tex] = q - w

For an ideal gas, the internal energy depends only on the temperature, so [tex]Δ_{U}[/tex] is zero if the final temperature is the same for both processes. Therefore, we can set [tex]Δ_{U}[/tex] to zero and solve for the difference in heat (q) between the two processes;

q(two-step) - q(one-step) = w(two-step) - w(one-step)

The work done by or on the gas can be calculated using the equation;

w = -P[tex]Δ_{V}[/tex]

where P is the external pressure, and [tex]Δ_{U}[/tex] is the change in volume. The negative sign indicates that work is done on the gas when it is compressed ([tex]Δ_{U}[/tex] < 0), and work is done by the gas when it expands ([tex]Δ_{U}[/tex] > 0).

For the two-step process, we can calculate the work done in two stages;

w(two-step) = -2.00 atm × (4.40 L - 2.20 L) - 2.50 atm × (2.20 L - 1.76 L)

= -3.32 atm L - 0.605 atm L

= -3.925 atm L

For the one-step process, we can calculate the work done in one step;

w(one-step) = -2.50 atm × (4.40 L - 1.76 L)

= -6.10 atm L

Substituting these values into the equation for the difference in heat, we get;

q(two-step) - q(one-step) = -3.925 atm L - (-6.10 atm L)

= 2.175 atm L

To convert this to joules, we need to multiply by the conversion factor for atm L to joules;

1 atm L = 101.3 J

Therefore; q(two-step) - q(one-step) = 2.175 atm L × 101.3 J/atm L

= 220.38 J

Therefore, the difference in heat between the two processes is 220.38 joules.

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Answer:

5l NO

2

at STP

No. of molecules=

22.4

5

mol=

22.4

5

×N

A

molecules

A) 5ℊ of H

2

(g)

No. of moles=

2

5

mol=

2

5

×N

A

molecules

B) 5l of CH

4

(g)

No. of moles of CH

4

=

22.4

5

mol=

22.4

5

N

A

molecules

C) 5 mol of O

2

=5N

A

O

2

molecules

D) 5×10

23

molecules of CO

2

(g)

Molecules of 5l NO

2

(g) at STP=5l of CH

4

(g) molecules at STP

Therefore, option B is correct.

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The number that is the same as the exponentiation given as follows: 2.5 × 10-³ is 0.0025.

Exponentiation is the process of calculating a power by multiplying together a number of equal factors, where the exponent specifies the number of factors to multiply.

For example, if 10 is multiplied three times, then it can be written as "10 raised to 3" which means 10³. In this case, 10 is the base, and 3 is the exponent.

Therefore, a number 0.0025 can be written in exponentiation as 2.5 × 10-³ by counting the number of zeros forward.

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Mescaline produces a wide range of psychoactive effects when ingested, including altered perception of reality, hallucinations, and euphoria. It is a powerful psychedelic drug that has been used for centuries by Native American tribes in spiritual ceremonies

Mescaline is a hallucinogenic alkaloid that is derived from the Peyote cactus. Mescaline is a complex organic molecule that can be synthesized in the laboratory from 3,4,5-trimethoxybenzyl bromide in two steps.The first step involves nucleophilic substitution using sodium cyanide, and the second step is a reduction using lithium aluminum hydride (LAH).Here's how mescaline can be synthesized from 3,4,5-trimethoxybenzyl bromide:Step 1: Nucleophilic substitution using sodium cyanideThe reaction of 3,4,5-trimethoxybenzyl bromide with sodium cyanide results in the formation of the nitrile derivative. NaCN serves as the nucleophile in this reaction, and it replaces the bromide ion.The mechanism for this reaction involves the following steps: A nucleophilic attack by the cyanide ion on the benzyl bromide. The carbon-bromine bond breaks, and the benzyl cation is formed. A second nucleophilic attack by the cyanide ion occurs on the benzyl cation, resulting in the formation of the nitrile derivative.Here's the reaction equation for this step:Step 2: Reduction using lithium aluminum hydrideThe next step is the reduction of the nitrile derivative using LAH. LAH serves as a strong reducing agent in this reaction and reduces the nitrile derivative to the amine. The mechanism for this reaction involves the following steps: A nucleophilic attack by LAH on the nitrile derivative. This results in the formation of an imine intermediate. The imine intermediate reacts with another LAH molecule, resulting in the formation of the amine.Here's the reaction equation for this step:Mescaline structure: Mescaline is a psychoactive compound that belongs to the phenethylamine class of alkaloids. The structure of mescaline is as follows: The molecule has three methoxy groups attached to the benzene ring, and it has an amine functional group. The molecule is a white crystalline powder that is soluble in water and alcohol.

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After the minimum point, the conductance typically increases again as the concentration of ions in the solution increases.

Without additional context, it is difficult to determine what exactly is being referred to by "this minimum point." However, in general, the conductance of an electrolyte solution depends on the concentration of ions in the solution. Before the minimum point, the conductance typically increases as the concentration of ions in the solution increases. This is because more ions are available to carry the electric current through the solution, leading to a higher conductance. At the minimum point, the conductance reaches a minimum value due to a balance between the effect of ion concentration and the effect of ion mobility. At this point, the concentration of ions in the solution is not high enough to support a high conductance, while the mobility of the ions is not low enough to reduce the conductance significantly. The ion concentration rises to a level where it can support a larger conductance than at the minimum point, which causes this increase in conductance to occur. The addition or removal of a solute, the reaction between various species in the solution, or any number of other events can cause the concentration of ions in the solution to change throughout the process of changing ion concentrations. The amount of accessible ions in the solution is affected by the change in concentration, and this in turn influences the conductance of the solution. In conclusion, the concentration of ions in an electrolyte solution affects the conductance of the solution. The conductance rises with increasing ion concentration, falls to a minimal value at a specific point, and then rises once more with increasing ion  concentration continues to increase. The ion concentration in the solution can change due to various factors, which can affect the conductance of the solution.

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In the following experiment, a coffee-cup calorimeter containing 100 mL of [tex]H_{ 2} O[/tex] is used. The initial temperature of the calorimeter is 23.0 ∘C. If 6.60 g of [tex]CaCl_{2}[/tex] is added to the calorimeter, Final temperature of the solution in the calorimeter = 11.

The first step in solving this problem is to calculate the number of moles of [tex]CaCl_{2}\\[/tex] added to the calorimeter.

Moles of [tex]CaCl_{2}[/tex] = mass of [tex]CaCl_{2}[/tex] / molar mass of [tex]CaCl_{2}[/tex]

Moles of[tex]CaCl_{2}[/tex] = 6.60 g / 110.98 g/mol (molar mass of [tex]CaCl_{2}[/tex]

Moles of[tex]CaCl_{2}[/tex] = 0.0594 mol

We can use the equation for heat transfer to find the change in temperature of the solution. q = mCsΔT, where q is the heat transferred, m is the mass of the solution, Cs is the specific heat of the solution, and ΔT is the change in temperature.

We know that the initial temperature of the calorimeter is 23.0 ∘C and the mass of the solution is 100 g (since the density of water is 1 g/mL). We can solve for ΔT: ΔT = q / mCs

To find q, we can use the enthalpy change of solution (ΔHsoln) and the number of moles of[tex]CaCl_{2}[/tex]added: q = ΔHsoln x moles of[tex]CaCl_{2}[/tex]

q = -82.8 kJ/mol x 0.0594 mol

q = -4.92 kJ

Now we can solve for ΔT: ΔT = (-4.92 kJ) / (100 g x 4.184 J/g⋅∘C)

ΔT = -11.8 ∘C

We can find the final temperature of the solution by adding the change in temperature to the initial temperature: Final temperature = 23.0 ∘C - 11.8 ∘C =11 ∘C.

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The energy associated with the formation of 2.55 g of 4He by the fusion of 3H and 1H is -2.982 x 10⁻¹⁰ J.

The given masses of the isotopes can be converted to kilograms using the conversion factor: 1 u = 1.661 x 10⁻²⁷ kg.

Mass of 4He = 2.55 g = 2.55 x 10⁻³ kg

Mass of 3H = 3.01605 u = 3.01605 x 1.661 x 10⁻²⁷ kg/u

= 5.0099 x 10⁻²⁷ kg

Mass of 1H = 1.00783 u = 1.00783 x 1.661 x 10⁻²⁷ kg/u

= 1.6737 x 10⁻²⁷ kg

The balanced equation for the fusion reaction is;

3H + 1H → 4He

The molar mass of 4He is 4.0026 g/mol, which can be converted to kg/mol using the conversion factor: 1 g/mol = 1 x 10⁻³ kg/mol.

Molar mass of 4He = 4.0026 g/mol = 4.0026 x 10⁻³ kg/mol

The number of moles of 4He formed can be calculated from its mass;

n(4He) = m(4He) / M(4He)

= 2.55 x 10⁻³ kg / 4.0026 x 10⁻³ kg/mol

= 0.638 mol

From the balanced equation, 3 moles of H atoms react with 1 mole of He atoms to form 1 mole of He atoms. Therefore, the number of moles of H atoms required for the reaction is;

n(H) = 3/4 x n(4He)

= 3/4 x 0.638 mol

= 0.479 mol

The energy released in the reaction can be calculated using the mass-energy equivalence equation;

E = Δm c²

where Δm is change in mass, c is the speed of light.

The change in mass is;

Δm = [3H + 1H - 4He] = [5.0099 x 10⁻²⁷ kg + 1.6737 x 10⁻²⁷kg - 4.0026 x 10⁻³ kg]

= -3.315 x 10⁻²⁷ kg (negative because mass is lost in the reaction)

The energy released is;

E = (-3.315 x 10⁻²⁷ kg) c²

= (-3.315 x 10⁻²⁷ kg) (2.998 x 10⁸ m/s)²

= -2.982 x 10⁻¹⁰ J

The negative sign indicates that energy is released in the reaction (exothermic reaction).

Therefore, the energy associated is -2.982 x 10⁻¹⁰ J.

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0.238 moles of NaN₃ are produced from 9.98 grams of N₂.

The moles of he mass of NaN₃ produced

The balanced equation for the reaction is:

2 NaN₃ → 2 Na + 3 N₂

The molar ratio between NaN₃ and N₂ is 2:3, which means that for every 2 moles of NaN₃, 3 moles of N₂ are produced.

The mole ratio is used to determine how many moles of NaN₃ are produced from 9.98 grams of N₂.

First, we need to convert the mass of N₂ to moles:

moles of N₂ = mass of N2 / molar mass of N₂

moles of N₂ = 9.98 g / 28.02 g/mol

moles of N₂ = 0.356 mol

moles of NaN₃ = (2/3) * moles of N₂

moles of NaN₃ = (2/3) * 0.356 mol

moles of NaN₃ = 0.238 mol

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A sp² hybrid orbital on C-1 overlaps with a sp hybrid orbital on C-2 to form the sigma bond between C-1 and C-2. This statement regarding the skeletal structure of the organic molecule true. The correct option is option A.

In general, molecules containing carbon (C) are referred to as organic compounds. Carbon atoms serve as the primary structural framework for the enormous diversity of naturally occurring compounds. Organic substances play a critical role in the existence of all life forms on Earth (and perhaps elsewhere in the universe). A sp² hybrid orbital on C-1 overlaps with a sp hybrid orbital on C-2 to form the sigma bond between C-1 and C-2.

Therefore, the correct option is option A.

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The final temperature of the water is 71.99°C.

The final temperature when 625 grams of water at 75.0°C loses 7.96 x 10⁴ J can be found using the specific heat capacity equation:

q = mcΔT

where q is the amount of heat transferred, m is the mass of the substance, c is the specific heat capacity, and ΔT is the change in temperature.

First, we need to determine the specific heat capacity of water, which is 4.18 J/g°C. Then we can rearrange the equation to solve for ΔT:

ΔT = q / (mc)

Substituting the given values, we get:

ΔT = (7.96 x 10⁴ J) / (625 g x 4.18 J/g°C)

ΔT = 3.01°C

Therefore, the final temperature is:

Tfinal = Tinitial - ΔT

Tfinal = 75.0°C - 3.01°C

Tfinal = 71.99°C

As a result, the water's ultimate temperature is 71.99°C.

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Answer:

A

Explanation:

If Hydrogen is H₂  There will be two silver

and is Carbon is C There will only be one gray

and if Oxygen is O₃ There will be three red

The climatic change in Earth's history that has resulted in glaciers is the cold climate.

During the last 2.6 million years, the Earth has experienced a series of ice ages, or periods of colder global climate, which have led to the growth of glaciers in regions with sufficient snowfall.

These colder periods are associated with changes in the Earth's orbit, tilt, and precession, which affect the amount and distribution of solar radiation received by the Earth. These climatic changes have had significant impacts on the Earth's surface and have influenced the evolution of life on our planet.

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Considering the definition of pH, the pH is 6.35 and the solution is acidic.

pH is the Hydrogen Potential and it is a measure of acidity or alkalinity. pH indicates the amount of hydrogen ions present in a solution or substance.

Mathematically, pH is calculated as the negative base 10 logarithm of the activity of hydrogen ions:

pH= - log [H⁺]

The numerical scale that measures the pH of substances includes the numbers from 0 to 14. The pH value 7 corresponds to neutral substances. Acidic substances are those with a pH lower than 7, while basic substances have a pH higher than 7.

In this case, being [H⁺]=4.5×10⁻⁷ M, you can replace this value in the definition of pH:

pH= -log (4.5×10⁻⁷ M)

Solving:

pH= 6.35

Finally, the pH is lower than 7, the solution is acidic.

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To create 100 mL of solution with a concentration of 1.5 M, 6.00 grams of NaOH are required.

The amount of NaOH needed to make 100. mL of solution with a concentration of 1.5 M can be calculated using the formula:

mass = molarity x volume x molar mass

where:

molarity = 1.5 M (given)

volume = 100. mL = 0.1 L (given)

molar mass of NaOH = 40.00 g/mol (from periodic table)

Substituting the values, we get:

mass = 1.5 mol/L x 0.1 L x 40.00 g/mol

mass = 6.00 g

Therefore, 6.00 grams of NaOH are needed to make 100. mL of solution with a concentration of 1.5 M.

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