please help, its sorta easy<3

The effect of pressure and temperature on the volume of a gas is included by Boyle's law and Charles's law respectively. Boyle's law states that the pressure of a given quantity of gas changes inversely with its volume at constant temperature.

Charles's law projects that the volume of a given amount of gas is seriously proportional to its temperature on the kelvin scale when the pressure is held constant.

In context to the given question we can evaluate the effect of pressure and temperature on volume using Boyle's law and Charles's law .

Let us consider that contain a gas with an initial volume V1, initial pressure P1 and initial temperature T1.

If we gradually increase the pressure to P2 while keeping the temperature constant at T1, then according to Boyle's law, the new volume V2 can be calculated is

V2 = (P1 x V1) / P2

Now,  we increase the temperature to T2 while keeping the pressure constant at P1, hence  Charles's law, the new volume V3 can be evaluated

V3 = (T2 / T1) x V1

Then, we conclude that both pressure and temperature have an effect on volume. Then, it is imperative to note that the effect of pressure is more significant than that of temperature. This is due to according to Boyle's law, pressure and volume are inversely proportional to each other.

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The balanced chemical equation is 2Pb(NO3)2 + 2KI 2KNO3 + PbI2 for the reaction between lead(II) nitrate and potassium iodide. Since lead(II) nitrate has a coefficient of 2, the equation must be balanced by adding 2 moles of lead(II) nitrate for every mole of potassium iodide.

The atom inventory technique, which counts the number of atoms of each element on both sides of the equation, can be used to ascertain this. For the equation to be balanced, the number of atoms on both sides must be equal.

In this instance, there are 4 atoms total—2 on each side of the nitrate molecule, 2 on each side of the potassium molecule, 1 on each side of the lead atom, and 2 on each side of the iodine molecule.

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[1]pb(no3)2+[1]Na2CrO4

[1]PbCrO4+[2]NaNO3

The number of grams of nitrogen gas that can be made from 13.3 moles of CuO is 124.04 grams.

Stoichiometry is the study and calculation of quantitative (measurable) relationships of the reactants and products in chemical reactions (chemical equations).

According to this question, 3 moles of CuO reacts to produce 1 mole of nitrogen gas. This means that 13.3 moles of CuO will produce 4.43 moles of Nitrogen gas.

4.43 moles of nitrogen gas is equivalent to 124.04 grams of nitrogen.

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Answer:

The mass of sodium sulfate formed by the comolete reaction of 120.0 grams of sodium hydroxide is 142.04 grams.

Explanation:

The balanced chemical equation for the reaction between sodium hydroxide and sulfuric acid is:

2NaOH + H2SO4 -> Na2SO4 + 2H2O

From the equation, we can see that 2 moles of sodium hydroxide react with 1 mole of sulfuric acid to form 1 mole of sodium sulfate. We can use this information, along with the molar masses of the compounds, to calculate the mass of sodium sulfate formed.

First, we need to convert the given mqss of sodium hydroxide to moles. The molar mass of sodium hydroxide is 40.00 g/mol, so:

Moles of NaOH = Mass of NaOH / Molar mass of NaOH

Moles of NaOH = 120.0 g / 40.00 g/mol

Moles of NaOH = 3.00 mol

Next, we can use the mole ratio from the balanced equation to calculate the moles of sodium sulfate formed:

Moles of Na2SO4 = Moles of NaOH / 2

Moles of Na2SO4 = 3.00 mol / 2

Moles of Na2SO4 = 1.50 mol

Finally, we can convert the moles of sodium sulfate to grams using its molqr mass of 142.04 g/mol:

Mass of Na2SO4 = Moles of Na2SO4 x Molar mass of Na2SO4

Mass of Na2SO4 = 1.50 mol x 142.04 g/mol

Mass of Na2SO4 = 213.06 g

Therefore, the mass of sodium sulfate formed by the complete reaction of 120.0 grams of sodium hydroxide is 213.06 grams.

Heat now refers to energy in motion. Prior to the creation of thermodynamic rules, heat was seen as a measure of the caloric, an invisible fluid that is contained in all matter. Here the heat absorbed is 135.3 J.

The amount of heat needed to raise a substance's temperature by one degree is known as its heat capacity. The quantity of heat energy needed to raise the temperature of a unit mass of any substance or matter by one degree Celsius is known as its specific heat capacity.

Mathematically amount of heat is given as:

Q= m c ΔT

c = specific heat capacity of iron =  0.451 J / g°C

q = 100 × 0.451 × 3 = 135.3 J

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NO is the most stable among the given compound. Thus, the most appropriate answer comes out to be option D.

The bond order is the number of bonds between two atoms. The higher the bond order the higher the stability of the compound.

It is calculated by  (number of bonding electrons - Number of the anti-bonding electron) * 0.5  

So, for C[tex]O_2[/tex] the bond order is 2

For [tex]C_2[/tex] ,

The number of electrons in the bonding orbital = 8.

Number of electrons in anti-bonding orbital = 4.

Bond order = (8- 4) * 0.5 = 2.

For [tex]O_2[/tex] ,

The number of electrons in the bonding orbital = 10.

Number of electrons in anti-bonding orbital = 6.

Bond order = (10 - 6) * 0.5  = 2.

For NO,

The number of electrons in the bonding orbital = 10.

Number of electrons in anti-bonding orbital = 5.

Bond order = (10 - 5) * 0.5  = 2.5

Thus, the most stable compound is NO as it has a 2.5 bond order.

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A mixture of ethane and ethene with a volume of 20 l reacts with 8 l of hydrogen. The percentage composition of the ethane and ethene are 71.4% and 26.5% respectively.

In chemistry, the term "percent composition" often refers to the percentage of the compound's entirety that each element makes up. For instance, the percent composition of every component would be as follows if you had an 80.0 g sample of a compound that contained 20.0 g element X & 60.0 g element Y

percentage composition of ethane = (20/20+8)×100=71.4%

percentage composition of ethene = (8/20+8)×100=26.5%

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The concentration of two reactants is decreased by the same amount. The change in the reaction rate depends on the balanced chemical equation. Option D is correct.

The effect of changing the concentration of a reactant on the rate of a reaction depends on the balanced chemical equation of the reaction.

In general, changing the concentration of a reactant will affect the rate of a reaction, but the magnitude and direction of this effect will depend on the stoichiometry of the reaction.

Thus, without knowing the balanced chemical equation and the rate law of the reaction, cannot determine a change in the concentration of the reactants will affect the rate of the reaction.

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The number of grams of sodium hydroxide that would be required to react with 400 grams of sulfuric acid is 326.53 grams.

Stoichiometry is the study and calculation of quantitative (measurable) relationships of the reactants and products in chemical reactions (chemical equations).

According to this question, sodium hydroxide reacts with sulfuric acid as follows:

[tex]2NaOH + H_[2]SO4 = Na_[2]SO_[4] + 2H_[2]O[/tex]

400 grams of sulfuric acid is equivalent to 4.08 moles.

Based on the above equation, 2 moles of [tex]NaOH[/tex] reacts with 1 moles of H_[2]SO4. This means that 4.08 × 2 = 8.16 moles of NaOH.

8.16 moles of [tex]NaOH[/tex] is equivalent to 326.53 grams.

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The volume of a 0.220 M HBr solution required to produce 0.0120 moles of HBr is 54.5 mL.

To calculate the volume of a 0.220 M HBr solution required to produce 0.0120 moles of HBr, we can use the following formula:

Volume (in liters) = moles / concentration

First, we need to convert the moles of HBr to liters:

0.0120 moles / (0.220 mol/L) = 0.0545 L

Next, we can convert liters to milliliters:

0.0545 L x 1000 mL/L = 54.5 mL

Therefore, the volume of a 0.220 M HBr solution required to produce 0.0120 moles of HBr is 54.5 mL.

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The total energy required to heat the water to the final temperature is 290,550.6 J.

The total energy required to heat the water to the final temperature is calculated as follows;

E = Q₁  + Q₂  + Q₃ + Q₄

E = mcΔθ + mf + mcΔθ₂ +  mLv

where;

The heat capacity of the water is calculated as;

E = 93 x 4.2 x (10)  +  334 x 93    +  93 x 4.2 x (121 - 0)  + 2240 x 93

E = 290,550.6 J

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There was no reaction because of the absence of water in the system.

Solid sodium bicarbonate, popularly known as baking soda, and solid acetic acid do react with one another, but water is necessary for the reaction to occur. There is no water present to aid in the reaction because both chemicals are dry solids at room temperature (25 degrees Celsius).

When solid sodium bicarbonate and solid acetic acid are combined, a reaction takes place that results in the production of carbon dioxide gas bubbles. If water were added to the mixture, this reaction would happen.

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The molar mass of the salt is 104 g/mol, so we can convert the molar solubility to grams per liter as follows:

[tex]8.54*10^{-5} mol/L * 104 g/mol = 8.87*10^{-3} g/L[/tex]

To calculate the solubility of the salt A2B, we first need to write the equilibrium equation for the dissolution of the salt:

A2B(s) ⇌ 2A+(aq) + B2-(aq)

The equilibrium constant for this reaction is the solubility product, Ksp, which is given as [tex]7.30*10^{-13}[/tex]

Ksp = [A+]^2[B2-]

Let x be the molar solubility of A2B, then the concentration of A+ and B2- ions will also be x.

Substituting these values in the Ksp expression, we get:

[tex]7.30*10^{-13} = x^2 (x)\\x^3 = 7.30*10^{-13}\\x = (7.30*10^{-13})^{1/3}[/tex]

x =[tex]8.54*10^{-5}[/tex] M

The molar mass of the salt is 104 g/mol, so we can convert the molar solubility to grams per liter as follows:

[tex]8.54*10^{-5} mol/L * 104 g/mol = 8.87*10^{-3} g/L[/tex]

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15.98 kJ of heat energy is required to melt 536.4 g of HBr.

To calculate the heat energy required to melt a given amount of HBr, we need to use the formula:

Q = n * ΔHf

First, we need to calculate the number of moles of HBr:

n = mass / molar mass

where the molar mass of HBr is 80.91 g/mol.

n = 536.4 g / 80.91 g/mol = 6.625 mol

Next, we can calculate the heat energy required to melt this amount of HBr:

Q = n * ΔHf = 6.625 mol * 2.41 kJ/mol = 15.98 kJ

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Approximately 48.64 grams of iron(III) oxide will be produced.

To solve this problem, we need to use the balanced chemical equation for the reaction between iron and water vapor:

[tex]3 Fe + 4 H_2O - Fe_3O_4 + 4 H_2[/tex]

From the equation, we can see that 3 moles of Fe react with 4 moles of [tex]H_2O[/tex] to produce 1 mole of [tex]Fe_3O_4[/tex] and 4 moles of [tex]H_2[/tex].

First, we need to calculate the number of moles of water vapor present:

[tex]n(H_2O) = PV/RT[/tex]

where P is the pressure in atm, V is the volume in L, R is the gas constant (0.08206 L·atm/K·mol), and T is the temperature in Kelvin.

Converting the temperature to Kelvin:

[tex]T = 50.2 + 273.15 = 323.35 K[/tex]

[tex]n(H_2O) = (0.121 atm)(5.50 L)/(0.08206 L·atm/K·mol)(323.35 K) \\= 0.840 mol[/tex]

Since Fe is in excess, we can assume that all the water vapor reacts to form [tex]Fe_3O_4[/tex] and [tex]H_2[/tex]. Therefore, the number of moles of [tex]Fe_3O_4[/tex] produced is equal to the number of moles of [tex]H_2O[/tex] consumed:

[tex]n(Fe_3O_4) = n(H_2O) * (1 mol Fe_3O_4/4 mol H_2O) \\= 0.840 mol * (1/4) \\= 0.210 mol[/tex]

Finally, we can use the molar mass of [tex]Fe_3O_4[/tex] to convert the number of moles to grams:

[tex]m(Fe_3O_4) = n(Fe_3O_4) * M(Fe_3O_4)[/tex]

where [tex]M(Fe_3O_4)[/tex] is the molar mass of [tex]Fe_3O_4[/tex]:

[tex]M(Fe_3O_4) = 3 * M(Fe) + 4 * M(O) = 3 * 55.845 g/mol + 4 * 15.999 g/mol \\= 231.532 g/mol[/tex]

[tex]m(Fe_3O_4) = 0.210 mol * 231.532 g/mol = 48.64 g[/tex]

Therefore, approximately 48.64 grams of iron(III) oxide will be produced.

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The volume of air in the bottle for each trial is as follows:

Trial 1: 33.61 mL, Trial 2: 34.04 mL, Trial 3: 34.09 mL, Trial 4: 35.30 mL, Trial 5: 35.12 mL, Trial 6: 35.15mL

Based on the data provided in parts C and D, we can fill out the first three rows of the table as follows:

Trial Mass of Bottle + Water (g) 70.85, 70.42, 70.37

Mass of Empty Bottle (g) 54.46, 54.46, 54.46

Mass of Water (g)                 16.39, 15.96, 15.91

Volume of Water (mL)           16.39, 15.96, 15.91

To calculate the volume of air in the bottle for each trial, we need to subtract the volume of water in the bottle from the total volume of the bottle. From part B, we know that the total volume of the bottle is 50 mL. Therefore, the volume of air for each trial is:

Trial 1: 50 mL - 16.39 mL = 33.61 mL

Trial 2: 50 mL - 15.96 mL = 34.04 mL

Trial 3: 50 mL - 15.91 mL = 34.09 mL

We can also calculate the volume of air for the remaining trials in the same way:

Trial 4: 50 mL - 14.70 mL = 35.30 mL

Trial 5: 50 mL - 14.88 mL = 35.12 mL

Trial 6: 50 mL - 14.85 mL = 35.15 mL

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The right response is C, which refers to a substance's melting or boiling point at a particular pressure. A phase diagram is a visual depiction of a substance's physical condition under various pressures and temperatures.

The melting or boiling point of a material at a certain pressure is represented by a specific point on the line of a phase diagram. This is so because the temperature at which a material transforms from one phase to another is known as the melting point or boiling point of that substance.

For instance, a material transforms from a solid to a liquid state when its temperature exceeds its melting point. Similar to this, when a substance's temperature hits its boiling point, it transforms from liquid to gaseous.

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The mass of water that can be heated from 15.0° C to 43.1° C by the addition of 1282J is 10.9grams.

The mass of water can be calculated using the following expression;

Q = mc∆T

Where;

According to this question, a sample of water can be heated from 15.0° C to 43.1° C by the addition of 1282J. The mass of the substance can be calculated as follows;

1282 = m × 4.184 × (43.1 - 15)

1282 = 117.57m

m = 10.9grams

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The theoretical yield of NH₃ produced under the given conditions is 11.75 g.

The balanced equation for the reaction between hydrogen (H₂) and nitrogen (N₂) to form ammonia (NH₃) is;

N₂ + 3H₂ → 2NH₃

To determine the theoretical yield of NH₃ produced from the given amounts of H₂ and N₂, we need to calculate the limiting reactant and then use stoichiometry to find the maximum amount of NH₃ which can be produced.

The molar masses of H₂, N₂, as well as NH₃ are;

H₂; 2.02 g/mol

N₂; 28.02 g/mol

NH₃; 17.03 g/mol

The number of the moles of each reactant will be calculated as;

moles of H₂ = mass of H₂ / molar mass of H₂ = 1.40 g / 2.02 g/mol = 0.693 mol

moles of N₂ = mass of N₂ / molar mass of N₂ = 9.66 g / 28.02 g/mol = 0.345 mol

To determine the limiting reactant, we need to compare the mole ratios of H₂ and N₂ in the balanced equation with the actual mole ratios of the reactants. The balanced equation shows that 1 mole of N₂ will reacts with 3 moles of H₂ to produce a 2 moles of NH₃. The actual mole ratio of N₂ to H₂ in the reaction mixture is;

moles of N₂ / moles of H₂ = 0.345 mol / 0.693 mol

= 0.498

This ratio is less than the required ratio of 1/3, which means that N₂ is the limiting reactant. This means that all the N₂ will be consumed in the reaction and the amount of NH₃ produced will depend on the amount of N₂ present.

Using the mole ratio from the balanced equation, we can calculate the theoretical yield of NH₃ that can be produced from the 0.345 mol of N₂;

moles of NH₃ = (0.345 mol N₂) × (2 mol NH3 / 1 mol N₂)

= 0.690 mol NH₃

The mass of this amount of NH₃ can be calculated as;

mass of NH₃ = moles of NH₃ × molar mass of NH₃ = 0.690 mol × 17.03 g/mol = 11.75 g

Therefore, the theoretical yield is 11.75 g.

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125 grams of zinc was reacted, therefore, 125 grams of zinc will produce 42.6 liters of hydrogen gas at STP, and the problem involves stoichiometry, which is the calculation of reactants and products in a chemical reaction.

The balanced chemical equation for the reaction of zinc and hydrochloric acid is:

Zn + 2HCl → ZnCl₂ + H₂

From the equation,

125 g Zn x (1 mole Zn/65.38 g) = 1.91 moles Zn

So, one can expect 1.91 moles of hydrogen gas to be produced.

Now, at STP (standard temperature and pressure), the volume of one mole of gas is 22.4 liters. Therefore, the volume of hydrogen gas produced at STP can be calculated as:

1.91 moles H2 x 22.4 L/mole = 42.6 L H2

Therefore, 125 grams of zinc will produce 42.6 liters of hydrogen gas at STP.

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The total pressure in Container C is [tex]5_{P}[/tex]/([tex]2_{V}[/tex]). If the initial pressure in Container A was doubled and Container B was reduced by one-half.

To solve this problem, we need to use combined gas law, which relates with pressure, volume, and temperature of a gas;

(P₁V₁)/T₁ = (P₂V₂)/T₂

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

Let's assume that the volume and temperature are constant in all three containers. Therefore, we can simplify the equation to;

P₁/P₂ = V₁/V₂

We can use this equation to solve for the final pressure in Container C.

First, let's calculate the new pressures in Containers A and B;

Container A; the initial pressure was doubled, so P₁ = [tex]2_{P}[/tex] and V₁ = V (since the volume is constant). Therefore, P₂ = P₁/(V₁/V₂) = [tex]2_{P}[/tex]/(1/2) = [tex]4_{P}[/tex].

Container B; the initial pressure was reduced by one-half, so P₁ = P/2 and V₁ = V (since the volume is constant). Therefore, P₂ = P₁/(V₁/V₂) = (P/2)/(1/2) = P.

Now that we have the new pressures in Containers A and B, we can use them to find the total pressure in Container C:

Container C; we are mixing equal volumes of gases from Containers A and B, so the total volume is [tex]2_{V}[/tex]. The total pressure is the sum of the partial pressures of the gases in Containers A and B, which are [tex]4_{P}[/tex] and P, respectively. Therefore, the total pressure in Container C is:

[tex]P_{total}[/tex] = ([tex]4_{P}[/tex] + P)/([tex]2_{V}[/tex])

= [tex]5_{P}[/tex]/([tex]2_{V}[/tex])

So, the final pressure in Container C is [tex]5_{P}[/tex]/([tex]2_{V}[/tex]).

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The temperature at which this substance would boil is 125.3°C.

The phase of the substance which is most dense is the solid phase.

The change of phase or change of state of a substance is the process whereby the substance change from one physical state to another on the application or removal of heat.

The processes of change of phase, when heat is added to a substance, include the following:

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Miscible liquids are separated via a kind of distillation called fractional distillation. he mixture is typically divided into component parts after a series of distillations and condensations.

Miscible liquids are separated via a kind of distillation called fractional distillation. The mixture is typically divided into component parts after a series of distillations and condensations. When the combination is heated to a specific temperature where some of the mixture begins to vaporise, the separation takes place.

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Answer:

CH3CH(OH)CN

Explanation:

This product is also known as acrylamide or prop-2-enamide. The reaction is a type of nucleophilic addition, in which the CN^- ion acts as a nucleophile and attacks the carbonyl carbon of CH3CH=O, forming a tetrahedral intermediate. Then, a proton transfer occurs, resulting in the formation of CH3CH(OH)CN and a water molecule. The mechanism of the reaction can be shown as follows:

The volume of 4.2 moles of [tex]CH_{4}[/tex] at 1.18 atm and 50.0°C is approximately 94.1 L.

To find the volume of the given amount of [tex]CH_{4}[/tex] at a specific temperature and pressure, we can use the Ideal Gas Law equation: PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the gas constant, and T is the temperature in Kelvin.

First, we need to convert the temperature from Celsius to Kelvin by adding 273.15. So, T = 50.0 + 273.15 = 323.15 K.

Next, we can substitute the given values into the Ideal Gas Law equation:

(1.18 atm) V = (4.2 mol) (0.0821 L·atm/mol·K) (323.15 K)

Simplifying the equation, we get:

V = (4.2 mol) (0.0821 L·atm/mol·K) (323.15 K) / (1.18 atm)

V ≈ 94.1 L

Therefore, the volume of 4.2 moles of [tex]CH_{4}[/tex] at 1.18 atm and 50.0°C is approximately 94.1 L.

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The [H₃O⁺] for the solution is less than 1 x 10⁻⁷ M, the solution is considered as basic.

A basic solution is one that has a pH greater than 7. It is also sometimes referred to as an alkaline solution. Basic solutions contain a lower concentration of hydrogen ions (H⁺) than hydroxide ions (OH⁻), meaning that they have a higher concentration of OH⁻.

We know that pH + pOH = 14 at 25°C.

Therefore, pH = 14 - pOH = 14 - 2.33

= 11.67.

Since pH = -log[H₃O⁺], we can rearrange to solve for [H₃O⁺]; [H₃O⁺] = [tex]10^{(-pH)}[/tex]

= [tex]10^{(-11.67)}[/tex] = 1.74 x 10⁻¹²) M.

Since the [H₃O⁺] concentration is less than 1 x 10⁻⁷ M, the solution is considered basic.

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An isotope contains 11 protons, 10 electrons, and 12 neutrons. ²³Na⁺ is the identity of the isotope. Therefore, the correct option is option A.

A chemical element's isotope is one of more than one species of atoms that share the same atomic number, place on the periodic table, and almost identical chemical activity, but differ in atomic mass and physical characteristics. There are a number of isotopes for each chemical element. The first step in identifying and labelling an atom is to count the protons within its nucleus. An isotope contains 11 protons, 10 electrons, and 12 neutrons. ²³Na⁺ is the identity of the isotope.

Therefore, the correct option is option A.

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Taking into account the reaction stoichiometry, 103.5 grams of Na⁺ is required to react with 4.5 moles of chloride.

In first place, the balanced reaction is:

Na⁺ + Cl⁻  → NaCl

By reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of moles of each compound participate in the reaction:

The molar mass of the compounds is:

By reaction stoichiometry, the following mass quantities of each compound participate in the reaction:

The following rule of three can be applied: If by reaction stoichiometry 1 mole of Cl⁻ react with 23 grams of Na⁺, 4.5 moles of Cl⁻ react with how many moles of Na⁺?

mass of Na⁺= (4.5 moles of Cl⁻×23 grams of Na⁺)÷ 1 moles of Cl⁻

mass of Na⁺= 103.5 grams

Finally, 103.5 grams of Na⁺ is required.

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