What is the average experimental volume per mole of carbon dioxide calculated for sodium carbonate and sodium bicarbonate?

Answer:

Disproportionation reactions involve a molecule being both oxidized and reduced simultaneously, while comproportionation reactions involve two different species reacting to form a single product with an intermediate oxidation state.

For example, the reaction of hydrogen peroxide (H2O2) with potassium permanganate (KMnO4) is a disproportionation reaction, which can be represented as:

For example, the reaction of sodium thiosulfate (Na2S2O3) with iodine (I2) is a comproportionation reaction, which can be represented as:

The balanced equation is:

2[tex]Fe^{3+}[/tex]+ 4[tex]H^{+}[/tex] + 2[tex]No_{2} ^{-}[/tex] → 2[tex]Fe^{2+}[/tex] + 2H2O + 2[tex]No_{3} ^{-}[/tex]

What is balanced equation?

To balance this redox reaction, we first need to identify which elements are being oxidized and reduced. In this case, the iron (Fe) is being reduced from a +3 oxidation state to a +2 oxidation state, while the nitrogen (N) in the nitrite ion (NO2-) is being oxidized from a +3 oxidation state to a +5 oxidation state in the nitrate ion (NO3-).

To balance the equation, we can follow these steps:

Write out the unbalanced equation:

[tex]Fe^{3+}[/tex] + [tex]No_{2} ^{-}[/tex] + H₂O → [tex]Fe^{2+}[/tex] +[tex]H^{+}[/tex] + [tex]No_{3} ^{-}[/tex]

Separate the equation into two half-reactions, one for the oxidation and one for the reduction:

Oxidation: [tex]No_{2} ^{-}[/tex] → [tex]No_{3} ^{-}[/tex]

Reduction: [tex]Fe^{3+}[/tex] → [tex]Fe^{2+}[/tex]

Balance the elements in each half-reaction by adding the appropriate number of electrons (e-):

Oxidation: 4[tex]H^{+}[/tex] + [tex]No_{2} ^{-}[/tex] + e- → [tex]No_{3} ^{-}[/tex] + 2H2O

Reduction: Fe³+ [tex]e^{-}[/tex] → [tex]Fe^{2+}[/tex]

Balance the number of electrons transferred in the two half-reactions by multiplying one or both of the half-reactions by a suitable coefficient:

Oxidation: 4[tex]H^{+}[/tex] + [tex]No_{2} ^{-}[/tex] + 2[tex]e^{-}[/tex] → [tex]No_{3} ^{-}[/tex] + 2H2O

Reduction: 2[tex]Fe^{3+}[/tex] + 2[tex]e^{-}[/tex] → 2[tex]Fe^{2+}[/tex]

Add the two half-reactions together and cancel out any common species on both sides of the equation:

2[tex]Fe^{3+}[/tex]+ 4[tex]H^{+}[/tex] + 2[tex]No_{2} ^{-}[/tex] → 2[tex]Fe^{2+}[/tex] + 2H2O + 2[tex]No_{3} ^{-}[/tex]

The balanced equation is:

2[tex]Fe^{3+}[/tex]+ 4[tex]H^{+}[/tex] + 2[tex]No_{2} ^{-}[/tex] → 2[tex]Fe^{2+}[/tex] + 2H2O + 2[tex]No_{3} ^{-}[/tex]

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Answer:

95.5 g/mol

Explanation:

To find the molar mass of the solute, we need to first calculate the number of moles of the solute in the solution.

Number of moles of solute = concentration x volume

We are given the concentration (0.400 m) and the volume (500 grams) of the solution, but we need to convert the mass of water to volume. We can do this using the density of water:

Density of water = 1 g/mL

Volume of water = mass of water / density of water = 500 g / 1 g/mL = 500 mL

Now we can calculate the number of moles of solute:

Number of moles of solute = 0.400 m x 500 mL = 200 mmol

Next, we need to find the mass of the solute:

Mass of solute = number of moles x molar mass

We are given the number of moles (200 mmol), so we can rearrange the equation to solve for the molar mass:

Molar mass = mass of solute / number of moles

Molar mass = 19.10 g / 200 mmol = 95.5 g/mol

Therefore, the molar mass of the solute is 95.5 g/mol.

Hopes this helps

Answer:

9mol N

Explanation:

The balanced chemical equation is:

N₂ + 3H₂ → 2NH₃

According to the stoichiometry of this reaction, 1 mole of N₂ reacts with 3 moles of H₂ to produce 2 moles of NH₃.

So, to produce 2 moles of NH₃, we need 1 mole of N₂.

Therefore, to produce 18 moles of NH₃, we would need:

18 mol NH₃ × (1 mol N₂ / 2 mol NH₃) = 9 mol N₂

Thus, 9 moles of nitrogen are required to produce 18 mol of NH₃.

Answer:

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Hence, four body systems are explained above.

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Answer:

Ethane(C2H6) reacts with oxygen gas to produce carbon dioxide and water.

The final balanced chemical equation is:

C₉H₂₀(l) + 14O₂(g) → 9CO₂(g) + 10H₂O(l)

A balanced chemical equation is a representation of a chemical reaction using chemical formulas and coefficients to show the reactants and products involved in the reaction.

To balance a chemical equation, coefficients are placed in front of the chemical formulas to ensure that the number of atoms of each element is equal on both sides of the equation. This results in a balanced equation that accurately represents the reactants and products involved in the reaction, and their relative amounts.

To balance the chemical equation, we need to make sure that the number of atoms of each element is the same on both sides of the equation.

Let's start by counting the number of atoms of each element in the reactants and products:

Reactants:

C₉H₂₀(l) → 9 carbon atoms (C), 20 hydrogen atoms (H)

O₂(g) → 2 oxygen atoms (O) per molecule, so 14 O₂ molecules give 28 oxygen atoms (O)

Products:

9CO₂(g) → 9 carbon atoms (C), 18 oxygen atoms (O)

10H₂O(l) → 20 hydrogen atoms (H), 10 oxygen atoms (O)

Now, we can see that the number of carbon and hydrogen atoms are already balanced, but we need to balance the oxygen atoms.

To do so, we can see that 14 O₂ molecules give us 28 oxygen atoms, which means we need 28/2 = 14 molecules of CO₂ and 14 molecules of H₂O to balance the equation.

The final balanced equation is:

C₉H₂₀(l) + 14O₂(g) → 9CO₂(g) + 10H₂O(l)

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Answer:

Explanation:

need points :>

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Layer 1 because it is closest to the surface

Answer:

the sodium hydroxide solution was 0.4 M.

Explanation:

To solve this problem, we can use the dilution formula:

M1V1 = M2V2

where M1 is the initial concentration, V1 is the initial volume, M2 is the final concentration, and V2 is the final volume.

We are given that 45.0 mL of a sodium hydroxide solution was diluted and the resulting volume is 180 mL with a concentration of 0.100 M. Let's use M1 for the initial concentration, V1 for the initial volume, M2 for the final concentration, and V2 for the final volume.

M2 = 0.100 M

V1 = 45.0 mL

V2 = 180 mL

We can rearrange the dilution formula to solve for M1:

M1 = (M2V2) / V1

Substituting the given values, we get:

M1 = (0.100 M x 180 mL) / 45.0 mL

M1 = 0.400 M

Therefore, the initial molarity of the sodium hydroxide solution was 0.4 M.

In comparison to the "P" alkalinity, the "T" alkalinity will be higher.

The "P" alkalinity of a sample is determined by the amount of hydrogen ions ([tex]H^+[/tex]) that are neutralized by the anions present. The "T" alkalinity of a sample is determined by the amount of hydroxide ions ([tex]OH^-[/tex]) that are neutralized by the cations present. In a sample where the only anion contributing to alkalinity is hydroxide ion, [tex]OH^-[/tex], the "P" alkalinity will be zero since there are no hydrogen ions ([tex]H^+[/tex]) to be neutralized. However, the "T" alkalinity will be equal to the amount of hydroxide ions ([tex]OH^-[/tex]) present since all of the [tex]OH^-[/tex] ions will be neutralized by the cations present. Therefore, in this case, the "T" alkalinity will be greater than the "P" alkalinity.

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The number of moles of CO contained in the 20.0 L tank at 93 °C and 4.52 atm is 3.01 moles (3rd option)

First, we shall list out the given parameters from the question. Details below:

We can obtain the number of mole in the tank as follow:

PV = nRT

4.52 × 20 = n × 0.0821 × 366

90.4 = n × 30.0486

Divide both sides by 30.0486

n = 90.4 / 30.0486

n = 3.01 moles

Thus, we can conclude that the number of mole of the gas is 3.01 moles (3rd option)

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The numerical value of Kc for this reaction is 0.0804 if a 2.00 mol sample of A is sealed in a 1.00 L flask and allowed to reach equilibrium with B and C. The equilibrium concentration of B is found to be 0.39 M.

The equilibrium constant expression for the reaction A (g) ⇔ B (g) + C (g) is:

Kc = [B] [C] / [A]

where [A], [B], and [C] are the molar concentrations of A, B, and C at equilibrium, respectively.

We are given that the equilibrium concentration of B is 0.39 M. However, we are not given the equilibrium concentration of A or C. To solve for Kc, we need to find the equilibrium concentrations of all three species.

Since the reaction is in a 1.00 L flask and we started with a 2.00 mol sample of A, the initial concentration of A is 2.00 M. At equilibrium, the concentration of A will be equal to (2.00 - [B]) M, and the concentration of C will also be equal to [B] M (because the stoichiometric coefficients for B and C are equal).

Substituting these values into the equilibrium constant expression, we get:

Kc = (0.39 M)^2 / (2.00 M - 0.39 M) = 0.0804

Therefore, the numerical value of Kc for this reaction is 0.0804.

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The percentage of iron in ferric oxide is 70%.

Molecular formula of ferric oxide is [tex]Fe_{2} O_{3}[/tex].

First, we will calculate the total mass of ferric oxide.

Molecular weight of oxygen = 16

Molecular weight of iron = 56

Therefore, total mass of oxygen=3 × 16 =48 g

and total mass of iron=2 × 56 =112 g

Percentage of metal in the chemical = total mass of the metal/ total mass of the compound× 100

Now, total mass of ferric oxide = 48+ 112 =160 g

Percentage of iron in the chemical = total mass of the iron/ total mass of ferric oxide × 100= 112160 × 100=70 %

Therefore, the percentage of iron in ferric oxide is 70%.

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The molarity of the sodium hydroxide solution is 0.09111 M.

What is Molarity?

Molarity is a measure of the concentration of a solution. It is defined as the number of moles of solute dissolved in one liter of solution.

Molarity is an important concept in chemistry because it allows us to quantify the amount of solute in a solution and to make predictions about the behavior of the solution in various chemical reactions.

First, we can calculate the number of moles of potassium hydrogen phthalate using its molecular weight:

moles of KHP = mass / molecular weight

moles of KHP = 0.600 g / 204.22 g/mol

moles of KHP = 0.002938 mol

Since one mole of KHP reacts with one mole of NaOH, we know that there are also 0.002938 moles of NaOH in the titrated solution. We can calculate the molarity of the NaOH solution using the formula:

molarity = moles of solute / volume of solution (in liters)

First, we need to convert the volume of the NaOH solution from milliliters to liters:

volume of NaOH solution = 32.21 mL = 0.03221 L

Now we can calculate the molarity of the NaOH solution:

molarity = 0.002938 mol / 0.03221 L

molarity = 0.09111 M

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To calculate the heat gained by the cold liquid, we need to use the formula: q = m * c * ΔT

where q is the heat gained or lost by the substance, m is the mass of the substance, c is the specific heat capacity of the substance, and ΔT is the change in temperature of the substance.

Let's assume that we have a cold liquid with a mass of 100 grams and an initial temperature of 10°C. We then add 50 grams of a hot solid that has been heated to 80°C. After the solid and liquid are allowed to equilibrate, the final temperature of the mixture is 20°C.

We can first calculate the heat lost by the hot solid using the same formula:

q = m * c * ΔT

q = 50 g * 0.385 J/g°C * (80°C - 20°C)

q = 15400 J

The negative sign indicates that the solid lost heat to the colder liquid.

To calculate the heat gained by the cold liquid, we can use the same formula and the final temperature of the mixture:

q = m * c * ΔT

q = 100 g * 4.184 J/g°C * (20°C - 10°C)

q = 4184 J

Therefore, the heat gained by the cold liquid is 4184 J.

The complete question is :

Use the equation qliquid = m × c × ΔT to calculate the heat gained by the cold liquid. Use the specific heat for the liquid you selected.

Use the equation qwater = m × c × ΔT to calculate the heat lost by the hot water. Show your work using the problem-solving method shown in previous rubrics.

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Answer:

the pressure of the gas is 1.67 atm when the volume increased to 4.5 L.

Explanation:

Assuming the temperature and the amount of gas remain constant (i.e., the process is isobaric):

Using Boyle's law, we can relate the initial pressure (P1) and volume (V1) to the final pressure (P2) and volume (V2):

P1V1 = P2V2

Plugging in the given values:

P1 = 1.50 atm

V1 = 5.00 L

V2 = 4.5 L

Solving for P2:

P2 = (P1V1)/V2 = (1.50 atm x 5.00 L)/4.5 L = 1.67 atm

Fe²⁺ is the reducing agent since in the given redox reaction it loses electrons which are transferred to another species.

A reducing agent, also known as a reductant, is a substance that causes a reduction in another substance by donating electrons or by removing oxygen from the substance being reduced. In other words, it is a substance that facilitates a chemical reaction by undergoing oxidation itself.

A reducing agent is typically an electron donor and is oxidized during the reaction. Examples of reducing agents include metals like zinc, hydrogen gas, and organic compounds like alcohols and sugars.

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For a 1-gram piece of magnesium, the amount of MgO produced by Mg is less than the amount produced by O2. Therefore, Mg is the limiting reactant and O2 is in excess.

For a 1000-gram piece of magnesium, the amount of MgO produced by Mg is still less than the amount produced by O2, even though there is a lot more Mg present. Therefore, Mg is still the limiting reactant and O2 is in excess.

To determine which reactant is limiting and which is in excess, we need to calculate the amount of product that each reactant can produce and compare them.

For a 1-gram piece of magnesium:

Moles of Mg = mass/Molar mass = 1 g/24.31 g/mol = 0.041 moles

Moles of O2 = volume x pressure/RT = (22.4 L at STP x 1 atm)/(0.0821 L·atm/mol·K x 273 K) = 1.0 moles

According to the balanced equation, 2 moles of Mg reacts with 1 mole of O2 to produce 2 moles of MgO

Therefore, the amount of MgO produced by 0.041 moles of Mg would be (2/2) x 0.041 moles = 0.041 moles

The amount of MgO produced by 1 mole of O2 would be (2/1) x 0.041 moles = 0.082 moles

We can see that the amount of MgO produced by Mg is less than the amount produced by O2. Therefore, Mg is the limiting reactant and O2 is in excess.

For a 1000-gram piece of magnesium:

Moles of Mg = mass/Molar mass = 1000 g/24.31 g/mol = 41.1 moles

Moles of O2 = volume x pressure/RT = (22.4 L at STP x 1 atm)/(0.0821 L·atm/mol·K x 273 K) = 1.0 moles

According to the balanced equation, 2 moles of Mg reacts with 1 mole of O2 to produce 2 moles of MgO

Therefore, the amount of MgO produced by 41.1 moles of Mg would be (2/2) x 41.1 moles = 41.1 moles

The amount of MgO produced by 1 mole of O2 would be (2/1) x 41.1 moles = 82.2 moles

We can see that the amount of MgO produced by Mg is still less than the amount produced by O2, even though there is a lot more Mg present. Therefore, Mg is still the limiting reactant and O2 is in excess.

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The probability that two heterozygous individuals will have a child together with the short earlobes phenotype is 15%.

The likelihood of the genotype and phenotype of the progeny can be calculated using a Punnett square. Let's refer to the long earlobe gene as "L" and the short earlobe allele as "l."

The probability of each genotype is:

LL = 1/4

Ll = 1/2

ll = 1/4

Now we need to take into account the 60% penetrance of the short earlobes genotype. This means that even if an individual has the genotype for short earlobes (ll), they may not actually exhibit the phenotype.

So the probability of having the short earlobes phenotype is:

P(short earlobes phenotype) = P(ll) x 0.6

P(ll) = 1/4

Substituting in:

P(short earlobes phenotype) = (1/4) x 0.6

P(short earlobes phenotype) = 0.15 or 15%

The probability that two heterozygous individuals will have a child together with the short earlobes phenotype is 15%.

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Answer:

D

The temperature an object would be if its molecules were completely still. Absolute zero is the lowest possible temperature that can be achieved, at which the particles of matter would have zero thermal energy and all molecular motion would stop.

The percentage yield of the reaction is 99.5%.

Percentage yield is the ratio of the actual yield of a reaction to the theoretical yield, expressed as a percentage. It is a measure of the efficiency of a reaction and indicates how much of the desired product was obtained relative to the amount that could have been produced under ideal conditions.

The yield of a chemical reaction can be affected by a variety of factors, including the purity and amount of the reactants, the conditions under which the reaction occurs (e.g. temperature, pressure, and reaction time), and the efficiency of the reaction process (e.g. whether the product is recovered efficiently).

Other factors that can affect yield include the presence of impurities or side reactions, the stability of the reactants and products, and the skill and experience of the person carrying out the reaction.

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Answer:

a. The chemical equation for the reaction between CaI2 and K2SO4 is:

CaI2 + K2SO4 → 2 KI + CaSO4

The solubility product constant (Ksp) for CaSO4 is 2.4 x 10^-5 at 25°C.

Using the Ksp expression for CaSO4, we can write:

Ksp = [Ca2+][SO42-]

Let x be the concentration of Ca2+ ions in the solution. Then, the concentration of SO42- ions will also be x, since the reaction is 1:1. Substituting into the Ksp expression, we get:

Ksp = x^2

Solving for x, we get:

x = sqrt(Ksp) = sqrt(2.4 x 10^-5) = 0.0049 M

Therefore, the minimum concentration of CaI2 needed to cause precipitation is 0.0049 M.

b. The chemical equation for the reaction between AgNO3 and RbCl is:

AgNO3 + RbCl → AgCl + RbNO3

The solubility product constant (Ksp) for AgCl is 1.8 x 10^-10 at 25°C.

Using the Ksp expression for AgCl, we can write:

Ksp = [Ag+][Cl-]

Let x be the concentration of Ag+ ions in the solution. Then, the concentration of Cl- ions will also be x, since the reaction is 1:1. Substituting into the Ksp expression, we get:

Ksp = x^2

Solving for x, we get:

x = sqrt(Ksp) = sqrt(1.8 x 10^-10) = 1.34 x 10^-5 M

Therefore, the minimum concentration of AgNO3 needed to cause precipitation is 1.34 x 10^-5 M.

Answer:

2.67 x 10^-4 M & 8.85 x 10^-8 M

Explanation:

The minimum concentration of the precipitating agent required to cause precipitation of the cation from the solution can be determined using the solubility product constant (Ksp) of the salt that would be formed. The Ksp is an equilibrium constant that represents the maximum amount of solid that can dissolve in water to form a saturated solution.

For part a, calcium iodide (CaI2) and potassium sulfate (K2SO4) are mixed. The reaction that occurs is:

CaI2(aq) + K2SO4(aq) → CaSO4(s) + 2KI(aq)

The solubility product constant for calcium sulfate (CaSO4) is 2.4 x 10^-5. Let x represent the minimum concentration of K2SO4 required to cause precipitation. The concentration of Ca2+ ions in solution is 9.0 x 10^-2 M. The Ksp expression for CaSO4 is:

Ksp = [Ca2+][SO42-]

Substituting the known values gives:

(2.4 x 10^-5) = (9.0 x 10^-2)(x)

Solving for x gives:

x = (2.4 x 10^-5)/(9.0 x 10^-2)

x = 2.67 x 10^-4 M

So, the minimum concentration of K2SO4 required to cause precipitation of CaSO4 is 2.67 x 10^-4 M.

For part b, silver nitrate (AgNO3) and rubidium chloride (RbCl) are mixed. The reaction that occurs is:

AgNO3(aq) + RbCl(aq) → AgCl(s) + RbNO3(aq)

The solubility product constant for silver chloride (AgCl) is 1.77 x 10^-10. Let y represent the minimum concentration of RbCl required to cause precipitation. The concentration of Ag+ ions in solution is 2.0 x 10^-3 M. The Ksp expression for AgCl is:

Ksp = [Ag+][Cl-]

Substituting the known values gives:

(1.77 x 10^-10) = (2.0 x 10^-3)(y)

Solving for y gives:

y = (1.77 x 10^-10)/(2.0 x 10^-3)

y = 8.85 x 10^-8 M

So, the minimum concentration of RbCl required to cause precipitation of AgCl is 8.85 x 10^-8 M.

Without more information about the specific experiment, it is difficult to provide a definitive answer to this question. However, based on the limited information provided, it seems that the experiment involves placing water samples from Detroit and Flint into separate beakers, stirring them, and then introducing a metal coil into each beaker.

What is Metal?

Metal is a type of element characterized by its shiny appearance, malleability, ductility, and ability to conduct heat and electricity. Metals are typically solid at room temperature (except for mercury, which is a liquid) and have a high melting and boiling point.

If the metal coil is made of a reactive metal such as iron, zinc, or aluminum, and the water contains dissolved oxygen and other ions, then the metal coil may undergo a chemical reaction with the water. This could result in the formation of metal ions and the release of hydrogen gas. The metal ions and other reaction products could then combine with other compounds in the water to form insoluble precipitates, which could appear as a cloudy or turbid solution.

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The buffer capacity for OH- is equal to the concentration of OH- added, which is [tex]1.87 × 10^-9[/tex] M.

The buffer capacity for OH- is given by the Henderson-Hasselbalch equation:

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

where pH is the desired pH change, pKa is the dissociation constant of the acid, [A-] is the concentration of the conjugate base, and [HA] is the concentration of the acid.

Rearranging the equation, we get:

[A-]/[HA] = [tex]10^(pH - pKa)[/tex]

In this case, CH3COOH is the acid, and NaCH3COO is its conjugate base. The dissociation constant of acetic acid (CH₃COOH) is 4.76. We can calculate the concentrations of CH₃COOH and NaCH₃COO as follows:

[CH₃COOH] = 1.0 M

[NaCH₃COO] = 0.575 M

To calculate the buffer capacity for OH-, we first need to find the pH of the buffer before adding OH-. We can use the Henderson-Hasselbalch equation to do this:

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

pH = 4.76 + log(0.575/1.0)

pH = 4.19

Now, if we add OH- to the buffer, it will react with CH₃COOH to form CH₃COO-, according to the following balanced equation:

CH₃COOH + OH- → CH₃COO- + H₂O

The concentration of OH- added to the buffer can be calculated using the pH change:

pH = 4.19 + 4.52 = 8.71

[OH-] = 10^-(pH) = 1.87 × [tex]10^-9[/tex]M

The change in concentration of CH3COOH and CH3COO- due to the addition of OH- can be calculated using stoichiometry:

Δ[CH₃COOH] = -[OH-]

Δ[CH₃COO-] = [OH-]

Therefore, the new concentrations of CH₃COOH and CH₃COO- after adding OH- are:

[CH₃COOH] = 1.0 - [OH-] = 1.0 - 1.87 ×[tex]10^-9[/tex] = 1.0 M

[CH₃COO-] = [NaCH₃COO] + [OH-] = 0.575 + 1.87 × [tex]10^-9[/tex] = 0.575 M

Using the Henderson-Hasselbalch equation, we can calculate the new pH of the buffer:

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

pH = 4.76 + log(0.575/1.0)

pH = 4.19

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Answer:

To lower the impacts of climate change, we can undertake the following activities:

Invest in low-carbon technologies: Utilize renewable energy sources such as solar panels, tidal energy, wind, wave, and geothermal power to reduce greenhouse gas emissions and minimize our reliance on fossil fuels.

Improve transportation: Opt for fuel-efficient cars, electric vehicles, and better public transport systems to reduce CO2 emissions and pollution.

Adopt sustainable diets: Shift towards plant-based diets, reduce meat and dairy consumption, and support sustainable farming practices to help mitigate climate change.

Restore nature: Plant trees, engage in rewilding schemes, and protect forests and oceans to increase their natural ability to absorb carbon dioxide.

Reduce consumption and waste: Minimize consumption of energy-intensive products, reduce plastic use, and adopt energy-efficient practices in homes, such as insulating walls and roofs and using heat pumps.

The molarity of the sodium hydroxide solution is 0.0911 M.

we need to calculate the number of moles of potassium hydrogen phthalate:

moles of KHP = mass / molar mass = 0.600 g / 204.22 g/mol = 0.00294 mol

Since one mole of KHP reacts with one mole of NaOH, we know that there are 0.00294 moles of NaOH used in the titration. We can use this information to calculate the molarity of the NaOH solution:

molarity of NaOH = moles of NaOH / volume of NaOH used in liters

We need to convert the volume of NaOH from milliliters to liters:

volume of NaOH = 32.21 mL = 0.03221 L

Now we can calculate the molarity of NaOH:

molarity of NaOH = 0.00294 mol / 0.03221 L = 0.0911 M

The molarity of the sodium hydroxide solution is 0.0911 M.

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Answer:

the molarity of the HCl is 1.17 M.

Explanation:

The balanced chemical equation for the reaction between HCl and NaOH is:

HCl + NaOH -> NaCl + H2O

From the equation, we can see that one mole of HCl reacts with one mole of NaOH to produce one mole of NaCl and one mole of water.

First, let's calculate the number of moles of NaOH used:

moles of NaOH = Molarity x Volume in liters

moles of NaOH = 0.650 M x 0.0900 L

moles of NaOH = 0.0585 mol

Since the reaction between HCl and NaOH is a 1:1 stoichiometry, the number of moles of HCl is also equal to 0.0585 mol.

Now, we can use the number of moles of HCl and the volume of HCl to calculate the molarity:

Molarity of HCl = moles of HCl / Volume in liters

Molarity of HCl = 0.0585 mol / 0.0500 L

Molarity of HCl = 1.17 M

Therefore, the molarity of the HCl is 1.17 M.