A solution of naf is added dropwise to a solution that is 0.0122 m in ba2 . when the concentration of f- exceeds ________ m, will precipitate. neglect volume changes. for , ksp = 1.7 x 10-6. 4. calculate the ph of a solution prepared by dissolving 0.270 mol of weak acid ha (ka = 1.77 x 10-4) and 0.260 mol of its conjugate base in water sufficient to yield 1.00 l of solution.

Fe is the limiting reactant.

The balanced chemical equation between iron and oxygen to produce iron (III) oxide is

4Fe(s) + 302(g) - ---> 2Fe2O3(s)

Mass of Fe = 227.8 g

Moles of Fe = 227.8gFe*Imol Fe/55.85g Fe = 4.079mol Fe

Mass of oxygen = 128 g

Moles of O2 = 128g02 * 1molo/32g02 = 4molO2

Calculating the limiting reactant: The reactant that produces the least amount of product will be the limiting reactant.

Mass of iron (III) oxide produced from Iron = 4.079mol Fe*2molFe2O3/4molFe*159.69g Fe2O3/1 mol Fe2O3 = 325.7gFe2O3

Mass of iron (III) oxide produced from oxygen = 4molO2 *2mol Fe2O3/3molO2*159.69gFe2O3/1mol Fe2O3 = 425.84gFe2O3

Iron (Fe) produces the least amount of the product iron (III) oxide. So, Fe is the limiting reactant.

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The product of reduction of ethyl 4-oxobutanoate with sodium borohydride in ethanol at room temperature for 30 minutes is ethyl 4- hydroxybutanoate .

Sodium borohydride is a relatively selective reducing agent  Ethanolic solutions of Sodium borohydride reduces aldehyde , and ketone , in the presence of acid chloride , ester , epoxide , lactones , acids , nitriles , nitro groups.

The sodium borohydride does not reduce ester group because sodium borohydride is not strong enough and the electrophilicity at carbony carbon of ester is not more as compare toaldehyde , and ketone

The product of reduction of ethyl 4-oxobutanoate with sodium borohydride in ethanol at room temperature for 30 minutes is ethyl 4- hydroxybutanoate .

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Pair of LiI and Ca(NO3)2 is soluble in water.

Here LiI stands for lithium iodide and Ca(NO3)2 stands for Calcium Nitrate.

Water's polarity, which is the unequal sharing of electrons among its atoms, is what allows it to be such a great solvent. Typically, water is held together through this polarity of partial positive and negative charges. Positive is attracted to negative, making a cohesive structure. When polar compounds or ions are added to water, they break up into smaller components, or dissolve, to become part of the solution. The water's partial charges attract different parts of the compound, making them soluble in water.

Polar Molecules

Similar to ions, polar molecules are able to dissolve in water. Like water, polar molecules have partial positive and negative charges. The attraction between opposite partial charges helps polar molecules dissolve in water.

Water doesn't have a full positive or negative charge; instead, it possesses a partial positive and partial negative charge. This is because oxygen is a lot more electronegative than hydrogen. This means that oxygen draws more electrons to it than the hydrogen does. Since more electrons are moving towards the oxygen, it has a slightly higher negative charge than water's hydrogen atoms.

Let's look at another simple example: a carbon-fluorine bond. Since fluorine is more electronegative than carbon, it draws more electrons over to it, making it partially negative.

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The delocalized cloud of π electrons in benzene is formed by the overlap of 6 p-orbitals.

So, option B is correct one.

In the case of hydrocarbons, delocalisation occurs in benzene rings , where a hexagon of six carbon atoms has decalized electrons spread over the whole ring.

All of the carbon atoms in the benzene rings are [tex]sp^{2}[/tex] orbitals around the ring produces a framework of six sigma bonds, while the unhybridized p-orbitals which are perpendicular to this plane over in side-to-side fashion to form three pi-bonds.

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The "break-even" interest rate for year n that equates the return on an n-period zero-coupon bond to that of an n - 1 - period zero-coupon bond rolled over into a one-year bond in year n is defined as the forward rate.

A forward rate is a specified price agreed by all parties involved for the delivery of a good at a specific date in the future. The use of forward rates can be speculative if a buyer believes the future price of a good will be greater than the current forward rate. Alternatively, sellers use forward rates to mitigate the risk that the future price of a good materially decreases.

Regardless of the prevailing spot rate at the time the forward rate meets maturity, the agreed-upon contract is executed at the forward rate. For example, on January 1st, the spot rate of a case of iceberg lettuce is $50. The restaurant and the farmer agree to the delivery of 100 cases of iceberg lettuce on July 1st at a forward rate of $55 per case. On July 1st, even if the price per case has decreased to $45/case or increased to $65/case, the contract will proceed at $55/case.

To extract the forward rate, we need the zero-coupon yield curve.

We are trying to find the future interest rate [tex]{\displaystyle r_{1,2}}{\displaystyle r_{1,2}}[/tex]  for time period  

[tex]{\displaystyle (t_{1},t_{2})}(t_1, t_2), {\displaystyle t_{1}}t_{1}[/tex]  and  [tex]{\displaystyle t_{2}}t_{2}[/tex]  expressed in years, given the rate  [tex]{\displaystyle r_{1}}r_{1}[/tex]  for

time period  [tex]{\displaystyle (0,t_{1})}(0, t_1)[/tex]  and rate [tex]{\displaystyle r_{2}}r_{2}[/tex]  for time period [tex]{\displaystyle (0,t_{2})}(0, t_2)[/tex].  To do this, we use the property that the proceeds from investing at rate  [tex]{\displaystyle r_{1}}r_{1}[/tex] for

time period [tex]{\displaystyle (0,t_{1})}(0, t_1)[/tex]  and then reinvesting those proceeds at rate

[tex]{\displaystyle r_{1,2}}{\displaystyle r_{1,2}}[/tex]  for time period  [tex]{\displaystyle (t_{1},t_{2})}(t_1, t_2)[/tex]  is equal to the proceeds from

investing at rate [tex]{\displaystyle r_{2}}r_{2}[/tex] for time period [tex]{\displaystyle (0,t_{2})}(0, t_2)[/tex].

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The partial pressure of gas C is 0.54atm. The correct option is D.

The pressure that one of the gases in a mixture would exert if it were in the same volume on its own.

The total pressure of A, B and C is 6.11 atm.

The partial pressure of A is 1.68 atm

The partial pressure of B is 3.89 atm.

Total pressure = sum of partial pressures of all the gasses in that mixture

6.11atm = 1.68atm + 3.89atm + Pc

Pc = 6.11atm - (1.68 + 3.89) = 0.54atm

Thus, the partial pressure of C is 0.54atm. The correct option is D.

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Hydroplaning occurs when water between the tires and the road builds until the tires begin to ride on a film of water, resulting in traction loss.

The motor vehicles or aircraft tires passing over wet pavement may lose contact with sufficient speed or water depth for a particular given tread design. The tires lose the braking friction required for stopping and start to hydroplane or aquaplane when they come into contact with water. Wet road surfaces can cause tires to hydroplane . This could result in loss of control and steering ability.

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Answer:

44.1 L

Explanation:

Since volume is being held constant, we can use the following variation of the Ideal Gas Law to find the new pressure.

[tex]\frac{P_1}{T_1N_1}=\frac{P_2}{T_2N_2}[/tex]

In the equation, "P₁", "T₁", and "N₁" represent the initial pressure, temperature, and moles. "P₂", "T₂", and "N₂" represent the final pressure, temperature, and moles. Your answer should have 3 sig figs to match the sig figs of the given values.

P₁ = 234 atm                                 P₂ = 68.0 atm

T₁ = 295 K                                     T₂ = 280 K

N₁ = 144 moles                              N₂ = ? moles

[tex]\frac{P_1}{T_1N_1}=\frac{P_2}{T_2N_2}[/tex]                                              <----- Equation

[tex]\frac{234 atm}{(295 K)(144 moles)}=\frac{68.0 atm}{(280 K)N_2}[/tex]                        <----- Insert values

[tex]\frac{234 atm}{42480}=\frac{68.0 atm}{(280 K)N_2}[/tex]                                      <----- Multiply 295 and 144

[tex]0.00551=\frac{68.0 atm}{(280 K)N_2}[/tex]                                    <----- Simplify left side

[tex]1.54=\frac{68.0 atm}{N_2}[/tex]                                            <----- Multiply both sides by 280

[tex](1.54)N_2={68.0 atm}[/tex]                                   <----- Multiply both sides by N₂

[tex]N_2 = 44.1L[/tex]                                                <----- Divide both sides by 1.54

There will be approximately 49.7 moles of air in the tank at the end of the dive.

The Ideal Gas Law, which asserts the following, can be used to solve this problem:

PV = nRT

Where:

P = pressure

V = volume

n = moles of gas

R = ideal gas constant

T = temperature

Let us first use the given data to obtain the initial amount of moles of air in the tank:

[tex]P_1[/tex]= 234 atm

V = 15.0 L

[tex]T_1[/tex] = 295 K, and

R = 0.0821 L/(Kmol) atm.

Using the ideal gas law equation, we can solve for [tex]n_1[/tex]:

[tex]n_1 = (P_1 * V) / (R * T_1)[/tex]

= (234 atm * 15.0 L) / (0.0821 L·atm/(K·mol) * 295 K)

≈ 189.6 moles

Using the new pressure and temperature, we can determine the exact amount of air in the tank:

P2=68.0 atm and T2=280 K

We can determine [tex]n_2[/tex] by applying the same formula:

[tex]n_2 = (P_2 * V) / (R * T_2)[/tex]

= (68.0 atm * 15.0 L) / (0.0821 L·atm/(K·mol) * 280 K)

≈ 49.7 moles

As a result, there will be approximately 49.7 moles of air in the tank at the end of the dive.

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The ka of the acid will be-

HA(aq)+OH−(aq)→A−(aq)+H2O(l)

You have the half-equivalence point when

[HA]=[A−]

it suggests that

log([HA][A−])=log(1)=0

As a result, it can be said that the pH of the solution and the pKa of the weak acid are equivalent at the half-equivalence point.

At the halfway point of equivalence: pH=pKa

The acid dissociation constant of the weak acid, Ka, determines the pKa. pKa=log(Ka), which indicates that Ka=10pKa.

Ka=10pH will be present when the two points are half equal.

Enter your value to determine Ka=105.67=2.1106.

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Answer:

.24 mol / liters

Explanation:

38. 2 ml = .0382 liters

.0382 liters times ( 0. 163 m KOH/1 liter) times ( 1 mol H2SO4/  1 mol koh) =

.006/.025 liters = .24 mol / liters

chegg

The conductivity of the solution in a reaction can be used to measure rates of reactions involving aqueous electrolytes. The correct option is D.

Conductivity is a property that tells if a material can pass heat and light through it. Conductor are hose material that pass the heat and plight through it. Those material that do not transfer the heat are called non-conductor.

The conductivity of a solution in a reaction. To measure the change in the rates of a solution that involve the aqueous electrolyte, the conductivity of the solution is measured.

Thus, the correct option is D. aqueous electrolytes.

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The question is incomplete. Your complete question is given below:

These are the options:

Gaseous substances

brightly colored substances

organic compounds

aqueous electrolytes

The pH at 25°c of a 0. 11 m solution of a weak acid that has ka = 9. 2 × 10−6 is 3.

The base dissociation constant is termed as Kb. Throughout a base split into ts constituent ions in water is determined by its base dissociation constant.

Kb = [H+] [A-]/[HA]

Now, let the concentration of [H+] = [A-] = x

Given,

Ka = 9.2 × 10−6.

Firstly we will calculate the value of the concentration of [H+]

pKa = x^2/(0.11-x)

9.2 × 10−6 = x^2/(0.11-x)

x^2 = 1.012 × 10−6

x = 1.002 × 10^−3.

The concentration of [H+] = [A-] = 1.002 × 10^−3.

Now, we will find pH as

pH = -log[H+]

pH = -log(1.002 × 10^−3)

pH = 3

Thus we calculated that the pH of the solution is 3.

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The final answer is 29.8 mL

Molar mass HCl = 36.5 g/mol

Mol HCl in 1.85 g = 1.85 g / 36.5 g/mol = 0.05068 mol

The molarity of the HCl solution is 0.0507 mol /L

Mol HCl in 50 mL = 50 mL / 1000 mL/L * 0.0507 mol /L = 0.002535 mol

This will require 0.002535 mol NaOH

1000 mL contains 0.0850 mol

The volume that contains 0.002535 mol

Volume = 0.002535 mol / 0.0850 mol /L *1000 mL/L = 29.8 mL

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The volume of 0.0850 M NaOH is required to titrate 25.0 ml of 0.0720 M is 21.176 ml.

Normality is defined as the number of equivalents per litre of the solution

It is given by

Normality = number of equivalents / 1 L of solution

Molarity is defined as the number of moles of solute present in one litre of solution

Molarity = moles of solute / 1 litre of solution

Relationship between molarity and normality

Normality = Molarity × Acidity or Basicity of a salt

Here, the Acidity Of NaOH is 1 and the basicity of HBr is also 1.

Thus, Normality = Molarity

We know, N₁V₁ = N₂V₂

We can also write it as M₁V₁ = M₂V₂

V1 = [tex]\frac{M_2V_2}{M_!}[/tex]

Substitute M₁ = 0.0850 M, M₂ = 0.0720 M, V₂= 25 ml

V1 = 21.176 ml

Hence, The volume of 0. 0850 m NaOH are required to titrate 25. 0 ml of 0. 0720 m is 21.176 ml.

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0.42 mol 7.3 L volume of O2 is produced when 28. 5 g of hydrogen peroxide (H2O2) decomposes to form water and oxygen at 150°C and 2. 0 atm

A moderate antiseptic used topically to treat minor burns, scrapes, and cuts is hydrogen peroxide. As a mouth rinse, it can be used to assist clear mucus or soothe minor oral irritation (caused, for example, by canker/cold sores, or gingivitis).

Stronger than chlorine, chlorine dioxide, and potassium permanganate, hydrogen peroxide is a potent oxidant.

Human blood is exposed to a hydrogen peroxide solution, which causes the peroxide to break down and release oxygen. Before transfusion, this process has been utilized to oxygenate banked blood.

When hydrogen peroxide comes into contact with skin, a chemical process takes place that causes hydrogen peroxide to turn white. When the solution oxidizes, oxygen bubbles that arise give the mixture a white, foamy appearance.

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Sucrose, a disaccharide (consisting of two monosaccharides), is table sugar.

A digestive enzyme called sucrase catalyses the breakdown of sucrose into its component sugars, fructose and glucose. One type, sucrase-isomaltase, is secreted at the brush boundary of the small intestine. Invertase, a form of the sucrase enzyme that is more frequently found in plants, also hydrolyzes sucrose but does it in a different way.

Maltose is converted to glucose by maltase. Succrase and lactase, respectively, break down other disaccharides like sucrose and lactose. Sucrase and lactase are enzymes that break down sucrose (also known as "table sugar") and lactose (sometimes known as "milk sugar"), respectively, into glucose and fructose.

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Alkali metals nitrates on heating give the corresponding nitrogen dioxide and oxygen.

These are the metals which forms alkali when dissolved in water. They are any of the soft, light, reactive metals of Group 1 of the periodic table.

Some examples of alkali metals are as follows:

A periodic table is a tabular chart of the chemical elements according to their atomic numbers so that elements with similar properties are in the same column.

So therefore, alkali metals nitrates on heating give the corresponding nitrogen dioxide and oxygen.

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A monosaccharide that consists of 5 carbon atoms, one of which is in a ketone group, is classified as a(n) ketopentose.

Ketone family

The suffix "-one" is added after locating the carbonyl group and, if necessary, designating it with a location number. Ketones are given their common names by first identifying the alkyl groups linked to the carbonyl (in alphabetical order), followed by the prefix "ketone." Several Regular Ketones. The three most significant ketones in terms of scale are acetone, methyl ethyl ketone, and cyclohexanone. Though less frequently than in general organic chemistry, they are still widespread in biochemistry. The simplest ketone is dimethyl ketone, or CH3COCH3, often known as acetone.

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The molarity of the diluted solution is 4.16 M

Given:

volume of water = 45ml

volume of k2so4 = 250ml

Molarity of k2so4 = 0.75 M

To Find:

molarity of the diluted solution

Solution: Molarity (M) is the amount of a substance in a certain volume of solution. Molarity is defined as the moles of a solute per liters of a solution. Molarity is also known as the molar concentration of a solution

M1V1 = M2V2

45*M1 = 250*0.75

M1 = 250*0.75/45

M1 = 4.16 M

So, Molarity of given solution is 4.16 M

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The primary sources of acidity in acid rain are the gases SO[tex]_{2}[/tex] and NO[tex]_{x}[/tex] So, the correct option is (c) SO[tex]_{2}[/tex] and NO[tex]_{x}[/tex]

A chemical process that starts when gases like sulphur dioxide and nitrogen oxides are discharged into the atmosphere leads to acid rain. These substances have the ability to ascend very far into the sky, where they combine and react with oxygen, water, and other chemicals to produce acid rain, an additional form of acidic pollution.

Acidic particles corrode metal and hasten the deterioration of paint and stone. Power plants are the primary sources of SO[tex]_{2}[/tex] . Burning fossil fuels produces nitrogen dioxide (like gasoline). Cars, lorries, and power plants are the biggest producers of NO[tex]_{2}[/tex].

Therefore, The primary sources of the acidity in acid rain are the gases SO[tex]_{2}[/tex] and NO[tex]_{x}[/tex].So, the correct option is (c) SO[tex]_{2}[/tex] and NO[tex]_{x}[/tex]

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The mass, in grams, of the chlorine gas will be 0.84 grams

At standard temperature and pressure, 22.4 L of gas has 1 mole of the gas in it.

This time, what we have is 265 mL of the gas at STP. 265 ml is equivalent to: 265/1000 = 0.265 L

If.

22.4 L = 1 mole

Then,

0.265 L = 0.265 x 1/22.4 = 0.01183 moles

This means that 0.01183 moles of chlorine is present in the gas at STP.

Mass of 0.01183 moles of chlorine = moles x molar mass.

The molar mass of chlorine gas is 71

Mass of the chlorine gas = 0.01183 x 71 = 0.84 grams

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A membrane is impermeable to charged molecules and does not allow the passage of ions because the charged molecules possess polarity and the membrane possesses hydrophobic interiors.

Charged hydrophilic molecules of all sizes, from small molecules to giant enzymes, can now be made more soluble via a technique called hydrophobic ion pairing. Hydrophobic molecules having hydrophilic moieties are ionically coupled with charged hydrophilic molecules, and the resulting uncharged complex is water-insoluble and will precipitate in aqueous conditions.

With polar solvents like water or alcohol, hydrophilic compounds can easily establish hydrogen bonds. Ionic (charged) groups with oxygen or nitrogen atoms make up the chemical structure of hydrophilic compounds. Typically, a substance's hydrophilicity is determined by its polarity. A molecule will be hydrophobic overall even if it has polar covalent bonds and these bonds are distributed symmetrically.

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The δs∘rxn  for the reaction [tex]2NO(g) + O_{2}[/tex] → [tex]2NO_{2} (g)[/tex] will be -146 J/K.

Entropy would be a measurable physical characteristic and a scientific notion that is frequently connected to a condition of disorder, unpredictability, or uncertainty.

Entropy would be a measurement of the system's unpredictability or disorder. The entropy increases as randomness do. It has broad properties as well as a state function. It has the unit [tex]JK^{-1} mol^{-1}[/tex].

Entropy of the reaction can be calculated by the reaction.

Δ[tex]S^{0} rxn[/tex] = 2 mol × [tex]S^{0} (NO_{2} (g) - 2 mol[/tex] × [tex]S^{0} NO (g)[/tex] - 1 mol × [tex]S^{0} (O_{2} )[/tex]

Δ[tex]S^{0} rxn[/tex]  = 2 mol × 240 J/mol.K - 2 mol × 210 J/mol.K-1 mol  ×205.2 J/mol.K

Δ[tex]S^{0} rxn[/tex]  = -146.8 J/K

Therefore, the δs∘rxn  for the reaction [tex]2NO(g) + O_{2}[/tex] → [tex]2NO_{2} (g)[/tex] will be -146 J/K.

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The statement that the concentration of carbon will not change if gaseous water is added to the system is false. The concentration of carbon will increase.

Concentration of a substance is the present of a substance in a per-unit square of an area. It is the total amount of solute present in the solution.

The equilibrium is C(s) + H₂O(g) < - > CO(g) + H₂(g), ΔH = 2296 J.

If we increase or add the gaseous water in the system, the amount of carbon will increase because the carbon is a gas, and increasing the pressure will increase the volume of the gas, at Constant temperature.

Thus, the statement is false.

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Explanation:

According to orbital hybridization theory, in ethene the C atoms are using sp2 hybrid orbitals. This leaves one unhybridized p orbital available to form the second bond (Pi bond) in the C-C double bonds.

One sp2 hybrid orbital from each C atom overlap to form one sigma bond between the two carbon atoms. One p orbital from each C atom then overlap to form a pi bond that completes the C-C double bond. The H atoms are bonded via sigma bonds to the C atoms when the H s orbitals overlap with the remaining sp2 hybrid orbitals from each C.

Answer:

A solution with a ph of 6 has 100 times as many hydrogen ions as a solution with a ph of 8.

Explanation:

At 7 on the pH scale, a solution is neutral. Any pH lower than 7 is acidic and any pH above 7 is basic.

Acidity is a measure of the concentration of hydrogen ions in a solution. The more acidic a solution, the more hydrogen ions it has. Therefore, if the pH is decreasing, the hydrogen ion concentration is increasing.

The pH scale increases by a factor of 10. This means that for every one unit change, the hydrogen ion concentration changes tenfold. Therefore, if the pH is decreasing by 2 units (from pH 8 to pH 6), the hydrogen ion concentration should increase 100-fold (10 x 10 = 100).

There are 0. 650 moles of sodium bromide can be produced from the reaction of 1. 03 moles of bromine gas with 0. 650 moles of sodium .

Calculation ,

To find the number of moles of sodium bromide we have to write balanced chemical equation first .

[tex]2Na (s)+ Br_{2}(g)[/tex] → 2NaBr (s)

from the equation we can conclude that 2 moles sodium react with one mole of bromine to form 2 mole of sodium bromide .

But  1. 03 moles of bromine gas ( that is equal to the number of moles required in the equation) with 0. 650 moles of sodium ( that is half of required mole ) is given .

So, sodium is present as limiting reactant .

So , the number of sodium bromide formed is equal to the number of moles of sodium atom.

Hence ,  the number of sodium bromide formed =  0. 650 moles

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The pH of the solution is the negative logarithm of a proton or the hydrogen ion concentration. The pH of 0.39 M acetic acid solution (CH₃COONa) is 2.58.

The pH has been said to be the hydrogen ion concentration that can also be given by the pOH.

Given,

The acid dissociation constant Ka = 1.8 × 10⁻⁵

Concentration of acetic acid (C) = 0.39 M

The hydrogen ion concentration from Ka and molar concentration are calculated as:

H⁺ = √ Ka × C

= √1.8 × 10⁻⁵ × 0.39

= √0.00000702

= 0.0026

Now, pH from hydrogen ion is calculated as,

pH = - log [H⁺]

= - log [0.0026]

= 2.58

Therefore, the pH of acetic acid is 2.58.

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Old and inefficient mining smokestacks contaminate the soil around abandoned mine sites.

A mine is an area from which natural resources are extracted such as coal, metals, etc. A mine is known to be the source of many heavy metals as well as acid.

As such the statement that does not represent the impact of mining services is that; "Old and inefficient mining smokestacks contaminate the soil around abandoned mine sites".

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Answer:

The reactivity series allows us to predict correctly how metals will react. Another importance is that so we can use displacement. We can spot a more reactive metal allowing us to displace a less reactive metal from a compound or an equation.

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Importance of Metal Reactivity Series are:

What are metal reactivity series?

Metal reactivity series is a chart of listing the metals in the order of decreasing reactivity.

See the below reaction:

K(s)        +            [tex]H_2O[/tex]   →           KOH(aq)              +                  1/2 [tex]H_2[/tex] (g)

Potassium      water              Potassium Hydroxide                Hydrogen

Lets see the reaction between Lead with Hydrochloric acid and Sulphuric acid.

Pb (s)      + 2HCl (aq)           →              [tex]PbCl_2[/tex] (aq)           +       [tex]H_2[/tex] (g)

Lead        Hydrochloric acid          Lead chloride                  Hydrogen

Pb (s)      +   [tex]2H_2SO_4[/tex]             →     [tex]PbSO_4[/tex]              +   [tex]SO_2[/tex]                   +   [tex]H_2O[/tex]

Lead         Sulphuric acid            Lead Sulphate      Sulphur dioxide    water

To learn more about Metal reactivity series:

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Answer:

instantaneous rate would be the term.