Calculate the pH of the solution that results from each of the following mixtures.

PART A---- 150. 0 mL of 0. 26 M HF with 230. 0 mL of 0. 32 M NaF The Ka of hydrofluoric acid is 6. 8 x 10−4. Express your answer using two decimal places.

PART B---- 170. 0 mL of 0. 11 M C2H5NH2 with 270. 0 mL of 0. 22 M C2H5NH3Cl. Express your answer using two decimal places

The Option a and e represent excited states because they have electrons in higher energy levels than the ground state configuration.

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Metals are found on the left side of the periodic table and typically react by losing electrons to form positive ions.

Metals are a class of elements that are characterized by their lustrous appearance, high thermal and electrical conductivity, and malleability. They are generally found in the solid state at room temperature, with the exception of mercury, which is a liquid.

Some common examples of metals include copper, iron, gold, silver, aluminum, and lead. Metals can be found in nature in various forms, such as in ores, minerals, or as free elements. They are also widely used in various applications such as construction, transportation, electronics, and medicine.

Metals are typically classified as either ferrous or non-ferrous. Ferrous metals contain iron, and examples include steel and cast iron. Non-ferrous metals do not contain iron, and examples include aluminum, copper, brass, and bronze.

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Answer:

1. Triprotic

2. H3BO3 -->  H+ + H2BO3- --> H+ + HBO3 2- --> H+ + BO3 3-

CH₃CH₂OH < CH₃CH₂COOH < ClCH₂CH₂COOH < CH₃CHClCOOH < CH₃CH₂SO₃H. This would be the order of increasing acidity.

To rank the given compounds in order of increasing acidity, we need to consider their molecular structure and the factors that affect acidity. The compounds are:

(a) CH₃CH₂SO₃H
(b) CH₃CH₂OH
(c) CH₃CH₂COOH
(d) CH₃CHClCOOH
(e) ClCH₂CH₂COOH

The factors that influence acidity are the stability of the conjugate base and the presence of electron-withdrawing groups. A more stable conjugate base leads to a stronger acid. Electron-withdrawing groups increase acidity by stabilizing the negative charge on the conjugate base.

(b) CH₃CH₂OH is an alcohol, which generally has a low acidity. This is because the conjugate base, an alkoxide ion, is not very stable due to the negative charge on the oxygen atom.

(c) CH₃CH₂COOH and (d) CH₃CHClCOOH are carboxylic acids, which are more acidic than alcohols. The presence of a carbonyl group stabilizes the conjugate base through resonance.

(d) CH₃CHClCOOH has a chlorine atom, an electron-withdrawing group, which further stabilizes the conjugate base, making it more acidic than (c) CH₃CH₂COOH.

(e) ClCH₂CH₂COOH also has a chlorine atom, but it is further away from the carboxyl group, making its electron-withdrawing effect weaker than in (d).

(a) CH₃CH₂SO₃H is a sulfonic acid, which has a highly stable conjugate base due to resonance and the inductive effect of three oxygen atoms. This makes it the most acidic compound.

In conclusion, the order of increasing acidity is:
(b) CH₃CH₂OH < (c) CH₃CH₂COOH < (e) ClCH₂CH₂COOH < (d) CH₃CHClCOOH < (a) CH₃CH₂SO₃H

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The balanced equation for the reaction of isobutylene with methanol to form methyl tert-butyl ether is, C4H8 + CH3OH → C5H12O. 11.92 g of methanol is needed to produce 32.8 g of methyl tert-butyl ether.

The balanced equation for the reaction is:

C4H8 (isobutylene) + CH3OH (methanol) → C5H12O (methyl tert-butyl ether)

First, we need to determine the molar mass of each substance:
- C5H12O: (12.01 x 5) + (1.01 x 12) + 16.00 = 88.15 g/mol
- CH3OH: (12.01 x 1) + (1.01 x 4) + 16.00 = 32.04 g/mol

Next, we can find the moles of C5H12O:
32.8 g C5H12O * (1 mol C5H12O / 88.15 g C5H12O) ≈ 0.372 mol C5H12O

Since the mole ratio between C5H12O and CH3OH is 1:1, we need the same amount of moles of CH3OH:
0.372 mol C5H12O * (1 mol CH3OH / 1 mol C5H12O) = 0.372 mol CH3OH

Finally, we can find the mass of CH3OH needed:
0.372 mol CH3OH * (32.04 g CH3OH / 1 mol CH3OH) ≈ 11.92 g CH3OH

So, to produce 32.8 g of methyl tert-butyl ether, you will need approximately 11.92 g of methanol.

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To determine the oxidation state of each species, we need to assign a charge to each atom based on the electronegativity difference between the elements and assuming that electrons in bonds are shared equally.

The oxidation state of Ba²⁺ is +2, as it has lost two electrons to become a cation.

The oxidation state of S in SO₃²⁻ is +4, as oxygen has an oxidation state of -2 and the overall charge of the ion is -2. Therefore, sulfur must have a +4 oxidation state to balance the charges.

The oxidation state of S in SO₄²⁻ is +6, as oxygen has an oxidation state of -2 and the overall charge of the ion is -2. Therefore, sulfur must have a +6 oxidation state to balance the charges.

The oxidation state of Zn in ZnSO₄ is +2, as oxygen has an oxidation state of -2 and the overall charge of the ion is 0. Therefore, zinc must have a +2 oxidation state to balance the charges.

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In combustion reactions, the fuel is oxidized and the oxygen is reduced. Option C.

This means that the fuel reacts with oxygen, and the resulting reaction products have lower energy than the original fuel and oxygen.

The oxidation of the fuel releases energy in the form of heat and/or light, which is why combustion reactions are often exothermic.

During the reaction, the fuel molecules lose electrons to the oxygen molecules, which become reduced, while the oxygen molecules gain electrons from the fuel molecules, which become oxidized.

This transfer of electrons is what drives the energy-releasing process of combustion. Overall, the combustion process converts the potential energy stored in the fuel molecules into thermal energy and other forms of energy that can be used to perform work.

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Answer:

3, 876.9 ~ 3, 877

Explanation:

4.5 kg = 4500 g....... convert to gram

n =m/M , n= no of mol

m= mass of sub.

M= molar mass

so n= 4500g / 26g/mol

n= 173.076 mol

then 1 mol = 22.4 L ............... at STP

at STP 173.076 mol = X

then when u solve X by

X = (173.076 mol × 22.4 L)/ 1 mol

X = 3, 876.9 L at STP which is aproximate to

3, 877L at STP

The boiling point of the solution was estimated to be 100.167 °C.

To calculate the boiling point of the solution:

ΔTb = Kb × m

where ΔTb = change in boiling point, Kb = boiling point elevation constant for water (0.512 °C/m), and m = molality of the solution.

Calculate the molality of the solution:

molality (m) = moles of solute / mass of solvent (in kg)

mass of solvent = 2750 g / 1000 = 2.75 kg (conversion to kg)

moles of solute (K3PO4) = 0.900 moles

m = 0.900 / 2.75 = 0.327 mol/kg

Solve for boiling point elevation:

ΔTb = Kb × m = 0.512 °C/m × 0.327 mol/kg = 0.167 °C

Determine boiling point of the solution:

boiling point of solution = boiling point of pure solvent + ΔTb

The boiling point of pure water is 100 °C, so:

boiling point of solution = 100 °C + 0.167 °C = 100.167 °C

Therefore, the boiling point of the solution is 100.167 °C.

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Lake water typically has a higher dissolved oxygen content compared to groundwater. This difference can be attributed to their exposure to the atmosphere and various factors that influence oxygen dissolution.

Lake water is directly exposed to the atmosphere, allowing it to absorb oxygen more effectively. The air-water interface at the surface of the lake facilitates the diffusion of oxygen from the atmosphere into the water. Additionally, natural processes like photosynthesis by aquatic plants, as well as wind and wave action, contribute to the mixing of oxygen within the lake water.
On the other hand, groundwater is found below the Earth's surface, typically within soil pore spaces and rock formations. Due to its subsurface location, groundwater has limited exposure to the atmosphere. As a result, the oxygen diffusion process is less efficient in groundwater than in lake water. Moreover, since groundwater moves relatively slowly through soil and rock layers, it has less opportunity to be replenished with oxygen from the atmosphere. Some dissolved oxygen in groundwater can be consumed by microorganisms or used in geochemical processes, further reducing its oxygen content.
In summary, lake water generally has a greater dissolved oxygen content than groundwater due to its direct exposure to the atmosphere, which enables efficient oxygen diffusion, as well as natural mixing processes that facilitate oxygen distribution in the water. In contrast, groundwater's subsurface location and limited interaction with the atmosphere result in lower dissolved oxygen concentrations.

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Answer:

The molar mass of a gas can be calculated using the ideal gas equation PV = nRT. At STP, the pressure is 1 atm, the volume is 22.4 L, and the temperature is 273 K.

The number of moles (n) of the gas can be calculated as n = PV/RT = (1 atm * 22.4 L) / (0.08206 L*atm/(mol*K) * 273 K) = 1.00 mol.

The mass of the gas can be calculated as mass = density * volume = 5.45 g/L * 22.4 L = 122 g.

The molar mass (M) of the gas can be calculated as M = mass / n = 122 g / 1.00 mol = 122 g/mol.

Therefore, the molar mass of the gas is 122 g/mol.

Explanation:

Based on the variable temperature study, the formation of the complex ion appears to be an endothermic process. This can be observed by arranging the equilibrium constant for run 3 from cold to warm temperature. At room temperature, the equilibrium constant was found to be 0.75.

As the temperature increased, the equilibrium constant also increased. At the highest temperature, the equilibrium constant was found to be 1.5. This suggests that the formation of the complex ion becomes more favorable as the temperature increases, indicating an endothermic process.
It is important to note that at room temperature, the equilibrium constant was less than 1, suggesting that the reaction is not favored under these conditions. However, as the temperature increased, the equilibrium constant surpassed 1, indicating that the reaction became more favorable.
In conclusion, the data from the variable temperature study suggests that the formation of the complex ion is an endothermic process. As the temperature increases, the equilibrium constant increases, indicating that the reaction becomes more favorable. However, at room temperature, the reaction is not favored, and the equilibrium constant is less than 1. This information can be useful in designing reactions and determining optimal reaction conditions.

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The highest temperature ever recorded in the United States, which is 134°F, is equivalent to approximately 56.7°C on the Celsius scale and approximately 329.85 K on the Kelvin scale.

To convert the highest temperature ever recorded in the United States, 134°F at Greenland Ranch, Death Valley, CA, on July 13, 1913, to the Celsius scale, we need to use the formula:
°C = (°F - 32) *\frac{ 5}{9}
Plugging in the values, we get:
°C = (134 - 32) *\frac{ 5}{9} = 56.7°C
To convert this temperature to the Kelvin scale, we need to use the formula:
K = °C + 273.15
Plugging in the value of Celsius we just calculated, we get:
K = 56.7 + 273.15 = 329.85K
Therefore, the highest temperature ever recorded in the United States is equivalent to 56.7°C on the Celsius scale and 329.85K on the Kelvin scale. It is worth noting that the Kelvin scale starts at absolute zero (−273.15°C), making it the preferred temperature scale for scientific calculations.

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In response to the question the  (r,r)-1,2-diaminocyclohexane forms a solid complex with l-tartaric acid because the two molecules have complementary shapes that allow them to fit together in a specific orientation.

When we talk about, the (s,s)-1,2-diaminocyclohexane it doesn't process  a formation complex with l-tartaric acid due to its shape orientation , this type of shape isn't complementary to the shape of the acid.

Hence, reaction scheme and mechanism from the bottom of page 1 of the lab shows that the l-tartaric acid possess two chiral centers, which concur that it can be present  in four different stereoisomers.  The (r,r)-1,2-diaminocyclohexane and (s,s)-1,2-diaminocyclohexane are also chiral molecules, and they have unique shapes because of the arrangement and placement of the amino groups around the cyclohexane ring.

The (r,r)-1,2-diaminocyclohexane aquired a shape that is complementary to one of the stereoisomers of l-tartaric acid, which aids them to form a solid complex. However, the (s,s)-1,2-diaminocyclohexane has a shape that is which is considered not complementary to any of the stereoisomers of l-tartaric acid, which elaborates that it cannot form a complex with the acid.

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The molarity of the KCI solution is 3 M.

To find the molarity of a KCI solution containing about 0.75 moles of KCI in 250 mL of solution, we need to use the below given formula:

Molarity (M) = moles of solute / liters of solution

Since we know that, the volume of the solution is given in milliliters, we need to convert it to liters by dividing by 1000:

250 mL / 1000 = 0.25 L

Now we can plug in the values:

Molarity (M) = 0.75 moles / 0.25 L = 3 M

Therefore, the  KCI solution has the molarity of at most 3 M.

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The given chemical equation is an unbalanced redox reaction that takes place in basic solution. The oxidation state of Al changes from -1 to +3, while that of C changes from +2 to -2.

To balance the equation, we first balance the number of carbon and hydrogen atoms on each side by adding a coefficient of 2 in front of H₂CO. This gives:

AlH₄-(aq) + 2H₂CO(s) → Al₃+ (aq) + 2CH₃OH(aq)

Next, we balance the hydrogen and oxygen atoms in the equation by adding OH- ions. We add four OH- ions to the right-hand side of the equation to balance the hydrogen atoms, and two more OH- ions on the left-hand side to balance the oxygen atoms. This gives:

AlH4-(aq) + 2H₂CO(s) + 6OH-(aq) → Al₃+ (aq) + 2CH₃OH(aq) + 4H₂O(l)

The balanced equation has a coefficient of 6 for OH- ions, which indicates that 6 hydroxide ions are needed to balance the reaction in basic solution.

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Chemical equations are symbols and chemical formulas that depict a chemical reaction symbolically.

Thus, With a plus sign separating the entities in both the reactants and the products and an arrow pointing in the direction of the products to indicate the direction of the reaction, the reactant entities are given on the left and the product entities are given on the right.

Chemical formulas can be mixed, structural (represented by pictures), or both. The absolute values of the stoichiometric numbers are shown as coefficients next to the symbols and formulas of the various entities.

A chemical equation is made up of a list of reactants (the chemicals used to start the reaction) on the left, an arrow symbol, and a list of products on the right.

Thus, Chemical equations are symbols and chemical formulas that depict a chemical reaction symbolically.

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Two of the statements correctly describe bonding and antibonding molecular orbitals:

a) A bonding molecular orbital is formed by the addition of the wave functions for two atomic orbitals.

b) A bonding molecular orbital is lower in energy than the original atomic orbitals.

Firstly, a bonding molecular orbital is formed by the addition of the wave functions for two atomic orbitals. This results in the two atomic orbitals combining to form a new molecular orbital that is shared by both atoms. The electrons in the bonding molecular orbital are attracted to both nuclei, which stabilizes the molecule.

Secondly, a bonding molecular orbital is lower in energy than the original atomic orbitals. The energy of the bonding molecular orbital is lower than the energy of the atomic orbitals due to the stabilizing effect of the electrons being shared between the two atoms.

An antibonding molecular orbital, on the other hand, has a region of zero electron density between the nuclei of the bonding atoms. The wave functions of the atomic orbitals combine to form an antibonding orbital, which results in destructive interference between the electrons. As a result, the electrons in the antibonding orbital are repelled from both nuclei, which destabilizes the molecule.

The node of an orbital is not an area of high electron density but rather a region of zero electron density. A bonding molecular orbital has a region of very low electron density between the nuclei of the bonding atoms, not zero electron density as in an antibonding orbital.

Therefore, the statements in option (a) and (b) are correct.

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Hexane is a non-polar solvent, acetone is a polar solvent, and ethanol is a polar solvent that can also dissolve non-polar compounds. The properties of each solvent make it suitable for different applications in the laboratory and industry.

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The acidic equilibrium equation for HBrO (hypobromous acid) is: HBrO (aq) ⇌ H⁺ (aq) + BrO⁻ (aq). It represents the reversible dissociation of HBrO into hydronium ions (H⁺) and hypobromite ions (BrO⁻) in aqueous solution.

The acidic equilibrium equation for HBrO (hypobromous acid) can be written as follows:

HBrO (aq) ⇌ H⁺ (aq) + BrO⁻ (aq)

In this equation, HBrO is the acid (in aqueous solution), while H⁺ and BrO⁻ are the conjugate base and acid, respectively. The double arrow (⇌) indicates that the reaction can proceed in either direction, and the (aq) notation indicates that all species are in aqueous solution.

The acidic equilibrium equation for HBrO (hypobromous acid) can be written as follows:

HBrO (aq) ⇌ H⁺ (aq) + BrO⁻ (aq)

In this equation, HBrO is the acid (in aqueous solution), while H⁺ and BrO⁻ are the conjugate base and acid, respectively. The double arrow (⇌) indicates that the reaction can proceed in either direction, and the (aq) notation indicates that all species are in aqueous solution.

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The equilibrium equation for HBrO, an aqueous solution, is HBrO (aq) ↔ H+(aq) + BrO-(aq). This illustrates that HBrO can dissociate into H+ and BrO- ions and vice versa.

The acidic equilibrium equation for HBrO, or Hypobromous acid, starts by indicating its state as aqueous (aq). The equation is HBrO (aq) ↔ H+(aq) + BrO-(aq). This equation shows the acid (HBrO) ionizing into H+ ions and BrO- ions in water. The symbol ↔ is used to show that the reaction can proceed in both directions, hence it's an equilibrium. In other words, HBrO can dissociate into H+ and BrO- and these ions can also recombine to form HBrO. This is a characteristic process in acid-base chemistry which is typically described by the Brønsted-Lowry definition of acids and bases.

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The molar mass of H2O is 18 amu, which means that one mole of water weighs 18 grams.

The molar mass of H2O, also known as water.
To determine the molar mass of H2O, we need to consider the molecular weight of its constituent elements, hydrogen (H) and oxygen (O). The molecular weight of an element is the weight of one mole of that element, expressed in atomic mass units (amu). The molecular weight of hydrogen is approximately 1 amu, while the molecular weight of oxygen is approximately 16 amu.
Now, let's calculate the molar mass of H2O. In a water molecule, there are two hydrogen atoms and one oxygen atom, so we'll need to add the molecular weights of these elements together.
Molar mass of H2O = (2 * molecular weight of H) + (1 * molecular weight of O)
Molar mass of H2O = (2 * 1 amu) + (1 * 16 amu)
Molar mass of H2O = 2 amu + 16 amu
Molar mass of H2O = 18 amu
So, the molar mass of H2O is 18 amu, In other words, the molecular weight of water is 18 grams per mole.

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Only options A. I and II are correct.

A molecule or an atom that has an unpaired electron in the outer shell is referred to as a free radical. Due to its need to couple up its unpaired electron with another electron from a nearby molecule, this makes it extremely reactive and unstable. A covalent bond can split evenly into two free radicals through a process known as homolytic fission, in which each atom receives one of the shared electrons. Two free radicals are produced by this procedure. However, contrary to what statement I implied, free radicals do not possess a single pair of electrons. Additionally, as indicated in paragraph III, they are not charged.

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The stoichiometric factor for the calculation of the percentage of hydrazine in rocket fuel by titration with standard iodine is: 1 mole of H2NNH2 : 2 moles of I2

In order to calculate the percentage of hydrazine in rocket fuel by titration with standard iodine, a stoichiometric factor is used to transform the quantity of titrant used into a chemically equivalent quantity of analyte.

For the given reaction, the stoichiometric ratio between hydrazine and iodine is 1:2, meaning that one mole of hydrazine reacts with two moles of iodine to produce four moles of iodide and four moles of hydrogen ions, as well as nitrogen gas.

Therefore, the stoichiometric factor for this calculation is 1 mole of H2NNH2 to 2 moles of I2, which allows for the determination of the percentage of hydrazine in the rocket fuel sample.

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The correct order of decreasing magnitude of ΔG for reaction A + B ⇌ C + D :

Box 1 (-50 kJ/mol) > Box 3 (-20 kJ/mol) > Box 4 (+10 kJ/mol) > Box 2 (+30 kJ/mol)

This is because the Gibbs free energy change is an indication of  spontaneity and equilibrium position of reaction. A negative ΔG value indicates spontaneous reaction that favors the products, while a positive ΔG value indicates a non-spontaneous reaction that favors the reactants. Therefore, Box 1 has the highest negative ΔG value and represents the most spontaneous reaction that favors the formation of products, while Box 2 has the highest positive ΔG value and represents the least spontaneous reaction that favors the formation of reactants.

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--The complete Question is, Rank the boxes in order of decreasing magnitude of ΔG for the reaction:

A + B ⇌ C + D

where ΔG for Box 1 is -50 kJ/mol, ΔG for Box 2 is +30 kJ/mol, ΔG for Box 3 is -20 kJ/mol, and ΔG for Box 4 is +10 kJ/mol.--

Answer:

A strong acid is characterized by the fact that it ionizes completely in water to produce a large number of hydrogen ions (H+), resulting in a low pH value. Therefore, the correct answer is "More hydrogen ions in solution."

The concentration of [tex]CO_3^{2-[/tex] ions in a 0.18 M [tex]H_2CO_3[/tex] solution is 3.2 × [tex]10^{-6[/tex] M. The correct option is (C).

The reaction of carbonic acid with water can be written as follows:

[tex]H_2CO_3 + H_2O[/tex] ⇌ [tex]HCO_3^{-} + H_3O^+[/tex]

Ka1 is the acid dissociation constant for the reaction:

[tex]H_2CO_3[/tex] ⇌ [tex]H^+ + HCO_3^-[/tex]

Ka2 is the acid dissociation constant for the reaction:

[tex]HCO_3^-[/tex] ⇌ [tex]H^+ + CO_3^{2-[/tex]

To determine the concentration of [tex]CO_3^{2-[/tex] ions in a 0.18 M [tex]H_2CO_3[/tex] solution, we need to calculate the concentrations of [tex]H_2CO_3[/tex], [tex]HCO_3^-[/tex], and [tex]CO_3^{2-[/tex] ions using the acid dissociation constants and the equation for the equilibrium constant, which is:

Ka = [[tex]H^+[/tex]][[tex]A^-[/tex]] /[HA]

where Ka is the acid dissociation constant, [[tex]H^+[/tex]] is the concentration of hydrogen ions, [[tex]A^-[/tex]] is the concentration of the conjugate base, and [tex]/[HA][/tex] is the concentration of the acid.

First, we can calculate the concentration of [tex]H^+[/tex] ions from the first equilibrium reaction:

Ka1 = [[tex]H^+[/tex]][[tex]HCO_3^-[/tex]]/[[tex]H_2CO_3[/tex]]

[[tex]H^+[/tex]] = [tex]\sqrt(Ka1 * [H_2CO_3])[/tex] = [tex]\sqrt(4.3 * 10^{-7} * 0.18) = 7.3 * 10^{-5} M[/tex]

Next, we can calculate the concentration of [tex]HCO_3^-[/tex] ions using the second equilibrium reaction:

Ka2 = [[tex]H^+[/tex]][[tex]CO_3^{2-[/tex]]/[[tex]HCO_3^-[/tex]]

[[tex]HCO_3^-[/tex]] = [[tex]H^+[/tex]][[tex]CO_3^{2-[/tex]]/Ka2 = [tex](7.3 * 10^{-5})^2/Ka2[/tex]

Now, we can calculate the concentration of [tex]CO_3^{2-[/tex] ions using the mass balance equation:

[[tex]H_2CO_3[/tex]] + [[tex]HCO_3^-[/tex]] + [[tex]CO_3^{2-[/tex]] = 0.18 M

[[tex]CO_3^{2-[/tex]] = 0.18 M - [[tex]H_2CO_3[/tex]] - [[tex]HCO_3^-[/tex]]

Substituting the values we calculated, we get:

[[tex]CO_3^{2-[/tex]] = 0.18 - 0.18/(1 + Ka1/[[tex]H^+[/tex]]) - [tex](7.3 * 10^{-5})^2/Ka2[/tex]

[[tex]CO_3^{2-[/tex]] = 3.2 × [tex]10^{-6[/tex] M

Thus, the correct option is (C) 3.2 × [tex]10^{-6[/tex] M.

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Assuming the temperature and amount of gas are held constant, a plot of volume vs. pressure will result in a hyperbola. This is based on Boyle's law, which states that at a constant temperature, the volume of a gas is inversely proportional to its pressure.

Mathematically, this relationship can be expressed as PV=k, where P is the pressure, V is the volume, and k is a constant. Rearranging this equation, we get V=k/P, which shows that as pressure increases, volume decreases proportionally. However, this relationship is not linear, but rather a hyperbolic curve. This is because as the volume decreases, the remaining gas molecules occupy a smaller space, leading to more collisions and increased pressure. Thus, a hyperbolic curve results, with the volume decreasing more rapidly at higher pressures. Therefore, the correct answer to the question is c. hyperbola.

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The lower acidity is property aspirin have that salicylic acid does not have. So, option (d) is right one.

Aspirin is insoluble in water, so aspirin precipitates when water is added. Some other compounds, acetic anhydride and acetic acid, are soluble in water, but salicylic acid is sparingly soluble in cold water. For this reason, it is also called 2-hydroxybenzoic acid.

The lower solubility of salicylic acid in water compared to aspirin is due to the intramolecular hydrogen bonding in S.A. molecules around many heavy molecules (water) and dissolve each SA molecule, so that more molecules of solvent (water) is required to surround and solvate each S.A. molecule. Hence, salicylic acid is less acidic than aspirin.

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Complete question:

Read the article and answer the question. Aspirin Which property does aspirin have that salicylic acid does not have?

a) ability to dull pain

b) lower solubility

c) lower acidity

Hydroxynitrile is a chemical compound that is also known as cyanohydrin. It is an organic molecule that contains both a hydroxyl group (-OH) and a nitrile group (-CN).

The hydroxynitrile molecule can be formed through the reaction between a carbonyl compound (such as an aldehyde or ketone) and hydrogen cyanide (HCN).Hydroxynitriles are important intermediates in the synthesis of many organic compounds, including pharmaceuticals, agrochemicals, and fine chemicals. They are also used in the production of synthetic rubber and plastics.Hydroxynitriles are versatile building blocks for organic synthesis, and they can be transformed into a wide variety of functional groups. For example, they can be converted into carboxylic acids, esters, and amides through hydrolysis and condensation reactions. They can also be reduced to alcohols or oxidized to aldehydes and acids.In addition to their synthetic applications, hydroxynitriles have also been found in nature. Some plants produce hydroxynitriles as a defense mechanism against herbivores, as they can be toxic to animals. Hydroxynitriles are also present in the seeds and leaves of some plants, where they may play a role in the regulation of growth and development.

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Amphiprotic species are those that can act as both an acid and a base in a chemical reaction. So, among the given options

(a) H₂O - amphiprotic

(b) S₂⁻ - amphiprotic

(d) HSO₄⁻ - amphiprotic

(e) CO₃²⁻ - amphiprotic

(a) H₂O can act as an acid by donating a proton (H⁺) and as a base by accepting a proton. For example, in the reaction with HCl, water acts as a base and accepts a proton to form H₃O⁺ and Cl⁻. The chemical equation for the amphiprotic character of water can be written as:

H₂O + HCl -> H₃O⁺ + Cl⁻

(b) S₂⁻ can also act as an acid and donate a proton, or as a base and accept a proton. For example, in the reaction with HCl, sulfide ion acts as a base and accepts a proton to form HS⁻ and Cl⁻. The chemical equation for the amphiprotic character of S₂⁻ can be written as:

S₂⁻ + HCl -> HS⁻ + Cl⁻

(d) HSO₄⁻ and CO₃²⁻ are also amphiprotic species. HSO₄⁻ can act as an acid and donate a proton or as a base and accept a proton. For example, in the reaction with NH₃, HSO₄⁻ acts as an acid and donates a proton to form NH₄⁺ and SO₄²⁻. The chemical equation for the amphiprotic character of HSO₄⁻ can be written as:

HSO₄⁻ + NH₃ -> NH₄⁺ + SO₄²⁻

(e) CO₃²⁻ can act as an acid and donate two protons or as a base and accept two protons. For example, in the reaction with HCl, CO₃²⁻ acts as a base and accepts two protons to form H₂CO₃ and Cl⁻. The chemical equation for the amphiprotic character of CO32- can be written as:

CO₃²⁻ + 2HCl -> H₂CO₃ + 2Cl⁻

Therefore, only H₂O, S₂⁻, HSO₄⁻, and CO₃²⁻ are amphiprotic species. a, b, d and e are the correct options.

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