what is the ph of a solution prepared by mixing 100. ml of 0.0500 m hcl with 300. ml of 0.500 m hno2? [ka(hno2)

The product shown in the structure can be generated from a Suzuki coupling reaction using the reactants Na2CO3, Pd(PPh3)4, SnBu3 and MgBr.

First, the palladium (Pd) catalyst is activated by the Na2CO3 under basic conditions, which then reacts with the organoboron compound SnBu3 to form a palladium-boron complex. This complex then reacts with the aryl halide, E, under mildly basic conditions and the reaction is accelerated by heating. The aryl halide is then replaced with the aryl Grignard, MgBr, which undergoes a transmetalation to give the desired product A.

In summary, the two reactants needed for the Suzuki coupling are Na2CO3, Pd(PPh3)4, SnBu3 and MgBr.  

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The equilibrium potential of a copper wire immersed in 0.0007 M CuSO₄ is 1.191 V.

To calculate the equilibrium potential of a copper wire immersed in 0.0007 M CuSO₄ solution, we need to use the Nernst equation. The Nernst equation is:

E = E° - (RT/nF) * ln(Q)

Where E is the equilibrium potential (in volts), E° is the standard electrode potential, R is the gas constant (8.314 J K-1 mol-1), T is the temperature (in Kelvin), n is the number of electrons transferred (2 in this case), F is Faraday’s constant (96485 C mol-1), and Q is the reaction quotient.

In this case, E° = 0.34 V, T = 298 K, n = 2, F = 96485 C mol-1, and Q = 0.0007 M CuSO4. Therefore, the equilibrium potential of the copper wire is:

E = 0.34 V - (8.314 J K-1 mol-1 * 298 K / (2 * 96485 C mol-1)) * ln(0.0007 M CuSO4)
E = 0.34 V - (-0.851 V)
E = 1.191 V

Therefore, the equilibrium potential of the copper wire immersed in 0.0007 M CuSO₄ solution at 25°C is 1.191 V.

Complete question:

Calculate the equilibrium potential of a copper wire immersed in 0.0007 M CuSO4 solution. The standard electrode potential for the reaction Cu2+ + 2e- = Cu0 at 25°C is 0.34 V (NHE).

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For precipitation to occur, the value of Qsp (the ion product constant) should be greater than the solubility product constant (Ksp).

Precipitation is the conversion of a dissolved substance into a solid, which then settles out of a solution. Precipitation occurs when a liquid solution is cooled or heated, causing it to become super-saturated with one or more solutes. A solution's super-saturation means that it contains more of a solute than it can contain at equilibrium.

A tiny seed crystal of the solute is added to the solution to kick off the precipitation. The seed crystal provides a template for the rest of the solute to nucleate and form a solid. For precipitation to occur, the value of Qsp (the ion product constant) should be greater than the solubility product constant (Ksp). When Qsp is greater than Ksp, the solution is supersaturated and precipitates are formed. If Qsp is less than Ksp, the solution is unsaturated and no precipitation occurs.

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In order to conduct PCR (polymerase chain reaction), the reagents included in the PCR bead should be template DNA, primers, DNA polymerase, dNTPs (deoxynucleotide triphosphates), and a buffer.

1. Template DNA is the target DNA that is to be amplified during the PCR process.

2. Primers are short pieces of single-stranded DNA that are complementary to the ends of the target DNA sequence.

3. DNA polymerase is an enzyme used to replicate the target DNA sequence.

4. dNTPs are the building blocks of DNA which include adenine, thymine, cytosine, and guanine.

5. Lastly, the buffer helps maintain the appropriate pH for the reaction to occur.

In conclusion, the PCR bead used in the experiment must have contained a template DNA, primers, DNA polymerase, dNTPs, and a buffer in order for PCR to take place.

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The Ultimate BOD ( Biochemical Oxygen Demand) of the waste is 81.3 mg/L when the 5-day BOD is 20 mg/L and the BOD rate coefficient is 0.15/day.

The Ultimate Biochemical Oxygen Demand (BOD) is defined as he quantity of oxygen required to stabilize or,

eliminate biodegradable organic matter in wastewater by the action of aerobic microorganisms under controlled laboratory conditions at a specified temperature over a period of time.

The 5-day BOD is measured by calculating the amount of oxygen consumed by microorganisms over a period of five days.

The Ultimate BOD of a waste can be determined by knowing the 5-day BOD and BOD rate coefficient. The following formula is used to determine the Ultimate BOD:

Ultimate BOD = 5-day BOD × [(e^(k×t))-1] / e^(k×t)Where k is the BOD rate coefficient and t is the time required to reach the Ultimate BOD.

The Ultimate BOD of the waste as follows: 5-day BOD = 20 mg/L k = 0.15/day t = ? Ultimate BOD = 5-day BOD × [(e^(k×t))-1] / e^(k×t) Ultimate BOD = 20 × [(e^(0.15×t))-1] / e^(0.15×t)

The Ultimate BOD is reached after 20 days. Ultimate BOD = 20 × [(e^(0.15×20))-1] / e^(0.15×20) Ultimate BOD = 81.3 mg/L

Therefore, the Ultimate BOD of the waste is 81.3 mg/L when the 5-day BOD is 20 mg/L and the BOD rate coefficient is 0.15/day. The coefficient is the numerical multiplier of a variable or quantity that follows a term or a factor.

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Answer: 15.0667

Explanation:

Answer:

1. Ans: B

Explanation: Pinakbet, which contains vegetables such as eggplants and kalabasa, are physically combined. Therefore, the pinakbet is an example of a mixture

2. Ans: A

Explanation: Heterogenous are different building blocks that are mixed UNEVENLY.

At the equivalence point of a titration between 24.6 mL of 0.389 M ethylamine, C2H5NH2, and 0.325 M hydroiodic acid, the pH is 0.

At the equivalence point of a titration between 24.6 mL of 0.389 M ethylamine, C2H5NH2, and 0.325 M hydroiodic acid, the pH is 0. The equation for the reaction is:

C2H5NH2 + HI → C2H5NH3+ + I-

The number of moles of hydroiodic acid, HI, needed to reach the equivalence point is equal to the number of moles of ethylamine, C2H5NH2. To calculate this, use the following equation:

Moles of HI = Moles of C2H5NH2

Volume of C2H5NH2 x Molarity of C2H5NH2 = Volume of HI x Molarity of HI

24.6 mL x 0.389 M = Volume of HI x 0.325 M

Volume of HI = 24.6 mL x 0.389 M / 0.325 M

Volume of HI = 30.53 mL

At the equivalence point, the pH of the solution is 0.

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Acetic acid has a ka of 1.80x10-5. The pH of a buffer solution made from 0.150 m hc2h3o2 and 0.530 m c2h3o2 is 4.76.

The pH of a buffer solution produced from 0.150 M HC2H3O2 and 0.530 M C2H3O2 is 4.76.

The following are the steps to solve the problem:

Acetic acid is a weak acid with the formula CH3COOH, which is also known as ethanoic acid.

HC2H3O2 is the molecular formula for this substance.

Acetic acid has a Ka of 1.8 x 10-5.

The ionization of acetic acid can be expressed as follows: CH3COOH + H2O ↔ H3O+ + CH3COO-

The ionization constant, Ka, is equal to the product of the concentration of H3O+ and CH3COO- ions divided by the concentration of CH3COOH.

Hence, Ka = ([H3O+] [CH3COO-])/[CH3COOH]

The Henderson-Hasselbalch equation is used to compute the pH of a buffer solution.

pH = pKa + log (base/acid), where pKa = -logKa.

In the equation, the base is C2H3O2-, and the acid is HC2H3O2.

Substituting the values in the equation, pH = -log1.8 x 10-5 + log(0.530/0.150) = 4.76.

Therefore, the pH of a buffer solution produced from 0.150 M HC2H3O2 and 0.530 M C2H3O2 is 4.76.

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the molality of solutes in the aqueous solution is 0.182 molal.

⇒m = (molality) = (Δ[tex]T_f[/tex]) / ([tex]K_f[/tex] × w2)

We have to calculate the molality of solutes in the solution by using the freezing-point depression constant and freezing-point depression of the aqueous solution.

Now, Substituting the given values, we get,

⇒ m = (Δ[tex]T_f[/tex]) / ([tex]K_f[/tex] × w2)

⇒ m = 0.34 / (1.86 × w2)

⇒ m = 0.182 molal

Therefore, the molality of solutes in the solution is 0.182 molal.

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(a) If the pressure is 10.0 atm, the temperature is 62.0 K.

(b) if the temperature is 225 k, the pressure is 36.3 atm.

a) In order to calculate the temperature, we need to use the ideal gas law, PV = nRT, where P is the pressure, V is the volume of the container, n is the number of moles of argon, R is the ideal gas constant, and T is the temperature.

We can calculate the number of moles, n, by using the molar mass of argon, which is 39.948 g/mol.

We have n = 186 g / 39.948 g/mol = 4.656 mol.

So we can plug in our values and solve for T:

T = (10.0 atm)(2.37 L) / (4.666 mol)(0.08206 L·atm/mol·K) = 62.0 K.

b) To calculate the pressure, we can again use the ideal gas law, PV = nRT. We know the values of n, R, and T from the previous question.

Since the volume of the container is given, we can plug in these values to solve for P:

P = (4.666 mol)(0.08206 L·atm/mol·K)(225 K) / 2.37 L = 36.3 atm.

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To summarize, at absolute zero, the energy needed for diffusion to occur is completely absent, and the molecules are completely frozen. So diffusion does not take place.

Diffusion is a process of net movement of molecules from an area of high concentration to an area of low concentration. The process of diffusion requires some form of energy, and at absolute zero, the energy is completely eliminated. This means that there would be no potential for molecules to move from a region of higher concentration to one of lower concentration. Therefore, diffusion does not occur at absolute zero.  At absolute zero, molecules stop vibrating and the atoms cease all motion. All molecular motion is frozen and stopped, so diffusion is not possible. Diffusion requires energy to move molecules, which is not available at absolute zero. The energy needed to drive molecules to move is not present, so molecules cannot move.

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The new volume of the ideal gas is 28.9 liters when heated from 268K to 343K and the pressure is increased from 5.16atm to 8.00atm.

This can be calculated using the Ideal Gas Law: PV = nRT, where P is pressure, V is volume, n is the number of moles, R is the ideal gas constant, and T is temperature.

First, calculate the number of moles of the ideal gas, n = PV/RT. Plug in the given values, and rearrange for n: n = (5.16atm)(35.0L)/(8.314L∙K⁻¹∙mol⁻¹)(268K) = 0.41mol.

Then, calculate the new volume of the gas, V = nRT/P. Plug in the given values, and rearrange for V: V = (0.41mol)(8.314L∙K⁻¹∙mol⁻¹)(343K)/(8.00atm) = 28.9L.

Therefore, the new volume of the ideal gas is 28.9 liters when heated from 268K to 343K and the pressure is increased from 5.16atm to 8.00atm.

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In order to prepare benzoic acid, the reactants that could be used are Na/NH2 and Reactant A Reactant B O3. In a chemical equation, a chemical reaction is described using chemical formulas and symbols.

Benzoic acid is a white crystalline powder that is somewhat soluble in water. It has a mild, sweet odor and a sour, astringent flavor. It is an organic acid that occurs naturally in several plants and animals' tissues. Benzoic acid is a colorless crystalline solid that is an important precursor for the synthesis of many other organic compounds.

Benzoic acid can be prepared by reacting a Grignard reagent with carbon dioxide, reducing the resulting carboxylic acid, or from toluene using the Perkin reaction. The reactants that could be used to prepare benzoic acid are: Na/NH2Reactant A Reactant B O3. The chemical equation for the preparation of benzoic acid from Na/NH2 is:

C6H5NO2 + 3H2 => C6H5COOH + 2NH3

The chemical equation for the preparation of benzoic acid from Reactant A Reactant B O3 is:

C6H5CH3 + 2O3 => C6H5COOH + H2O + O2

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There are 567.8 mmHg in 75.7 kpa.

Pressure is the amount of force that is applied over a given area divided by the size of this area.

The units of pressure are as follows:

In SI units, pressure is measured in pascals where;

1 kPa = 760/101.325 = 7.5 mmHg

Hence, 75.7kpa is equal to 567.8 mmHg

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An electrolyte solution is one that contains ions. The correct option is d.

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When determining the energy effect of a chemical reaction, the system is/are reactants and products and the surroundings are everything outside the system.

When determining the energy effect of a chemical reaction, the system and surroundings are involved. The system refers to the reactants and products involved in the chemical reaction, whereas the surroundings refer to everything else outside the system, including the temperature, pressure, and any other factors that can affect the reaction.

The energy effect of a chemical reaction can be determined by calculating the difference between the energy of the products and the energy of the reactants. This difference is known as the energy change or the enthalpy change of the reaction.

If the energy change is positive, it means that the reaction is endothermic, and energy is absorbed from the surroundings. This results in a decrease in the temperature of the surroundings.

On the other hand, if the energy change is negative, it means that the reaction is exothermic, and energy is released into the surroundings. This results in an increase in the temperature of the surroundings.

It is important to note that the energy effect of a chemical reaction can also be affected by external factors such as pressure, temperature, and the presence of a catalyst.

In conclusion, the system is the reactants, and the products and surroundings are factors like temperature and pressure, i.e., everything outside the system.

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Potassium hydrogen phthalate (KHP) is used to standardize the NaOH solution instead of measuring out a known mass of solid NaOH to be dissolved in water because KHP has several benefits that make it a better option.

Potassium hydrogen phthalate (KHP) is a crystalline powder that has a chemical formula of KHC8H4O4. It is a primary standard for acid-base titrations, meaning that its molar mass and purity are known to a high degree of accuracy.

KHP is stable when exposed to air and is easy to obtain.4. KHP is inexpensive in comparison to other primary standards, such as sodium carbonate, which is costly and difficult to prepare.KHP reacts with NaOH in a 1:1 molar ratio, so one mole of KHP reacts with one mole of NaOH. Because the amount of KHP used in the titration is known, the concentration of the NaOH solution can be determined mathematically.

A secondary standard, such as NaOH, can be standardized using KHP, which is a primary standard. As a result, it is preferred to use KHP to standardize NaOH rather than measuring out a known mass of solid NaOH to be dissolved in water.

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The ratio of [HCO₃⁻] to [HCO₂H] in an acetate buffer is 5.01.

The ratio of [HCO₃⁻] to [HCO₂H] (formic acid) in an acetate buffer at pH 4.50 is determined by the Henderson-Hasselbalch equation:

pH = pKa + log ([HCO₃⁻]/[HCO₂H]).
[HCO₃⁻]/[HCO₂H] = 10^(pH-pKa)
= 10^(4.50 - 3.80)
= 5.01


To further understand the buffering capacity of an acetate buffer, we must first understand the role of formic acid and bicarbonate in an acetate buffer.

Formic acid is an organic acid and bicarbonate is a salt of carbonic acid. Both of these species can form and break down as needed to maintain the pH of the buffer.

As the pH of the buffer is increased, the formic acid will break down, forming more bicarbonate.

On the other hand, as the pH of the buffer is decreased, more formic acid will form, resulting in fewer bicarbonate ions.

The buffering capacity of an acetate buffer is dependent on the relative concentrations of formic acid and bicarbonate ions, and these concentrations can vary depending on the pH of the buffer.

In summary, the ratio of [HCO₃⁻] to [HCO₂H] is found to be 5.01 in an acetate buffer at pH 4.50.

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If the scientist wishes to increase the heat of the boiling water as much as possible, she can increase the heat source's temperature or increase the amount of heat being supplied to the water.

This can be done by increasing the flame on the stove, increasing the power of the heating element, or adding more heat from an external source. However, it is important to note that the boiling point of water is 100°C at standard atmospheric pressure, so the water will not get any hotter than that unless the pressure is increased. Also, caution should be taken when working with boiling water to prevent burns or accidents.

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The pKₐ of a 0.109M solution of a weak acid (HA) is 4.50 and pKₐ arranged to 4 decimal places is 4.5000.

The pH of a solution of a weak acid can be used to determine the pKₐ of the acid. The pKa is the negative logarithm of the acid dissociation constant (Kₐ). In other words, pKₐ = -log Kₐ.

To calculate the pKₐ of a weak acid, we can use the following equation:

pKₐ = pH + log (base concentration/acid concentration).

In this case, we are given a 0.109 m solution of a weak acid (HA) with a pH of 4.50.

To calculate the pKₐ, we need to know the concentration of the acid and the concentration of the base. Since we do not know the concentrations, we can assume that the acid and base concentrations are equal and equal to 0.109 m.

Using the equation above, we can calculate the pKₐ as follows:

pKₐ = 4.50 + log (0.109 / 0.109) = 4.50 + 0 = 4.50.

Therefore, the pKₐ of the weak acid (HA) comes out to be 4.50, and rearranging it to four decimal places gives the answer as 4.5000.

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Quantitative inference: The energy of a wave is instantly proportional to the square of its amplitude. This means that if the amplitude of a wave is doubled from 2 to 4, the wave's energy will increase by a factor of four (2^2 = 4). This can be expressed mathematically as E2 = (A2/A1)^2xE1. Qualitative inference: Increasing the amplitude of a wave from 2 to 4 would result in a wave with higher energy. This means that the wave would be more intense and capable of doing more work.

Amplitude refers to the highest displacement of a wave from its resting position. In other words, it is the distance from the highest point (crest) to the lowest point (trough) of a wave, measured in distance units, such as meters or centimetres.

Yes, increasing the amplitude of a wave increases its energy. The energy of a wave is directly proportional to the square of its amplitude. This means that if the amplitude of a wave is doubled, its energy will increase by a factor of four, and if the amplitude is tripled, the energy will increase by a factor of nine.

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a) 3.21 kg of uranium would be consumed in a year to run a 100 W light bulb for a year, assuming a thermal to electric efficiency of 30% using 235u.

b)  35 kg of coal would be required to run a 100 W light bulb for a year

We have;

Thermal to electric efficiency = 30%

Power of the light bulb = 100 W

Thermal output of coal = 25 GJ/tona)

Uranium that would be consumed in a year to run a 100 W light bulb for a year:

Energy consumed in a year by the light bulb = 100 W × 24 hours/day × 365 days/year

= 876,000 Wh

= 876 kWh

Electric energy produced from the thermal energy = (Thermal to electric efficiency / 100) × Energy consumed

electric energy produced from the thermal energy = (30 / 100) × 876 kWh

= 262.8 kWh

Amount of uranium consumed in a year = Electric energy produced from the thermal energy / Energy density of uranium

= 262.8 kWh / 81.8 GJ/t

= 0.00321 t

= 3.21 kg

Therefore 3.21 kg of uranium would be consumed in a year to run a 100 W light bulb for a year.

b) Coal that would be required to run a 100 W light bulb for a year given a thermal output of 25 GJ/ton:

Energy consumed in a year by the light bulb = 100 W × 24 hours/day × 365 days/year

= 876,000 Wh

= 876 kWh

Electric energy produced from the thermal energy = (Thermal to electric efficiency / 100) × Energy consumedElectric

energy produced from the thermal energy = (30 / 100) × 876 kWh = 262.8 kWh

Amount of coal required = Thermal energy required / Thermal output of coal

The thermal energy required = Electric energy produced from the thermal energy / (Thermal to electric efficiency / 100)

Thermal energy required = 262.8 kWh / (30 / 100) = 876 kWh

Amount of coal required = 876 kWh / 25 GJ/ton = 0.035 t = 35 kg

35 kg of coal would be required to run a 100 W light bulb for a year given a thermal output of 25 GJ/ton.

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The chemical equation for the ion pairing of Sr2+ (aq) and C2O42- (aq) leading to their soluble ion pair is given by the following chemical equation: Sr2+ (aq) + C2O42- (aq) ⇌ SrC2O4 (s).

Here, Sr2+ (aq) is an aqueous solution of strontium ions and C2O42- (aq) is an aqueous solution of oxalate ions. When these two solutions are mixed, they undergo a reaction to form a precipitate of strontium oxalate (SrC2O4) which is a soluble ion pair.

The reaction is reversible because the soluble ion pair can dissociate into its constituent ions under certain conditions. The solubility of the ion pair is determined by the equilibrium constant (Ksp) of the reaction which is given by the following equation: Ksp = [Sr2+][C2O42-] where [Sr2+] and [C2O42-] are the concentrations of strontium ions and oxalate ions in the solution, respectively.

Thus, the chemical equation for the ion pairing of Sr2+ (aq) and C2O42- (aq) leading to their soluble ion pair is: Sr2+ (aq) + C2O42- (aq) ⇌ SrC2O4 (s).

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The pH of the buffer prepared using 0.750 moles of HCN and 0.250 moles of NaCN dissolved into 2.00l of the solution is 9.31.

a) To determine the pH of the buffer, we need to first calculate the concentration of the acid and its conjugate base.

HCN is a weak acid and NACN is its conjugate base. The equation for the dissociation of HCN is:

HCN + H2O ⇌ H3O+ + CN-

The equilibrium constant for this reaction is Ka = [H3O+][CN-]/[HCN].

The concentration of HCN is 0.750 moles/2.00 L = 0.375 M

The concentration of CN- is 0.250 moles/2.00 L = 0.125 M

Therefore, Ka = (x)(x)/(0.375-x)

where x is the concentration of H3O+ and is assumed to be very small compared to 0.375.

Solving for x, we get x = 4.9 x 10^-10 M

Therefore, the pH of the buffer is pH = -log[H3O+]

pH = -log(4.9 x 10^-10)

pH = 9.31

b) The buffer has a higher capacity for additions of acid because it is made up of a weak acid and its conjugate base. The weak acid can neutralize added base, and the conjugate base can absorb added H3O+.

c) The pH will change by 1 pH unit when the amount of NaOH added is equal to the amount of HCN present in the buffer.

The moles of HCN in the buffer is 0.750 moles.

The reaction between NaOH and HCN is:

NaOH + HCN → NaCN + H2O

For every mole of HCN, we need one mole of NaOH to neutralize it.

Therefore, the amount of NaOH needed to change the pH by 1 unit is 0.750 moles.

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The concentration of sulfuric acid that remains after neutralization is 0.056 M.

To find out what concentration of sulfuric acid remains after neutralization, you will need to use the balanced equation for the reaction:

H2SO4 + 2KOH → K2SO4 + 2H2O

First, you will need to determine the moles of each reactant in the solution.

Moles can be determined using the formula:

moles = concentration x volume

In this case:

moles of H2SO4 = 0.850 L x 0.400 M = 0.34 mol

moles of KOH = 0.800 L x 0.250 M = 0.2 mol

Since the reaction is a 1:2 ratio, you will need to determine which reactant is limiting the reaction.

To do this, compare the mole ratios of the reactants:

0.34 mol H2SO4 : 0.2 mol KOH = 1.7 : 1

Since the ratio of H2SO4 to KOH is greater than 1:2, KOH is the limiting reactant. Therefore, all of the KOH is used up in the reaction, leaving some H2SO4 unreacted.

To find the amount of H2SO4 remaining, you will need to use the mole ratio of H2SO4 to KOH.

Since 2 moles of KOH react with 1 mole of H2SO4, you can use the mole ratio:

0.2 mol KOH x (1 mol H2SO4 / 2 mol KOH) = 0.1 mol H2SO4 remaining

Finally, you can determine the concentration of the H2SO4 remaining:

concentration = moles / volume

concentration = 0.1 mol / (0.850 L + 0.800 L)

concentration = 0.056 M

Therefore, the concentration of sulfuric acid that remains after neutralization is 0.056 M.

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To determine the volume of NaOH used to reach the equivalence point of the titration using the titration data, we need to find the point where the acid and base are neutralized.

At this point, the moles of acid and base are equal, and this is called the equivalence point.To find the volume of NaOH used at the equivalence point, we can use the following

Steps:1. Plot the titration data on a graph of pH versus volume of NaOH added.

Steps:2. Identify the point where the pH changes abruptly. This is the equivalence point.

Steps:3. Determine the volume of NaOH added at the equivalence point by reading the volume from the graph.

Steps:4. Compare the volume of NaOH used at the equivalence point of the titration with the predicted volume calculated above.The extent of agreement with the predicted volume can be assessed by calculating the percent error.

The percent error is calculated using the formula:

                                      Percent error = [(experimental value - theoretical value) / theoretical value] x 100

If the percent error is small, then the agreement is good. If the percent error is large, then there is a significant difference between the predicted and experimental values.

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The molecular equation for the gas evolution reaction between aqueous hydrobromic acid (HBr) and aqueous lithium sulfite (Li2SO3) is as follows:  2 HBr (aq) + [tex]Li_{2} So_{3}[/tex] (aq) → 2 LiBr (aq) + [tex]H_{2} So_{3}[/tex] (aq)

In this reaction, hydrobromic acid (HBr) reacts with lithium sulfite ([tex]Li_{2} So_{3}[/tex]) to form lithium bromide (LiBr) and sulfurous acid ([tex]H_{2} So_{3}[/tex]). The sulfurous acid is unstable and decomposes into water( [tex]H_{2o[/tex]) and sulfur dioxide gas ([tex]So_{2}[/tex]):

[tex]H_{2} So_{3}[/tex] (aq) → [tex]H_{2} 0[/tex]l) + [tex]So_{2}[/tex] (g)

The overall reaction is:

2 HBr (aq) + [tex]Li_{2} So_{3}[/tex] (aq) → 2 LiBr (aq) + [tex]H_{2} o[/tex] (l) + [tex]So_{2}[/tex] (g)

In this gas evolution reaction, the mixing of the two aqueous solutions results in the formation of a new compound, lithium bromide, which remains dissolved in the solution. The other product, sulfurous acid, decomposes into water and sulfur dioxide gas, which is released as bubbles in the solution. This release of gas is the characteristic feature of gas evolution reactions.

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A diatomic molecule consists of two atoms of the same element or two atoms of different elements, bonded together with a covalent bond, or in some cases, two lone pairs of electrons. So all statements are true.

A diatomic molecule is a molecule made up of two atoms of the same element or two atoms of different elements bonded together. The bonding of the atoms is usually done through a covalent bond, meaning that electrons are shared between the two atoms in order to create a stable arrangement.  In the case of two atoms of different elements, each atom has a different electronegativity, resulting in the formation of a polar covalent bond. This means that the electrons will be pulled closer to one atom than the other, resulting in an overall dipole moment for the molecule. In the case of two atoms of the same element, a nonpolar covalent bond is formed. This means that the electrons are shared equally between the two atoms and no dipole moment is formed. In some cases, two lone pairs of electrons may be present instead of a covalent bond. This results in a molecule with a larger overall dipole moment.

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