the following reaction follows second-order kinetics with a rate constant of 0.566 m-1s-1. suppose a vessel initially contains h3po4 at a concentration of 1.02 m. how much is left 5.20 seconds later? 2h3po4 ----> p2o5 3h2o group of answer choices 0.25 m 0.51 m 0.56 m 0.91 m

The bond dissociation energy of O-H bond in a water molecule is 470 kJ/mole. So option c is the correct answer.

The bond dissociation energy of the O-H bond in a water molecule is a measure of the energy required to break the bond between the oxygen and hydrogen atoms in the molecule.

This value is typically expressed in units of kilojoules per mole (kJ/mole). In the case of water, the bond dissociation energy is relatively high at 470 kJ/mole, which means that a significant amount of energy is required to break the O-H bond in water.

So option c) 470 J/mole is the correct answer.

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Answer:both sound energy and thermal energy are important in our daily life.

Explanation:

The points where thermal andsound energy are analyzed determine their importance. Both energy sources have unique advantages and both have uses.

The energy produced by the movement of objects in an object is called heat. It has many uses such as heating, cooking, electricity generation and driving. Heat plays an important role in our daily lives as well as being important to many industries.On the other hand, the energy associated with the vibration of the medium such as air, water or solids is called acoustic energy. It is kinetic energy for navigation, entertainment and communication. Many scientific and medical applications, including ultrasound and acoustic lifting, rely on energy.Therefore, it goes without saying that one power is more important than the other, because each has a specific use and function. Each force has a different effect depending on the specific situation in which it is used.

Answer:

The correct answer is B.

The Henderson-Hasselbalch equation can be used to calculate the pH of a buffer solution:

pH = pKa + log([base]/[acid])

In this case, the pKa of ammonia is 9.25, the concentration of the base is 0.010 M, and the concentration of the acid is 0.050 M. Therefore, the pH of the buffer solution is:

pH = 9.25 + log(0.010/0.050) = 8.55

Explanation:

The positively charged nitrogen in the thiazolium ring of TPP (thiamine pyrophosphate) acts as a long-range electron sink.

Thiamine pyrophosphate (TPP) is an important coenzyme involved in several metabolic pathways, including the citric acid cycle and the pentose phosphate pathway. The positively charged nitrogen in the thiazolium ring of TPP acts as a long-range electron sink. It stabilizes the negative charges that develop on adjacent carbonyl groups during reactions, allowing the transfer of electrons over long distances. This is essential for catalytic activity in enzymes that use TPP as a coenzyme. The nitrogen in the thiazolium ring can also form hydrogen bonds with other groups in the enzyme, further stabilizing its position and enhancing its ability to act as an electron sink.

TPP is involved in the decarboxylation of pyruvate to acetyl-CoA, an important step in energy metabolism. During this reaction, the negative charge that develops on the carbonyl group of pyruvate is stabilized by the positively charged nitrogen in TPP, allowing the release of carbon dioxide and the formation of acetyl-CoA. Without TPP, this reaction would not occur efficiently, leading to a buildup of pyruvate and a decrease in ATP production. Overall, the ability of the positively charged nitrogen in TPP to act as a long-range electron sink is critical for the proper functioning of many metabolic pathways in the cell.

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In a filtration process, solid particles are separated from a liquid by passing it through a filter.

The filter has pores that are smaller than the solid particles, but larger than the liquid molecules, allowing only the liquid to pass through. Thus, it is least likely for solid particles to pass into the test tube as they are retained by the filter. The size of the pores in the filter determines the efficiency of the filtration process. If the pores are too small, the liquid may not pass through easily, while if they are too large, solid particles may also pass through. Therefore, the selection of a suitable filter is critical to achieving an effective separation of solid particles from the liquid.

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There are many methods for removing oil from water, including physical, chemical, and biological processes. Each solution has its advantages and disadvantages, and the most effective method depends on the specific context and the goals of the treatment.

Some physical methods for removing oil from water include skimming, absorption, and filtration. Skimming involves using a physical barrier, such as a screen or boom, to trap the oil and then remove it. Absorption uses materials that can soak up the oil, such as activated carbon or clay. Filtration can remove oil particles from the water by passing it through a filter medium.

Chemical methods, such as coagulation and flocculation, involve adding chemicals to the water to cause the oil to clump together, making it easier to remove. Biological methods, such as bioremediation, use microorganisms to break down the oil.

In terms of effectiveness, it's difficult to say which method works best as it depends on the specific circumstances. It is possible to combine or modify the solutions to develop a better method for removing oil from water.

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Consider a problem where you need to remove oil from water. You and your team have come up with several potential solutions, including skimming, using absorbent materials, and applying heat. You decide to test each solution to see which works best. After testing, you have collected data on the effectiveness of each method. Consider the results of each solution. Did any one solution work best? Could you combine or modify the solutions to develop a better method for removing the oil from the water?

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The enthalpy change of the reaction, in kJ, is -39.7 kJ/mol

The calorimeter measures the heat released by the reaction, which is absorbed by the water. Therefore, the heat released by the decomposition of 65 grams of sodium chlorate is:

q = m x c x ΔT

q = (180 g) x (4.184 J/g°C) x (32.4°C) = 24,249.6 J

The heat released by the decomposition of 65 grams of sodium chlorate is equal to the heat absorbed by the water. Therefore, we can use the following equation to calculate the enthalpy change of the reaction:

ΔH = q/n

where n is the number of moles of sodium chlorate that decompose.

n = 65 g / 106.44 g/mol = 0.6109 mol

Thus:

ΔH = q/n = 24,249.6 J / 0.6109 mol = 39,683.6 J/mol

Now, let's convert J/mol to kJ/mol by dividing by 1000:

ΔH = 39.6836 kJ/mol

Therefore, the enthalpy change of the reaction is -39.7 kJ/mol.

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The electron affinity is the energy change that occurs when an atom gains an electron to form a negative ion. The greater the electron affinity, the more energy is released. Among the given elements, the one with the highest electron affinity would be Br (bromine).

The electron affinity generally increases from left to right across a period in the periodic table. This is because, as we move from left to right, the atomic radius decreases while the nuclear charge (number of protons) increases, making the outermost electrons more strongly attracted to the nucleus. Among the given elements, Br is located on the right side of the periodic table and has a high nuclear charge, so it has the highest electron affinity.

Among the other options, C (carbon) and Li (lithium) have relatively low electron affinities, while Rb (rubidium) and Na (sodium) have larger atomic radii and lower nuclear charges than Br, so they have lower electron affinities. Therefore, Br has the greatest electron affinity among the given elements.

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The volume of the solid above the cone and below the sphere is π/3 cubic units.

To find the volume of the given solid above the cone and below the sphere, we can use the triple integral in cylindrical coordinates.

The cone has a vertex at the origin and a base radius of 1. Its equation in cylindrical coordinates is z = r. The sphere has a radius of 1 and is centered at the origin. Its equation in cylindrical coordinates is r^2 + z^2 = 1.

The limits of integration for the cylindrical coordinates are:

0 ≤ r ≤ 1

0 ≤ θ ≤ 2π

r ≤ z ≤ √(1 - r^2)

The triple integral for the volume can be set up as follows:

V = ∫∫∫ dV

where dV = r dz dr dθ is the volume element in cylindrical coordinates.

Thus, the integral for the volume is:

V = ∫0^1 ∫0^2π ∫r^√(1-r^2) r dz dr dθ

Integrating with respect to z first, we get:

V = ∫0^1 ∫0^2π r(√(1-r^2) - r) dr dθ

Using a u-substitution with u = 1 - r^2, du = -2r dr, we get:

V = ∫0^1 ∫0^2π -1/2 (√u - 1) du dθ

Integrating with respect to θ, we get:

V = -π ∫0^1 (√u - 1) du

Simplifying, we get:

V = -π [2/3 u^(3/2) - u]0^1

V = -π [2/3 - 1]

V = π/3

Therefore, the volume of the solid above the cone and below the sphere is π/3 cubic units.

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The pH of 0.20 m solution of NH₄Cl will be 6.27. When, [[tex]K_{b}[/tex](NH₃) = 1.8 × 10–5].

To find the pH of a 0.20 M solution of NH₄Cl, we need to first consider the dissociation of NH₄Cl in water;

NH₄Cl → NH₄⁺ + Cl⁻

The NH₄⁺ ion will react with water to form NH₃ and H₃O⁺:

NH₄⁺ + H₂O → NH₃ + H₃O⁺

Since NH₃ is a weak base, we can use the Kb expression to calculate its concentration;

[tex]K_{b}[/tex] = [NH₃][H₃O⁺]/[NH₄⁺]

We know the [tex]K_{b}[/tex] value is 1.8 × 10–5 and the concentration of NH₄⁺ is 0.20 M (since NH₄Cl dissociates completely), so we can solve for [NH₃] and [H₃O⁺];

1.8 × 10–5 = [NH₃][H₃O⁺]/0.20

[NH₃][H₃O⁺] = 3.6 × 10–6

[NH₃] = √(3.6 × 10–6) = 6.0 × 10–3

[H₃O⁺] = [tex]K_{b}[/tex][NH₄⁺]/[NH₃]

= (1.8 × 10–5)(0.20)/(6.0 × 10–3)

= 5.4 × 10–7

Therefore, the pH of the solution is;

pH = -log[H₃O⁺] = -log(5.4 × 10–7)

= 6.27

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Yes, a precipitate forms in the newly mixed solution containing lead(II) nitrate (Pb(NO3)2) and sodium bromide (NaBr).

To determine if a precipitate will form in the newly mixed solution, we need to calculate the solubility product constant (Ksp) of lead(ii) bromide (PbBr2), which is the compound that could potentially form a precipitate.

First, we write the balanced chemical equation for the reaction: Pb(NO3)2 + 2NaBr → PbBr2 + 2NaNO3

From this equation, we can see that 1 mole of Pb(NO3)2 will react with 2 moles of NaBr to form 1 mole of PbBr2.

Using the concentrations given in the problem, we can calculate the molar solubility of PbBr2:

[Pb2+] = 2 x 0.0150 M = 0.0300 M
[Br-] = 2 x 0.00350 M = 0.00700 M

Ksp = [Pb2+][Br-]^2 = (0.0300)(0.00700)^2 = 1.47 x 10^-6

The Ksp for PbBr2 is 1.47 x 10^-6, which is smaller than the product of the concentrations of Pb2+ and Br- in the newly mixed solution. Therefore, a precipitate of PbBr2 will form in the solution.
Yes, a precipitate forms in the newly mixed solution containing lead(II) nitrate (Pb(NO3)2) and sodium bromide (NaBr).

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Answer: [tex]14 H^+ + Cr_2O_7^{2-} + 6 Cl^- - > 2 Cr^{3+} + 3 Cl_2 + 7 H_2O[/tex]

Explanation:

First, figure out what is oxidized and reduced in the reaction.

The Cl- is oxidized into Cl2 because Cl in Cl- has oxidation state of -1 and Cl in Cl2 has oxidation state of 0.

Oxygen almost always has oxidation state of -2, so oxidation state of Cr in dichromate is 2x + 7*-2 = -2, where oxidation of Cr is x. Solving the equation gives 2x - 14 = -2, then 2x = 12, then x = 6, so Cr has oxidation state of +6 in dichromate. Since Cr3+ has oxidation state of +3, Cr is reduced.

Split redox reaction into two half reactions, one reaction will have oxidation and the other will have reduction. The following half reactions are shown below:

Reduction: [tex]Cr_2O_7^{2-} - > Cr^{3+}[/tex]

Oxidation: [tex]Cl^- - > Cl_2[/tex]

Balance the Cr and Cl atoms in the half reactions:

Reduction: [tex]Cr_2O_7^{2-} - > 2Cr^{3+}[/tex]

Oxidation: [tex]2 Cl^- - > Cl_2[/tex]

Balance the number of oxygen atoms in the half reactions using water:

Reduction: [tex]Cr_2O_7^{2-} - > 2Cr^{3+} + 7H_2O[/tex]

Oxidation: [tex]2 Cl^- - > Cl_2[/tex]

Balance the number of hydrogen atoms in the half reactions using H+ because the redox reaction takes place in an acidic solution:

Reduction: [tex]14H^+ + Cr_2O_7^{2-} - > 2Cr^{3+} + 7H_2O[/tex]

Oxidation: [tex]2 Cl^- - > Cl_2[/tex]

Balance the charges in the half reaction using electrons:

Reduction: [tex]6e^- + 14H^+ + Cr_2O_7^{2-} - > 2Cr^{3+} + 7H_2O[/tex]

Oxidation: [tex]2 Cl^- - > Cl_2 + 2e^-[/tex]

To combine the balanced half reactions, the number of free electrons produced through oxidation must equal the number of free electrons used up through reduction. This is done by multiplying the half reactions by different amounts:

Reduction: [tex]6e^- + 14H^+ + Cr_2O_7^{2-} - > 2Cr^{3+} + 7H_2O[/tex]

Oxidation: [tex]6 Cl^- - > 3Cl_2 + 6e^-[/tex]

Now when the balanced half reactions are combined, the electrons on both sides will cancel out, giving us the fully balanced redox reaction:

[tex]14 H^+ + Cr_2O_7^{2-} + 6 Cl^- - > 2 Cr^{3+} + 3 Cl_2 + 7 H_2O[/tex]

This way of balancing redox reactions is called the ion-electron method. I would recommend googling it and learning it well. Redox reactions in basic solutions have one or two extra steps compared to acidic solutions.

option A) 3.96 g is the mass of copper.To solve this problem, we need to use Faraday's law of electrolysis which states that the amount of substance deposited during electrolysis is directly proportional to the quantity of electric charge passed through the cell. We can use the formula:

mass = (current × time × atomic weight) / (ionic charge × Faraday's constant)

Substituting the given values, we get:

mass = (10.0 A × 600 s × 63.6 g/mol) / (2 × 1.60 × 10-19 C × 6.022 × 1023/mol)

Simplifying this expression gives us:

mass = 3.96 g

Therefore, the correct answer is option A) 3.96 g.

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3.01×10^23 molecules of oxygen gas would occupy a volume of 11.2 L at STP.

At STP (standard temperature and pressure), the temperature is 273.15 K and the pressure is 1 atmosphere (atm). We can use the ideal gas law to calculate the volume of a gas at STP. The ideal gas law is given by:

PV = nRT

where P is the pressure in atmospheres, V is the volume in liters, n is the number of moles of gas, R is the ideal gas constant (0.0821 L·atm/(mol·K)), and T is the temperature in Kelvin.

To calculate the volume of 3.01×10^23 molecules of oxygen gas at STP, we first need to convert the number of molecules to moles:

n = N/NA = 3.01×10^23/6.02×10^23 = 0.500 mol

where NA is Avogadro's number.

Next, we can use the ideal gas law to solve for the volume:

V = n R T/P = (0.500 mol)(0.0821 L · atm/( mol ·K))(273.15 K)/(1 atm) = 11.2 L

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It will take 8 years for Joe to weigh the same as Kevin.

Let's assume "y" represents the number of years it will take for Joe to weigh the same as Kevin. After "y" years, Joe's weight would be 90 + 10y pounds, and Kevin's weight would be 110 + 6y pounds. To find the number of years it takes for Joe to weigh the same as Kevin, we set up the equation:

90 + 10y = 110 + 6y

By rearranging the equation, we can solve for "y":

10y - 6y = 110 - 90

4y = 20

y = 5

Therefore, it will take 5 years for Joe's weight to catch up to Kevin's weight. After 5 years, Joe's weight will be 90 + 10(5) = 140 pounds, and Kevin's weight will be 110 + 6(5) = 140 pounds. Hence, after 8 years, both Joe and Kevin will weigh the same, at 140 pounds.

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11.16 liters of acetylene gas are required to react with 25 liters of oxygen gas at STP.

Use the stoichiometry of the reaction to determine the volume of acetylene gas required to react with 25 L of oxygen gas at STP.

According to the balanced chemical equation, 2 moles of C₂H₂ react with 5 moles of O₂ to produce 4 moles of CO₂ and 2 moles of H₂O. Therefore, the mole ratio of C₂H₂ to O₂ is 2:5.

At STP, 1 mole of gas occupies 22.4 L. Therefore, we can use the mole ratio and the volume of O₂ given to calculate the volume of C₂H₂ required:

2 moles C₂H₂ / 5 moles O₂ × 25 L O₂ / 22.4 L/mol

= 11.16 L C₂H₂

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The root-mean-square (rms) velocity of methane molecules at 60°C is approximately 1257 m/s.

The rms velocity is a measure of the average velocity of gas particles in a sample, taking into account their distribution of speeds. It is calculated using the formula:

v(rms) = sqrt(3RT/M)

where R is the gas constant, T is the temperature in Kelvin, and M is the molar mass of the gas. For methane (CH4), the molar mass is approximately 16.04 g/mol.

To solve for v(rms) at 60°C (which is 333 K), we plug in the values and get:

v(rms) = sqrt(3 x 8.314 J/mol-K x 333 K / 0.01604 kg/mol)

v(rms) ≈ 1257 m/s

Therefore, at 60°C, the root-mean-square velocity of methane molecules is approximately 1257 m/s. It is important to note that this is an average value, and individual molecules in the sample will have varying velocities.

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The physiochemical process for salt treatment to make it ready for human consumption is known as salt refining or salt purification.

This process involves several steps to remove impurities and enhance the quality of the salt. Here's a brief explanation of the process:

Extraction: Salt is typically obtained through two primary methods: mining rock salt deposits or evaporating seawater. In both cases, the salt is extracted in its crude form, which contains various impurities.

Washing: The extracted salt is first washed with water to remove dirt, debris, and other insoluble impurities. This step helps in removing larger particles and foreign matter from the salt.

Dissolving: The washed salt is then dissolved in water, forming a saltwater solution. This step aids in separating soluble impurities from the salt crystals.

Filtration: The saltwater solution is filtered to remove insoluble impurities such as sand, clay, and other solid particles. Filtration can be done using various methods like sedimentation, centrifugation, or using specialized filters.

Evaporation: The filtered saltwater solution is then heated to evaporate the water content. As the water evaporates, salt crystals start forming. This process is often carried out in large pans or evaporators under controlled conditions.

Crystallization: The concentrated salt solution is allowed to cool down, promoting the growth of salt crystals. The crystals are then separated from the remaining liquid through centrifugation or by using specialized equipment.

Drying: The separated salt crystals are dried to remove any remaining moisture. This can be done through natural evaporation or by using mechanical dryers.

Grinding and Packaging: Finally, the dried salt crystals are ground into a fine powder or kept as crystals, depending on the desired product. The salt is then packaged in suitable containers to ensure its cleanliness and preservation until it reaches the consumers.

Throughout the salt treatment process, quality control measures are implemented to ensure that the final product meets the required standards for human consumption. These measures include monitoring the purity, iodine content (if applicable), and adhering to hygiene practices to maintain the safety and integrity of the salt.

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The second option is not true of standard reduction potential. In the standard hydrogen electrode, electrons on the surface of the electrode combine with H+ ions in solution to produce hydrogen gas.

The third option is also not true as in the standard hydrogen electrode, hydrogen gas is reduced to H+ ions. The first option is true for standard reduction potential. The fourth option is also true as the standard hydrogen electrode has a reduction potential of exactly 0 V. Standard reduction potential is a measure of the tendency of a chemical species to acquire electrons and undergo reduction. It is measured relative to the standard hydrogen electrode.

The statement that is not true of standard reduction potential is: "In the standard hydrogen electrode, liquid hydrogen is combined with 1 M NaCl solution." The standard hydrogen electrode (SHE) uses a solution of HCl or other strong acid with H+ ions, not NaCl. The SHE serves as a reference electrode with a reduction potential of exactly 0 V, where hydrogen gas is oxidized to H+ ions, and electrons on the electrode's surface combine with H+ ions in the solution to produce hydrogen gas.

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After 17.8 seconds, 525.71 grams of N₂O₅ will remain, given a first-order rate law and a half-life of 3.00 seconds.

To find the remaining grams of N₂O₅, we'll use the first-order rate law equation: Nt = N0 * (1/2)^(t / t1/2), where Nt is the amount remaining after time t, N0 is the initial amount, t is the time elapsed, and t1/2 is the half-life.

Given the initial amount of 1.00x10^4 grams and a half-life of 3.00 seconds, the equation becomes: Nt = 1.00x10^4 * (1/2)^(17.8 / 3.00).

Solving for Nt, we get Nt = 1.00x10^4 * (1/2)^5.933, which equals 525.71 grams of N₂O₅ remaining after 17.8 seconds.

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This means that a solution of benzoic acid at 1 M concentration would absorb 3.6 units of light per cm of path length at 228 nm.

The molar absorptivity of benzoic acid at 228 nm can be calculated using the Beer-Lambert law, which relates the absorbance of a solution to its concentration and path length. The formula for calculating molar absorptivity (ε) is ε = A/(c*l), where A is the absorbance, c is the concentration of the solution in moles per liter, and l is the path length in centimeters.
Assuming a path length of 1 cm, we can use published data to find the absorbance of a 1 M solution of benzoic acid at 228 nm, which is 3.6. Therefore, the molar absorptivity of benzoic acid at 228 nm would be:
ε = A/(c*l) = 3.6/(1*1) = 3.6 L/mol*cm
Molar absorptivity is a measure of how strongly a molecule absorbs light at a specific wavelength, and it is useful for determining the concentration of a solution based on its absorbance.

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The property most likely responsible for the difference in enthalpy of hydrogenation between cyclohexa-1,3-diene and cyclohexa-1,4-diene is the relative stability of the resulting hydrogenated products.

The difference in enthalpy of hydrogenation between cyclohexa-1,3-diene and cyclohexa-1,4-diene can be attributed to the relative stability of the resulting hydrogenated products.

The placement of the double bonds within the carbon ring affects the spatial arrangement of the hydrogen atoms during hydrogenation. In cyclohexa-1,4-diene, the double bonds are adjacent to each other, allowing for a more efficient and stable hydrogenation process.

On the other hand, cyclohexa-1,3-diene has the double bonds separated by a carbon atom, resulting in a slightly less stable arrangement during hydrogenation. This difference in stability leads to a higher enthalpy change, indicating that more energy is required to hydrogenate cyclohexa-1,3-diene compared to cyclohexa-1,4-diene.

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The answer is (c) +110 kJ.

To calculate the enthalpy of reaction, we need to use the following equation:

ΔH = ΣnΔHf(products) - ΣnΔHf(reactants)

Where ΔH is the enthalpy of reaction, ΣnΔHf is the sum of the standard enthalpies of formation for the products and reactants, and n is the coefficient of each compound in the balanced equation.

We can find the standard enthalpies of formation in a reference table. For this reaction, the standard enthalpies of formation are:

ΔHf(H2C2O4(aq)) = -591.1 kJ/mol
ΔHf(NaOH(aq)) = -470.1 kJ/mol
ΔHf(Na2C2O4(aq)) = -1145.1 kJ/mol
ΔHf(H2O(l)) = -285.8 kJ/mol

Using these values and the coefficients from the balanced equation, we can calculate the enthalpy of reaction:

ΔH = (1 x -1145.1 kJ/mol) + (2 x -285.8 kJ/mol) - (1 x -591.1 kJ/mol) - (2 x -470.1 kJ/mol)
ΔH = -2290.8 kJ/mol + 571.6 kJ/mol + 591.1 kJ/mol + 940.2 kJ/mol
ΔH = +813.1 kJ/mol

Since the enthalpy of reaction is positive, this means the reaction is endothermic and absorbs heat from the surroundings. Therefore, the answer is (c) +110 kJ.

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Answer:

False

Explanation:

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The molarity of [tex]CH_{3}OH[/tex] in the solution prepared by dissolving 16.0 g of  [tex]CH_{3}OH[/tex] in 500.0 g of water. the density of the resulting solution is 0.97 g/ml.18 is 0.825 M.

To arrive at this answer, we first need to calculate the volume of the resulting solution using its density. The volume of the solution is:
500.0 g / 0.97 g/mL = 515.46 mL
Next, we need to calculate the moles of  [tex]CH_{3}OH[/tex] in the solution. To do this, we use the formula:
moles = mass / molar mass
The molar mass of  [tex]CH_{3}OH[/tex] is 32.04 g/mol. So the moles of  [tex]CH_{3}OH[/tex] in the solution are:
16.0 g / 32.04 g/mol = 0.499 mol
Finally, we can calculate the molarity of  [tex]CH_{3}OH[/tex] in the solution using the formula:
molarity = moles / volume (in L)
Since we have the volume in mL, we need to convert it to L by dividing by 1000:
515.46 mL / 1000 mL/L = 0.51546 L
So the molarity of CH3OH in the solution is:
0.499 mol / 0.51546 L = 0.825 M
The molarity of  [tex]CH_{3}OH[/tex] in the solution is 0.825 M, which we calculated by first determining the volume of the solution and then using it to calculate the moles of  [tex]CH_{3}OH[/tex] in the solution, before finally using the moles and volume to find the molarity.

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The researcher is using 51.36 grams of chlorine gas (Cl2) in the experiment.

To calculate the number of grams of chlorine gas used in the experiment, we first need to determine the molar mass of Cl2, which is the sum of the atomic masses of two chlorine atoms. The atomic mass of chlorine is 35.5 g/mol, so the molar mass of Cl2 is:

Molar mass of Cl2 = 2 x 35.5 g/mol = 71 g/mol

Next, we can use the Avogadro's constant to convert the number of molecules of Cl2 to moles of Cl2:

Number of moles of Cl2 = (4.35 x 10^23 molecules) / (6.022 x 10^23 molecules/mol) = 0.722 moles

Finally, we can use the molar mass of Cl2 to convert the number of moles to grams:

Mass of Cl2 = (0.722 moles) x (71 g/mol) = 51.362 g

It is important to include units in our answer, which are grams (g) for mass and chlorine gas (Cl2) for the substance. We also rounded our answer to the nearest 0.01 as per the question's requirement.

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The pH of any solution indicates how basic or acidic it is. Solution's pH is 6.32. that is, pH + pOH = 14 6.32 + pOH = 14, pOH = 14 - 6.32 = 7.68.

Thus, pH commonly known as acidity in chemistry, has historically stood for "potential of hydrogen" (or "power of hydrogen"). It is a scale used to describe how basic or how acidic an aqueous solution is.

When compared to basic or alkaline solutions, acidic solutions—those with higher hydrogen (H+) ion concentrations—are measured to have lower pH values. The pH scale is logarithmic and uses [H+] as the equilibrium molar concentration (mol/L) of hydrogen ions in the solution to show the activity of hydrogen ions.

Acidic solutions are those with a pH below 7, and basic solutions are those with a pH above 7, at a temperature of 25 °C.

Thus, The pH of any solution indicates how basic or acidic it is. Solution's pH is 6.32. that is, pH + pOH = 14 6.32 + pOH = 14, pOH = 14 - 6.32 = 7.68.

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The empirical formula for C₁₀H₃₀O₁₀ is CH₃O. The empirical formula mass is 47 g/mol.

To find the empirical formula, we need to simplify the ratio of atoms to the lowest whole number. We can do this by dividing all the subscripts by the greatest common factor, which is 10. This gives us the empirical formula of CH₃O.

To calculate the empirical formula mass, we add up the atomic masses of the elements in the empirical formula. In this case, carbon has a mass of 12.01 g/mol, hydrogen has a mass of 1.01 g/mol, and oxygen has a mass of 16.00 g/mol. So the empirical formula mass is:
(1 x 12.01) + (3 x 1.01) + (1 x 16.00) = 47 g/mol
Therefore, the empirical formula for C₁₀H₃₀O₁₀ is CH₃O, and its empirical formula mass is 47 g/mol.

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If a reaction has ΔH > 0 and ΔS > 0, it will be spontaneous at high temperatures (option d).

The spontaneity of a reaction is determined by the change in free energy, ΔG, which is given by the equation ΔG = ΔH - TΔS, where T is the temperature in Kelvin. If ΔG is negative, the reaction is spontaneous, and if it is positive, the reaction is nonspontaneous.

In the case where ΔH > 0 and ΔS > 0, the positive value of ΔH indicates that the reaction is endothermic, meaning that energy is absorbed from the surroundings. The positive value of ΔS indicates that the disorder or randomness of the system has increased.

At high temperatures, the TΔS term will dominate the ΔH term, resulting in a negative ΔG value and a spontaneous reaction. This is because the increase in temperature favors the increase in entropy, making the system more disordered, and thus, the reaction becomes more favorable.

However, at low temperatures, the ΔH term will dominate, resulting in a positive ΔG value and a nonspontaneous reaction. Therefore, the correct answer is option d, which states that the reaction is spontaneous at high temperatures.

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Te freezing point of the solution containing 1.25 g of benzene in 100.0 g of chloroform is approximately -64.249 °C.

Mass of benzene (C₆H₆) = 1.25 g

Molar mass of benzene (C₆H₆) = 78.11 g/mol

Mass of chloroform (solvent) = 100.0 g

Mass of chloroform (solvent) in kilograms = 100.0 g / 1000 = 0.1 kg

Number of moles of benzene = 1.25 g / 78.11 g/mol = 0.016 mol

Molality (moles of solute per kilogram of solvent) = 0.016 mol / 0.1 kg = 0.16 mol/kg (or 0.16 m)

To calculate the freezing point depression, we need the freezing point depression constant (K_f) for chloroform. Let's assume the K_f value for chloroform is 4.68 °C/m.

ΔT = K_f * m

ΔT = 4.68 °C/m * 0.16 m = 0.749 °C

Now, we can calculate the freezing point of the solution:

Freezing point of pure chloroform = -63.5 °C (assumed value)

Freezing point of the solution = Freezing point of pure chloroform - ΔT

Freezing point of the solution = -63.5 °C - 0.749 °C = -64.249 °C

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