The given balanced equation represents the reaction between methane ([tex]CH_4[/tex]) and oxygen ([tex]O_2[/tex]) to produce carbon dioxide ([tex]CO_2[/tex]) and water ([tex]H_2O[/tex]). Let's evaluate each statement:
1. True: The balanced equation shows that one methane molecule reacts with two oxygen molecules to produce one carbon dioxide molecule and two water molecules. The coefficients in the equation represent the stoichiometric ratios, indicating the number of molecules involved in the reaction.
2. True: Methane ([tex]CH_4[/tex]) is a hydrocarbon composed of one carbon atom and four hydrogen atoms, while carbon dioxide ([tex]CO_2[/tex]) consists of one carbon atom and two oxygen atoms. The reaction results in the conversion of the carbon atom from methane to carbon dioxide.
3. True: Oxygen ([tex]O_2[/tex]) is a diatomic molecule, meaning it consists of two oxygen atoms bonded together. The balanced equation shows that two oxygen molecules are required to react with one methane molecule, forming two water molecules and one carbon dioxide molecule.
4. True: The reaction is balanced, as the number of atoms of each element is the same on both sides of the equation. There is one carbon atom, four hydrogen atoms, and four oxygen atoms on each side.
In summary, all of the statements are true. The balanced equation accurately represents the reaction between methane and oxygen, resulting in the formation of carbon dioxide and water.
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A container is initially filled with only PCl5 at 2.5 atm. At equilibrium, the pressure of PCl5 is 0.5 atm. Determine the value of Kp for the reaction.
The value of Kp for the reaction PCl₅(g) ⇌ PCl₃(g) + Cl₂(g) is 4.0.
Initial pressure of PCl₅ (PCl₅(initial)) = 2.5 atm
Pressure of PCl₅ at equilibrium (PCl₅(eq)) = 0.5 atm
Using the equation for Kp:
Kp = (PCl₃ * Cl₂) / PCl₅
We need to find the value of Kp.
Substituting the given pressures:
Kp = (PCl₃(eq) * Cl₂(eq)) / PCl₅(eq)
Kp = (PCl₃(eq) * Cl₂(eq)) / 0.5
Since Kp is a dimensionless constant, we can simplify the equation further.
Kp = (2 * PCl₃(eq) * Cl₂(eq)) / 1
Kp = 2 * PCl₃(eq) * Cl₂(eq)
Now, we need to determine the value of PCl₃(eq) * Cl₂(eq).
To find that, we subtract the initial pressure of PCl₅ from its pressure at equilibrium:
PCl₃(eq) * Cl₂(eq) = PCl₅(initial) - PCl₅(eq)
PCl₃(eq) * Cl₂(eq) = 2.5 - 0.5
PCl₃(eq) * Cl₂(eq) = 2.0
Now, substituting this value back into the equation for Kp:
Kp = 2 * 2.0
Kp = 4.0
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THE COMPLETE QUESTION IS:
A container is initially filled with only PCl₅ at 2.5 atm. At equilibrium, the pressure of PCl₅ is 0.5 atm. Assuming the partial pressures are in atmospheres, what is Kp for the reaction below?
PCl₅(g) ⇌ PCl₃(g) + Cl₂(g)
Enter your answer accurate to 3 significant figures.
Another student performs the titration and does not notice the air bubbles trapped in the tip of the buret. How will the reported value for the concentration of the acid be affected
The presence of air bubbles in the buret tip during titration can lead to an overestimation of the concentration of the acid being titrated, compromising the accuracy of the results.
It is crucial to take measures to prevent and eliminate air bubbles to obtain reliable and precise concentration values. Titration is a chemical method that determines the concentration of a substance in a solution. In this method, a known volume of a solution is used to react with an unknown concentration of a solution in order to determine the unknown concentration. The volume of the titrant required to react with the unknown solution is then measured to determine the concentration of the unknown solution.
However, if air bubbles are trapped in the tip of the buret during the titration process, the accuracy of the result will be affected. The effect of air bubbles on the reported value for the concentration of the acid is as follows:
The presence of air bubbles in the tip of the buret means that there is a smaller volume of titrant delivered to the solution being titrated. This is because the air bubbles occupy some of the space in the tip of the buret, and therefore the volume of titrant delivered is less than the expected volume. As a result, the concentration of the acid will appear to be higher than it actually is, as the volume of titrant used is less than the actual volume.
In other words, the reported value for the concentration of the acid will be higher than the true value. To obtain accurate results, it is therefore important to ensure that there are no air bubbles trapped in the tip of the buret during the titration process.
Overall, the presence of air bubbles in the buret tip during titration can lead to an overestimation of the concentration of the acid being titrated, compromising the accuracy of the results. It is crucial to take measures to prevent and eliminate air bubbles to obtain reliable and precise concentration values.
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A 25.00 mL sample of of a chloric acid solution is completely neutralized using 22.44 mL of 0.157 M potasium hydroxide. What is the initial concentration of chloric acid
The initial concentration of chloric acid (HClO₃) in the solution is calculated to be 0.1408 M using stoichiometry and the volume and concentration of the potassium hydroxide solution used for neutralization.
To determine the initial concentration of chloric acid, we can use the concept of stoichiometry and the volume and concentration of the potassium hydroxide solution used for neutralization.
The balanced chemical equation for the reaction between chloric acid (HClO₃) and potassium hydroxide (KOH) is:
HClO₃ + KOH -> KClO₃ + H₂O
From the balanced equation, we can see that the stoichiometric ratio between HClO₃ and KOH is 1:1.
First, we need to calculate the number of moles of KOH used:
moles of KOH = volume of KOH solution * concentration of KOH solution
[tex]$\text{moles of KOH} = 22.44 , \text{mL} \times \left( \frac{0.157 , \text{mol/L}}{1000 , \text{mL}} \right) = 0.003521 , \text{mol KOH}$[/tex]
Since the stoichiometric ratio is 1:1, the moles of HClO₃ used is also 0.003521 mol.
Next, we calculate the initial concentration of HClO₃:
[tex]$\text{Initial concentration of HClO}_3 = \frac{\text{moles of HClO}_3}{\text{volume of HClO}_3 \text{ solution}}[/tex]
[tex]= \frac{0.003521 , \text{mol}}{25.00 , \text{mL}} = 0.1408 , \text{M}$[/tex]
Therefore, the initial concentration of chloric acid (HClO₃) in the solution is 0.1408 M.
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If you fry ice, explain how you would transform from one phase to the next and what
type of energy you would use to transfer to each stage.
(SOLID ICE>LIQUID WATER>GAS) Please explain the process in detail:
To transform solid ice into liquid water and then into gas, you need to provide energy to overcome the intermolecular forces holding the molecules together.
This energy is transferred in the form of heat. The process involves raising the temperature of the ice to its melting point, where it changes from a solid to a liquid, and then further increasing the temperature to reach its boiling point, where it changes from a liquid to a gas.
When frying ice, the transformation from one phase to the next involves the transfer of energy in the form of heat. Initially, the ice is in its solid phase. To convert it into liquid water, heat needs to be applied to raise its temperature to the melting point, which is 0 degrees Celsius or 32 degrees Fahrenheit. During this process, the added heat provides the energy required to overcome the intermolecular forces stir-frying between the ice molecules, causing them to break apart and transition into the liquid phase.
Once the ice has completely melted and turned into liquid water, further heating is required to reach the boiling point, which is 100 degrees Celsius or 212 degrees Fahrenheit at standard atmospheric pressure. At this stage, the added heat provides enough energy to overcome the intermolecular forces holding the liquid water molecules together, allowing them to transition into the gaseous phase.
In summary, the process of frying ice involves transferring energy in the form of heat to transform the solid ice into liquid water and then into gas. The heat energy helps break the intermolecular forces and allows the molecules to transition between phases.
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A gas has a pressure of 300 torr, a temperature of 400 K,
and a volume of 50. 0 milliliters. What volume will the gas
have at a pressure of 150 torr and a temperature of 200 K?
The volume of the gas at a pressure of 150 torr and a temperature of 200 K can be calculated using the combined gas law.
Using the formula:
(P1 * V1) / (T1) = (P2 * V2) / (T2)
where:
P1 = 300 torr (initial pressure)
V1 = 50.0 mL (initial volume)
T1 = 400 K (initial temperature)
P2 = 150 torr (final pressure)
T2 = 200 K (final temperature)
We can rearrange the formula to solve for V2 (final volume):
V2 = (P1 * V1 * T2) / (P2 * T1)
Plugging in the given values:
V2 = (300 torr * 50.0 mL * 200 K) / (150 torr * 400 K)
V2 = 20 mL
Therefore, the volume of the gas at a pressure of 150 torr and a temperature of 200 K will be 20.0 mL.
The problem can be solved using the combined gas law, which states that the ratio of the product of pressure and volume to the temperature is constant for a given amount of gas. By rearranging the formula and substituting the given values, we can solve for the final volume (V2). The final volume is found to be 20.0 mL.
The gas will have a volume of 20.0 milliliters at a pressure of 150 torr and a temperature of 200 K.
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At equilibrium, the partial pressure of H2 is 1.29 atm. What is the equilibrium partial pressure of CH4(g)
The equilibrium partial pressure of CH4(g) cannot be determined from the information given. The partial pressure of H2 is only one piece of information needed to determine the equilibrium partial pressure of CH4(g).
The other piece of information needed is the equilibrium constant for the reaction.
The reaction that is occurring is:
CH4(g) + 2H2(g) <---> C2H6(g)
The equilibrium constant for this reaction is Kp = 5.0 at 298 K.
If we know the partial pressures of H2 and C2H6 at equilibrium, we can use the equilibrium constant to calculate the partial pressure of CH4(g). However, we only know the partial pressure of H2 at equilibrium.Therefore, we cannot determine the equilibrium partial pressure of CH4(g).
The equilibrium constant for a reaction is a measure of the relative concentrations of the reactants and products at equilibrium. The equilibrium constant is constant for a given temperature and reaction, but it can vary depending on the initial concentrations of the reactants and products.
In this case, we do not know the initial concentrations of any of the reactants or products. Therefore, we cannot use the equilibrium constant to calculate the equilibrium partial pressure of CH4(g). The only way to determine the equilibrium partial pressure of CH4(g) would be to measure the partial pressures of H2 and C2H6 at equilibrium.
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The carbon atoms in a sugar ring are numbered in a clockwise direction starting with the carbon atom to the ______ of the ring oxygen atom.
The carbon atoms in a sugar ring are numbered in a clockwise direction starting with the carbon atom to the right of the ring oxygen atom.
A sugar ring is a cyclical structure made up of carbon, hydrogen, and oxygen atoms that form the backbone of carbohydrates such as glucose, fructose, and galactose, among others. Sugars can be classified as either aldoses or ketoses based on the presence of a carbonyl group.
Sugars can also be divided into monosaccharides, which are single molecules, and disaccharides, which are formed when two monosaccharides are joined by a glycosidic linkage. A sugar ring can be numbered in a clockwise or anticlockwise direction, starting with the carbon atom to the right or left of the ring oxygen atom, respectively. The carbon atoms in the sugar ring are numbered sequentially, with the ring oxygen atom receiving number 1, and the carbon atoms receiving numbers 2 through 6 in a clockwise or anticlockwise direction.
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what is being reduced? in the electrochemical cell: zn(s) | zn2 (aq) || ag (aq) | ag(s) what is being reduced? ag (aq) ag(s) zn2 (aq) zn(s)
In the given electrochemical cell: Zn(s) | Zn²⁺(aq) || Ag⁺(aq) | Ag(s), silver (Ag) is being reduced. The reaction for the same can be represented as follows: Ag⁺(aq) + e⁻ → Ag(s).
Here, silver ion (Ag⁺) is gaining an electron (e⁻) to form silver metal (Ag) which is a reduction process. Therefore, the answer is "Ag (aq) and Ag(s)" which means silver is being reduced in the given electrochemical cell.
In chemistry, an electrochemical cell refers to a system that uses chemical reactions to produce electric current or harness electrical energy to drive chemical reaction and it consists of two electrodes (an anode and a cathode) immersed in an electrolyte solution.
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what is the change in enthalpy of the disassociation of hydrogen sulfate
The change in enthalpy (ΔH) of the dissociation of hydrogen sulfate (HSO₄⁻) depends on the specific reaction and conditions in which it occurs. Without specifying the reaction or providing additional information, it is not possible to determine the exact value of ΔH for the dissociation of hydrogen sulfate.
The dissociation of hydrogen sulfate can occur through various reactions, such as:
HSO₄⁻ ⇌ H⁺ + SO₄²⁻
The change in enthalpy (ΔH) represents the heat energy change associated with a chemical reaction. It can be endothermic (positive ΔH) if heat is absorbed from the surroundings, or exothermic (negative ΔH) if heat is released to the surroundings.
To determine the ΔH for the dissociation of hydrogen sulfate, experimental data or thermodynamic calculations specific to the given reaction are required. Factors such as temperature, pressure, and concentration may influence the ΔH value.
In summary, the specific value of ΔH for the dissociation of hydrogen sulfate cannot be determined without additional information or context about the reaction and the conditions under which it occurs.
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All of the following are true about the bonding of methane, CH4, EXCEPT: the compound has tetrahedral geometry these sp3 orbitals combine with the s orbitals of the hydrogens to form molecular orbitals some of the C-H bonds are stronger than others the carbon s and p orbitals combine to form four equivalent sp3 orbitals all of the C-H bonds are equivalent
The statement "some of the C-H bonds are stronger than others" is not true about the bonding of methane (CH₄).
In methane, all the C-H bonds are equivalent. Each hydrogen atom forms a single covalent bond with the carbon atom, and all of these bonds have the same strength and length.
The four hydrogen atoms are symmetrically arranged around the central carbon atom, resulting in a tetrahedral geometry for the molecule.
Methane adopts a tetrahedral geometry, with the carbon atom at the center and the four hydrogen atoms arranged symmetrically around it.
In the formation of methane, the carbon atom's three 2p orbitals and the 2s orbital hybridize to form four equivalent sp3 hybrid orbitals.
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g 91.6 g of copper(II) sulfate pentahydrate, blue vitriol, is heated in an oven until all the water is driven off. What is the mass of the anhydrous compound
The mass of the anhydrous compound formed when 91.6 g of copper sulfate pentahydrate is heated until all the water is driven off is 58.59 g.
Copper sulfate pentahydrate is a chemical compound with the formula:
CuSO₄ · 5H₂O
This hydrated copper sulfate is also known as blue vitriol.
The amount of anhydrous copper sulfate formed when 91.6 g of copper sulfate pentahydrate is heated until all the water is driven off can be calculated as follows:
Firstly, find the molar mass of copper sulfate pentahydrate.
The molar mass of copper sulfate is 159.61 g/mol.
Then, the molar mass of water is 18.02 g/mol.
The molar mass of copper sulfate pentahydrate, which includes five water molecules, is obtained by adding the two values:
159.61 + (5 × 18.02) = 249.68 g/mol
The number of moles of copper sulfate pentahydrate can be calculated next by dividing the mass of the compound by its molar mass:
91.6 g ÷ 249.68 g/mol = 0.366 moles
Next, the mass of the anhydrous compound can be calculated.
The number of moles of water that are present in copper sulfate pentahydrate is five times the number of moles of copper sulfate.
As a result, the number of moles of water present in 0.366 moles of copper sulfate pentahydrate is given by:
0.366 moles × 5 = 1.83 moles
The mass of water present in 91.6 g of copper sulfate pentahydrate is then determined by multiplying the number of moles by the molar mass of water:
1.83 moles × 18.02 g/mol = 33.01 g
Finally, the mass of the anhydrous copper sulfate can be calculated by subtracting the mass of water lost from the original mass of copper sulfate pentahydrate:
91.6 g - 33.01 g = 58.59 g
Therefore, the mass of the anhydrous compound formed when 91.6 g of copper sulfate pentahydrate is heated until all the water is driven off is 58.59 g.
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Consider the carbon monoxide molecule. Draw a valid Lewis structure for this molecule and use the Lewis structure to answer the following questions. Group of answer choices What is the total, valence electron count for the molecule
To draw the Lewis structure for the carbon monoxide (CO) molecule, we need to consider the valence electrons of each atom.
Carbon (C) has 4 valence electrons, while oxygen (O) has 6 valence electrons. The total valence electron count for the CO molecule can be calculated by summing the valence electrons of each atom and taking into account the molecular formula:
Total valence electron count = Number of valence electrons of carbon + Number of valence electrons of oxygen
Total valence electron count = 4 (valence electrons of carbon) + 6 (valence electro
Total valence electron count = 10
Therefore, the total valence electron count for the carbon monoxide (CO) molecule is 10.
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The vapor pressure of water is 20 degrees C is 17.5 torr. A 21.6% solution of the nonvolatile solute ethylene glycol, C2H4(OH)2, in water is prepared. Calculate the vapor pressure of the solution to three significant figures
The vapor pressure of the solution is approximately 13.4 torr.
The vapor pressure of a solution can be calculated using Raoult's law, which states that the vapor pressure of a solution is directly proportional to the mole fraction of the solvent. In this case, water is the solvent and ethylene glycol is the solute.
Given that the vapor pressure of pure water at 20°C is 17.5 torr, we can calculate the mole fraction of water in the solution. Since the solution is 21.6% ethylene glycol, the mole fraction of water (solvent) is 1 - 0.216 = 0.784.
Using Raoult's law, the vapor pressure of the solution is:
vapor pressure of solution = mole fraction of solvent * vapor pressure of pure solvent
vapor pressure of solution = 0.784 * 17.5 torr
vapor pressure of solution ≈ 13.7 torr
Rounding to three significant figures, the vapor pressure of the solution is approximately 13.4 torr.
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Which of the following acids would you use to prepare a buffer with a pH of 3.52 Oa.Hydrocyanic acidpKa=9.21 Points out of 8.00 ObHypoiodous acid(pKa=11.6) O c.None of these would be appropriate for a buffer solution with pH of 3.52 Od.Butyric acid(pKa=4.82) PFlag question Oe.Citric acidpKa=3.13
To prepare a buffer with a pH of 3.52, the appropriate choice would be Butyric acid (pKa = 4.82). The correct option is D.
In a buffer solution, the acid and its conjugate base work together to resist changes in pH upon addition of acid or base. Since the desired pH is lower than the pKa value of Butyric acid, it indicates that the acid will be predominantly in its undissociated form and can act as an effective buffer.
Hydrocyanic acid and Hypoiodous acid have pKa values higher than the desired pH, making them unsuitable choices for preparing a buffer with a pH of 3.52. Citric acid has a pKa value of 3.13, which is close to the desired pH, but Butyric acid is a better choice as its pKa is closer to the target pH.
The correct option for preparing a buffer with a pH of 3.52 is option D: Butyric acid (pKa = 4.82).
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A buffer solution with a pH of 3.52 would require an acid with a pKa close to this value. Therefore, the appropriate acid to use to prepare the buffer would be Butyric acid (pKa=4.82).
Buffer solution: A buffer solution is a solution that can resist a change in pH when a small amount of acid or base is added to it. The pH of the buffer solution changes by a small amount when a small amount of acid or base is added to it because it contains both a weak acid and its conjugate base. pH is the measure of how acidic or basic a solution is. The pH scale ranges from 0 to 14. A solution with a pH less than 7 is acidic, while one with a pH greater than 7 is basic. A solution with a pH of 7 is neutral. Acids are substances that contribute hydrogen ions (H+) to a solution and have a pH of less than 7. Bases are substances that contribute hydroxide ions (OH-) to a solution and have a pH greater than 7.pKa valueA pKa value is a measure of the acidity of a molecule. It represents the pH at which the molecule is half-protonated and half-deprotonated. The lower the pKa value, the stronger the acid. An acid with a pKa value close to the pH of the buffer solution will be a suitable candidate for making the buffer solution. A buffer with a pH of 3.52 would require an acid with a pKa close to this value.
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Because of the position of arsenic in the periodic table, arsenate (AsO4-) is chemically similar to inorganic phosphate and is used by phosphate-requiring enzymes as an alternate substrate. However, organic arsenates are quite unstable and spontaneously hydrolyze. Arsenate is known to inhibit glycolysis. Indentify the target enzyme, and explain the mechanism of inhibition.
The target enzyme inhibited by arsenate in glycolysis is the enzyme phosphofructok---inase
Is arsenic an inhibitor?
An important stage in glycolysis is the conversion of fructose-6-phosphate to fructose-1,6-bisphosphate, which is catalyzed by phosphofruc---tokinase, a crucial regulatory enzyme in the glycolytic system.
The structural resemblance of arsenate to inorganic phosphate is thought to be the mechanism of inhibition. In order to complete the process, phosphofructokinase binds to the substrate fructose-6-phosphate and needs inorganic phosphate as a cofactor. However, because and arsenate have comparable chemical characteristics, arsenate can also attach to the enzyme in its place. An arsenate-enzyme complex is created as a result.
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If a solution containing 30.91 g of mercury(II) chlorate is allowed to react completely with a solution containing 9.718 g of sodium dichromate, how many grams of solid precipitate will form? How many grams of the reactant in excess will remain after the reaction?
The balanced chemical equation for the reaction between mercury(II) chlorate and sodium dichromate is:3Hg(ClO3)2 + Na2Cr2O7 → 3HgCrO4 + NaClO3 + NaClFirst, we can calculate the amount of solid precipitate formed using stoichiometry.
The molar mass of Hg(ClO3)2 is 386.61 g/mol.The molar mass of Na2Cr2O7 is 297.80 g/mol.Therefore, the number of moles of Hg(ClO3)2 is:30.91 g ÷ 386.61 g/mol = 0.08 molAnd the number of moles of Na2Cr2O7 is:9.718 g ÷ 297.80 g/mol = 0.03 mol
Using the stoichiometric coefficients, we can see that 3 moles of Hg(ClO3)2 react with 1 mole of Na2Cr2O7 to form 3 moles of HgCrO4.
Therefore, the limiting reactant is Na2Cr2O7 because it forms fewer moles of the product. The amount of Hg(ClO3)2 used is 1/3 of the amount of Na2Cr2O7 used.So, 0.03 mol of Na2Cr2O7 would require 3 × 0.03 mol = 0.09 mol of Hg(ClO3)2 to react completely.
Since only 0.08 mol of Hg(ClO3)2 is available, it is the limiting reactant.So, all of the 0.03 mol of Na2Cr2O7 reacts to form 0.03 × 3 mol = 0.09 mol of HgCrO4.The molar mass of HgCrO4 is 396.66 g/mol.Therefore, the mass of HgCrO4 formed is:0.09 mol × 396.66 g/mol = 35.70 gSo, 35.70 g of solid precipitate will form.
Now, we can calculate the amount of reactant in excess. 0.08 mol of Hg(ClO3)2 reacted with all of the Na2Cr2O7. So, the amount of Hg(ClO3)2 in excess is:0.08 mol - 0 mol = 0.08 molThe mass of Hg(ClO3)2 in excess is:0.08 mol × 386.61 g/mol = 30.93 gTherefore, 30.93 grams of the reactant in excess will remain after the reaction.
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The reaction between 30.91 grams of mercury(II) chlorate and 9.718 grams of sodium dichromate will produce 30.35 grams of solid precipitate (Hg2Cl2) and leave 2.86 grams of sodium dichromate in excess.
To determine the grams of solid precipitate formed and the grams of the reactant in excess, we first need to write the balanced equation for the reaction between mercury(II) chlorate (Hg(ClO3)2) and sodium dichromate (Na2Cr2O7). The balanced equation is as follows:
3 Hg(ClO3)2 + Cr2O7^2- + 14 Na+ → 3 Hg2Cl2 + 2 Cr3+ + 7 NaClO3
According to the balanced equation, 3 moles of mercury(II) chlorate react with 1 mole of sodium dichromate to produce 3 moles of solid mercury(I) chloride (Hg2Cl2).
Calculating the grams of solid precipitate formed (Hg2Cl2):
To determine the grams of solid precipitate formed, we need to calculate the number of moles of mercury(II) chlorate and sodium dichromate, and then use the stoichiometry of the balanced equation to find the number of moles of solid precipitate.
Finally, we can convert the moles to grams using the molar mass of Hg2Cl2.
Molar mass of Hg2Cl2 = 2 * (atomic mass of Hg) + atomic mass of Cl
Molar mass of Hg2Cl2 = 2 * (200.59 g/mol) + 35.45 g/mol
Molar mass of Hg2Cl2 = 436.63 g/mol
Number of moles of Hg(ClO3)2 = mass / molar mass
Number of moles of Hg(ClO3)2 = 30.91 g / (2 * (200.59 g/mol) + 3 * 16.00 g/mol + 2 * 35.45 g/mol)
Number of moles of Hg(ClO3)2 = 30.91 g / 444.22 g/mol
Number of moles of Hg(ClO3)2 = 0.0696 mol
Number of moles of Na2Cr2O7 = mass / molar mass
Number of moles of Na2Cr2O7 = 9.718 g / (2 * 22.99 g/mol + 2 * 52.00 g/mol + 7 * 16.00 g/mol)
Number of moles of Na2Cr2O7 = 9.718 g / 294.95 g/mol
Number of moles of Na2Cr2O7 = 0.0329 mol
From the balanced equation, we can see that the mole ratio of Hg(ClO3)2 to Hg2Cl2 is 3:3. Therefore, the number of moles of Hg2Cl2 formed will be equal to the number of moles of Hg(ClO3)2.
Number of moles of Hg2Cl2 formed = 0.0696 mol
Mass of Hg2Cl2 formed = number of moles * molar mass
Mass of Hg2Cl2 formed = 0.0696 mol * 436.63 g/mol
Mass of Hg2Cl2 formed = 30.35 g
Therefore, 30.35 grams of solid precipitate (Hg2Cl2) will form.
Calculating the grams of the reactant in excess (Na2Cr2O7):
To determine the grams of the reactant in excess, we need to compare the stoichiometry of the balanced equation to see which reactant is left over.
Since the mole ratio of Hg(ClO3)2 to Na2Cr2O7 is 3:1, we can calculate the number of moles of Hg(ClO3)2 that react with the available amount of Na2Cr2O7, and then subtract that from the initial moles of Na2Cr2O7 to find the remaining moles.
Finally, we can convert the remaining moles to grams using the molar mass of Na2Cr2O7
Number of moles of Na2Cr2O7 reacted = (number of moles of Hg(ClO3)2) / 3
Number of moles of Na2Cr2O7 reacted = 0.0696 mol / 3
Number of moles of Na2Cr2O7 reacted = 0.0232 mol
Number of moles of Na2Cr2O7 remaining = (initial moles of Na2Cr2O7) - (moles of Na2Cr2O7 reacted)
Number of moles of Na2Cr2O7 remainin = 0.0329 mol - 0.0232 mol
Number of moles of Na2Cr2O7 remaining = 0.0097 mol
Mass of Na2Cr2O7 remaining = number of moles * molar mass
Mass of Na2Cr2O7 remaining = 0.0097 mol * (2 * 22.99 g/mol + 2 * 52.00 g/mol + 7 * 16.00 g/mol)
Mass of Na2Cr2O7 remaining = 0.0097 mol * 294.95 g/mol
Mass of Na2Cr2O7 remaining = 2.86 g
Therefore, 2.86 grams of sodium dichromate (Na2Cr2O7) will remain after the reaction.
The reaction between 30.91 grams of mercury(II) chlorate and 9.718 grams of sodium dichromate will produce 30.35 grams of solid precipitate (Hg2Cl2) and leave 2.86 grams of sodium dichromate in excess.
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The solubility of any gas in a liquid can always be increased by ________ Select one: a. increasing the temperature. b. using a solvent that can form hydrogen bonds. c. using a solvent that is nonpolar. d. decreasing the pressure of the gas above the solvent. e. increasing the pressure of the gas above the solvent.
The solubility of any gas in a liquid can always be increased by increasing the pressure of the gas above the solvent. So, the correct option is d.
The solubility of gases in a liquid is affected by many factors, including temperature, pressure, and the nature of the solvent. Gas solubility decreases as temperature increases since an increase in temperature increases the kinetic energy of the gas molecules. The increase in kinetic energy provides energy for the gas molecules to overcome the forces that hold them in the solution.
The solubility of any gas in a liquid can be increased by increasing the pressure of the gas above the solvent, which pushes more gas molecules into the liquid, leading to an increase in gas solubility. This is known as Henry's law. The nature of the solvent is also important in gas solubility. Gases dissolve best in solvents that have a similar polarity. Therefore, solvents that can form hydrogen bonds, such as water, are better at dissolving polar gases, whereas nonpolar solvents, such as oil, are better at dissolving nonpolar gases. Therefore, the correct option is d.
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Proteins often have well-defined structure that is essential to their function. At a high enough T, proteins denature and no longer function. What do the enthalpy and entropy look like in denaturization processes?
Protein denaturation is accompanied by changes in both enthalpy and entropy.
How are enthalpy and entropy affected during protein denaturation?During denaturation, the enthalpy and entropy of the protein are altered. Enthalpy refers to the heat energy involved in the breaking of intra- and intermolecular bonds within the protein structure. In denaturation, an increase in enthalpy is observed as the bonds holding the protein's three-dimensional structure together are disrupted. This increase in enthalpy is due to the energy required to break these bonds.
Simultaneously, the entropy of the system also changes during denaturation. Entropy is a measure of the disorder or randomness in a system. In the folded state, proteins have a lower entropy because their atoms are arranged in an ordered manner. However, during denaturation, the protein's structure becomes more disordered, leading to an increase in entropy.
The balance between enthalpy and entropy determines the overall stability of the protein structure. When the temperature rises, the increase in enthalpy due to bond disruption can overcome the decrease in entropy, leading to denaturation. As a result, the protein loses its functional conformation and may become inactive or form aggregates.
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A 200-L insulated tank contains nitrogen gas at 200 kPa, 300 K. A line with nitrogen at 500 K, 500 kPa adds 40% more mass to the tank with a flow through a valve. Use constant specific heats to find the final temperature and the exergy destruction.
To find the final temperature and the exergy destruction when the line with nitrogen at 500 K and 500 kPa adds 40% more mass to the insulated tank containing nitrogen gas at 300 K and 200 kPa, we need to consider the conservation of mass and energy.
First, let's calculate the initial mass of nitrogen in the tank using the ideal gas law:
PV = nRT
Where:
P = Pressure
V = Volume
n = Number of moles
R = Ideal gas constant
T = Temperature
Initial pressure (P1) = 200 kPa
Initial volume (V1) = 200 L
Initial temperature (T1) = 300 K
Using the ideal gas law, we can find the initial number of moles (n1) of nitrogen:
n1 = (P1 * V1) / (R * T1)
Next, we need to calculate the additional mass of nitrogen that is added to the tank. Since the line adds 40% more mass, the final mass of nitrogen (m2) will be:
m2 = 1.4 * m1
where m1 is the initial mass of nitrogen in the tank
Now, we can calculate the final number of moles (n2) of nitrogen using the final mass:
n2 = m2 / M
where M is the molar mass of nitrogen.
Next, we can calculate the final pressure (P2) using the ideal gas law:
P2 = (n1 * R * T1 + n2 * R * T2) / V1
where T2 is the final temperature.
We can rearrange the equation to solve for the final temperature:
T2 = (P2 * V1 - n1 * R * T1) / (n2 * R)
Finally, we can calculate the exergy destruction using the equation:
Exergy destruction = (m2 * Cp * (T1 - T2)) + (m2 * R * (T1 * ln(P1/P2)))
where Cp is the specific heat capacity at constant pressure and R is the specific gas constant.
By plugging in the known values and evaluating the equations, we can determine the final temperature and the exergy destruction.
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Ethyl trichloroacetate is significantly more reactive toward hydrolysis than ethyl acetate. Explain this observation.
Ethyl trichloro acetate is significantly more reactive toward hydrolysis than ethyl acetate because of the polar nature of the functional group. Ethyl trichloro acetate is more reactive towards hydrolysis than ethyl acetate because of the polar nature of the functional group.
The reaction between an ester and water is known as hydrolysis. An ester is a derivative of carboxylic acid, which contains a carbonyl group that is linked to an alkyl or an aryl group through an oxygen atom. The hydrolysis of esters involves the addition of water to the ester's carbonyl group, resulting in the formation of a carboxylic acid and an alcohol.The reactivity of esters toward hydrolysis is determined by the alkyl or aryl group linked to the carbonyl group, as well as the functional group. The higher the electron density of the carbonyl group, the more reactive the ester will be toward hydrolysis.
Ethyl trichloro acetate has a more polar carbonyl group than ethyl acetate. Trichloro acetate has a chlorine atom that draws electron density away from the carbonyl group, reducing the electron density of the carbonyl group, making it more electrophilic and reactive towards nucleophiles like water. Therefore, ethyl trichloro acetate is significantly more reactive towards hydrolysis than ethyl acetate.
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a) The eluent on the left is _____________________ than acetone. [ Select ]Copyright 2021. Govindarajoo, G. Rutgers, The State University of New Jerseyb) Acetone is _______________________ to overcome intermolecular attractions between spotted sample and adsorbent than the eluent on the left
a) The eluent on the left is less polar than acetone.
b) Acetone is better able to overcome intermolecular attractions between the spotted sample and adsorbent than the eluent on the left.
a) The eluent on the left is described as being "less polar" than acetone. Polar solvents have a higher ability to dissolve polar substances due to their polarity resulting from a difference in electronegativity between the atoms. Acetone, being a polar solvent, has a higher polarity compared to the eluent on the left.
b) In chromatography, the eluent is the solvent or mixture of solvents used to carry the sample through the stationary phase (adsorbent). The eluent needs to have sufficient strength to overcome intermolecular attractions between the sample and the adsorbent. Acetone, being a relatively polar and volatile solvent, is more effective in breaking the intermolecular attractions and carrying the sample along the stationary phase compared to the eluent on the left, which is described as less polar. This allows for better separation and migration of the components of the sample during chromatographic analysis.
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A group of 4 stars form at the same time in a cluster. Star A is the mass of our Sun, Star B is 4 solar masses, Star C is 0. 4 solar masses and Star D is 12 solar masses. How many stars are left on the main sequence after 8 billion years?
0, 1, 2, 3, or 4?
This leaves only star B with 4 solar masses on the main sequence, and as per the life expectancy of a star like this, it leaves only 1 star remaining on the main sequence after 8 billion years. Therefore, the answer is one.
A group of four stars is formed at the same time in a cluster. The stars are A, B, C, and D, with masses of the sun, four solar masses, 0.4 solar masses, and 12 solar masses, respectively. We need to find out the number of stars that remain on the main sequence after eight billion years. Answer: One star remains on the main sequence after 8 billion years. :
Stars A and C are less massive than our Sun; they will stay on the main sequence for longer than our Sun and will still be there after 8 billion years. Star D, on the other hand, will die as a supernova long before 8 billion years. This leaves only star B with 4 solar masses on the main sequence, and as per the life expectancy of a star like this, it leaves only 1 star remaining on the main sequence after 8 billion years. Therefore, the answer is one.
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1. Define the following terms: a) Avogadro's number (NA) b) Formula mass c) Molar mass (MM) d) Mole (mol) 3. See the example in Introduction for set-up of calculations. Calculate the number of a) moles of F b) atoms of F in 32.5 g of fluorine gas, F2. 4. Which of the following represents the least number of molecules? Show all work. a) 20.0 g of H20 b) 77.0 g of CHA c) 68.0 g of CaH2
Avogadro's number (NA) Avogadro's number (NA) refers to the number of particles in one mole of a substance. It has a value of 6.022 × 1023 particles/mol.b) Formula mass Formula mass refers to the sum of the atomic masses of all the atoms in a molecule.
It is expressed in atomic mass units (amu).c) Molar mass (MM)Molar mass (MM) refers to the mass of one mole of a substance. It is expressed in grams/mole.d) Mole (mol)Mole (mol) refers to the quantity of a substance that contains as many particles (atoms, molecules or ions) as there are in exactly 12.000 g of pure 12C. It is also known as the Avogadro constant. 3. Example for set-up of calculations: The molar mass of CO2 is 44.01 g/mol.How many moles are in 132.03 g of CO2?n(CO2) = mass (CO2) / molar mass (CO2)n(CO2) = 132.03 g / 44.01 g/moln(CO2) = 3.00 moles(a) moles of FThe molar mass of F2 is 38.00 g/mol.
How many moles are in 32.5 g of F2?n(F2) = mass (F2) / molar mass (F2)n(F2) = 32.5 g / 38.00 g/moln(F2) = 0.855 moles(b) atoms of FThe number of atoms of F in 0.855 moles of F2 is:n = NA × n(F2)n = (6.022 × 1023 particles/mol) × (0.855 mol)n = 5.14 × 1023 atoms of F4. The least number of molecules is represented by option (b) 77.0 g of CHA.How many moles of CHA are in 77.0 g of CHA n = mass (CHA) / molar mass (CHA)n = 77.0 g / molar mass (CHA)The molar mass of CHA is 30.08 g/mol.n = 77.0 g / 30.08 g/moln = 2.56 molNow, we can find the number of molecules of CHA by using Avogadro's number:The number of molecules = 2.56 mol × 6.022 × 1023 molecules/mol= 1.54 × 1024 molecules of CHA
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Explain, in terms of temperature and volume, why cool water entered into the flask when it was placed into the cool-water bath
In terms of temperature and volume, the cool water entered into the flask when it was placed into the cool-water bath the water level in the flask will rise.
This is due to the contraction of gases when they are cooled, resulting in a decrease in volume at a constant pressure. When the flask is placed in a cool-water bath, the air inside it cools and contracts, resulting in a decrease in volume. This generates a vacuum in the flask, which pulls water up the stem, increasing the volume. Since the flask is in equilibrium with the water bath's temperature, the temperature of the water and the flask is the same.
Furthermore, since the pressure within the flask is less than that outside, the flask will continue to fill with water until the pressure within it equals that outside, which is essentially atmospheric pressure, at which point the water will stop rising. The relationship between temperature, volume, and pressure is important to understand to comprehend the physics of how gases function. So therefore when cool water is poured into a flask placed in a cool-water bath, the water level in the flask will rise.
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A 0.050 M sample of hydrogen iodide is decomposed into hydrogen and iodine, according to the equation above. The [HI], ln[HI], and 1/[HI] vs. time graphs are plotted. (The equations are given for the best-fit straight line for the data, and R 2 value indicates the strength of correlation.) a. Write the rate law for this reaction. Justify your response using the graphs provided. b. Find the value of the rate constant, k. Include units. c. At what time will [HI] be 0.001 M
Locate the point where the concentration of HI is 0.001 M on the y-axis, and then find the corresponding time on the x-axis.
What are the main characteristics of cloud computing and how does it differ from traditional on-premises infrastructure?To determine the rate law for the reaction, we need to analyze the graphs provided.
The rate law can be determined by examining the initial rates of the reaction under different conditions.
From the graphs, we can observe that the ln[HI] vs. time graph shows a linear relationship, indicating a first-order reaction. Therefore, the rate law for this reaction is:
Rate = k[HI]
To find the value of the rate constant (k), we can use the slope of the ln[HI] vs. time graph.
The slope of a linear graph of ln[A] vs. time is equal to -k. From the graph, we can determine the slope and obtain the value of k.
The units of k depend on the overall reaction order and the units of concentration used.
Without specific information on the units, it is not possible to provide the exact units for k.
To determine the time at which [HI] will be 0.001 M, we can use the 1/[HI] vs. time graph.
The time at which [HI] will be 0.001 M can be read from the graph.
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A chemist adds 190.0mL of a 13.0/molL silver(I) perchlorate AgClO4 solution to a reaction flask. Calculate the mass in kilograms of silver(I) perchlorate the chemist has added to the flask.
The mass of solute of silver(I) perchlorate in kilograms is 0.354 kg.
Here are the steps involved in calculating the mass of silver(I) perchlorate in kilograms:
First, we need to determine the number of moles of silver(I) perchlorate in the solution. We can do this by multiplying the volume of the solution (in liters) by the concentration of the solution (in moles per liter).
volume = 190.0 mL
concentration = 13.0 mol/L
number of moles = volume * concentration
= 190.0 mL * 13.0 mol/L
= 2470.0 μmol
Once we know the number of moles of silver(I) perchlorate, we can calculate its mass by multiplying the number of moles by the molar mass of silver(I) perchlorate.
molar mass of silver(I) perchlorate = 143.32 g/mol
mass = number of moles * molar mass
= 2470.0 μmol * 143.32 g/mol
= 354.0 g
Finally, we need to convert the mass from grams to kilograms.
1 kg = 1000 g
mass in kg = mass in g / 1000 g
= 354.0 g / 1000 g
= 0.354 kg
Therefore, the mass of silver(I) perchlorate in kilograms is 0.354 kg.
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An ionic salt contains an anion relatively large compared to the size of its cation. Which ion is typically arranged in a close-packed array
In an ionic salt where the anion is relatively large compared to the size of its cation, the cation is typically arranged in a close-packed array.
This arrangement allows for efficient packing of the smaller cations around the larger anions, maximizing the attractive electrostatic interactions between them.
Close-packed arrays are characterized by the arrangement of spheres in a way that maximizes the packing density.
The two common types of close-packed structures are face-centered cubic (FCC) and hexagonal close-packed (HCP). In both structures, the spheres are arranged in layers, and each sphere is surrounded by its nearest neighbors.
In the case of an ionic salt with a large anion and a smaller cation, the cation will fit better within the interstitial spaces between the anions.
This allows the cations to form a close-packed arrangement around the anions, maximizing the efficiency of the packing.
Therefore, the cation is typically arranged in a close-packed array.
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Which of the following compounds is insoluble in water? Hg2Cl2 Li2SO4 K2CO3 BaS All of these compounds are soluble in water.
All of these compounds are soluble in water.
When determining the solubility of compounds in water, we need to consider the solubility rules. According to the general solubility rules, most chloride (Cl-) salts are soluble in water, including Hg2Cl2 (mercury(I) chloride). Therefore, Hg2Cl2 is soluble in water.
Li2SO4 (lithium sulfate) is also soluble in water. Sulfates (SO4²⁻) are generally soluble, except for a few exceptions such as barium sulfate (BaSO4) and calcium sulfate (CaSO4). Lithium sulfate does not fall into the exception category, so it is soluble in water.
K2CO3 (potassium carbonate) is also soluble in water. Carbonates (CO3²⁻) are generally soluble, except for a few exceptions such as calcium carbonate (CaCO3) and lead(II) carbonate (PbCO3). Potassium carbonate does not fall into the exception category, so it is soluble in water.
BaS (barium sulfide) is also soluble in water. Sulfides (S²⁻) are generally insoluble, except for a few exceptions such as alkali metal (Group 1) sulfides and alkaline earth metal (Group 2) sulfides. Barium sulfide is an alkaline earth metal sulfide, so it is soluble in water.
Based on the solubility rules and the information provided, all of the compounds Hg2Cl2, Li2SO4, K2CO3, and BaS are soluble in water. Therefore, the correct answer is that all of these compounds are soluble in water.
All of the compounds Hg2Cl2, Li2SO4, K2CO3, and BaS are soluble in water according to the solubility rules. It is important to refer to solubility rules or solubility tables to determine the solubility of specific compounds in water or other solvents.
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Out of the list of compounds provided (Hg2Cl2, Li2SO4, K2CO3, and BaS), Hg2Cl2 (Mercury (I) Chloride) is most accurately characterized as insoluble in water. This is based on established solubility rules, which state that although most chloride salts are soluble, Hg2Cl2 is a commonly recognized exception with very low solubility.
Explanation:The question asks which of the provided compounds is insoluble in water. The compounds listed are Hg2Cl2 (Mercury (I) Chloride), Li2SO4 (Lithium Sulfate), K2CO3 (Potassium Carbonate), and BaS (Barium Sulfide). According to Solubility Rules, most sulfate salts are soluble in water with a few exceptions; Li2SO4 is therefore soluble. Most carbonate salts are also soluble, making K2CO3 soluble as well. However, Hg2Cl2 is an exception to the generality that chloride salts are soluble, it has low solubility, and is therefore largely insoluble in water. Lastly, sulfide salts such as BaS are typically insoluble. But between BaS and Hg2Cl2, the latter is more commonly known to be insoluble. So, out of the provided compounds, Hg2Cl2 is most accurately characterized as insoluble in water.
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Which factors have the greatest effect on the rate of a chemical reaction betweenAgNO3(aq) and Cu(s)
The rate of a chemical reaction between AgNO3(aq) and Cu(s) is influenced by several factors like surface area, temperature, concentration, prescence of catalyst and stirrring or agitation.
Surface area: The rate of the reaction is directly proportional to the surface area of the reactants. By increasing the surface area of the solid Cu (e.g., by using powdered Cu instead of a solid piece), more Cu atoms are exposed to the AgNO3 solution, leading to increased collision frequency and a faster reaction rate.
Temperature: Increasing the temperature generally increases the reaction rate. Higher temperatures provide more kinetic energy to the reactant particles, leading to more frequent and energetic collisions, which results in a faster reaction rate.
Concentration: The concentration of reactants affects the reaction rate. Higher concentrations of AgNO3(aq) increase the likelihood of collisions between Ag+ ions and Cu atoms, resulting in a faster reaction rate. Similarly, increasing the concentration of Cu(s) would also increase the number of available Cu atoms for reaction, leading to an increased rate.
Presence of a catalyst: The addition of a catalyst can significantly speed up the reaction rate without being consumed in the process.
While a catalyst does not directly participate in the reaction, it provides an alternative reaction pathway with lower activation energy, making it easier for the reaction to occur.
Stirring or agitation: Stirring or agitating the reaction mixture enhances the rate of reaction by ensuring uniform mixing and maintaining a higher concentration of reactants at the reaction interface.
This promotes effective collisions between Ag+ ions and Cu atoms, leading to a faster reaction rate.
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Given the reaction C(g) + 2H(g) + 2F(g) → CH2F2(g) What is the heat of reaction, ΔH, in kJ at 25 °C? Bond Energies (kJ.mol-1) at 25 °C C-H 414 C-F 486 H-H 435 F-F 159 H-F 569 a. ΔH= +900 kJ b. ΔH= -900 kJ c. ΔΗ = 465 kJ d. ΔH= -1800 kJ
The heat of reaction, ΔH, for the given reaction C(g) + 2H(g) + 2F(g) → CH2F2(g) at 25 °C is 392 kJ/mol.
To determine the heat of reaction, ΔH, for the given reaction, we need to calculate the energy change associated with breaking the bonds in the reactant molecules and forming the bonds in the product molecule.
Given the bond energies at 25 °C:
C-H: 414 kJ/mol
C-F: 486 kJ/mol
H-H: 435 kJ/mol
F-F: 159 kJ/mol
H-F: 569 kJ/mol
Let's calculate the energy change for breaking the bonds in the reactant molecules:
2(C-H) + 2(H-H) + 2(F-F)
2(414 kJ/mol) + 2(435 kJ/mol) + 2(159 kJ/mol)
= 828 kJ/mol + 870 kJ/mol + 318 kJ/mol
= 2016 kJ/mol
Next, let's calculate the energy change for forming the bonds in the product molecule:
1([tex]CH_2[/tex]) + 2(F-H)
1(486 kJ/mol) + 2(569 kJ/mol)
= 486 kJ/mol + 1138 kJ/mol
= 1624 kJ/mol
Now, we can calculate the overall energy change for the reaction by subtracting the energy change associated with breaking the bonds from the energy change associated with forming the bonds:
ΔH = Energy of bonds broken - Energy of bonds formed
ΔH = 2016 kJ/mol - 1624 kJ/mol
ΔH = 392 kJ/mol
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