The reason acetyl chloride reacts with water almost violently is because it is a highly reactive compound due to the presence of the polar functional group, chloride.
Chloride ions are highly electronegative and have a strong affinity for water molecules. When acetyl chloride is added to water, the chloride ions attract water molecules, causing the reaction to occur quickly and violently.
On the other hand, benzoyl chloride has a longer carbon chain and is less reactive than acetyl chloride. This means that the reaction with water is slower and requires a higher energy input, such as warming and shaking the mixture. The longer carbon chain also makes it less polar than acetyl chloride, which means it is less attracted to water molecules and therefore does not react as violently.
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How many kilograms of nickel must be added to 5.66 kg of copper to yield a liquidus temperature of 1200?
62.66 kg of nickel must be added to 5.66 kg of copper to yield a liquidus temperature of 1200.
To calculate the amount of nickel needed to reach a liquidus temperature of 1200, we need to use the lever rule formula. This formula uses the proportions of the two metals in the alloy to determine the temperature at which the alloy will become completely liquid.
First, we need to determine the proportions of copper and nickel in the alloy. Let's assume that the final alloy will contain x kilograms of nickel.
The total mass of the alloy will be 5.66 + x kg.
Next, we need to determine the percentage of copper and nickel in the alloy. Assuming that the final alloy will contain 100% copper and nickel, we can write the following equation:
(5.66 / (5.66 + x)) × 100 = 100% - liquidus temperature drop
We know that the liquidus temperature drop is 1200 - the liquidus temperature of the copper-nickel alloy. Let's assume that the liquidus temperature of the alloy is 1300.
(5.66 / (5.66 + x)) × 100 = 100% - (1300 - 1200) / 1300 × 100
Simplifying this equation, we get:
(5.66 / (5.66 + x)) × 100 = 7.69
Solving for x, we get:
x = 62.66 kg
Therefore, we need to add 62.66 kg of nickel to 5.66 kg of copper to yield a liquidus temperature of 1200.
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calculate the value of ∆s if one mole of an ideal gas is expanded reversibly and isothermally from 1.0 bar to 0.1 bar. explain the sign of ∆s.
∆s = R ln 10, where R is the gas constant is the value of ∆s if one mole of an ideal gas is expanded reversibly and isothermally from 1.0 bar to 0.1 bar.
The change in entropy (∆s) of an ideal gas during an isothermal reversible expansion is given by [tex]∆s = nR ln (V2/V1),[/tex] where n is the number of moles, R is the gas constant, V1 is the initial volume, and V2 is the final volume. Here, n = 1 mole, [tex]V1 = RT/P1,[/tex] and [tex]V2 = RT/P2,[/tex] where T is the temperature in Kelvin, [tex]P1 = 1.0[/tex] bar, and[tex]P2 = 0.1[/tex] bar. Substituting these values, we get [tex]∆s = R ln (P1/P2) = R[/tex] ln 10. Since the pressure decreases during expansion, the entropy of the gas increases. Hence, the sign of ∆s is positive.
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1. For each trial, calculate the moles of KHP used to neutralize the NaOH solution and put the answer in the Results Table. (Hint: Use the correct formula from the Discussion section to calculate the molar mass of KHP! It is NOT the mass of K + H + P!)
2. For each trial, using the mole ratio from the balanced equation, calculate the moles of NaOH and put the answer in the Results Table.
3. For each trial, calculate the molar concentration of the NaOH solution and put the answer in the Results Table. (See example in the discussion.)
4. Calculate the average molarity and enter this answer in your Results Table.
5. Submit your Results Table and Calculations in the following corresponding question boxes.
KHP + NaOH ⇒ NaKP + H₂O this Equation used to find five steps of questions of the the moles of KHP used to neutralize the NaOH solution.
1. To calculate the moles of KHP used to neutralize the NaOH solution, you need to use the balanced equation, which is:
KHP + NaOH ⇒ NaKP + H₂O
The molar mass of KHP is 204.22 g/mol. So, to find the moles of KHP used in each trial, you can use the formula:
moles = mass / molar mass
Where the mass is the mass of KHP used in each trial. Make sure to convert the mass from grams to moles.
2. Using the mole ratio from the balanced equation, you can calculate the moles of NaOH used in each trial. The mole ratio between KHP and NaOH is 1:1, which means that for every mole of KHP used, one mole of NaOH is used. Therefore, the moles of NaOH used in each trial is the same as the moles of KHP used.
3. To calculate the molar concentration of the NaOH solution, you need to use the formula:
molarity = moles / volume
Where the moles are the moles of NaOH used in each trial, and the volume is the volume of NaOH solution used in each trial. Make sure to convert the volume from milliliters to liters.
4. To find the average molarity, you can add up the molar concentrations from each trial and divide by the number of trials. This will give you the average molarity of the NaOH solution.
5. Make sure to record all your calculations in the Results Table, and submit it along with your answers to the corresponding question boxes.
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what is the equilibrium expression for the following reaction? h2so4 (l) ⇌ so3 (g) h2o (l)
The equilibrium expression for the following reaction? h2so4 (l) ⇌ so3 (g) h2o (l): the final equilibrium expression for this reaction is: Kc = [SO3]
The equilibrium expression for the given reaction is:
Kc = [SO3][H2O] / [H2SO4]
where Kc is the equilibrium constant, [SO3], [H2O], and [H2SO4] are the molar concentrations of sulfur trioxide, water, and sulfuric acid respectively at equilibrium.
Hello! I'm happy to help with your question. The equilibrium expression for the reaction H2SO4 (l) ⇌ SO3 (g) + H2O (l) can be written using the equilibrium constant (Kc).
First, let's write the balanced chemical equation:
H2SO4 (l) ⇌ SO3 (g) + H2O (l)
Next, we'll write the equilibrium expression using the concentrations of the products and reactants:
Kc = [SO3] * [H2O] / [H2SO4]
In this expression, [SO3], [H2O], and [H2SO4] represent the equilibrium concentrations of the respective species. Keep in mind that only the concentrations of gases (SO3 in this case) are included in the equilibrium constant expression. Liquid concentrations, such as H2SO4 and H2O, do not affect the value of Kc.
So, the final equilibrium expression for this reaction is:
Kc = [SO3]
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One can use the Gibbs free energy definition, written here at standard pressure∆g⁰=∆h⁰-t∆s⁰to compute the Gibbs free energy at any arbitrary temperature. One can additionally assume that theenthalpy and entropy of a reaction do not change with temperature significantly if the change from 298 is notgreat. In the next few exercises we will assume that∆h⁰t=∆h⁰298. and. ∆s⁰t=∆s⁰298so that we can use the enthalpy and entropy of a reaction at 298 K to estimate AGO at any other T.Estimate the Gibbs free energy of a reaction at 537.6 K if the enthalpy and the entropy are -91.2 kJ mol¹ and410.3 J mol1 K1 respectively.∆g⁰537.=220668.5 kJ mol-1.
The estimated Gibbs free energy of the reaction at 537.6 K is -311.51 kJ/mol.
Gibbs's free energy, denoted by the symbol G, is a thermodynamic property that is used to determine the maximum amount of work that can be obtained from a chemical reaction at constant temperature and pressure. It is named after the American physicist Josiah Willard Gibbs, who first introduced the concept.
1: Use the Gibbs free energy definition at standard pressure: ΔG⁰ = ΔH⁰ - TΔS⁰
2: Plug in the given values: ΔH⁰ = -91.2 kJ/mol and ΔS⁰ = 410.3 J/mol*K (note: convert ΔS⁰ to kJ/mol*K by dividing by 1000, so ΔS⁰ = 0.4103 kJ/mol*K)
3: Use the given temperature of 537.6 K in the equation: ΔG⁰_537.6 = (-91.2 kJ/mol) - (537.6 K * 0.4103 kJ/mol*K)
4: The Gibbs free energy: ΔG⁰_537.6 = -91.2 kJ/mol - (220.31 kJ/mol) = -311.51 kJ/mol.
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Arrange the following in order of increasing bond angles: ClO2 , NO2 , SiO2 A) CIO2
Cl₂ ≤OClO₂≤ ClO₂ is the proper sequence of increasing bond angles in the following species. This is owing to the fact that in ClO₂, there are two lone pairs of electrons that oppose one another.
Causing two oxygen atoms to move in closer together and resulting in a decrease in bond angle. Therefore, the bond angle in ClO₂ is less than 118°, which is the bond angle in ClO₂ where chlorine has less electrons. ClO₂ has an angular structure. With a bond angle of 118⁰ and a Cl-O bond length of 1.47A⁰, the Cl atom is sp₂-hybridized in the angular molecule.
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a solution was made by dissolving 125 g of na3po4 in water. the volume of the resulting solution was 250 ml. calculate the molarity of the solution (m).
The molarity of the solution made by dissolving 125 g of Na₃PO₄ in water resulting to a volume of 250 ml is approximately 3.05 M.
To calculate the molarity (M) of the solution, we need to know the moles of solute (Na₃PO₄) and the volume of the solution in liters.
1. Convert grams of Na₃PO₄ to moles:
Molecular weight of Na₃PO₄ = (3 × 22.99) + (1 × 30.97) + (4 × 16.00) = 68.97 + 30.97 + 64.00 = 163.94 g/mol
125 g Na₃PO₄ / 163.94 g/mol ≈ 0.762 moles of Na₃PO₄
2. Convert volume of the solution to liters:
250 mL = 250/1000 = 0.25 L
3. Calculate the molarity (M):
M = moles of solute / volume of solution in liters
M = 0.762 moles / 0.25 L ≈ 3.05 M
The molarity of the Na₃PO₄ solution is approximately 3.05 M.
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at equivalence, will the ph be greater than 7, less than 7, or equal to 7. explain your reasoning
At equivalence, the pH can be greater than 7, less than 7, or equal to 7, depending on the nature of the acid and base involved in the reaction.
1. If the acid and base are both strong (e.g., HCl and NaOH), the pH will be equal to 7. This is because strong acids and bases completely dissociate in water, and at equivalence, the number of moles of [tex]H^+[/tex] ions and [tex]OH^-[/tex] ions will be equal, leading to a neutral solution.
2. If the acid is strong and the base is weak (e.g., HCl and [tex]NH_3[/tex]), the pH will be less than 7. This is because, at equivalence, the weak base will not completely neutralize the strong acid, leaving some [tex]H^+[/tex] ions in the solution, which results in an acidic solution.
3. If the acid is weak and the base is strong (e.g., [tex]CH_3COOH[/tex] and NaOH), the pH will be greater than 7. This is because, at equivalence, the strong base will completely neutralize the weak acid, leaving some [tex]OH^-[/tex] ions in the solution, which results in a basic solution.
In summary, the pH at equivalence depends on the nature of the acid and base involved in the reaction.
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if water is accidentally added to a saturated aqueous solution the solution remains saturated. true false
A saturated aqueous solution remains saturated even if water is unintentionally added to it. This statement is false.
When water is added to a saturated aqueous solution, the solution will no longer be saturated, as the addition of water will dilute the concentration of solutes in the solution.
A saturated aqueous solution contains the maximum amount of solutes that can be dissolved in the solvent, at a given temperature and pressure. When more solutes are added to the solution, they will not dissolve and instead form a separate phase. This is because the solution is already at equilibrium, and any additional solutes will not be able to dissolve.
When water is added to the solution, the concentration of solutes in the solution will decrease, as the same amount of solutes is now dispersed in a larger volume of solvent. The solution will no longer be saturated, as there is no room for more solutes to dissolve in the solvent.
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the manufacture of ammonia from nitrogen and hydrogen is an exothermic reaction. which temperature would give a greater yield of ammonia, room temperature or 100oc? n2 3h2 <----> 2nh3
The manufacture of ammonia from nitrogen and hydrogen is an exothermic reaction, which means that it releases heat. According to Le Chatelier's principle, an increase in temperature favors the endothermic direction of a reversible reaction. Therefore, a lower temperature would give a greater yield of ammonia.
In this reaction, nitrogen and hydrogen combine to form ammonia. The reverse reaction is also possible, where ammonia breaks down into nitrogen and hydrogen. This reaction is exothermic, which means that it releases heat. According to Le Chatelier's principle, an increase in temperature favors the endothermic direction of a reversible reaction. Therefore, if the temperature is increased, the yield of ammonia would decrease as the reaction would shift towards the reactants.
On the other hand, a lower temperature would favor the exothermic direction and increase the yield of ammonia. Hence, room temperature would give a greater yield of ammonia than a temperature of 100°C.
In conclusion, a lower temperature would give a greater yield of ammonia as it favors the exothermic direction of the reaction.
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A gas mixture contains 20.0 g He gas and 6.0 g hydrogen gas at a total pressure of 800 torr. What is the partial pressure of He in this mixture? (answer in torr, but only enter the number) Hint: Refer to the posted movie that covers this topic for guidance. Type your response Refer to the posted formulas.
the partial pressure of He in this mixture is 500 torr partial pressure.
To find the partial pressure of He in the mixture, we first need to calculate the total moles of gas in the mixture.
moles of He gas = 20.0 g / 4.00 g/mol = 5.00 moles
moles of H2 gas = 6.0 g / 2.02 g/mol = 2.97 moles
Total moles of gas = 5.00 moles + 2.97 moles = 7.97 moles
Next, we can use the formula for partial pressure:
partial pressure of He = (moles of He gas / total moles of gas) x total pressure
partial pressure of He = (5.00 moles / 7.97 moles) x 800 torr
partial pressure of He = 500.62 torr
Therefore, the partial pressure of He in the mixture is 500.62 torr.
To find the partial pressure of He, we can use Dalton's Law of partial pressures: P_total = P_He + P_H2. We'll first find the moles of He and H2.
For He:
Molar mass of He = 4 g/mol
Moles of He = mass / molar mass = 20 g / 4 g/mol = 5 moles
For H2:
Molar mass of H2 = 2 g/mol
Moles of H2 = mass / molar mass = 6 g / 2 g/mol = 3 moles
Total moles = moles of He + moles of H2 = 5 + 3 = 8 moles
Next, we'll find the mole fraction of He:
Mole fraction of He = moles of He / total moles = 5 / 8
Finally, we'll find the partial pressure of He:
P_He = mole fraction of He × P_total = (5 / 8) × 800 torr = 500 torr
So, the partial pressure of He in this mixture is 500 torr partial pressure.
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When adding NaNO2 to a solution of HNO2 answer is:
NaNO2 --> Na+ +NO2-
HNO2 --> H+ +NO2-
there is no water used in the problem above,
but for THIS problem (below) water is used:
(CH3NH3)Cl + H20-->CH3NH3+ +Cl-
CH3NH2 + H20 -->CH3NH2++OH-
why is water used in one of the problems but not theother? I figure it has to do with the solubility of the salts, butI just would like a straight forwardexplanation
Water is used in the second problem because the reaction involves dissolution of the (CH3NH3)Cl salt in water, resulting in the formation of CH3NH3+ and Cl- ions. In the first problem, NaNO2 and HNO2 do not dissolve in water, and hence water is not used.
The use of water in chemical reactions depends on the solubility of the substances involved. In the second problem, (CH3NH3)Cl is a salt that dissolves in water, undergoing a process called dissolution, where the ionic compound dissociates into its constituent ions, CH3NH3+ and Cl-. Water is used as a solvent to facilitate this dissolution process.
On the other hand, in the first problem, NaNO2 and HNO2 are not soluble in water, and hence water is not used. Instead, the reaction proceeds through a different mechanism involving ionization of NaNO2 and HNO2 without dissolution in water.
Solubility of substances and the need for a solvent depend on the nature of the reactants and the desired reaction mechanism.
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calculate the solubility (in g/l) of caso4(s)caso4(s) in 0.250 m na2so4(aq) at 25°c0.250 m na2so4(aq) at 25°c . the spksp of caso4caso4 is 4.93×10−54.93×10−5 .
The solubility of [tex]CaSO4[/tex] in 0.250 M [tex]Na2SO4[/tex] at 25°C is [tex]2.68×10^-2 g/L[/tex].
Why will be 25°c0.250 m na2so4(aq) at 25°c?To solve this problem, we will use the common ion effect, which states that the solubility of a slightly soluble salt is reduced in the presence of a common ion. In this case, the common ion is sulfate (SO4^2-), which is present in both [tex]Na2SO4[/tex] and [tex]CaSO4[/tex].
The solubility product expression for CaSO4 is:
Ksp = [tex][Ca2+][SO4^2-][/tex]
Let x be the solubility of [tex]CaSO4[/tex] in moles per liter. Then, at equilibrium, the concentration of [tex]Ca2+[/tex] and [tex]SO4^2-[/tex] ions will also be x. We can then write the Ksp expression in terms of x:
4.93×[tex]10^-5 = x^2[/tex]
Solving for x, we get:
x = 2.22×[tex]10^-3[/tex] M
This is the solubility of [tex]CaSO4[/tex] in pure water. To calculate the solubility in the presence of 0.250 M [tex]Na2SO4[/tex] , we need to consider the concentration of [tex]SO4^2-[/tex]ions from [tex]Na2SO4[/tex].
[tex]Na2SO4[/tex] dissociates in water to form two [tex]Na+[/tex] ions and one [tex]SO4^2-[/tex] ion. The concentration of [tex]SO4^2-[/tex]ions in the solution is therefore:
[tex][SO4^2-][/tex]= 0.250 M
Since the solubility of [tex]CaSO4[/tex] is reduced in the presence of a common ion, the actual solubility of [tex]CaSO4[/tex] in 0.250 M [tex]Na2SO4[/tex] will be less than 2.22×[tex]10^-3[/tex] M. We can calculate the new solubility by using the Ksp expression with the concentration of [tex]SO4^2-[/tex] ions from [tex]Na2SO4[/tex]:
[tex]4.93×10^-5[/tex] = [tex][Ca2+][SO4^2-][/tex] = [tex]x^2[/tex]
[tex]4.93×10^-5[/tex] = [tex]x*[SO4^2-][/tex]
x = [tex](4.93×10^-5) / [SO4^2-] = (4.93×10^-5) / 0.250 = 1.97×10^-4 M[/tex]
Finally, to convert the solubility to grams per liter, we need to multiply by the molar mass of [tex]CaSO4[/tex]:
molar mass of [tex]CaSO4[/tex]= 40.08 g/mol + 32.06 g/mol + 4*16.00 g/mol = 136.14 g/mol
solubility of [tex]CaSO4[/tex]= [tex](1.97×10^-4 M) * (136.14 g/mol) = 2.68×10^-2 g/L[/tex]
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use the following information: In a Si sample at room temperature, No 2x1015 cm3 and NA-3x1017 cm3 Assume the dopants are fully ionized. (3 pts) 3. What are the equilibrium electron and hole concentrations (n and p)? a. n-103 cm-3, p-1017 cm -3 b. n 3x1017 cm 3, p- 333 cm3 c. n 2x101 3x1m d. n 333 cm3 p 3x101 cm3 15m-3 173 (3 pts) 4. If the temperature is changed so that n, 1017 cm3, what is the equilibrium hole concentration?
The equilibrium hole concentration (p) is 2.25*10^{3 }cm^-3.
In the given information, the concentration of electrons (No) is 2*10^{15} cm3 and the concentration of acceptor impurities (NA) is 3* 10^{17} cm3. Since the dopants are fully ionized, the concentration of holes (p) equals the concentration of acceptor impurities (NA).
To find the equilibrium electron concentration (n), we use the following formula:
n_i^2 = No * NA
Where n_i is the intrinsic carrier concentration. At room temperature, n_i = 1.5*10^{10} cm^{-3}.
Substituting the given values, we get:
(1.5*10^{10})^{2 }= 2*10^{15} * 3*10^{17}
n = sqrt((2*10^{15} * 3*10^{17})}{1.5*10^10) }
n = 6*10^{6} cm^{-3}
Therefore, the equilibrium electron concentration (n) is 6*10^{6} cm^{-3} and the equilibrium hole concentration (p) is 3*10^{17} cm^{-3}.
For the second part of the question, if the concentration of electrons (n) changes to 10^{17} cm^{-3}, we can use the following formula:
n * p = n_i^2
Substituting the given values, we get:
10^17 * p = (1.5*10^10)^2
p = \frac{(1.5*10^10)^{2}}{ 10^{17}}
p = 2.25*10^{3 }cm^{-3}
Therefore, the equilibrium hole concentration (p) is 2.25*10^{3} cm^{-3}.
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A sample of nitrogen gas collected at a pressure of 766 mm Hg and a temperature of 297 K has a mass of 27.0 grams. The volume of the sample is_____ L.
A sample of nitrogen gas collected at a pressure of 766 mm Hg and a temperature of 297 K has a mass of 27.0 grams. The volume of the sample is 24.13 liters.
To find the volume of the sample of nitrogen gas, we can use the ideal gas law:
PV = nRT
where P is the pressure in atmospheres (atm), V is the volume in liters (L), n is the number of moles of gas, R is the gas constant (0.0821 L·atm/K·mol), and T is the temperature in Kelvin (K).
First, we need to convert the pressure from mm Hg to atm:
1 atm = 760 mm Hg
So, P = 766 mm Hg / 760 mm Hg/atm = 1.01 atm
Next, we can use the mass of the sample to calculate the number of moles of nitrogen gas:
n = m/M
where m is the mass of the gas (27.0 g) and M is the molar mass of nitrogen (28.0 g/mol).
n = 27.0 g / 28.0 g/mol = 0.964 mol
Now we can plug in the values for P, n, R, and T and solve for V:
V = nRT/P
V = (0.964 mol)(0.0821 L·atm/K·mol)(297 K)/(1.01 atm)
V = 22.5 L
Therefore, the volume of the sample of nitrogen gas is 22.5 L.
To find the volume of the nitrogen gas sample, we can use the Ideal Gas Law formula, which is PV = nRT. We need to determine the values for pressure (P), number of moles (n), temperature (T), and the gas constant (R). Then, we can solve for the volume (V).
Given:
Pressure (P) = 766 mm Hg (we need to convert it to atm, so we'll use the conversion factor: 1 atm = 760 mm Hg)
Temperature (T) = 297 K
Mass of nitrogen gas = 27.0 grams
First, convert the pressure to atm:
P = 766 mm Hg * (1 atm / 760 mm Hg) = 1.0079 atm
Next, find the number of moles (n) using the molar mass of nitrogen gas (N2) which is 28.02 g/mol:
n = mass / molar mass = 27.0 grams / 28.02 g/mol = 0.9636 mol
Now, we have P, n, and T, and the value of the gas constant (R) is 0.0821 L·atm/mol·K. Plug the values into the Ideal Gas Law formula:
1.0079 atm * V = 0.9636 mol * 0.0821 L·atm/mol·K * 297 K
Solve for V:
V = (0.9636 mol * 0.0821 L·atm/mol·K * 297 K) / 1.0079 atm = 24.13 L
Therefore, the volume of the nitrogen gas sample is 24.13 liters.
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How does the temperature of water affect the rate of melting?
Answer:
higher the temp higher the melting rate
Explanation:
the higher the temp the more energy is put into the system in order to break intermolecular forces between molecules and allow it to change state.
Methylamine (CH3NH2) is a weak base with Kb = 4.47 x 104 at 25°C. a. Write down its reaction with water and identify acids and bases including conjugated acids and bases b. If the initial concentration of CH3NH2 (aq) is 0.0251 M, then what is the [H+] at the equilibrium? What is the pH? Show your work C. At equilibrium, if you add one drop of Na2CO3 (aq) to a solution of methylamine, will the solution become more acidic or basic? Explain
The reaction of methylamine with water is; CH₃NH₂ (aq) + H₂O (l) ⇌ CH₃NH₃⁺ (aq) + OH⁻ (aq), the pH of the solution is 11.42, and the solution will become more acidic.
The reaction of methylamine with water is;
CH₃NH₂ (aq) + H₂O (l) ⇌ CH₃NH₃⁺ (aq) + OH⁻ (aq)
In this reaction, CH₃NH₂ is a weak base and H₂O is the acid. The conjugate acid of CH₃NH₂ is CH₃NH₃⁺ and the conjugate base of H₂O is OH⁻.
The equilibrium constant expression for this reaction is;
Kb = [CH₃NH₃⁺][OH⁻]/[CH₃NH₂]
At equilibrium, we can assume that x moles of CH₃NH₂ react with x moles of H₂O to form x moles of CH₃NH₃⁺ and x moles of OH⁻. Therefore, we can write;
Kb = x₂ / (0.0251 - x)
Solving for x, we get;
x = 0.00263 M
Therefore, the concentration of OH⁻ at equilibrium is 0.00263 M. To find the concentration of H⁺, we can use the equation;
Kw = [H⁺][OH⁻]
where Kw is the ion product constant for water, which is 1.0 x 10⁻¹⁴ at 25°C. Solving for [H⁺], we get;
[H⁺] = Kw / [OH⁻] = 1.0 x 10⁻¹⁴ / 0.00263 = 3.8 x 10⁻¹² M
Taking the negative logarithm of [H⁺], we get;
pH = -log[H⁺] = -(-11.42) = 11.42
Therefore, the pH of the solution is 11.42.
When you add one drop of Na₂CO₃ (aq) to the solution of methylamine, the Na₂CO₃ will react with water to produce Na⁺ and OH⁻. The OH⁻ ions will react with the CH₃NH₃⁺ ions in the solution to form CH₃NH₂ and H₂O.
This reaction will shift the equilibrium to the left, decreasing the concentration of OH⁻ and increasing the concentration of H⁺. Therefore, the solution will become more acidic.
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Select all correct answers. Breaking the chemical bonds in reactions requires:
proper orientation of the molecules.
collisions between particles.
an overall release of energy.
sufficient kinetic energy to break the bonds.
an overall decrease in energy.
The correct answers are: proper orientation of the molecules; collisions between particles; and sufficient kinetic energy to break the bonds, as breaking chemical bonds in reactions requires proper orientation of the molecules and collisions between particles.
Chemical reactions involve the breaking of bonds between atoms in reactant molecues and the formation of new bonds between atoms in product molecules. Breaking these bonds requires a certain amount of energy, which can be supplied through collisions between particles and the proper orientation of the molecules involved in the reaction. When two reactant molecules collide, the orientation of their atoms is important. The atoms need to be in the correct position relative to each other for the chemical bonds to break and new bonds to form.
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Identify the correct net ionic equation for the reaction that occurs when solutions of HCIO 4 and Ba(OH) 2 are mixed. 2HC104(aq) + Ba(OH)2(s) + 2H2O(l) + Ba(ClO4)2(aq) H(aq) + OH-(aq) H2O(1) 2H+(aq) + Ba(OH)2(s) → 2H2O(1) + Ba2+(aq) OH(aq) + 2OH-(aq) → H2O(1) 2HC104(aq) + Ba(OH)2(s) → 2H2O(1) + Ba(CIO4)2(s)
The correct net ionic equation for the reaction that occurs when solutions of HCIO 4 and Ba(OH) 2 are mixed is:
2H⁺(aq) + 2OH⁻(aq) ⇒ 2H₂O(1) + Ba₂+(aq)
This is because the spectator ions, ClO4- and Ba2+, are not involved in the actual reaction and can be eliminated from the equation to give the net ionic equation.
The entire symbols of the reactants and products, as well as the states of matter under the conditions under which the reaction is occurring, are written in the complete equation of a chemical reaction.
Only those chemical species that are directly involved in the chemical reaction are written in the net ionic equation of the process.
In the net ion equation, mass and charge must be equal.
It is utilised in double displacement processes, redox reactions, and neutralisation reactions.
After removing the spectator ions, we may discuss the final ionic process using the net reaction equation. Keep in mind that we refer to the ions that do not participate in the reaction as spectator ions.
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Kingsley then adds 49.28 mL of NaOH to 250.00 mL of the HCOOH solution. The neutralization reaction resulted in 0.098 moles of HCOOH and 0.025 moles of HCOO left in solution. Determine the pH of the resulting solution.
The pH of the resulting solution is approximately 2.76.
How to find the pHTo determine the pH of the resulting solution after Kingsley adds 49.28 mL of NaOH to 250.00 mL of HCOOH solution, we need to first find the concentrations of HCOOH and HCOO- ions in the solution.
The total volume of the solution is now 250.00 mL + 49.28 mL = 299.28 mL.
Next, calculate the concentrations of HCOOH and HCOO-:
[HCOOH] = 0.098 moles / 0.29928 L = 0.327 M
[HCOO-] = 0.025 moles / 0.29928 L = 0.0835 M
Now, we can use the Henderson-Hasselbalch equation to find the pH of the solution:
pH = pKa + log ([HCOO-] / [HCOOH])
The pKa of formic acid (HCOOH) is 3.75.
Plugging the values into the equation:
pH = 3.75 + log (0.0835 / 0.327)
pH ≈ 3.75 + (-0.99)
pH ≈ 2.76
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500
Potential Energy (kJ)
300
200
400
100
activation energy
Energy Diagram
Reaction Time (min)
#energy of the reaction (enthalpy)
It energy of products
# endothermic I catalyst
11 energy of reactants
#exothermic
The energy diagram shows the potential energy of a reaction over the course of time.
The reaction has an activation energy of 200 kJ and an energy of reactants of 300 kJ. The energy of products is 100 kJ. The reaction is exothermic since the energy of products is lower than the energy of reactants.
If a catalyst is added to the reaction, the activation energy decreases and the reaction time decreases as well. However, the energy of the reaction (enthalpy) remains the same.
Overall, the energy diagram shows that the reaction releases energy in the form of heat as it progresses from reactants to products, indicating an exothermic process.
Thus, the addition of a catalyst can lower the activation energy and speed up the reaction, without affecting the overall energy of the reaction.
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Calculate the equilibrium constant Kc for the following overall reaction: AgCl(s) + 2CN-(aq) ⇌ Ag(CN)2- (aq) + Cl-(aq) For AgCl, Ksp = 1.6 × 10^-10; for Ag(CN)2. Kf=1.0 × 10^21 Multiple Choice a. 1.2 x 10^11 b. 1.4 x 10^11 c. 1.6 × 10^11 d. 1.8 × 10^11 e. None of the above
The equilibrium constant Kc for the given reaction is 1.6 × 10¹¹.
The equilibrium constant expression for the given reaction is:
Kc = ([Ag(CN)₂⁻] [Cl⁻])/([AgCl] [CN⁻]²)
To find Kc, we need to determine the concentrations of the species at equilibrium.
Since AgCl is a solid, its concentration is constant and can be assumed to be 1 (or any other convenient value). Let x be the concentration of Ag(CN)₂⁻ at equilibrium, then the concentrations of Cl⁻ and CN⁻ are also equal to x, as two moles of CN⁻ react with one mole of AgCl to form one mole of Ag(CN)₂⁻ and one mole of Cl⁻.
The solubility product expression for AgCl is:
Ksp = [Ag⁺] [Cl⁻]
Since [Ag⁺] is negligible compared to [CN⁻] in the presence of excess CN⁻, we can assume that [Cl⁻] = [AgCl] = 1. Therefore:
Ksp = 1 = [Ag⁺] [Cl⁻] = [Ag⁺]
Substituting the concentrations of the species into the equilibrium constant expression, we get:
Kc = ([Ag(CN)₂⁻] [Cl⁻])/([AgCl] [CN⁻]²) = (x²)/(1 x x²) = x
The formation constant expression for Ag(CN)₂⁻ is:
Kf = ([Ag(CN)₂⁻])/([Ag⁺] [CN⁻]²)
Substituting [Ag⁺] = 1 and solving for [Ag(CN)₂⁻], we get:
[Ag(CN)₂⁻] = Kf [Ag⁺] [CN⁻]² = 1.0 × 10⁻²¹ × 1 × x² = 1.0 × 10⁻²¹ x²
Substituting this expression for [Ag(CN)₂⁻] into the equilibrium constant expression, we get:
Kc = ([Ag(CN)₂⁻] [Cl⁻])/([AgCl] [CN⁻]²) = (1.0 × 10⁻²¹ x² x)/(1 x x²) = 1.6 × 10¹¹
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: For the reaction below: 1. NaN3 2. H20, heat a Draw the organic product.
The reaction of NaN3 and H2O, upon heating, results in the formation of NaNH2 and HN3. There is no organic product formed in this reaction as neither of the reactants is an organic compound, and the products are inorganic compounds.
NaN3 is sodium azide, which is a common reagent used in organic synthesis for the preparation of primary amines, among other things. HN3 is hydrazoic acid, which is a weak acid and a highly toxic and explosive compound. NaNH2 is sodium amide, which is a strong base used in organic synthesis for deprotonation reactions.
The reaction between NaN3 and H2O is an example of an inorganic reaction that is important in the preparation of inorganic compounds and is not relevant to organic synthesis.
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The reaction of NaN3 and H2O, upon heating, results in the formation of NaNH2 and HN3. There is no organic product formed in this reaction as neither of the reactants is an organic compound, and the products are inorganic compounds.
NaN3 is sodium azide, which is a common reagent used in organic synthesis for the preparation of primary amines, among other things. HN3 is hydrazoic acid, which is a weak acid and a highly toxic and explosive compound. NaNH2 is sodium amide, which is a strong base used in organic synthesis for deprotonation reactions.
The reaction between NaN3 and H2O is an example of an inorganic reaction that is important in the preparation of inorganic compounds and is not relevant to organic synthesis.
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in a 1-butanol molecule, what part of the molecule is described as hydrophobic?
In a 1-butanol molecule, the hydrophobic part is the butyl group (-CH₂CH₂CH₂CH₃), which is a long, nonpolar chain of carbon and hydrogen atoms.
The hydroxyl group (-OH) at the other end of the molecule is hydrophilic, as it is polar and can form hydrogen bonds with water molecules. The hydrophobic butyl group, on the other hand, tends to repel water and interact more favorably with other hydrophobic molecules.
To expand further, the term "hydrophobic" refers to a molecule or part of a molecule that tends to repel water and other polar substances. This is because hydrophobic substances are typically nonpolar or have a low polarity, meaning they have no or very few electrically charged or partially charged areas.
Water, on the other hand, is a polar molecule, meaning it has a partial positive charge on one end and a partial negative charge on the other. Polar substances like water interact favorably with other polar molecules and are repelled by nonpolar molecules.
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The above reaction scheme presents one possible synthesis of diphenhydramine. Work out the synthesis on a separate sheet of paper, and then draw the structure of compound h.
The structure of compound H, which is diphenhydramine, is given below.
What is chemical reaction?The process of changing one or more substances into new ones with different chemical and physical properties is known as a chemical reaction. The atoms of the reactants are rearranged to create new compounds or molecules during a chemical reaction, which causes some chemical bonds to break and new ones to form.
Different types of chemical reactions exist, including synthesis, decomposition, combustion, acid-base, and redox reactions.
H H
| |
H---C---N(CH3)---C---H
| | | |
H C C H
| |
C C
| |
H H
The molecule consists of two phenyl rings attached to a central carbon atom, which is bonded to a nitrogen atom and two methyl groups.
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What is the final ph if 0.03 mol hcl is added to .500l of a buffer solution that is .024 m nh3 and .20 m nh4cl?
The final pH of the buffer solution is 9.13 after adding 0.03 mol HCl to 0.500 L of a buffer solution containing 0.024 M NH₃ and 0.20 M NH₄Cl.
To calculate the final pH, we will use the Henderson-Hasselbalch equation: pH = pKa + log([A-]/[HA]). First, find the pKa value of the conjugate acid NH₄⁺, which is 9.25. Next, calculate the moles of NH₃ and NH₄Cl present in the buffer solution by multiplying their molarity by the volume (0.5 L).
Then, subtract the moles of HCl added from the moles of NH₃ and add them to the moles of NH₄Cl to find the new concentrations of NH₃ and NH₄Cl. Finally, plug the new concentrations into the Henderson-Hasselbalch equation to find the final pH.
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if a buffer solution is 0.190 m in a weak base ( b=6.8×10−5) and 0.530 m in its conjugate acid, what is the ph?
The pH of the buffer solution is slightly basic due to the presence of a weak base.
To determine the pH of the buffer solution, we need to use the Henderson-Hasselbalch equation: pH = pKa + log ([A-]/[HA]), where pKa is the acid dissociation constant of the weak acid, [A-] is the concentration of the conjugate base, and [HA] is the concentration of the weak acid.
In this case, the weak base and its conjugate acid act as the acid and base components of the buffer, respectively. The pKa can be calculated using the expression pKa = -log(Ka), where Ka is the equilibrium constant for the dissociation of the weak acid.
Plugging in the given values, we get a pKa of 9.17. Then, we can substitute the concentrations of the weak base and its conjugate acid into the Henderson-Hasselbalch equation and solve for the pH, which turns out to be 9.59.
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calculate the equilibrium constant of the reaction n2(g) 3 h2(g) ⇌ 2 nh3(g) at 25°c, given that δg°’ = -32.90 kj/mol
The equilibrium constant of the reaction is approximately 2.6 x [tex]10^5[/tex] at 25°C.
The equilibrium constant (K) of the reaction can be calculated using the equation:
ΔG° = -RTlnK
where ΔG° is the standard free energy change, R is the gas constant, T is the temperature in Kelvin, and ln is the natural logarithm.
Given that ΔG°’ = -32.90 kJ/mol and the temperature is 25°C (298.15 K), we can solve for K as follows:
ΔG° = -RTlnK
-32.90 kJ/mol = -(8.314 J/mol•K)(298.15 K) lnK
lnK = -32.90 kJ/mol / (-8.314 J/mol•K)(298.15 K)
lnK = 12.23
K = [tex]e^{(lnK)[/tex]
K =[tex]e^{(12.23)[/tex]
K ≈ 2.6 x [tex]10^5[/tex]
Therefore, the equilibrium constant of the reaction is approximately 2.6 x [tex]10^5[/tex] at 25°C.
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what monoalkylation product(s) would you obtain if benzene were alkylated with n-butyl chloride
The mono-alkylation product(s) obtained when benzene is alkylated with n-butyl chloride are:
The mono-alkylation product obtained from the reaction of benzene with n-butyl chloride is n-butylbenzene. This reaction occurs through the Friedel-Crafts alkylation process:
1. Formation of electrophile: n-Butyl chloride reacts with a Lewis acid catalyst, such as aluminum chloride (AlCl3), to form an electrophile, n-butyl carbocation (C4H9+).
2. Electrophilic attack: The electrophile (n-butyl carbocation) attacks the benzene ring, breaking one of the pi bonds in the aromatic ring.
3. Intermediate formation: A carbocation intermediate is formed, with the n-butyl group attached to the benzene ring.
4. Deprotonation: The carbocation intermediate loses a proton (H+), and the pi bond is restored to regenerate the aromatic character of the benzene ring.
The final product is n-butylbenzene, which is the mono-alkylation product of benzene with n-butyl chloride.
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when a student mixes 50 ml of 1.0 M HCL and 50 ml of 1.0M NaOH in a coffee-cup calorimeter, the temperature of the resultant solution increases from 21.0 C to 27.5 C. Calculate the enthalpy change for the reaction in kJ/mol HCL. For this problem, assume that the calorimeter loses only a negligible quantity of heat, the total volume of the solution is 100 ml, the density of the solution is 1.0 g/ml, and its specific heat is 4.18 J/g-K.
The balanced chemical equation for the reaction between HCl and NaOH is: the enthalpy change for the reaction between 1.0 M HCl and 1.0 M NaOH is -54.22 kJ/mol HCl (since the reaction is exothermic).
HCl(aq) + NaOH(aq) → NaCl(aq) + [tex]H_{2} O[/tex](l)
First, we need to calculate the amount of heat released or absorbed by the reaction using the formula:
q = m·C·ΔT
where q is the heat absorbed or released by the reaction, m is the mass of the solution, C is the specific heat of the solution, and ΔT is the change in temperature of the solution. Since the total volume of the solution is 100 mL and the density is 1.0 g/mL, the mass of the solution is:
m = 100 mL × 1.0 g/mL = 100 g
The specific heat of the solution is given as 4.18 J/g-K. The change in temperature is:
ΔT = 27.5°C - 21.0°C = 6.5°C
Therefore, the amount of heat released or absorbed by the reaction is:
q = 100 g × 4.18 J/g-K × 6.5°C = 2,711 J
Next, we need to convert the amount of heat to the enthalpy change for the reaction per mole of HCl. Since we mixed 50 mL of 1.0 M HCl with 50 mL of 1.0 M NaOH, we have 0.05 moles of HCl in the solution. Therefore, the enthalpy change per mole of HCl is:
ΔH = q / n
where n is the number of moles of HCl. Therefore,
ΔH = 2,711 J / 0.05 mol = 54,220 J/mol
To express the result in kJ/mol, we need to divide by 1000:
ΔH = 54.22 kJ/mol
Therefore, the enthalpy change for the reaction between 1.0 M HCl and 1.0 M NaOH is -54.22 kJ/mol HCl (since the reaction is exothermic).
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