Nitrogen gas will effuse about 2.16 times faster than xenon gas under these conditions.
The rate of effusion of a gas will be inversely proportional to square root of its molar mass. This is known as Graham's law of effusion. Mathematically, we will express this relationship as:
rate of effusion ∝ 1/√(molar mass)
Let's assume that the temperature and pressure of the gases are kept constant. The molar mass of nitrogen gas (N₂) is 28 g/mol, and the molar mass of xenon gas (Xe) is 131 g/mol. Therefore, the ratio of their molar masses will be;
molar mass of Xe / molar mass of N₂ = 131 g/mol / 28 g/mol = 4.68
According to Graham's law of effusion, the ratio of their effusion rates is the inverse of the ratio of their square roots of their molar masses. This can be expressed as;
rate of effusion of N₂ / rate of effusion of Xe = √(molar mass of Xe) / √(molar mass of N₂) = √4.68 = 2.16
Therefore, 2.16 times faster will the nitrogen effuse.
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A container containing 5.75L of a gas is collected at 115 K and then allowed to expand to 25 L. What must the new temperature be to maintain the same pressure?
Charles's Law-
[tex]\:\:\:\:\:\: \:\:\:\:\:\:\star\longrightarrow\sf \underline{\dfrac{V_1}{T_1}=\dfrac{V_2}{T_2}}\\[/tex]
Where:-
V₁ = Initial volumeT₁ = Initial temperatureV₂ = Final volumeT₂ = Final temperatureAs per question, we are given that -
V₁=5.75L T₁ = 115KV₂ =25 LNow that we have obtained all the required values, so we can put them into the formula and solve for T₂ :-
[tex]\:\:\:\:\:\: \:\:\:\:\:\:\star\longrightarrow\sf \underline{\dfrac{V_1}{T_1}=\dfrac{V_2}{T_2}}\\[/tex]
[tex]\:\:\:\:\:\: \:\:\:\:\:\:\longrightarrow\sf \underline{\dfrac{T_2}{V_2}=\dfrac{T_1}{V_1}}\\[/tex]
[tex]\:\:\:\:\:\: \:\:\:\:\:\:\longrightarrow\sf \underline{T_2=\dfrac{T_1}{V_1} \times V_2}\\[/tex]
[tex]\:\:\:\:\:\: \:\:\:\:\:\:\longrightarrow\sf T_2=\cancel{\dfrac{115}{5.75} }\times 25\\[/tex]
[tex]\:\:\:\:\:\: \:\:\:\:\:\:\longrightarrow\sf T_2=20 \times 25\\[/tex]
[tex]\:\:\:\:\:\: \:\:\:\:\:\:\longrightarrow\sf T_2=500\:K\\[/tex]
[tex]\:\:\:\:\:\: \:\:\:\:\:\:\longrightarrow\sf T_2=(500 -273)°C\\[/tex]
[tex]\:\:\:\:\:\: \:\:\:\:\:\:\longrightarrow\sf \underline{T_2=227\:°C}\\[/tex]
Therefore, the new temperature will become 500K or, 227°C to maintain the same pressure.
the rate constant of a certain reaction is known to obey the arrhenius equation, and to have an activation energy . if the rate constant of this reaction is at , what will the rate constant be at ? round your answer to significant digits.
The rate constant is 3.0 x [tex]10^{4} M^{-1} S^{-1}[/tex] which is calculated by using the Arrhenius equation.
We can use the Arrhenius equation to find [tex](E_{a} / R)[/tex][([tex](1/ T_{2} ) - ( 1/ T_{1} )[/tex]] separately,
We have, Activation energy, [tex]E_{a}[/tex] = 30.0 kJ/mole = 30000 J/mole
R = 8.314 J/mole .K
Temperature T2 = 201.0C = 201.0 + 273 = 474 K
T1 = 172 C = 172 + 273 = 445 K
Hence,
{ [tex](E_{a} / R)[/tex][([tex](1/ T_{2} ) - ( 1/ T_{1} )[/tex] } = (30000/8.314) x [(1/474) – (1/445)]
[tex](E_{a} / R)[/tex][([tex](1/ T_{2} ) - ( 1/ T_{1} )[/tex] = – 0.4961
Let us use this value and Rate constant [tex]K_{2}[/tex] = 5.0 x 104[tex]M^{-1} S^{-1}[/tex] and have to solve it for [tex]K_{1}[/tex]
[tex]K_{1}[/tex] / (5.0 x 104) = e –0.4961.
[tex]K_{1}[/tex] / (5.0 x 104) = 0.6089
[tex]K_{1}[/tex] = 5.0 x 104 x 0.6089
[tex]K_{1}[/tex] = 3.0 x [tex]10^{4} M^{-1} S^{-1}[/tex]
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The complete question is,
The rate constant of a certain reaction is known to obey the Arrhenius equation, and to have an activation energy is 30.0 kJ/mole. if the rate constant of this reaction is at 5.0 x 104[tex]M^{-1} S^{-1}[/tex], what will the rate constant be at 172 C? round your answer to significant digits.
Starting from benzene and any other needed starting materials/reagents, show how to make following compounds. a) ptert butylchlorobenzene b) 1-phenylcyclopentene c) m-bromonitrobenzene d) p-bromonitrobenzene e) o-bromonitrobenzene f) p-toluenesulfonic acid
To synthesize the following compounds, use these reactions- Friedel-Crafts alkylation, elimination reaction, Friedel-Crafts halogenation, nitration and sulfonation.
a) p-tert-butylchlorobenzene:,
1. Benzene + tert-butyl chloride + AlCl3 (Friedel-Crafts alkylation) → p-tert-butylbenzene
2. p-tert-butylbenzene + Cl2 + AlCl3 (Friedel-Crafts halogenation) → p-tert-butylchlorobenzene
b) 1-phenylcyclopentene:
1. Benzene + 1,5-dibromopentane (Friedel-Crafts alkylation) → 1-phenylcyclopentyl bromide
2. 1-phenylcyclopentyl bromide + KOH (elimination reaction) → 1-phenylcyclopentene
c) m-bromonitrobenzene:
1. Benzene + HNO3 + H2SO4 (nitration) → nitrobenzene
2. Nitrobenzene + Br2 + AlCl3 (Friedel-Crafts halogenation) → m-bromonitrobenzene
d) p-bromonitrobenzene:
1. Benzene + HNO3 + H2SO4 (nitration) → nitrobenzene
2. Nitrobenzene + Br2 + FeBr3 (halogenation) → p-bromonitrobenzene
e) o-bromonitrobenzene:
1. Benzene + Br2 + AlCl3 (Friedel-Crafts halogenation) → bromobenzene
2. Bromobenzene + HNO3 + H2SO4 (nitration) → o-bromonitrobenzene
f) p-toluenesulfonic acid:
1. Benzene + CH3Cl + AlCl3 (Friedel-Crafts alkylation) → toluene
2. Toluene + H2SO4 (sulfonation) → p-toluenesulfonic acid
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does the mass of the pt(s) electrode increase, decrease, or remain the same as the cell operates? justify your answer.
The mass of the Pt(s) electrode remains the equal because the Pt does now not react and no Cu atoms could be deposited at the Pt electrode.
Mass is a fundamental physical property that refers to the amount of matter in an object. It is a measure of the number of atoms, molecules, or particles that make up an object or substance. Mass is typically measured in grams (g) or kilograms (kg), and it is different from weight, which is a measure of the gravitational force exerted on an object.
In chemical reactions, the mass of reactants must be equal to the mass of the products, according to the law of conservation of mass. This means that the total mass of all substances involved in a chemical reaction remains constant, regardless of any physical or chemical changes that may occur. The mass of an atom is typically expressed in atomic mass units (amu), which are defined relative to the mass of a carbon-12 atom. The mass of a molecule is the sum of the masses of all its constituent atoms.
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what two carbonyl compounds would be required in order to synthesize the compound below by a crossed aldol condensation?
The compound shown below can be synthesized by a crossed aldol condensation between propanal and benzaldehyde.
A crossed aldol condensation involves the reaction of two different carbonyl compounds, where the enolate ion of one carbonyl compound attacks the carbonyl group of another carbonyl compound. In this case, propanal and benzaldehyde are the two carbonyl compounds required to synthesize the given compound.
The enolate ion of propanal attacks the carbonyl group of benzaldehyde to form an intermediate, which then undergoes a dehydration step to yield the final product. This reaction requires the use of a strong base, such as NaOH or KOH, to generate the enolate ion, and can be promoted by heating or by using a catalyst, such as piperidine or pyrrolidine. The crossed aldol condensation is a useful method for the synthesis of complex molecules, and is widely used in organic synthesis.
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--The complete question is, What two carbonyl compounds(as shown in image) would be required in order to synthesize the compound below by a crossed aldol condensation?--
100 points I don't know how to balance and don't know what type each are, pls help!
Explanation:
1. Count each type of atom in reactants and products. ...
2. Place coefficients, as needed, in front of the symbols or formulas to increase the number of atoms or molecules of the substances. ...
3. Repeat steps 1 and 2 until the equation is balanced.
what is the ph of a solution prepared by mixing 30.00 ml of 0.10 m ch3co2h with 30.00 ml of 0.020 m ch3co2na? assume that the volume of the solutions are additive and that ka
pH of solution prepared by mixing 30.00 ml of 0.10 m CH₃CO₂H with 30.00 ml of 0.020 m CH₃CO₂Na is 2.42
CH₃CO₂Na after mixing. Since the volumes are additive, the total volume of the solution is 60.00 mL. The initial concentration of CH₃CO₂H is 0.10 M, and after mixing with CH₃CO₂Na, some of it will react with the Na⁺ ions to form the weak acid's conjugate base CH₃CO₂⁻,
CH₃CO₂H + H₂O ⇌ H₃O+ + CH₃CO₂⁻
Initial: 0.10 M 0 0
Change: -x +x +x
Equilibrium: 0.10 - x x x
Similarly, the initial concentration of CH₃CO₂Na is 0.020 M, and it will dissociate in water to form CH₃CO₂⁻ and Na⁺ ions,
CH₃CO₂Na + H₂O ⇌ CH₃CO₂⁻ + Na⁺
Initial: 0.020 M 0 0
Change: 0 +x +x
Equilibrium:0.020 x x
Since CH₃CO₂⁻ is a common ion in both equilibria, we can use the expression for the acid dissociation constant, Ka, to find the concentration of H₃O⁺:
Ka = [H₃O⁺][CH₃CO₂⁻]/[CH₃CO₂H]
Substituting the equilibrium concentrations into the expression:
1.8 x 10^-5 = [H₃O⁺][x]/(0.10 - x)
Assuming that x is small compared to 0.10 M, we can simplify the expression:
1.8 x 10^-5 = x^2/(0.10)
Solving for x,
x = 0.00378 M
Therefore, the concentration of H₃O⁺ is 0.00378 M, and the pH of the solution is,
pH = -log[H3O+] = -log(0.00378) = 2.42
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--The complete question is, what is the ph of a solution prepared by mixing 30.00 ml of 0.10 m CH₃CO₂Hwith 30.00 ml of 0.020 m CH₃CO₂Na?--
100 POINTS!! PLEASE HELP!!!
Propane (C3H8) burns in oxygen to form CO2 and H2O according to the following equation. How many grams of O2 are required to burn 2.56 x 1022 propane molecules?
the balanced equation is:
C3H8 + 5O2 --> 3CO2 + 4H2O
2.56 x 10²² propane molecules must be burned with 6.82 grams of oxygen.
According to the following equation, how many propane molecules burn in oxygen to produce CO2 and H2O?The following is the balanced equation for propane combustion:
[tex]C3H8 + O2 = 3CO2 + 4H2O[/tex]
Hence, we require 5 oxygen molecules for every molecule of propane.
We must multiply the quantity of propane molecules by the ratio of oxygen molecules to propane molecules in order to determine how many oxygen molecules are needed to burn 2.56 x 1022 propane molecules.
[tex]O2[/tex] to [tex]C3H8[/tex] Ratio: 5:1
The necessary number of O2 molecules is (5/1) times 2.56, which equals 1.28 x 10²³.
So, using the molar mass of oxygen, we can convert the quantity of oxygen molecules to grams.
1 mole of [tex]O2[/tex] = 32 g
1.28 x 10²³ molecules of O2 = (1.28 x 10²³/ 6.022 x 10²³) moles of O2
Mass of [tex]O2[/tex] = (1.28 x 10²³/ 6.022 x 10²³) x 32 g.
Mass of [tex]O2[/tex] = 6.82 grams.
Hence, 6.82 grams of [tex]O2[/tex] are required to burn 2.56 x 10²² propane molecules.
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the heat capacity of a sample should be divided by which quantity to find the specific heat capacity?temperature in degrees celsius
Specific heat capacity is a property of a material that represents the amount of heat required to change the temperature of one kilogram of the substance by one degree Celsius.
It is expressed in units of joules per kilogram per degree Celsius (J/kg°C).To find the specific heat capacity of a sample, you should divide its heat capacity by its mass. This distinction is important because specific heat capacity allows for the comparison of different materials' abilities to store thermal energy.
For example, if you have two different materials with the same mass, the one with the higher specific heat capacity will require more energy to heat it up by the same temperature difference. This will provide you with the amount of energy required to change the temperature of one kilogram of the substance by one degree Celsius.
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which action will use up one of the components and destroy the buffer solution containing 0.500 mol hf and 0.500 mol kf? group of answer choices addition 0.500 mol koh will use up one of the components and destroy the buffer solution addition 0.005 mol hf will use up one of the components and destroy the buffer solution addition 0.005 mol hcl will use up one of the components and destroy the buffer solution addition 0.500 mol kcl will use up one of the components and destroy the buffer solution addition 0.500 mol kf will use up one of the components and destroy the buffer solution addition 0.005 mol koh will use up one of the components and destroy the buffer solution
When a 0.005 mol HCl is added to the buffer solution containing 0.500 mol HF and 0.500 mol KF, it will use up one of the components and destroy the buffer solution.
A buffer solution is a chemical mixture that resists changes in pH by neutralizing small amounts of added acids or bases. It contains a weak acid and its conjugate base or a weak base and its conjugate acid.
Buffer solutions are necessary for many chemical processes since pH changes can dramatically affect the behavior of chemical compounds.
Buffer solutions can be prepared using a wide range of chemical compounds, and the exact composition of the solution is determined by the desired pH range and the concentration of the buffer components.
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A student discovers a solution in an unmarked beaker. An equal amount of strong acid is added to the beaker. The products are salt and water.
What was the pH of the unmarked beaker before adding the acid?
hurry!!
In this case, since the unmarked beaker contained a strong base, we can expect its pH to be greater than 7.
.
To determine the pH of the unmarked beaker before adding the acid, we need to understand the pH scale. The pH scale ranges from 0 to 14, with a pH of 7 being neutral, below 7 being acidic, and above 7 being basic.
Since we know the solution in the unmarked beaker was a strong base, it is likely to have a high pH value.
Common strong bases include sodium hydroxide (NaOH) and potassium hydroxide (KOH). The pH of these strong bases in a concentrated solution is typically between 12 and 14.
However, without additional information about the specific strong base and its concentration, it is impossible to give an exact pH value for the unmarked beaker.
In summary, we can conclude that the pH of the unmarked beaker before adding the acid was above 7, indicating a strong basic solution, but we would need more information to determine the exact pH value.
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on a paper chromatogram, the solvent front traveled 6.0 cm and the sample spot traveld 4.2 cm. what is the rf for that sample?
The distance travelled by the chromatogram substance divided by the solvent front's travel distance yields the Rf (retention factor) value. Hence, the sample's Rf value is close to 0.7.
It is a dimensionless quantity that aids in the identification and description of substances separated by chromatography. A different technique that may be used to determine if a material is pure or impure is paper chromatography. Moreover, it may be used to separate mixtures of soluble chemicals in order to identify the components of the mixture.
They are frequently coloured materials like food colouring, ink, dye, or plant pigments. The Rf value in this situation may be determined using the formula below: Rf is equal to the sum of the distances covered by the substance front and the solvent front.
Rf = 4.2 cm ÷ 6.0 cm
Rf ≈ 0.7
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potassium fluoride is added to water at a temperature of 298 k. if the initial concentration of that potassium fluoride in water is 0.251 m, then what is the ph of this solution? facts you may need: kw
If the initial concentration of that potassium fluoride in water is 0.251 m, then The pH of the mixture is roughly 9.61.
Potassium fluoride is an ionic compound that dissociates into potassium cations [tex](K^+)[/tex] and fluoride anions [tex](F^-)[/tex] when it is dissolved in water. The dissociation equation is:
[tex]KF(s)[/tex] → [tex]K^+(aq) + F^-(aq)[/tex]
The equilibrium constant for this dissociation reaction is denoted by [tex]K_{sp}[/tex], and its value depends on the temperature of the solution.
At 298 K, the value of [tex]K_{sp}[/tex] for potassium fluoride is approximately [tex]1.76 * 10^{-10}[/tex]. This means that only a small fraction of the potassium fluoride molecules will dissociate into ions in water, and the majority of the molecules will remain in their undissociated form.
To calculate the pH of the solution, we need to consider the effect of the fluoride ions on the acidity of the water. Fluoride ions are weak bases that can react with water molecules to form hydrofluoric acid (HF) and hydroxide ions (OH-):
[tex]F^-(aq) + H_2O(l)[/tex] ⇌ [tex]HF(aq) + OH^-(aq)[/tex]
The equilibrium constant for this reaction is denoted by [tex]K_b[/tex] and its value is [tex]3.5 * 10^{-11}[/tex] at 298 K.
To determine the pH of the solution, we need to find the concentration of hydroxide ions in the solution. This can be calculated from the concentration of fluoride ions using the [tex]K_b[/tex] value and the equilibrium constant expression:
[tex]K_b = \frac{[HF][OH^-]}{[F^-]}[/tex]
⇒ [tex][{OH}^-] = K_b * \frac{[F^-]}{[HF]}[/tex]
The concentration of hydrofluoric acid can be calculated from the concentration of fluoride ions using the dissociation constant for HF, which is denoted by Ka and has a value of [tex]7.2 * 10^{-4}[/tex] at 298 K.
[tex]K_a = \frac{[H^+][F^-]}{[HF]}[/tex]
⇒ [tex][H^+] = K_a * \frac{[HF]}{[F^-]}[/tex]
Substituting the given values, we get:
[tex][F^-] = 0.251 M[/tex]
[tex]K_b = 3.5 * 10^{-11}[/tex]
[tex]K_a = 7.2 * 10^{-4}[/tex]
[tex][HF] = \frac{[F^-] * K_a}{[H^+]}[/tex]
⇒ [tex][H^+] = \frac{[F^-] * K_a}{[HF]}[/tex]
⇒ [tex][H^+] = \frac{[F^-] * K_a}{(\frac{[F^-] * K_a}{[H^+]})}[/tex]
⇒ [tex][H^+] = \frac{(K_b * [F^-])}{[HF]}[/tex]
⇒ [tex][H^+] = \frac{(3.5 * 10^{-11}) * (0.251)}{(\frac{[F^-] * K_a}{[HF]})}[/tex]
⇒ [tex][H^+] = 2.45 * 10^{-10} M[/tex]
[tex]pH = -log[H^+][/tex]
⇒ [tex]pH = -log(2.45 * 10^{-10})[/tex]
⇒ [tex]pH = 9.61[/tex]
Therefore, the pH of the solution is approximately 9.61.
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the acid - dissociation constant for benzoic acid, hc 7 h 5 o 2 , is 6.3 x 10 - 5 . calculate the equilibrium conc entrations of h 3 o , c 7 h 5 o 2 - , and hc 7 h 5 o 2 in the solution if the initial concentration of the acid is 0.050m
The equilibrium concentration of C₆H₅COOH is 0,06294 M which is calculated by using the balanced chemical reaction.
The balanced chemical reaction can be written as,
C₆H₅COOH(aq.) ⇄ H⁺(aq.) + C₆H₅COO⁻(aq.).
Ka(C₆H₅COOH) is 6,3·10⁻⁵.
c(C₆H₅COOH) is 6,3·10⁻² M.
[H⁺] = [C₆H₅COO⁻] = x which is called as equilibrium concentration.
The equilibrium concentration of the reaction may be defined as the ratio of concentrations of the substances on the right side of the reaction to the concentrations of those on the left side of the reaction which is equals a constant appropriate for that specific chemical reaction.
[C₆H₅COOH] = 0,063 M - x.
Ka = [H⁺] · [C₆H₅COO⁻] / [C₆H₅COOH].
0,000063 = x² / 0,063 M - x.
By solving the equation we get,
x = 0,00006 M.
[C₆H₅COOH] = 0,063 M - 0,00006 M.
[C₆H₅COOH] = 0,06294 M
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The melting point of a solid is 24.9OC. As heat is added to melt the solid, what happens to the particles?
The motion of the particles increases.
The particles move closer together.
The motion of particles decreases.
The particles move farther apart.
Explanation:
As heat is added to melt the solid, the motion of the particles increases. This is because the heat energy increases the kinetic energy of the particles, causing them to vibrate more and move faster. Eventually, the increased motion overcomes the forces holding the particles together in a solid lattice structure, and the solid melts into a liquid.
the nitration of bromobenzene occurs as shown in the scheme. following the reaction, several purification steps are performed where sodium bicarbonate, dichloromethane, sodium chloride, and ethanol are used. identify the purpose of each of these by answering the given questions. after the reaction has gone to completion, what is the purpose of adding saturated sodium bicarbonate to the reaction mixture? it is used to neutralize any remaining acid. it is used to absorb excess heat generated by the exothermic reaction. it causes excess bromine to be given off as a gas. it is used to extract the bromonitrobenzene from the reaction mixture. why is dichloromethane added next to the reaction mixture? it is used to crystallize the bromonitrobenzene. it neutralizes excess acid in the reaction mixture. it is used as the drying agent to remove water. it is used to extract the bromonitrobenzene from the acid mixture. when washing the aqueous sodium bicarbonate layer with dichloromethane, which layer contains the product? the product is in the aqueous sodium bicarbonate layer. the product is insoluble in both layers and precipitates out of solution. the product is in the dichloromethane layer. the product is an emulsion between the two layers. after removing the dichloromethane layer, why is it rinsed with saturated sodium chloride? it is used as the drying agent to remove any excess water. it causes the bromonitrobenzene to precipitate out of solution. it is used to remove any water soluble impurities from the dichloromethane layer. it is used to crystallize the bromonitrobenzene. what is the best way to remove the last remaining traces of water from the dichloromethane layer? using decantation adding a drying agent using distillation by evaporation what is the 95% ethanol used for after obtaining the solid bromonitrobenzene? it is used as the solvent for recrystallization. it is used as an additional extraction solvent. it is used to remove dichloromethane from the bromonitrobenzene. it is used to remove more water from the dichloromethane.
Decantation and distillation by evaporation can also be used.95% ethanol is used as the solvent for recrystallization after obtaining the solid bromonitrobenzene. This ensures that the bromonitrobenzene is purified to obtain a higher yield of the desired product.
As a question-answering bot, when answering questions on the platform Brainly, I should always be factually accurate, professional, and friendly while being concise and not providing extraneous amounts of detail. Additionally, I should use the following terms in my answer to this particular question: nitration of bromobenzene, sodium bicarbonate, dichloromethane, sodium chloride, and ethanol. Also, I will be identifying the purpose of each of these by answering the given questions. After the reaction has gone to completion, the purpose of adding saturated sodium bicarbonate to the reaction mixture is that it is used to neutralize any remaining acid. Why is dichloromethane added next to the reaction mixture? Dichloromethane is added to the reaction mixture to extract the bromonitrobenzene from the acid mixture. When washing the aqueous sodium bicarbonate layer with dichloromethane, the layer containing the product is the dichloromethane layer. After removing the dichloromethane layer, the layer is rinsed with saturated sodium chloride to remove any water-soluble impurities from the dichloromethane layer. The best way to remove the last remaining traces of water from the dichloromethane layer is by using a drying agent.
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tech a says that the two-way catalytic convertor was the first type of catalytic convertor designed. tech b says that it converts hydrocarbons and carbon monoxide to carbon dioxide and water. who is correct?
Both tech A and tech B are partially correct because as per tech A the two-way catalytic converter was the first type of converter which was introduced in the mid-1970s and as per tech B it converts hydrocarbons and carbon monoxide to carbon dioxide and water to reduce these two pollutants and is not effective in reducing Nitrogen Oxides.
The process of converting CO and HC to CO2 and H2O takes place through a chemical reaction that occurs on a catalytic surface within the two-way catalytic converter. The catalytic surface is typically made up of precious metals, such as platinum, palladium, or rhodium, which act as catalysts to speed up the chemical reaction.
The two-way catalytic converter converts two types of harmful emission, whereas a three-way catalytic converter, which was developed later converts three types of emissions-CO, HC, and NO into less harmful substances. It is also used as a catalyst to promote chemical reactions that convert harmful substances in CO and water vapour which helps in reducing harmful gases in vehicles and improving the air quality.
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a solution made with of phosporic acid ( , see below), dissolved in of solution, was titrated with . how many of solution are necessary to reach a of ?
To solve this titration problem, we can use the following formula:
M1V1 = M2V2
where M1 is the concentration of the phosphoric acid solution, V1 is the volume of the phosphoric acid solution, M2 is the concentration of the titrant (the solution being added during the titration), and V2 is the volume of the titrant required to reach the endpoint of the titration.
We can start by calculating the number of moles of phosphoric acid in the initial solution:
moles of H3PO4 = (0.20 mol/L) x (0.500 L) = 0.100 mol
Next, we can use the balanced chemical equation for the reaction between phosphoric acid and sodium hydroxide to determine the stoichiometry of the titration:
H3PO4 + 3 NaOH → Na3PO4 + 3 H2O
The equation shows that for every one mole of phosphoric acid, three moles of sodium hydroxide are required to reach the endpoint of the titration.
Since the desired pH is not provided, we will assume that the endpoint of the titration is pH 7, which is close to the neutral pH of water.
At pH 7, sodium hydroxide is completely neutralized and the solution contains only sodium phosphate. The balanced chemical equation for the reaction between phosphoric acid and sodium hydroxide shows that one mole of phosphoric acid reacts with one mole of sodium phosphate:
H3PO4 + Na3PO4 → 3 NaH2PO4
Therefore, to reach the endpoint of the titration, we need three times the number of moles of sodium hydroxide as moles of phosphoric acid:
moles of NaOH = 3 x moles of H3PO4 = 3 x 0.100 mol = 0.300 mol
Finally, we can use the concentration and number of moles of sodium hydroxide to calculate the volume required to reach the endpoint of the titration:
V2 = moles of NaOH / M2
Assuming the concentration of the titrant sodium hydroxide is 0.100 mol/L (which is commonly used for titrations), we have:
V2 = 0.300 mol / 0.100 mol/L = 3.00 L
Therefore, we need 3.00 L of the sodium hydroxide solution to reach the endpoint of the titration and achieve a pH of 7.
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after a lemon is squeezed into a bottle of water, the new concentration of H3O is 0.00006M. what is the pH of the water lemon mixture?
Answer:
Assuming that the lemon juice is the only source of H3O+ ions in the water, we can use the formula for pH to calculate the pH of the water-lemon mixture:
pH = -log[H3O+]
First, we need to determine the concentration of H3O+ in the solution. We are given that the concentration of H3O+ after the lemon is squeezed into the water is 0.00006 M. Therefore:
[H3O+] = 0.00006 M
Now, we can substitute this value into the pH formula:
pH = -log(0.00006)
pH = 4.22
Therefore, the pH of the water-lemon mixture is 4.22.
GEN CHEM2 PLEASE HELP
The concentration of the hydronium ion, [H₃O⁺] of 0.28 M calcium hydroxide, Ca(OH)₂ solution is 1.8×10⁻¹⁴ M (option A)
How do i determine the value of [H₃O⁺]?First, we shall determine the [OH⁻] of the solution. Details below:
Ca(OH)₂(aq) <=> Ca²⁺(aq) + 2OH⁻(aq)
From the above equation,
1 mole of Ca(OH)₂ is contains 2 mole ofOH⁻
Therefore,
0.28 M Ca(OH)₂ will contain = 0.28 × 2 = 0.56 M OH⁻
Finally, we shall determine the hydronium, ion [H₃O⁺] of the solution. Details below:
Concentration of hydroxide ion, [OH⁻] = 0.56 MConcentration of hydronium, ion [H₃O⁺] = ?[H₃O⁺] × [OH⁻] = 10¯¹⁴
[H₃O⁺] × 0.56 = 10¯¹⁴
Divide both side by 0.56
[H₃O⁺] = 10¯¹⁴ / 0.56
[H₃O⁺] = 1.8×10⁻¹⁴ M
Thus, hydronium, ion [H₃O⁺] of the solution is 1.8×10⁻¹⁴ M (option A)
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g anode is the electrode where oxidation occurs while the cathode is the electrode where reduction occurs group of answer choices true false
It is true that the g anode is indeed the electrode where oxidation occurs while the cathode is the electrode where reduction occurs.
Electrons are moved from one species to another in redox reactions. When a reaction occurs spontaneously, energy is released that can be put to good use. The process must be divided into the oxidation reaction and the reduction reaction to capture this energy. A wire is used to move the electrons from one side of the reactions to the other after they have been placed into two separate containers. A voltaic/galvanic cell is produced as a result.
Electrons are transported from one species to the other during a redox reaction. The energy that can be used for work is released if the reaction is spontaneous.
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some contact lens solutions use boric acid buffers to resist ph change. over what ph range does boric acid have good buffering capacity? explain
For boric acid, with a pKa of approximately 9.24, the good buffering capacity occurs over the pH range of 8.24 to 10.24.
Boric acid is a weak acid often used as a buffering agent in contact lens solutions to resist pH changes. Its good buffering capacity is due to its ability to maintain a relatively constant pH level when small amounts of acids or bases are added to the solution. To understand the pH range over which boric acid has a good buffering capacity, we need to look at its acid dissociation constant (Ka) and the Henderson-Hasselbalch equation.
The Ka value for boric acid is approximately 5.8 x 10^-10. The pKa, which is the negative logarithm of the Ka, is calculated as follows:
pKa = -log(Ka) ≈ 9.24
The Henderson-Hasselbalch equation relates the pH, pKa, and the ratio of the concentrations of the conjugate base ([A-]) to the weak acid ([HA]) in a solution:
pH = pKa + log ([A-]/[HA])
A good buffering capacity is achieved when the pH of the solution is within one unit of the pKa value. This is because the ratio of the conjugate base to the weak acid ([A-]/[HA]) is near 1, meaning both the acid and its conjugate base are present in similar amounts, and the buffer can effectively neutralize added acids or bases.
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when the skeleton equation above is balanced and all coefficients reduced to their lowest whole-number terms, what is the coefficient for h ?
The coefficient of h in the balanced chemical equation above is 2.
When the skeleton equation above is balanced and all coefficients reduced to their lowest whole-number terms, the coefficient for h is 2.What is a skeleton equation?
A skeleton equation is an equation in which the reactants and products are listed using their chemical formulas but not their stoichiometric coefficients. A balanced chemical equation is one in which the stoichiometric coefficients for reactants and products are included in such a way that the law of conservation of mass is fulfilled,
and the number of atoms of each element present in the reactants equals the number of atoms of that element present in the products.For instance, a typical chemical equation, which shows the chemical reactants and products, can be represented as:
2 NaOH (aq) + H2SO4 (aq) → Na2SO4 (aq) + 2 H2O
(l)Here, the reaction involves two moles of sodium hydroxide, one mole of sulfuric acid, one mole of sodium sulfate, and two moles of water. To balance this equation,
the coefficients should be added as:2 NaOH (aq) + H2SO4 (aq) → Na2SO4 (aq) + 2 H2O (l)The balanced chemical equation represents the actual stoichiometric ratios of reactants and products in the chemical reaction.
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which method should be used when stirring the contents of the calorimeter? select one: steady the thermometer and gently swirl the entire calorimeter stir the contents using the thermometer the contents of the calorimeter do not need to be stirred place a stir bar in the calorimeter
When stirring the contents of the calorimeter, the method that should be used is to steady the thermometer and gently swirl the entire calorimeter.
Describe the calorimeter The amount of heat absorbed or released by a chemical reaction or physical change is measured using a calorimeter. A reaction or change takes place in the inner compartment of the calorimeter, while the outer compartment acts as an insulator. Due to its excellent insulation, any heat generated or lost during the reaction or change can be measured.
A thermometer is placed through a hole at the top of a calorimeter to gauge the temperature inside. The correct way to stir the calorimeter's contents is to hold the thermometer firm while gently swirling the entire calorimeter.
The thermometer should not be used to stir the calorimeter's contents, nor should a stir bar be put within. The temperature inside the calorimeter is distributed uniformly throughout the entire mixture thanks to the gently whirling technique.
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Write a paragraph on how buffer solutions play an important role in regulating blood acidity.
PLEASE HELP
Answer:
Buffer solutions are essential in regulating blood acidity because they help maintain a stable pH level in the body. The pH of blood needs to be tightly controlled within a narrow range of 7.35 to 7.45 for optimal physiological functioning. Buffers work by absorbing excess hydrogen ions (H+) or hydroxide ions (OH-) that may be introduced into the bloodstream from various metabolic reactions. For example, if the blood becomes too acidic due to an increase in H+ ions, buffer systems such as the bicarbonate-carbonic acid system can neutralize the excess H+ ions and restore the pH balance. Conversely, if the blood becomes too alkaline due to an increase in OH- ions, buffer systems can absorb the excess OH- ions and bring the pH back to normal. In this way, buffer solutions help maintain the delicate acid-base balance crucial to the proper functioning of our bodies.
What is the boiling point in ºC of a 3.6 molal solution of ethylene glycol in benzene?
The boiling point of the 3.6 molal solution of ethylene glycol in benzene is 89.208 ºC.
What is Boiling Point?
Boiling point is the temperature at which the vapor pressure of a liquid equals the external pressure, usually atmospheric pressure. At the boiling point, bubbles of vapor are formed throughout the liquid, and the liquid changes phase to become a gas.
The boiling point elevation of a solution is given by the formula:
ΔTb = Kb x molality
Where ΔTb is the boiling point elevation, Kb is the ebullioscopic constant of the solvent, and molality is the molal concentration of the solute.
For benzene, Kb = 2.53 ºC/m.
Since the molality of the ethylene glycol solution is 3.6 molal, we can calculate the boiling point elevation as:
ΔTb = 2.53 ºC/m x 3.6 molal = 9.108 ºC
This means that the boiling point of the solution is 9.108 ºC higher than the boiling point of pure benzene.
To find the boiling point of the solution, we need to add this value to the boiling point of pure benzene, which is 80.1 ºC. Therefore:
Boiling point of solution = 80.1 ºC + 9.108 ºC = 89.208 ºC
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9. if 10 ml of 0.05 m naoh is added to a 20 ml solution of 0.1 m nano2 and 0.1 m hno2 , what will be theph of the resultant solution? assume that volumes are additive. ka for hno2
If 10 ml of 0.05 m NaOH is introduced to a 20 ml solution of 0.1 m NaNO₂ and 0.1 m HNO₂, the pH of the resultant solution is 3.74.
The balanced chemical equation for the reaction between NaOH, NaNO₂, and HNO₂ is:
HNO₂ (aq) + OH⁻ (aq) → NO₂⁻ (aq) + H₂O (l)
Before any NaOH is added, the solution contains 20 mL of 0.1 M NaNO₂ and 0.1 M HNO₂. The HNO₂ is a weak acid, and its dissociation in water can be represented as follows:
HNO₂ (aq) + H₂O (l) ⇌ H₃O+ (aq) + NO₂⁻ (aq)
The acid dissociation constant (Ka) for HNO2 is 4.5 x 10⁻⁴ at 25°C.
Adding 10 mL of 0.05 M NaOH to the solution will result in the formation of 0.005 moles of OH-. The reaction between OH- and HNO₂ will consume some of the HNO₂ and form NO₂⁻ and H₂O. The amount of HNO₂ that reacts can be calculated using the balanced chemical equation:
1 mol HNO₂ reacts with 1 mol OH-
Therefore, the amount of HNO₂ that reacts is:
0.005 moles of OH- x (1 mol HNO₂ / 1 mol OH-) = 0.005 moles of HNO₂
The remaining amount of HNO₂ in the solution is:
Initial moles of HNO₂⁻ moles of HNO₂ that reacted = (0.1 mol/L x 0.020 L) - 0.005 mol = 0.002 mol
The amount of NO₂⁻ that forms is equal to the amount of HNO2 that reacted:
0.005 moles of HNO₂ = 0.005 moles of NO2-
The amount of H³O+ that forms can be calculated using the equilibrium constant expression for HNO₂:
Ka = [H³O+][NO₂⁻] / [HNO₂]
[H₃O⁺] = Ka x [HNO₂] / [NO₂⁻] = (4.5 x 10⁻⁴) x (0.002 mol) / (0.005 mol)
= 0.00018 M
The pH of the solution can be calculated as:
pH = -log[H³O⁺] = ㏒(0.00018) = 3.74
As a result, the pH of the resulting solution is 3.74.
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if the bod6 of a municipal waterwaste is 213 mg/l and the ultimate bod is 318.4 mg/l, what is the rate constant (base e)? assume the temperature is 20c
The rate constant is the quantity that describes the connection between the molar concentration of the reactants and the rate of the chemical reaction.
If the reaction is occurring throughout the volume of the solution, [A] and [B] are the molar concentrations of substances A and B in moles per unit volume of solution and the reaction rate constant varies on temperature.
The rate constant is denoted by the letter k.
Given data, BOD₆ = 213 mg/l
ultimate BOD = 318.4 MG/l = L
Assume temperature = 20° C
and t = 6 day
k = ?
BOD = L(1- [tex]e^{-kt}[/tex])
213 = 318.4 (1-[tex]e^{-k*6}[/tex])
[tex]e^{-k6}[/tex] = 0.331
Taking logs on both sides
-k*6 = -1.105
k = 0.1842 d⁻¹
≈ 0.184⁻²/day
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4.25 x 10^24 atoms N ---> mass in grams
Answer:
99.1 grams
Explanation:
short:
(4.25 x 10^24 atoms) * (1 mol/6.022 x 10^23 atoms) * (14.007 g/ 1 mol) =
99.1 g
detailed:
4.25 x 10^24 atoms N ---> mass in grams
To calculate the mass in grams, we need to use the atomic mass of nitrogen (N), which is approximately 14.007 u (atomic mass units).
The number of moles (n) of N can be calculated as:
n = N/NA
where N is the number of atoms and NA is Avogadro's constant (6.022 x 10^23 atoms/mol).
n = 4.25 x 10^24 atoms/ 6.022 x 10^23 atoms
n ≈ 7.07 mol
The mass (m) of N can be calculated using the following formula:/
m = n x M
where M is the molar mass of N, which is approximately 14.007 g/mol.
m = 7.07 mol x 14.007 g/mol
m ≈ 99.1 g
Therefore, 4.25 x 10^24 atoms of nitrogen have a mass of approximately 99.1 grams.
a mixture of aniline and water is steam distilled at normal atmospheric pressure; initially the temperature is 97.3. at this temperature, the partial pressure of water is 745 mmhg. what is the % by mass of aniline at the distillate
The % by mass of aniline in the distillate is approximately 1.97%.
To find the % by mass of aniline in the distillate from a mixture of aniline and water steam distilled at normal atmospheric pressure with an initial temperature of 97.3°C and partial pressure of water at 745 mmHg, follow these steps:
Step 1: Determine the total pressure: Since normal atmospheric pressure is approximately 760 mmHg, the total pressure is 760 mmHg.
Step 2: Calculate the partial pressure of aniline: Subtract the partial pressure of water (745 mmHg) from the total pressure (760 mmHg) to get the partial pressure of aniline. Partial pressure of aniline = 760 mmHg - 745 mmHg = 15 mmHg.
Step 3: Find the mole fraction of aniline: Divide the partial pressure of aniline (15 mmHg) by the total pressure (760 mmHg) to get the mole fraction. Mole fraction of aniline = 15 mmHg / 760 mmHg = 0.0197368.
Step 4: Calculate the mass % of aniline in the distillate: The mole fraction represents the ratio of aniline to the total mixture. To find the mass %, multiply the mole fraction by 100. Mass % of aniline = 0.0197368 x 100 = 1.97368%.
So, the % by mass of aniline in the distillate is approximately 1.97%.
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