Complete question: An acid which could not be prepared by the reaction of an organic halide with cyanide ion followed by acid hydrolysis of the nitrile is: A) propanoic acid B) phenylacetic acid C) acetic acid D) (CH3)3CCO2H E) CH3(CH2)14CO2H
Answer: the correct option is option D ((CH3)3CCO2H).
Explanation: An acid which could not be prepared by the reaction of an organic halide with cyanide ion followed by acid hydrolysis of the nitrile is Pivalic acid. Pivalic acid with the molecular formula of CH3)3CCO2H is rather prepared by hydrocarboxylation of isobutene via the Koch reaction.
Final answer:
Phenylacetic acid cannot be prepared by the SN₂ reaction of an organic halide with cyanide ion followed by acid hydrolysis of the nitrile due to the reaction's failure with aryl halides. Instead, such reactions often lead to E₂ elimination or no reaction at all. Option B
Explanation:
The question pertains to the inability to synthesize certain carboxylic acids by the reaction of an organic halide with cyanide ion followed by acid hydrolysis of the resulting nitrile. Specifically, it's impossible to use this method to prepare an acid which can't be derived from a primary or secondary alkyl halide due to the limitations of the SN₂ reaction mechanism.
To synthesize acetic acid, for example, one may treat methyl bromide with cyanide ion to form acetonitrile, which upon hydrolysis yields acetic acid. However, phenylacetic acid, which requires a phenylacetyl halide precursor, cannot be formed this way because the reaction of cyanide ions with aryl halides does not proceed through an SN₂ mechanism. Instead, an E₂ elimination often occurs or, typically, there is no reaction due to the stability of the benzene ring and the partial double bond character of the C-X bond.
Therefore, phenylacetic acid could not be prepared using the reaction of an organic halide with cyanide ion followed by acid hydrolysis of the nitrile, while acetic acid and propanoic acid can be synthesized using this method. The correct answer to the student's question is phenylacetic acid.
My grandmother left me some silver years ago. I opened the box containing the silver and it had tarnished. Using the equation below, if I had 300 g of silver that had tarnished and 175g H2S, which would be the limiting reactant? 2 Ag(s) + H2S(g) → Ag2S(s) + H2(g) A. hydrogen B. silver C. There is no limiting reactant in this case D. hydrogen sulfide E. silver sulfide
Answer:
B
Explanation:
The limiting reagent in a chemical reaction is the one that is consumed completely in the course of the reaction. It dictates the extent to which the reaction would proceed as the completion of the reaction is based solely on it.
There are several ways to determine the limiting reagent. The easiest way to do this that works is to divide the number of moles of each reagent by their stoichiometric coefficient in the balanced equation. The reactant with the least value is the limiting reagent.
Thus, we need to calculate the number of moles of silver and hydrogen sulphide. To do this, we simply divide the masses by the relative atomic masses or molecular masses.
The atomic mass of silver is 108.
The number of moles of silver is =
300/108 = 2.8
We now divide this by the stoichiometric coefficient: 2.8/2 = 1.4
We do same for hydrogen sulphide. The molar mass of hydrogen sulphide is 34g/mol
The number of moles is thus 175/34 = 5.15
We now divide this by stoichiometric coefficient = 5.15/1 = 5.15
We can see that silver has less number of the value and hence it is the limiting reactant.
The use of fertilizers in agriculture has significantly altered several nutrient cycles including:
Answer:
potassium, nitrogen and phosphorous cycle
Explanation:
A fertilizer is a substance which is applied on the plants by farmers to increase the supply of nutrients for the plants. Fertilizers have known to be toxic in many ways such as they alter the potassium, nitrogen and phosphorus cycles. Nitrogen, potassium and phosphorus are present in abundant amounts in the fertilizers. Draining of these fertilizers into rivers and ponds is toxic for the aquatic life. Hence, the use of fertilizers disrupts the natural cycles and is toxic for many aquatic plants and animals.
In the development of atomic models, it was realized that the atom is mostly empty. Consider a model for the hydrogen atom where its nucleus is a sphere with a radius of roughly 10−15 m, and assume the electron orbits in a circle with a radius of roughly 10−10 m. In order to get a better sense for the emptiness of the atom, choose an object and estimate its width. This object will be your "nucleus". How far away would the "electron" be located away from your "nucleus"?
Answer:
See below
Explanation:
Lets choose a baseball ball which has to have a minimum of diameter of 1.43 inches ( 1.43 in x 2.54 cm/in = 3.63 cm )
ratio electron/nucleus = 10−10 m / 10−15 m = 100000
distance "e" = 3.63 cm x 100000 = 363000 cm ( 3630 m or 3.630 km )
Imagine to bat a baseball 3.6 km away !
The second messenger cyclic AMP (cAMP) is synthesized from ATP by the activity of the enzyme adenylyl cyclase. Cyclic AMP, in turn, activates protein kinase A (PKA), also called cAMP‑dependent protein kinase, which is responsible for most of the effects of cAMP within the cell. Determine the correct steps in the activation of PKA, and then place them in the correct order, starting after the adenylyl cyclase reaction.
The correct steps in the activation of PKA after the adenylyl cyclase reaction are: cAMP binding to and activating the regulatory subunits of PKA, the catalytic subunits of PKA phosphorylating target proteins, and PKA being inactivated when cAMP is hydrolyzed by phosphodiesterase.
Explanation:The correct steps in the activation of PKA, after the adenylyl cyclase reaction, are:
cAMP binds to and activates the regulatory subunits of PKA, causing them to release the catalytic subunits.The catalytic subunits of PKA phosphorylate target proteins, leading to their activation or inhibition.PKA is inactivated when cAMP is hydrolyzed by phosphodiesterase, reducing the levels of cAMP in the cell.Overall, the activation of PKA by cAMP allows for the regulation of various cellular processes, including metabolism, gene expression, and cell signaling.
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The correct steps in the activation of PKA after the adenylyl cyclase reaction are: adenylyl cyclase converts ATP to cAMP, cAMP binds to and activates Protein Kinase A (PKA), and activated PKA phosphorylates serine and threonine residues of target proteins, activating them.
Explanation:The correct steps in the activation of PKA after the adenylyl cyclase reaction are:
Adenylyl cyclase converts ATP to cAMP.cAMP binds to and activates Protein Kinase A (PKA).Activated PKA phosphorylates serine and threonine residues of target proteins, activating them.These steps occur in the cAMP second messenger system and are responsible for the effects of cAMP within the cell.
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ethanol is used as fuel for cars. Explain how energy is obtained from ethanol to fuel a car.
Answer:
the gas it runs on it
Explanation:
A chemist prepares a solution of copper(II) sulfate CuSO4 by measuring out 79.g of copper(II) sulfate into a 500.mL volumetric flask and filling the flask to the mark with water. Calculate the concentration in /molL of the chemist's copper(II) sulfate solution. Be sure your answer has the correct number of significant digits.
Answer:
The concentration of the copper(II) sulfate solution is 0.99 M
Explanation:
Step 1: Data given
Mass of copper(II) sulfate = 79 grams
Volume of the flask = 500 mL
Molar mass copper(II) sulfate = 159.61 g/mol
Step 2: Calculate moles copper(II) sulfate
Moles CuSO4 = Mass CuSO4 / molar mass CuSO4
Moles CuSO4 = 79.0 grams / 159.61 g/mol
Moles CuSO4 = 0.495 moles
Step 3: Calculate concentration
Concentration CuSO4 = moles / volume
Concentration CuSO4 = 0.495 moles / 0.5 L
Concentration = 0.99 M
The concentration of the copper(II) sulfate solution is 0.99 M
2) You are trying to determine a TLC solvent system which will separate the compounds X, Y, and Z. You ran the compounds on a TLC plate using hexanes/ethyl acetate 95:5 as the eluting solvent and obtained the chromatogram below. How could you change the solvent system to give better separation of these three compounds?
Answer:
Answer is in the explanation.
Explanation:
Thin layer chromatography is a chromatographic technique used to separate the components of a mixture using a thin stationary phase supported by an inert backing and a mobile phase. The separation principle is in the different affinities between the components of the mixture and the stationary or mobile phase.
The affinity in mobile phase could be improved changing the polarity of this phase. In this case, you could change proportion of hexane/ethyl acetate to change polarity of mobile phase and the affinity of the different compounds to mobile or stationary phase.
I hope it helps!
Thin-layer chromatography with its adjustable hexane/ethyl acetate ratio, serves as a precise and versatile tool for optimizing separation conditions and exploring diverse interactions in chromatography.
Thin-layer chromatography (TLC) is a chromatographic method that segregates mixture components via a slender stationary phase on an inert support and a mobile phase. The separation hin-ges on distinct affinities between mixture constituents and the stationary or mobile phase. Modifying the polarity of the mobile phase can enhance its affinity.
By adjusting the hexane/ethyl acetate ratio, the mobile phase's polarity transforms, influencing compounds' interactions with the stationary and mobile phases. This dynamic shift in affinity leads to differential migration rates, facilitating component separation.
TLC stands as a versatile tool in analytical chemistry, where subtle adjustments in solvent composition yield nuanced variations in separation patterns.
Fine-tuning the hexane/ethyl acetate proportions allows for targeted optimization of separation conditions, offering a precise means to explore and exploit the diverse interactions governing the chromatographic process.
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The percentage of water in an unknown hydrated salt is to bedetermined by weighing a sample of the salt, heating it to driveoff water, cooling it to room temperature, and re-weighing. Whichprocedural mistake would result in determining a percentage ofwater that is too low?
Procedural Mistakes
I. heating the sample in aclosed, rather than an open, container
II. re-weighing thesample before it has cooled to room temperature
A. I only
B. II only
C. both I andII
D. neither I norII
Answer:
B
Explanation:
If the sample is reweighed after it is removed from the furnace without being cooled to room temperature, its percentage water content will be too low as there would be almost no water if crystalization at such high temperature. However, when it is cooled to room temperature, the actual percentage of water contained in the sample can be accurately determined by weighing.
Final answer:
The procedural mistake that would result in determining a percentage of water that is too low is heating the sample in a closed container. Mistake I prevents the water from fully evaporating, which would cause an undercalculation of the water content.
Explanation:
When determining the percentage of water in a hydrated salt through heating, two procedural mistakes could lead to an inaccurate calculation of a lower percentage of water than the actual value. The correct answer to the given options is A. I only.
Heating the sample in a closed container (Mistake I) could prevent all the water from escaping, meaning that some water may remain inside, leading to an underestimation of the percentage of water in the salt.
In contrast, re-weighing the sample before it has cooled to room temperature (Mistake II) would not cause an underestimation but rather an overestimation, as the sample would weigh more due to the warmth. Therefore, Mistake II would not result in a lower percentage but rather potentially a higher percentage if the heat impacted the scale's reading.
Place the following compounds in order of increasing strength of intermolecular forces.
a. CH3CH3 CH3(CH2)8CH3 CH3CH2CH2CH3
b. CH3CH3 < CH3(CH2)8CH3 < CH3CH2CH2CH3
c. CH3CH2CH2CH3 < CH3CH3 < CH3(CH2)8CH3
d. CH3(CH2)8CH3 < CH3CH3 < CH3CH2CH2CH3
e. CH3CH3 < CH3CH2CH 2CH3 < CH3(CH2)8CH3
f. CH3(CH2)8CH3 < CH3CH2CH2CH3 < CH3CH3
Answer:
CH3(CH2)8CH3 > CH3CH2CH2CH3 > CH3CH3 (Option f)
Explanation:
Larger and heavier atoms and molecules exhibit stronger dispersion forces than smaller and lighter ones. In a larger atom or molecule, the valence electrons are, on average, farther from the nuclei than in a smaller atom or molecule. They are less tightly held and can more easily form temporary dipoles.
CH3CH3 has a molar mass of 30.07 g/mol
CH3(CH2)8CH3 has a molar mass of 142.28 g/mol
CH3CH2CH2CH3 has a molar mass of 58.12 g/mol
CH3(CH2)8CH3 > CH3CH2CH2CH3 > CH3CH3 (Option f)
The compounds should be ordered by increasing strength of their London dispersion forces, which relate to their molecular size. The correct order is CH3CH3 < CH3CH2CH2CH3 < CH3(CH2)8CH3, reflecting the increasing number of electrons and molecule size from ethane, to butane, to decane. Option e) is correct.
When placing the compounds CH3CH3, CH3(CH2)8CH3, and CH3CH2CH2CH3 in order of increasing strength of intermolecular forces (IMFs), it is important to consider the types of intermolecular forces present and the size of the molecules. All three compounds are nonpolar and primarily exhibit London dispersion forces. Because dispersion forces increase with the number of electrons and, thus, with the size of the molecule, the compound with the longest carbon chain will have the strongest intermolecular forces.
Thus, the correct ordering from weakest to strongest intermolecular forces is:
CH3CH3 (Ethane) - smallest molecule with the fewest electrons and hydrocarbonsCH3CH2CH2CH3 (Butane) - larger molecule with more electrons and hydrocarbons than ethaneCH3(CH2)8CH3 (Decane) - largest molecule with the most electrons and hydrocarbons, thus the strongest dispersion forcesTherefore, the correct answer is e. CH3CH3 < CH3CH2CH2CH3 < CH3(CH2)8CH3.
Nitroglycerin is a dangerous powerful explosive that violently decomposes when it is shaken or dropped. The Swedish chemist Alfred Nobel (1833-1896) founded the Nobel Prizes with a fortune he made by inventing dynamite, a mixture of nitroglycerin and inert ingredients that was safe to handle. (1) Write a balanced chemical equation, including physical state symbols, for the decomposition of liquid nitroglycerin ( C3H5NO33 ) into gaseous dinitrogen, gaseous dioxygen, gaseous water and gaseous carbon dioxide. (2) Suppose 41.0L of carbon dioxide gas are produced by this reaction, at a temperature of −14.0°C and pressure of exactly 1atm . Calculate the mass of nitroglycerin that must have reacted. Round your answer to 3 significant digits.
Answer:
a. [tex]4 C_3H_5N_3O_9 (l)\rightarrow 6N_2 (g) + O_2 (g) + 10 H_2O (g) + 12 CO_2 (g)[/tex]
b. 146.0 g
Explanation:
Question 1 (a). Just as the problem states, liquid nitroglycerin decomposes into nitrogen gas [tex]N_2[/tex], oxygen gas [tex]O_2[/tex], water vapor [tex]H_2O[/tex] and carbon dioxide [tex]CO_2[/tex]. Let's write the decomposition of nitroglycerin into these 4 components:
[tex]C_3H_5N_3O_9 (l)\rightarrow N_2 (g) + O_2 (g) + H_2O (g) + CO_2 (g)[/tex]
Now we need to balance the equation. Firstly, notice we have 3 carbon atoms on the left and 1 on the right, so let's multiply carbon dioxide by 3:
[tex]C_3H_5N_3O_9 (l)\rightarrow N_2 (g) + O_2 (g) + H_2O (g) + 3 CO_2 (g)[/tex]
Now, we have 3 nitrogen atoms on the left and 2 on the right, so let's multiply nitrogen on the right by [tex]\frac{3}{2}[/tex]:
[tex]C_3H_5N_3O_9 (l)\rightarrow \frac{3}{2}N_2 (g) + O_2 (g) + H_2O (g) + 3 CO_2 (g)[/tex]
We have 5 hydrogen atoms on the left, 2 on the right, so let's multiply the right-hand side by [tex]\frac{5}{2}[/tex]:
[tex]C_3H_5N_3O_9 (l)\rightarrow \frac{3}{2}N_2 (g) + O_2 (g) + \frac{5}{2} H_2O (g) + 3 CO_2 (g)[/tex]
Finally, count the oxygen atoms. We have a total of 9 on the left. On the right we have (excluding oxygen molecule):
[tex]\frac{5}{2} + 6 = 8.5[/tex]
This leaves [tex]9 - 8.5 = 0.5 = \frac{1}{2}[/tex] of oxygen. Since oxygen is diatomic, we need to take one fourth of it to get one half in total:
[tex]C_3H_5N_3O_9 (l)\rightarrow \frac{3}{2}N_2 (g) + \frac{1}{4} O_2 (g) + \frac{5}{2} H_2O (g) + 3 CO_2 (g)[/tex]
To make it look neater without fractional coefficients, multiply both sides by 4:
[tex]4 C_3H_5N_3O_9 (l)\rightarrow 6N_2 (g) + O_2 (g) + 10 H_2O (g) + 12 CO_2 (g)[/tex]
Question 2 (b). Now we can make use of the balanced chemical equation and apply it for the context of this separate problem. We're given the following variables:
[tex]V_{CO_2} = 41.0 L[/tex]
[tex]T = -14.0^oC + 273.15 K = 259.15 K[/tex]
[tex]p = 1 atm[/tex]
Firstly, we may find moles of carbon dioxide produced using the ideal gas law [tex]pV = nRT[/tex].
Rearranging for moles, that is, dividing both sides by RT (here R is the ideal gas law constant):
[tex]n_{CO_2} = \frac{pV_{CO_2}}{RT} = \frac{1 atm\cdot 41.0 L}{0.08206 \frac{L atm}{mol K}\cdot 259.15 K} = 1.928 mol[/tex]
According to the stoichiometry of the balanced chemical equation:
[tex]4 C_3H_5N_3O_9 (l)\rightarrow 6N_2 (g) + O_2 (g) + 10 H_2O (g) + 12 CO_2 (g)[/tex]
4 moles of nitroglycerin (ng) produce 12 moles of carbon dioxide. From here we can find moles o nitroglycerin knowing that:
[tex]\frac{n_{ng}}{4} = \frac{n_{CO_2}}{12} \therefore n_{ng} = \frac{4}{12}n_{CO_2} = \frac{1}{3}\cdot 1.928 mol = 0.6427 mol[/tex]
Multiplying the number of moles of nitroglycerin by its molar mass will yield the mass of nitroglycerin decomposed:
[tex]m_{ng} = n_{ng}\cdot M_{ng} = 0.6427 mol\cdot 227.09 g/mol = 146.0 g[/tex]
Why do the metals Co, Rh, and Lr form octahedral complexes (rather than tetrahedral or square planar complexes)? Hint: Look at the placement in the periodic table and the respective electron configurations.
Explanation:
Octahedral complexes will be favoured over tetrahedral ones because:
It is more favourable to form six bonds rather than four
The crystal field stabilisation energy is usually greater for octahedral than tetrahedral complexes.
The transition metals Co, Rh, Lr are in group 9 of d block and they have 3d, 4d, and 5d orbitals respectively
Metals like Co, Rh, and Lr form octahedral complexes due to their electron configurations and the stability achieved with six ligands. This coordination results in a geometry that minimizes electron repulsions and maximizes stability.
Transition metals like Co (Cobalt), Rh (Rhodium), and Lr (Lawrencium) typically form octahedral complexes due to their electron configurations and positions in the periodic table. These metals have available d-orbitals that can accommodate six ligands, resulting in a coordination number of six which is most stable in an octahedral geometry.
For instance, Co³⁺ has an electron configuration that favors the formation of octahedral complexes due to the stabilization of its d-electrons in a ligand field that splits the d-orbitals into two energy levels, with four orbitals at lower energy and two at higher energy. Similarly, Rh³⁺ and Lr³+ configurations also favor six-coordinate octahedral structures, providing maximum separation of electron pairs and minimizing electron-electron repulsions.
Examples:
[Co(NH₃)₆]³⁺: An example of an octahedral complex where the cobalt ion is bonded to six ammonia ligands.
[RhCl₆]³⁺: An octahedral complex where the rhodium ion is coordinated by six chloride ligands.
• Which of the following elements are most likely to act as 9 acceptor impurities in germanium? (a) bromine, (b) gallium, (c) sil- icon, (d) phosphorus, (e) magnesium • Which of the following elements are most likely to serve 10 as donor impurities in germanium? (a) bromine, (b) gallium, (c) sil- icon, (d) phosphorus, (e) magnesium
Answer:
Phosphorus can act as an acceptor while galium can act as a donor
Explanation:
In electronics, pentavalent atoms are used as acceptors while trivalent atoms are used donors. Hence the answers provided above.
10-1Copyright © 2016 McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.1.Chlorination of pentane gives a mixture of isomers having the molecular formula C5H11Cl. The percentage of 1-chloropentane is 22%. Assuming the secondary hydrogens in pentane are equally reactive to monochlorination, what is the percentage of 3-chloropentane in the mixture?
Answer:
26%
Explanation:
Chlorination is a reaction that substitutes an atom of hydrogen for an atom of chlorine in a hydrocarbon. Pentane has 5 carbons and 12 hydrogens, and the chlorination can happen at the carbons 1,2 or 3. If it happens at carbon 4, the structure will be the same as carbon 2, and if it happens at carbon 5, will be the same as carbon 1.
Then, the percentage of 2-chloropentane and 3-chloropentane is 100 - 22 = 78%. As stated above, 4 hydrogens can be substituted to form 2-chloropentane (two at carbon 2, and two at carbon 4), and only two for 3-chloropentane. So, the percentage of secondary hydrogens to form the structure wanted is:
2/6 = 1/3
Thus, the percentage of it is:
(1/3)*78% = 26%
reddit The amount of NE released by sympathetic nerve terminals will be most strongly influenced by a change in which of the following? Alpha receptor sensitivity Alpha receptor density Extracellular [Ca2+] COMT activity
Answer: Extracellular [Ca2+]
Explanation:
The sensitivity and density of the alpha receptors serve to enhance the response to the release of norepinephrine (NE) . However, they do not exert a strong influence as the concentration of calcium ions on the amount of norepinephrine (NE) released by sympathic nerve terminals.
The release of neurotransmitters depends more on either an external or internal stimulus.This results in an action potential which on reaching a nerve terminal, results in the opening of Ca²⁺ channels in the neuronal membrane. Because the extracellular concentration of Ca²⁺ is greater than the intracellular Ca²⁺ concentration, Ca²⁺ flows into the nerve terminal. This triggers a series of events that cause the vesicles containing norepinephrine (NE) to fuse with the plasma membrane and release norepinephrine (NE) into the synapse. The higher the action potential, the higher the Ca²⁺ flow into the terminals resulting in higher amount of norepinephrine (NE) into the synapse, and vice versa.
Catechol-O-methyltransferase (COMT) is one of several enzymes that degrade catecholamines such as dopamine, epinephrine, and norepinephrine. It serves a regulatory purpose to lower the concentration of norepinephrine upon its release from nerve terminals.
What is the conjugate acid of each of the following? What is the conjugate base of each?
(a) OH−
(b) H2O
(c) HCO3 −
(d) NH3
(e) HSO4 −
(f) H2O2
(g) HS−
(h) H5N2+
The specie formed when an acid looses a proton is its conjugate base.
According to the Brownstead - Lowry definition, an acid donates a proton while a base accepts a proton. The specie formed when an acid looses a proton is its conjugate base.
The conjugate acids of the following Brownstead - Lowry bases are;
[tex]OH^-[/tex] ----> [tex]H2O[/tex]
[tex]H2O[/tex]----->[tex]H3O^+[/tex]
[tex]NH3[/tex] -----> [tex]NH4^+[/tex]
[tex]HS^-[/tex]----> [tex]H2S[/tex]
The conjugate bases of the following acids are shown;
[tex]H2O2[/tex] -----> [tex]HO2^-[/tex]
[tex]H5N2+[/tex] ------> [tex]H4N2[/tex]
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This is a high school chemistry question about identifying conjugate acids and bases. The conjugate acid of a base is formed when the base accepts a proton, while a conjugate base is formed when an acid donates a proton.
Explanation:When a base accepts a proton (H+), it becomes a conjugate acid, and when an acid donates a proton, it becomes a conjugate base.
For the given compounds:
(a) OH−: Conjugate acid is H2O; Conjugate base doesn't exist as OH- is a base.(b) H2O: Conjugate acid is H3O+; Conjugate base is OH-.(c) HCO3− : Conjugate acid is H2CO3; Conjugate base is CO3^2-.(d) NH3 : Conjugate acid is NH4+; Conjugate base doesn't exist as NH3 is a base.(e) HSO4− : Conjugate acid is H2SO4; Conjugate base is SO4^2-.(f) H2O2 : Conjugate acid is H3O2+; Conjugate base is HO2-.(g) HS− : Conjugate acid is H2S; Conjugate base is S^2-.(h) H5N2+ : Conjugate acid doesn't exist as H5N2+ is an acid; Conjugate base is H4N2.Learn more about Conjugate Acids and Bases here:https://brainly.com/question/36337476
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In an ionic compound, the size of the ions affects the internuclear distance (the distance between the centers of adjacent ions), which affects lattice energy (a measure of the attractive force holding those ions together). Based on ion sizes, rank these compounds of their expected lattice energy..
Note: Many sources define lattice energies as negative values. Please rank by magnitude and ignore the sign. |Lattice energy| = absolute value of the lattice energy.
a. RbCl ,b. RbBr ,c. Rbl ,d. RbF
Answer:
b
Explanation:
bc
Calculate the change in the entropy of the system and also the change in the entropy of the surroundings, and the resulting total change in entropy, when a sample of nitrogen gas of mass 14 g at 298 K and 1.00 bar doubles its volume in (a) an isothermal reversible expansion, (b) an isothermal irreversible expansion against Pext = 0, and (c) an adiabatic reversible expansion.
Answer:
(a) Δ[tex]S_{sys}[/tex] = 2.881 J/K; Δ[tex]S_{sur}[/tex] = -2.881 J/K; total change in entropy = 0
(b)Δ[tex]S_{sys}[/tex] = 2.881 J/K; Δ[tex]S_{sur}[/tex] = 0 ; total change in entropy = 2.881 J/K
(c) Δ[tex]S_{sys}[/tex] = 0 ; Δ[tex]S_{sur}[/tex] = 0 ; total change in entropy = 0
Explanation:
In the given problem, we need to calculate the change in the entropy of the system and also the change in the entropy of the surroundings, and the resulting total change in entropy, when a sample of nitrogen gas of mass 14 g at 298 K and 1.00 bar doubles its volume. We have the following variable:
mass (m) = 14 g
Temperature = 298 K
Pressure = 1.00 bar
Initial volume = [tex]V_{1}[/tex]
Final volume = [tex]V_{2}[/tex] = [tex]2V_{1}[/tex]
(a) Change in entropy of the system Δ[tex]S_{sys}[/tex] = [tex]nRIn\frac{V_{2} }{V_{1} }[/tex]
where R = 8.314 J/(mol*K)
n = number of moles = mass/molar mass = 14/ 28 = 0.5 moles
Δ[tex]S_{sys}[/tex] = 0.5*8.314*ln2 = 2.881 J/K
Change in entropy of the surrounding Δ[tex]S_{sur}[/tex] = -2.881 J/K
Therefore, for a reversible process, the total change in entropy = Δ[tex]S_{sys}[/tex]+Δ[tex]S_{sur}[/tex] = 2.881 - 2.881 = 0
(b) Because entropy is a state function, we use the same procedure as in part (a). Thus, Δ[tex]S_{sys}[/tex] = 2.881 J/K
Since surrounding does not change in this process Δ[tex]S_{sur}[/tex] = 0.
total change in entropy = Δ[tex]S_{sys}[/tex]+Δ[tex]S_{sur}[/tex] = 2.881 - 0 = 2.88 J/K
(c) For an adiabatic reversible expansion, q(rev) = 0, thus:
Δ[tex]S_{sys}[/tex] = 0
Since heat energy is not transferred from the system to the surrounding
Δ[tex]S_{sur}[/tex] = 0
total change in entropy = Δ[tex]S_{sys}[/tex]+Δ[tex]S_{sur}[/tex] = 0
Choose the statement below that is TRUE. A solution will form between two substances if solute-solvent interactions are of comparable strength to the solute-solute and solvent-solvent interactions. A solution will form between two substances if the solute-solvent interactions are small enough to be overcome by the solute-solute and solvent- solvent interactions. A solution will form between two substances if the solute-solute interactions are strong enough to overcome the solvent-solvent interactions. A solution will form between two substances only if the solvent-solvent interactions are weak enough to overcome the solute-solvent interactions. None of these are true.
Answer:
Non of these are true
Explanation:
To form a solution, solute-solvent interaction must exceed solute-solute and solvent-solvent interaction. Hence a new attraction leading to solvation of the solid in the solvent is set up.
What were the concentrations of the solutions (zinc solution, copper solution, and salt bridge)? Were the concentrations consistent with those of standard state conditions? Explain your answer.
Answer:
1. 1M for zinc and copper solutions. No concentration was needed for the salt bridge.
2. See explanation.
Explanation:
Hello,
1. In this case, and considering the experiment you performed, the concentration of both copper solution and zinc solution were measured as 1 M, while the concentration of salt bridge was not taken into account.
2. However, as the reaction proceeded, the transfer of anions allowed the total charges in solution to remain neutral until the end, thus, the net movement of anions produced a concentration gradient between the solutions. It means that, after a while the net concentration of both anions and cations in the zinc oxidation section became greater than in the copper reduction section, making the concentration gradient opposite to the movement of anions. In such a way, as anions moved, cations moved to the right as well.
Best regards.
The concentrations of the zinc, copper, and salt bridge solutions are dependent on the amount of solute dissolved in the solvent. Standard state conditions, used as a reference point, include specific constant values for pressure, temperature, and concentration. Factors such as temperature and ion product of water can significantly affect a solution's properties and how they align to these standard conditions.
Explanation:The question asked about the concentrations of different solutions in standard state conditions. Based on the information provided, it's difficult to conclusively determine the concentration of each solution as the specific concentrations are not detailed. However, you can generally understand that a solution's concentration describes the amount of solute dissolved in a solvent. For example, for Zn(OH)₂(s) in a solution buffered at a pH of 11.45, this suggests a certain quantity of Zn (zinc) ions in the solution.
The standard state condition refers to a specific set of conditions including pressure (1 bar), temperature (298.15 K), and concentration (1M). For example, Zn(s) + 2HCl(aq) ZnCl₂ (aq) + H₂(g) details a reaction involving zinc and hydrochloric acid producing zinc chloride and hydrogen gas. This standardized condition forms the basis for the determination of properties such as enthalpy, entropy, and Gibbs free energy. Therefore, whether the solution is consistent with the standard state condition would be based on if the pressure, temperature, and concentration are adhering to these standard parameters.
Many factors such as temperature, ion product of water, solutes in the solutions can affect a solution's properties and its alignment to standard state conditions. Thus, it's important that these factors are closely monitored and regulated in a reaction setup.
Learn more about Solution Concentrations here:https://brainly.com/question/10720472
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Given that E°red = -1.66 V for Al3+/ Al at 25°C, find E° and E for the concentration cell expressed using shorthand notation below.
Al(s) Al3+(1.0 × 10-5 M) Al3+(0.100 M) Al(s)
A) E° = 0.00 V and E = +0.24 V
B) E° = 0.00 V and E = +0.12 V
C) E° = -1.66 V and E = -1.42 V
D) E° = -1.66 V and E = -1.54 V
The concentration cell involving Al3+ has a standard cell potential (E°) of -1.66 V, and the cell potential (E) is approximately -1.54 V. The correct option is D: E° = -1.66 V and E = -1.54 V.
To determine the standard cell potential (E°) and the cell potential (E) for the given concentration cell, we can use the Nernst equation and the standard reduction potential (E°red) for the half-reaction involved.
The shorthand notation for the concentration cell is given as:
[tex]\[ \text{Al(s) | Al}^{3+} (1.0 \times 10^{-5} \, \text{M}) \, || \, \text{Al}^{3+} (0.100 \, \text{M}) \, | \, \text{Al(s)} \][/tex]
Firstly, let's identify the two half-reactions occurring in the cell. The half-reaction for the left side (anode) is [tex]\(\text{Al} \rightarrow \text{Al}^{3+} + 3e^-\)[/tex], and for the right side (cathode), it is [tex]\(\text{Al}^{3+} + 3e^- \rightarrow \text{Al}\)[/tex].
The standard cell potential (E°) can be calculated using the standard reduction potentials [tex](\(E°_{\text{red}}\))[/tex] of the half-reactions:
[tex]\[ E° = E°_{\text{cathode}} - E°_{\text{anode}} \][/tex]
Given[tex]\(E°_{\text{red}} = -1.66 \, \text{V}\) for \(\text{Al}^{3+}/\text{Al}\)[/tex], we can substitute this into the equation to find E°. However, we need to consider the fact that the half-reaction for the anode is reversed in the standard reduction potential table. Therefore, \(E°_{\text{anode}} = -(-1.66 \, \text{V}) = 1.66 \, \text{V}\).
[tex]\[ E° = 0 - 1.66 \, \text{V} = -1.66 \, \text{V} \][/tex]
Now, to find the cell potential (E), we use the Nernst equation:
[tex]\[ E = E° - \frac{0.0592}{n} \log\left(\frac{[\text{Al}^{3+}]_{\text{cathode}}}{[\text{Al}^{3+}]_{\text{anode}}}\right) \][/tex]
Here, n is the number of electrons transferred (which is 3 for both half-reactions), and[tex]\([X]_{\text{cathode}}\[/tex] ) and [tex]\([X]_{\text{anode}}\)[/tex]are the concentrations of X at the cathode and anode, respectively.
For the given cell, plugging in the values:
[tex]\[ E = -1.66 - \frac{0.0592}{3} \log\left(\frac{0.100}{1.0 \times 10^{-5}}\right) \][/tex]
Calculating this gives approximately -1.54 V.
Therefore, the correct answer is option D: [tex]\(E° = -1.66 \, \text{V}\) and \(E = -1.54 \, \text{V}\)[/tex].
Complete question :- Given that E°red = -1.66 V for Al3+/ Al at 25°C, find E° and E for the concentration cell expressed using shorthand notation below.
Al(s) | Al3+(1.0 × 10-5 M) || Al3+(0.100 M) | Al(s)
A) E° = 0.00 V and E = +0.24 V
B) E° = 0.00 V and E = +0.12 V
C) E° = -1.66 V and E = -1.42 V
D) E° = -1.66 V and E = -1.54 V
The composition of a compound used to make polyvinyl chloride (PVC) is 38.4% C, 4.8% H and 56.8% Cl by mass. It took 7.73 min for a given volume of the compound to effuse through a porous plug, but it took only 6.18 min for the same amount of Ar to diffuse at the same temperature and pressure. What is the molecular formula of the compound?
Answer:
C₂H₃Cl
Explanation:
We can calculate the compound's molar mass using the data given by the problem and Graham's law:
Rate₁/Rate₂ = [tex]\sqrt{\frac{M_{2}}{M_{1}} }[/tex]
In this case the subscript 1 refers to the compound and 2 refers to Ar.
Keeping in mind that Rate = volume/time, and that the volume is the same for both compounds, we can rewrite the equation as:
Time₂/Time₁ = [tex]\sqrt{\frac{M_{2}}{M_{1}} }[/tex]
6.18/7.73 = [tex]\sqrt{\frac{39.95}{M_{1}} }[/tex]
M₁ = 62.5 g/mol
Now we determine the molecular formula by using the elemental % analysis:
Assuming we have 1 mol of the compound:
C ⇒ 62.5 g * 38.4/100 = 24 g C = 2 mol C
H ⇒ 62.5 g * 4.8/100 = 3 g H = 3 mol H
C ⇒ 62.5 g * 56.8/100 = 35.5 g Cl = 1 mol Cl
Thus the molecular formula is C₂H₃Cl
Rose bengal is a chromophore used in biological staining that has an absorption maximum at 559.1 nm and several other shorter wavelength absorption bands in the ultraviolet and visible regions of the spectrum when dissolved in ethanol. What is the energy difference, in kilojoules per mole, between the absorption maximum at 559.1 nm and a band at 263.5nm?
the energy difference is negative, it means that the band at 263.5 nm has a lower energy than the absorption maximum at 559.1 nm.
To calculate the energy difference between two absorption bands, we can use the equation:
[tex]\[ \Delta E = \frac{hc}{\lambda} \][/tex]
Where:
- [tex]\( \Delta E \)[/tex] is the energy difference in joules (J)
- [tex]\( h \)[/tex] is Planck's constant [tex](\(6.62607015 \times 10^{-34} \, \text{J} \cdot \text{s}\))[/tex]
- c is the speed of light in a vacuum [tex](\(3.00 \times 10^8 \, \text{m/s}\))[/tex]
- [tex]\( \lambda \)[/tex] is the wavelength in meters (m)
First, we need to convert the wavelengths from nanometers (nm) to meters (m):
[tex]\[ \lambda_1 = 559.1 \, \text{nm} \times \frac{1 \, \text{m}}{10^9 \, \text{nm}} = 5.591 \times 10^{-7} \, \text{m} \][/tex]
[tex]\[ \lambda_2 = 263.5 \, \text{nm} \times \frac{1 \, \text{m}}{10^9 \, \text{nm}} = 2.635 \times 10^{-7} \, \text{m} \][/tex]
Now, we can calculate the energy difference (\( \Delta E \)):
[tex]\[ \Delta E = \frac{(6.62607015 \times 10^{-34} \, \text{J} \cdot \text{s})(3.00 \times 10^8 \, \text{m/s})}{5.591 \times 10^{-7} \, \text{m}} - \frac{(6.62607015 \times 10^{-34} \, \text{J} \cdot \text{s})(3.00 \times 10^8 \, \text{m/s})}{2.635 \times 10^{-7} \, \text{m}} \][/tex]
[tex]\[ \Delta E = (3.00 \times 10^8 \, \text{m/s})\left(\frac{6.62607015 \times 10^{-34} \, \text{J} \cdot \text{s}}{5.591 \times 10^{-7} \, \text{m}} - \frac{6.62607015 \times 10^{-34} \, \text{J} \cdot \text{s}}{2.635 \times 10^{-7} \, \text{m}}\right) \][/tex]
[tex]\[ \Delta E = (3.00 \times 10^8 \, \text{m/s})(2.3723 \times 10^{-19} \, \text{J} - 2.5161 \times 10^{-19} \, \text{J}) \][/tex]
[tex]\[ \Delta E \approx (3.00 \times 10^8 \, \text{m/s})(-0.1438 \times 10^{-19} \, \text{J}) \][/tex]
[tex]\[ \Delta E \approx -4.314 \times 10^{-12} \, \text{J} \][/tex]
[tex]\[ \Delta E \approx -4314 \, \text{pJ} \][/tex]
Since the energy difference is negative, it means that the band at 263.5 nm has a lower energy than the absorption maximum at 559.1 nm.
The energy difference between the absorption maximum at 559.1 nm and the band at 263.5 nm is 240 kJ/mol.
To calculate the energy difference between the absorption maximum at 559.1 nm and a band at 263.5 nm in kilojoules per mole, follow these steps:
Step 1. Convert Wavelengths to Energy:
Use the equation [tex]\( E = \frac{hc}{\lambda} \)[/tex], where:
(E) is the photon's energy.
Planck's constant [tex](\(6.626 \times 10^{-34}\) J.s)[/tex] is represented as (h).
The speed of light is represented as (c) = [tex](3.00 \times 10^8\)[/tex] m/s.
The wavelength in meters is [tex](\lambda \))[/tex].
Step 2. Calculate Energy for 559.1 nm:
[tex]\[ \lambda_1 = 559.1 \text{ nm} = 559.1 \times 10^{-9} \text{ m} \][/tex]
[tex]\[ E_1 = \frac{6.626 \times 10^{-34} \times 3.00 \times 10^8}{559.1 \times 10^{-9}} = 3.556 \times 10^{-19} \text{ J} \][/tex]
Step 3. Calculate Energy for 263.5 nm:
[tex]\[ \lambda_2 = 263.5 \text{ nm} = 263.5 \times 10^{-9} \text{ m} \][/tex]
[tex]\[ E_2 = \frac{6.626 \times 10^{-34} \times 3.00 \times 10^8}{263.5 \times 10^{-9}} = 7.552 \times 10^{-19} \text{ J} \][/tex]
Step 4. Find the Energy Difference:
[tex]\[ \Delta E = E_2 - E_1 = 7.552 \times 10^{-19} \text{ J} - 3.556 \times 10^{-19} \text{ J} = 3.996 \times 10^{-19} \text{ J} \][/tex]
Step 5. Convert to Kilojoules per Mole:
[tex]\[ 1 \text{ photon} = 3.996 \times 10^{-19} \text{ J} \][/tex]
[tex]\[ 1 \text{ mole of photons} = 3.996 \times 10^{-19} \text{ J} \times 6.022 \times 10^{23} \text{ photons/mol} \][/tex]
[tex]\[ \Delta E_{\text{mol}} = 2.406 \times 10^{5} \text{ J/mol} = 240.6 \text{ kJ/mol} \][/tex]
So, the energy difference between the absorption maximum at 559.1 nm and the band at 263.5 nm is 240.6 kJ/mol.
A fixed amount of gas at 25.0°C occupies a volume of 10.0 L when the pressure is 667 torr. Use Boyle's law to calculate the pressure (torr) when the volume is reduced to 7.88 L at a constant temperature of 25.0°C.a. 846 torrb. 526 torrc. 0.118 torrd. 1.11 torre. 5.26 × 104 torr
Answer:
The pressure, when the volume is reduced to 7.88L, is 846 torr (option A)
Explanation:
Step 1: Data given
The temperature of a gas = 25.0°C
AT 25 °C the gas occupies a volume of 10.0L and a pressure of 667 torr.
The volume reduces to 7.88 L but the temperature stays constant.
Step 2: Boyle's law
(P1*V1)/T1 = (P2*V2)/T2
⇒ Since the temperature stays constant, we can simplify to:
P1*V1 = P2*V2
⇒ with P1 = the initial pressure 667 torr
⇒ with V1 = the initial volume = 10.0 L
⇒ with P2 = the final pressure = TO BE DETERMINED
⇒ with V2 = the final volume = 7.88L
P2 = (P1*V1)/V2
P2 = (667*10.0)/7.88
P2 = 846 torr
The pressure, when the volume is reduced to 7.88L, is 846 torr (option A)
Final answer:
Using Boyle's law, the new pressure of the gas when its volume decreases from 10.0 L to 7.88 L (with initial pressure 667 torr) is calculated to be 846 torr.
Explanation:
To calculate the new pressure of a gas when its volume changes, we can use Boyle's law, which states that for a fixed amount of gas at constant temperature, the pressure of the gas is inversely proportional to its volume.
This means if the volume decreases, the pressure increases, and vice versa, as long as the temperature and the amount of gas remain unchanged.
When the volume of the gas decreases from 10.0 L to 7.88 L and the initial pressure is 667 torr, we set up Boyle's law as follows: P1 * V1 = P2 * V2
Substituting the known values gives: 667 torr * 10.0 L = P2 * 7.88 L
Assuming no typographical errors, to solve for P2 (the new pressure), we rearrange the equation:
P2 = (667 torr * 10.0 L) / 7.88 L
After performing the division, we find: P2 = 846 torr.
Therefore, the answer is (a) 846 torr, which aligns with Boyle's expectation that a decrease in volume leads to an increase in pressure.
Electricity is the flow of electrons. The following questions relate to how electricity is quantified.1. Electrons are charged particles. The amount of charge that passes per unit time is calleda. potentialb. currentc. voltage2. The driving force for the electrons (i.e., the reason they are flowing in the first place) is measured bya. chargeb. currentc. potential3. Charge is measured ina. amperes (A)b. coulombs (C)c. joules (J)d. volts (V)4. Current is measured ina. amperes (A)b. coulombs (C)c. joules (J)d. volts (V)5. Potential is measured ina. amperes (A)b. coulombs (C)c. joules (J)d. volts (V)
Answer:
The correct option are:
1. (b) current
2. (c) potential
3. (b) coulombs (C)
4. (a) amperes (A)
5. (d) volts (V)
Explanation:
Electric charge is a property of a particle, like electron and proton. The subatomic particles, electrons and protons are negatively and positively charged particles, respectively.
The electric charge of a particle can be measured in coulomb, denoted by C.
Current or electric current is defined as the net flow of the electric charges through a given region, per unit time. Electric current can be measured in ampere, denoted by A.
Potential or Electric potential is described as the work done to displace the electric charges from one point to another. Potential is the driving force for the movement of charges and can be calculated in volt, denoted by V.
The flow of charge per unit time is current, measured in amperes (A), while the driving force for electron flow is electric potential, measured in volts (V), and charge itself is quantified in coulombs (C).
Understanding Electrical Quantities
The amount of charge that passes per unit time is called current, and is measured in amperes (A). The driving force that causes electrons to flow, known as the electric potential, is measured in volts (V). Charge itself is measured in coulombs (C).
Answering the questions provided:
The amount of charge that passes per unit time is called current.
The driving force for the electrons is measured by potential.
Charge is measured in coulombs (C).
Current is measured in amperes (A).
Potential is measured in volts (V).
What is Δn for the following equation in relating Kc to Kp?SO3(g) + NO(g) ↔ SO2(g) + NO2(g)12-20-1
Answer:
0
Explanation:
The relation between Kp and Kc is given below:
[tex]K_p= K_c\times (RT)^{\Delta n}[/tex]
Where,
Kp is the pressure equilibrium constant
Kc is the molar equilibrium constant
R is gas constant
T is the temperature in Kelvins
Δn = (No. of moles of gaseous products)-(No. of moles of gaseous reactants)
For the first equilibrium reaction:
[tex]SO_3_{(g)}+NO_{(g)}\rightleftharpoons SO_2_{(g)}+NO_2_{(g)}[/tex]
Δn = (2)-(2) = 0
Thus, Kp is:
[tex]K_p= Kc\times \times (RT)^{0}[/tex]
[tex]K_p= Kc[/tex]
Final answer:
The Δn represents the difference in moles of gaseous products and reactants for a reaction when relating Kc to Kp. For the reaction SO3(g) + NO(g) ↔ SO2(g) + NO2(g), Δn would be 0. Kp is related to Kc by the equation Kp = Kc(RT)Δn.
Explanation:
The question relates to the concept of the reaction quotient (Δn) when relating the equilibrium constants Kc (equilibrium constant in terms of concentration) to Kp (equilibrium constant in terms of partial pressure) for a given chemical reaction involving gases. The value of Δn is the difference in the sum of the moles of gaseous products and the sum of the moles of gaseous reactants in a balanced chemical equation. In the example given:
[tex]SO_{3}[/tex] (g) + NO(g) ↔ [tex]SO_{2}[/tex] (g) + [tex]NO_{2}[/tex] (g), Δn would be (1 + 1) - (1 + 1) = 0.
1[tex]N_{2}[/tex](g) + [tex]2H_{2} O[/tex](g) ↔ 2NO(g) + [tex]2H_{2}[/tex] (g), Δn would be (2 + 2) - (1 + 2) = 1.
To relate Kc and Kp, we use the equation Kp = Kc(RT)Δn, where R is the gas constant and T is the temperature in Kelvins. In cases where Δn is zero, as in the first equation, Kp will be equal to Kc because (RT)0 equals 1.
Which one of the following represents a decrease in entropy?
a) The evaporation of perfume.
b) The sublimation of carbon dioxide.
c) The melting of ice.
d) The condensation of steam on a kitchen window.
e) It is not possible to have a decrease in entropy
Answer:
D
Explanation:
There are principally three states of matter. These are the solid, liquid and gaseous states. The gaseous state has the highest degree of disorderliness as gas particles can move randomly while the solid state has the highest level of compactness.
Hence, we need to be adequately fed with information as regards the phase change to know if entropy has decreased or increased.
A. is wrong
Evaporation is a change of state to the gaseous state meaning there is an increased entropy.
B. is wrong
Sublimation is a change of state which means a solid substance like iodine or naphthalene changes its state directly to the gaseous state. There is an increased entropy here too.
C is wrong
Melting of ice means going from ice block to liquid water. This is synonymous to going from the solid state to the liquid state which is an increased entropy
D is correct
Condensation involves going from the gaseous state to the liquid state. This means going from a less ordered state to a more ordered state. This is accompanied by an entropy decrease.
E is wrong
While there are some processes that increase entropy, we also have some process that decrease entropy.
Final answer:
The process that represents a decrease in entropy is the condensation of steam on a kitchen window. This is because it involves the transition from a disordered gas state to a more ordered liquid state.
Explanation:
The question asks which of the given scenarios represents a decrease in entropy. Entropy, in simple terms, is a measure of the disorder or randomness of a system. A decrease in entropy means the system is becoming more ordered.
The evaporation of perfume - This process increases entropy as the perfume molecules spread out.The sublimation of carbon dioxide - Sublimation, the process of transitioning directly from a solid to a gas phase, results in an increase in entropy due to the increased dispersion of molecules.The melting of ice - This process increases entropy because the water molecules move from a highly ordered solid state to a less ordered liquid state.The condensation of steam on a kitchen window - This process decreases entropy. When steam (gas) condenses into liquid water on a window, it transitions from a highly disordered state to a more ordered state. Therefore, this is the correct answer.It is not possible to have a decrease in entropy - This statement is not accurate as there are processes, such as condensation, that can result in a decrease in entropy.
Thus, the correct answer is d) The condensation of steam on a kitchen window.
Which of the following sequences ranks the structures below in order of increasing acidity?
A) 1 < 2 < 3
B) 2 < 3 < 1
C) 3 < 1 < 2
D) 2 < 1 < 3
Answer:
Hi, no structure in the question was given to the answer the question.
Explanation:
Certain molecules or groups, when attached to an organic compound can make it become more acidic or less acidic. These groups are classified into 2
Electron withdrawing groups e.g the halogens F, Cl, Br and I. Fluorine is the best withdrawing group, next to chlorine and the least Iodine. Electron donating groups e.g Alkyls and Phenyl groupsElectron withdrawing groups makes the carboxylic more acidic. Also, the closer the electron withdrawing groups to the proton in the acid also makes the acid more acidic. This electron withdrawing group allows the electron density around the oxygen atom to be decreased.
Electron donating group makes the carboxylic acid less acidic. The more alkyl groups in a carboxylic acid, the less acidic the acid becomes. This allows the electron density around the oxygen atom in to be increased.
Heavy metal ions like lead(II) can be precipitated from laboratory wastewater by adding sodium sulfide, Na2S. Will all the lead be removed from 14.0 mL of 6.30×10-3 M Pb(NO3)2 upon addition of 13.1 mL of 0.0121 M Na2S? If all the lead is removed, how many moles of lead is this? If not, how many moles of Pb remain?
Answer:
All the 8.82*10^-5 moles of lead present is removed.
Explanation:
The total amount of lead II ion present is less than the amount of sulphide ion present and they react in 1:1 ratio so all the lead II ions are removed from the solution. See attached image.
One way in which the useful metal copper is produced is by dissolving the mineral azurite, which contains copper (II) carbonate, in concentrated sulfuric acid. The sulfuric acid reacts with the copper (II) carbonate to produce a blue solution of copper (II) sulfate. Scrap iron is then added to this solution, and pure copper metal precipitates out because of the following chemical reaction: Fe(s) + CuSO4 (aq) rightarrow Cu (s) + FeSO4 (aq) Suppose an industrial quality-control chemist analyzes a sample from a copper processing plant in the following way. He adds powdered iron to a 250. mL copper (II) sulfate sample from the plant until no more copper will precipitate. He then washes, dries, and weighs the precipitate, and finds that it has a mass of 96. mg . Calculate the original concentration of copper (II) sulfate in the sample. Be sure your answer has the correct number of significant digits.
Answer:
6,04x10⁻³M
Explanation:
For the reaction:
Fe(s) + CuSO₄(aq) → Cu(s) + FeSO₄(aq)
The precipitate of Cu(s) weights 96,0 mg. In moles:
Moles of Cu(s):
0,096g×(1mol/63,546g) = 1,51x10⁻³ moles of Cu(s). If you see the balanced equation 1 mole of CuSO₄ produce 1 mole of Cu(s). That means moles of CuSO₄ are the same of Cu(s), 1,51x10⁻³ moles of CuSO₄
As volume of the solution is 250 mL, 0,250L, the molar concentration of the original solution is:
1,51x10⁻³ moles of CuSO₄ / 0,250L = 6,04x10⁻³M
I hope it helps!
Which of the following provides an instantaneous measure of radioactivity:1 Scintillation counter2. Geiger counter3. Film-badge dosimeterB onlyB and C onlyA and B onlyA, B, and C
Answer:
A and B only
Explanation:
Both Scintillation counter and Geiger counter are used for instantaneous measures of radioactivity.
Both are used to detect and quantify the amount of radiation. GM counter can detect all kind of radiation that alpha, beta and gamma radiation. Whereas Scintillation counter can detect only ionization radiation.