In the given reactions, F2 and O2 act as oxidizing agents by being reduced, while H2 and Mg serve as reducing agents by being oxidized. In the complex reaction with MnO4- and Fe2+, MnO4- is the oxidizing agent and Fe2+ is the reducing agent.
Explanation:Classification of Oxidizing and Reducing Agents
In the redox reaction F2 + H2 → 2HF, the molecule F2 is being reduced (gains electrons) and thus is the oxidizing agent. Conversely, H2 is being oxidized (loses electrons) and so it acts as the reducing agent.
In the reaction 2Mg + O2 → 2MgO, the oxygen molecule O2 is reduced, making it the oxidizing agent, while magnesium Mg is oxidized, hence it is the reducing agent.
For the more complex redox reaction, 5Fe2+ + 8H+ + MnO4- → 5Fe3+ + Mn2+ + 4H2O, the permanganate ion MnO4- is reduced and is the oxidizing agent. Fe2+ is oxidized and is the reducing agent. The H+ ion is neither an oxidizing agent nor a reducing agent in this reaction.
Calculate the shortest wavelength of light capable of dissociating the Br–I bond in one molecule of iodine monobromide if the bond energy, or bond dissociation energy, is 179 kJ/mol .
Answer: 6.7*10^-7 m
Explanation:
The full explanation is shown in the image attached. The energy of the photon is obtained by dividing the bond energy by the Avogadro's number. Using the Plank's equation, we cash obtain the frequency or wavelength of radiation required by substituting into the given equation appropriately.
Final answer:
To calculate the shortest wavelength of light capable of dissociating the Br-I bond, the bond energy given (179 kJ/mol) is first converted to Joules per photon. Then, applying the equation E = hc / λ with Planck's constant and the speed of light gives the shortest wavelength needed to break the bond.
Explanation:
The question asks to calculate the shortest wavelength of light capable of dissociating the Br–I bond in iodine monobromide, given that the bond energy is 179 kJ/mol. To solve this, we apply the equation relating the energy of a photon (E) to its wavelength (λ), using Planck's constant (h) and the speed of light (c).
The energy of the photon needed can be calculated using the equation E = hc / λ, where h = 6.626 x 10-34 J·s (Planck's constant) and c = 3.00 x 108 m/s (speed of light). First, convert the bond dissociation energy from kJ/mol to Joules (J) by multiplying by 1000 and dividing by Avogadro's number (6.022 x 1023 mol-1), giving the energy required per molecule.
Finally, rearrange the equation to solve for λ: λ = hc / E. Plugging in the values, we find the shortest wavelength of light capable of breaking the Br–I bond. Remember, since the question gives the bond dissociation energy in kJ/mol, conversion to Joules is necessary for the calculation to proceed correctly.
An aqueous solution is saturated with both a solid and a gas at 5 ∘C∘C. What is likely to happen if the solution is heated to 85 ∘C∘C ? View Available Hint(s)
Here is the complete question
An aqueous solution is saturated with both a solid and a gas at 5 °C. What is likely to happen if the solution is heated to 85 °C ?
View Available Hint(s)
a.) Some gas will bubble out of solution and more solid will dissolve.
b.) Some gas will bubble out of the solution and some solid will precipitate out of the solution.
c.) Some solid will precipitate out of solution.
d.) More gas will dissolve and more of the solid will dissolve.
Answer:
a.) Some gas will bubble out of solution and more solid will dissolve.
Explanation:
Temperature increase usually increases the dissolution of solids in liquids. From the question; Some gas will be bubble out of the solution as the temperature is being increased to 85 °C because that aqueous solution is saturated(i.e equal amount of solute and solvent in the solution) with both solid and gas at 5 °C, but when the solution is heated to 85 °C, the solution becomes supersaturated( i.e the solute is now more at the given temperature than the solvent).
Answer:
As the temperature increases, the solubility of the solid increases and the solubility of the gas decreases. When the solution that is saturated between a solid and a gas at 5 ° C and heated to 85 ° C, the gas comes out of the solution first.
Explanation:
A saturated solution is one that has the maximum amount of solute that is dissolved. An unsaturated solution is one that has a low amount of solute compared to the saturated solution. According to Henry's law, the solubility of a gas at a specific temperature is directly proportional to its partial pressure, that is
C ∝ p
C = kp
Where
p is the partial pressure
k ia a proportionality constant
C is the concentration of the gas
a. Lithium and sodium are the most similar because they are both________ elements located in the same__________ , and therefore have similar properties.b. Nitrogen and oxygen are not the most similar because although they are both______ elements, are each located in a different________
Answer:
Lithium and sodium are the most similar because they are both alkali elements located in the same group, and therefore have similar properties.
Nitrogen and oxygen are not the most similar because although they are both non metals elements, are each located in a different group
Explanation:
Li and Na are both alkali elements from group 1 that shares some similities. The both can be obtained by the water hydrolysis. These are common reactions:
Metal from group 1 + H₂O → Base + H₂
Metal from group 1 + O₂ → oxides
Metal from group 1 + group 17 → ionic halides
Both form cations with 1+ charge, they can release only 1 e-
N is an element from group 15 and O, from group 16. They are both non metal.
Nitrogen can make a variety of oxides.
They react in water to produce nitric acid:
N₂O₃ + H₂O → 2HNO₃
N₂O₅ + H₂O → 2HNO₃
It has an anion with -3, as oxidation state. (Nitride)
The N with H, makes a well known hidride → ammonia
N₂ + 3H₂ → 2NH₃
The Oxygen also makes a well known hidride → water
2H₂ + O₂ → 2H₂O
Both are covalent hidrides.
N can have many oxidation's states. O always acts with -2 except for the peroxydes, with -1. O can have a great power of oxidation, that N does not have.
O₂ always acts as a reactant, at combustion reactions.
Lithium and sodium are similar as they are both alkali elements, found in the same group of the periodic table, leading to similar properties. Nitrogen and oxygen, while both nonmetals, are in separate groups, yielding different properties.
Explanation:a. Lithium and sodium are the most similar because they are both alkali elements located in the same group, and therefore have similar properties. Lithium and sodium both belong to the alkali metals group in the periodic table. They share similar chemical behaviors, as they have only one electron in a valence s subshell outside a filled set of inner shells. This fact influences their reactivity and the compounds they can form.
b. Nitrogen and oxygen are not the most similar because although they are both nonmetal elements, they are each located in a different group. Even though nitrogen and oxygen are close to each other in the periodic table, they do not share the same group, hence they have different properties and different numbers of valence electrons.
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Write a full set of quantum numbers for the following:
(a) The outermost electron in an Rb atom
(b) The electron gained when an S⁻ ion becomes an S²⁻ ion
(c) The electron lost when an Ag atom ionizes
(d) The electron gained when an F⁻ ion forms from an F atom
The full set of quantum numbers varies depending on the electron being considered in each element or ion, with the numbers consisting of the principal quantum number (n), angular momentum quantum number (l), magnetic quantum number (m_l), and spin quantum number (m_s).
Quantum numbers describe values of conserved quantities in the dynamics of a quantum system, which in this case are the electrons in various elements or ions.
(a) The outermost electron in an Rb (Rubidium) atom will have the quantum numbers: n = 5, l = 0, ml = 0, ms = +1/2 or -1/2.(b) The electron gained when an S⁻ ion becomes an S²⁻ ion will have the quantum numbers: n = 3, l = 2, ml = -2 to 2 (any of these for one electron), ms = +1/2 or -1/2.(c) The electron lost when an Ag atom ionizes (silver) will have the quantum numbers: n = 5, l = 0, ml = 0, ms = +1/2 or -1/2 (since the last electron in the 5s subshell is lost).(d) The electron gained when an F⁻ ion forms from an F atom will have the quantum numbers: n = 2, l = 1, ml = -1, 0, or 1 (for an additional p electron), ms = +1/2 or -1/2.Calculate the amount of heat required to completely sublime 66.0 gg of solid dry ice (CO2)(CO2) at its sublimation temperature. The heat of sublimation for carbon dioxide is 32.3 kJ/molkJ/mol. Express your answer in kilojoules. nothing kJkJ
Answer:
48.5 kJ
Explanation:
Let's consider the sublimation of carbon dioxide at its sublimation temperature, that is, its change from the solid to the gaseous state.
CO₂(s) → CO₂(g)
The molar mass of carbon dioxide is 44.01 g/mol. The moles corresponding to 66.0 g are:
66.0 g × (1 mol/44.01 g) = 1.50 mol
The heat of sublimation for carbon dioxide is 32.3 kJ/mol. The heat required to sublimate 1.50 moles of carbon dioxide is:
1.50 mol × (32.3 kJ/mol) = 48.5 kJ
The tarnish that forms on objects made of silver is solid silver sulfide; it can be removed by reacting it with aluminum metal to produce silver metal and solid aluminum sulfide. How many moles of the excess reactant remain unreacted when the reaction is over if 5 moles of silver sulfide react with 8 moles of aluminum metal? Hint: Write a balanced chemical equation first. Enter to 1 decimal place.
When 5 moles of silver sulfide react with 8 moles of aluminum metal, there is an excess of aluminum. After the reaction is complete, 3.3 moles of the excess aluminum remain unreacted.
Explanation:The balanced chemical equation for the reaction between solid silver sulfide (Ag2S) and aluminum metal (Al) is:
3Ag2S + 2Al → 6Ag + Al2S3
Based on this equation, we can see that every 3 moles of Ag2S react with 2 moles of Al to produce 6 moles of Ag and 1 mole of Al2S3.
In the given question, 5 moles of Ag2S react with 8 moles of Al. Therefore, we have an excess of Al. To determine the moles of excess Al remaining unreacted, we can set up a ratio:
(8 moles Al reacted) / (2 moles Al required to react with 3 moles Ag2S) = x moles Ag2S / 5 moles Ag2S
Simplifying this ratio, we find:
x = (8 moles Al / 2) × (5 moles Ag2S / 3 moles Al)
x = 20/6 = 3.3 moles
Therefore, 3.3 moles of the excess reactant (Al) remain unreacted when the reaction is over.
Final answer:
After writing a balanced chemical equation for the reaction between silver sulfide and aluminum, we determine that 4.7 moles of aluminum remain unreacted when 5 moles of silver sulfide react with 8 moles of aluminum.
Explanation:
To determine the number of moles of the excess reactant that remain unreacted, we first need to write a balanced chemical equation for the reaction between silver sulfide and aluminum metal. Here is the balanced equation:
3 Ag₂S (s) + 2 Al (s) → 6 Ag (s) + Al₂S₃(s)
Using the balanced equation, we see that 3 moles of silver sulfide react with 2 moles of aluminum. Therefore, if we had 5 moles of silver sulfide, we would need 2/3 × 5 = 10/3 moles of aluminum to react completely with the silver sulfide.
Since 8 moles of aluminum were originally present, we subtract the amount of aluminum that reacted to find the excess:
8 moles Al - 10/3 moles Al = 14/3 moles Al
Thus, 14/3 moles or 4.7 moles of aluminum remain unreacted.
Palladium (Pd; Z 46) is diamagnetic. Draw partial orbital diagrams to show which of the following electron configurations is consistent with this fact:
(a) [Kr] 5s²4d⁸
(b) [Kr] 4d¹⁰
(c) [Kr] 5s¹4d⁹
Answer : The electron configurations consistent with this fact is, (b) [Kr] 4d¹⁰
Explanation :
Electronic configuration : It is defined as the representation of electrons around the nucleus of an atom.
Number of electrons in an atom are determined by the electronic configuration.
Paramagnetic compounds : They have unpaired electrons.
Diamagnetic compounds : They have no unpaired electrons that means all are paired.
The given electron configurations of Palladium are:
(a) [Kr] 5s²4d⁸
In this, there are 2 electrons in 's' orbital and 8 electrons in 'd' orbital. From the partial orbital diagrams we conclude that 's' orbital are paired but 'd' orbital are not paired. So, this configuration shows paramagnetic.
(b) [Kr] 4d¹⁰
In this, there are 10 electrons in 'd' orbital. From the partial orbital diagrams we conclude that electrons in 'd' orbital are paired. So, this configuration shows diamagnetic.
(c) [Kr] 5s¹4d⁹
In this, there are 1 electron in 's' orbital and 9 electrons in 'd' orbital. From the partial orbital diagrams we conclude that 's' orbital and 'd' orbital are not paired. So, this configuration shows paramagnetic.
The correct electron configuration for a diamagnetic substance like Palladium (Pd) is [Kr] 4d¹⁰. This is because diamagnetic substances like Palladium need to have all their electron orbitals fully filled, thereby having no unpaired electrons.
Explanation:The subject of this question is about the electron configuration of Palladium (Pd; Z 46), which is a diamagnetic element. Diamagnetic substances have no unpaired electrons and are not attracted to a magnetic field. Thus, to be consistent with this fact, the electron configuration of Palladium must not have any unpaired electrons.
(a) [Kr] 5s²4d⁸: This configuration would imply there are unpaired electrons in the 4d orbital, which contradicts the fact that Palladium is diamagnetic.
(b) [Kr] 4d¹⁰: This configuration correctly states that all the orbitals are filled, including the 5s orbital before the 4d orbital. Therefore, the correct electron configuration for Palladium, a diamagnetic element, is [Kr] 4d¹⁰.
(c) [Kr] 5s¹4d⁹: The proposed configuration would also suggest an unpaired electron exists in the 4d orbital, which contradicts Palladium being diamagnetic.
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Whenever dry nitrogen from a portable cylinder is used in service and installation practice, what item is of most importance in consideration of safety?
Answer:
a relief valve is inserted in the downstream line from the pressure regulator.
Explanation:
Whenever dry nitrogen from a portable cylinder is used in service and installation practice the item of most importance in consideration of safety is a relief valve is inserted in the downstream line from the pressure regulator.
The most important item to consider for safety when using dry nitrogen from a portable cylinder in service and installation practice is a pressure regulator, which controls the flow and release of the gas. It is also crucial to inspect the system for leaks regularly.
Explanation:When using dry nitrogen from a portable cylinder in service and installation practice, the most important item to consider for safety is a pressure regulator. A pressure regulator is used to control the flow and release of nitrogen gas from the cylinder. It ensures that the pressure does not exceed safe limits and reduces the risk of an explosion or other accidents.
Additionally, it is crucial to always check for leaks in the system before using dry nitrogen. Leaks can result in the accumulation of nitrogen gas, which can displace oxygen in the air and lead to asphyxiation. Therefore, regularly inspecting the system and using appropriate leak detection methods, such as soapy water or a leak detection solution, is essential for safety.
In summary, when working with dry nitrogen from a portable cylinder, the two most important safety considerations are the use of a pressure regulator and regular inspection for leaks.
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At a certain temperature this reaction follows first-order kinetics with a rate constant of 0.184 s^-1 ? Suppose a vessel contains Cl_2 O_5 at a concentration of 1.16 M. Calculate the concentration of Cl_2 O_5 in the vessel 5.70 seconds later.
To calculate the concentration of Cl2O5 in the vessel 5.70 seconds later, use the first-order rate law equation and the given rate constant and initial concentration.
Explanation:To calculate the concentration of Cl2O5 in the vessel 5.70 seconds later, we can use the first-order rate law equation.
The rate law equation for a first-order reaction is: ln([A]t/[A]0) = -kt
Where [A]t is the concentration at time t, [A]0 is the initial concentration, k is the rate constant, and t is the time.
In this case, we know the rate constant (k) is 0.184 s^-1 and the initial concentration ([A]0) is 1.16 M. We need to find the concentration at 5.70 seconds ([A]t).
Plugging in the values: ln([A]t/1.16) = -0.184 * 5.70
Solving for [A]t, we find that the concentration of Cl2O5 in the vessel 5.70 seconds later is approximately 0.64 M.
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Given data: Rate constant (k) = 0.184 s⁻¹
Initial concentration (Cl₂O₅) = 1.16 M
Time (t) = 5.70 s
To find the concentration of Cl₂O₅ after 5.70 seconds
We know that the first-order integrated rate law equation is:
ln([A]t/[A]0) = -where [A]t and [A]0 are the concentrations of reactant at time 't' and initial time '0' respectively. We can rearrange this equation to find the concentration of the reactant at any time 't':[A]t = [A]0 × e^(-kt)Putting the values in the equation, we get:
[Cl₂O₅]5.70 = 1.16 M × e^(-0.184 s⁻¹ × 5.70 s)
[Cl₂O₅]5.70 = 0.501
Therefore, the concentration of Cl₂O₅ in the vessel 5.70 seconds later is 0.501 M.
If one wished to obtain 0.050 moles of isopentyl alcohol, how many milliliters should one obtain? Enter only the number to two significant figures.
Answer:
1100 millimeters
Explanation:
1 mole of isopentyl alcohol = 22.4L = 22.4×1000 mL = 22,400L
0.050 moles of isopentyl alcohol = 0.050 × 22,400mL = 1120mL = 1100mL (to two significant figures)
In an aqueous solution containing Ni (II) and Ni (IV) salts, which cation would you expect to be the more strongly hydrated? Why?
Answer:
Well in comparison of Ni (II) and Ni (IV), Ni (IV) is a stronger vation with the +4 charge so it will attract the more oxygen ions in the aqueous solution. That is why Ni (IV) will be more strongly hydrated.
Explanation:
Final answer:
In an aqueous solution, the Ni (IV) cation is more strongly hydrated than the Ni (II) cation due to its higher charge density, which attracts more water molecules into its hydration sphere.
Explanation:
In an aqueous solution containing Ni (II) and Ni (IV) salts, the Ni (IV) cation is expected to be more strongly hydrated. This outcome is attributed primarily to the charge density. The Ni (IV) has a higher charge (+4) compared to Ni (II) which has a +2 charge. In terms of hydration, water molecules, which act as dipoles, are more strongly attracted to ions with a higher charge density. This means that the Ni (IV) ion, with its higher charge, attracts and binds water molecules more strongly than the Ni (II) ion.
This strong attraction results in a greater degree of hydration for the Ni (IV) ion as it pulls more water molecules into its hydration sphere. This process is crucial in understanding the properties of solutions, especially in predicting the behavior of ions in biological and chemical systems.
Solid silver chloride, AgCI(s), decomposes to its elements when exposed to sunlight Write a balanced equation for this reaction with correct formulas for the products. What is the coefficient in front of AgCI(s) when the equation is properly balanced? 01 03 02 04
Answer: The coefficient in front of AgCl when the equation is properly balanced is 2.
Explanation:
According to the law of conservation of mass, mass can neither be created nor be destroyed. Thus the mass of products has to be equal to the mass of reactants. The number of atoms of each element has to be same on reactant and product side. Thus chemical equations are balanced.
Decomposition is a type of chemical reaction in which one reactant gives two or more than two products.
Decomposition of silver chloride is represented as:
[tex]2AgCl(s)\rightarrow 2Ag(s)+Cl_2(g)[/tex]
Thus the coefficient in front of AgCl when the equation is properly balanced is 2.
Answer:
the coefficient in front of AgCl(s) when the equation is properly balanced is 2.
Explanation:
The decomposition of solid silver chloride (AgCl) into its elements when exposed to sunlight can be represented by the following balanced chemical equation:
2AgCl(s) → 2Ag(s) + Cl2(g)
In this balanced equation:
The coefficient in front of AgCl(s) is 2.
The coefficient in front of Ag(s) is 2.
The coefficient in front of Cl2(g) is also 2.
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Which of the following solutions will have the lowest freezing point? Input the appropriate letter. A. 35.0 g of C3H8O in 250.0 g of ethanol (C2H5OH) B. 35.0 g of C4H10O in 250.0 g of ethanol (C2H5OH) C. 35.0 g of C2H6O2 in 250.0 g of ethanol (C2H5OH)
Answer:
Solution with 35.0 g of [tex]C_3H_8O[/tex] in 250.0 g of ethanol will have lowest freezing point
Explanation:
[tex]\Delta T_f=K_f\times m[/tex]
where,
[tex]\Delta T_f[/tex] =depression in freezing point =
[tex]K_f[/tex] = freezing point constant
m = molality
As we can see that higher the molality of the solution more will depression in freezing point of the solution and hence lower will the freezing point of solution.
[tex]Molality=\frac{moles}{\text{mass of solvent in kg}}[/tex]
A. 35.0 g of [tex]C_3H_8O[/tex] in 250.0 g of ethanol.
Moles of [tex]C_3H_8O[/tex]=[tex]\frac{35.0 g}{60 g/mol}=0.5833 mol[/tex]
Mass of solvent i.e. ethanol = 250.0 g = 0.25 kg (1 g = 0.001 kg)
[tex]m=\frac{0.5833 mol}{0.25 kg}=2.33 m[/tex]
B. 35.0 g of [tex]C4H_{10}O[/tex] in 250.0 g of ethanol
Moles of [tex]C_4H_{10}O[/tex][tex]=[tex]\frac{35.0 g}{74 g/mol}=0.4730 mol[/tex]
Mass of solvent i.e. ethanol = 250.0 g = 0.25 kg (1 g = 0.001 kg)
[tex]m'=\frac{0.4730 mol}{0.25 kg}=1.89 m[/tex]
C. 35.0 g of [tex]C_2H_{6}O_2[/tex] in 250.0 g of ethanol
Moles of [tex]C_2H_{6}O_2[/tex]=[tex]\frac{35.0 g}{62g/mol}=0.5645 mol[/tex]
Mass of solvent i.e. ethanol = 250.0 g = 0.25 kg (1 g = 0.001 kg)
[tex]m''=\frac{0.5645 mol}{0.25 kg}=2.26 m[/tex]
[tex]m>m'''>m''[/tex]
Solution with 35.0 g of [tex]C_3H_8O[/tex] in 250.0 g of ethanol will have lowest freezing point
The volume density of atoms for a bcc lattice is 5 x 1026 m-3. Assume that the atoms are hard spheres with each atom touching its nearest neighbors. Determine the lattice constant and effective radius of the atom.
Explanation:
It is known that for a body centered cubic unit cell there are 2 atoms per unit cell.
This means that volume occupied by 2 atoms is equal to volume of the unit cell.
So, according to the volume density
[tex]5 \times 10^{26} atoms = 1 [tex]m^{3}[/tex]
2 atoms = [tex]\frac{1 m^{3}}{5 \times 10^{26} atoms} \times 2 atoms[/tex]
= [tex]4 \times 10^{-27} m^{3}[/tex]
Formula for volume of a cube is [tex]a^{3}[/tex]. Therefore,
Volume of the cube = [tex]4 \times 10^{-27} m^{3}[/tex]
As lattice constant (a) = [tex](4 \times 10^{-27} m^{3})^{\frac{1}{3}}[/tex]
= [tex]1.59 \times 10^{-9} m[/tex]
Therefore, the value of lattice constant is [tex]1.59 \times 10^{-9} m[/tex].
And, for bcc unit cell the value of radius is as follows.
r = [tex]\frac{\sqrt{3}}{4}a[/tex]
Hence, effective radius of the atom is calculated as follows.
r = [tex]\frac{\sqrt{3}}{4}a[/tex]
= [tex]\frac{\sqrt{3}}{4} \times 1.59 \times 10^{-9} m[/tex]
= [tex]6.9 \times 10^{-10} m[/tex]
Hence, the value of effective radius of the atom is [tex]6.9 \times 10^{-10} m[/tex].
Pepsinogen is the inactive protease secreted by the chief cells in the stomach; this enzyme is converted to the active form called ______ in the presence of HCl. The primary role of this enzyme is the digestion of proteins in the stomach.a. Bicarbonateb. Pepsinc. Hydrochloric acid
Answer:
b. Pepsin
Explanation:
In the stomach, when dygestion takes place,pepsinogen is secreted in the stomach for the dygestion of proteins, which are catalyzed to aminoacids for the use of the body.
When the pepsinogen gets in contact with the HCl it converts to pepsin, which is the actived form of the pepsinogen.
Pepsinogen is converted into its active form, called Pepsin, under acidic conditions rendered by Hydrochloric acid (HCl). Pepsin, once activated, aids in protein digestion.
Explanation:Pepsinogen, an inactive protease, is secreted by the chief cells located in the stomach. Under the acidic condition produced by the presence of Hydrochloric acid (HCl), pepsinogen is converted into its active form, known as Pepsin. Pepsin plays an integral role in the process of digestion, specifically assisting in the breakdown of proteins in the stomach.
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Phosgene, a poisonous gas, when heated will decompose into carbon monoxide and chlorine in a reversible reaction: COCl2 (g) <-----> CO (g) + Cl2 When 2.00 mol of phosgene is put into an empty 1.00 L flask and 395 ˚C and allowed to come to equilibrium, the final mixture contains 0.0398 mol of chlorine. Find Keq. Group of answer choices
Answer:
8.08 × 10⁻⁴
Explanation:
Let's consider the following reaction.
COCl₂(g) ⇄ CO (g) + Cl₂(g)
The initial concentration of phosgene is:
M = 2.00 mol / 1.00 L = 2.00 M
We can find the final concentrations using an ICE chart.
COCl₂(g) ⇄ CO (g) + Cl₂(g)
I 2.00 0 0
C -x +x +x
E 2.00 -x x x
The equilibrium concentration of Cl₂, x, is 0.0398 mol / 1.00 L = 0.0398 M.
The concentrations at equilibrium are:
[COCl₂] = 2.00 -x = 1.96 M
[CO] = [Cl₂] = 0.0398 M
The equilibrium constant (Keq) is:
Keq = [CO].[Cl₂]/[COCl₂]
Keq = (0.0398)²/1.96
Keq = 8.08 × 10⁻⁴
In a certain acidic solution at 25 ∘C, [H+] is 100 times greater than [OH −]. What is the value for [OH −] for the solution?
Answer: The value of [tex][OH^-][/tex] for the solution is [tex]10^{-6}M[/tex]
Explanation:
To calculate the concentration of hydroxide ion for the solution, we use the equation:
[tex][H^+]\times [OH^-]=10^{-14}[/tex]
We are given:
[tex][H^+]=100\times [OH^-][/tex]
Putting values in above equation, we get:
[tex]100\times [OH^-]\times [OH^-]=10^{-14}[/tex]
[tex][OH^-]^2=\frac{10^{-14}}{100}[/tex]
[tex][OH^-]=\sqrt{10^{-12}}[/tex]
[tex][OH^-]=10^{-6}M[/tex]
Hence, the value of [tex][OH^-][/tex] for the solution is [tex]10^{-6}M[/tex]
A certain metal M crystallizes in a lattice described by a body-centered cubic (bcc) unit cell. The lattice constant a has been measured by X-ray crystallography to be 409. Calculate the radius of an atom of M.
To calculate the radius of an atom in a body-centered cubic (BCC) structure, we can use the formula: radius (r) = √(3/4) * a/2, where a is the lattice constant.
Explanation:In a body-centered cubic (BCC) unit cell, the atoms in the corners do not touch each other but contact the atom in the center. The unit cell contains two atoms: one-eighth of an atom at each of the eight corners and one atom in the center. An atom in a BCC structure has a coordination number of eight. To calculate the radius of an atom in a BCC structure, we can use the formula:
Radius (r) = √(3/4) * a/2
where a is the lattice constant. Plugging in the given value of the lattice constant as 409, we can calculate the radius of the atom of metal M in a BCC structure using this formula.
Sucralfate (molecular weight: 2087 g/mol) is a major component of Carafate®, which is prescribed for the treatment of gastrointestinal ulcers. The recommended dose for a duodenal ulcer is 1.0g taken orally in 10.0 mL (2 teaspoons) of a solution four times a day. Calculate the molarity of the solution.a. 0.0005 Mb. 0.048 Mc. 0.02087 Md. 0.010 M
Answer:
Option b. 0.048 M
Explanation:
We have the molecular weight and the mass, from sulcralfate.
Let's convert the mass in g, to moles
1 g . 1 mol / 2087 g = 4.79×10⁻⁴ moles.
Molarity is mol /L
Let's convert the volume of solution in L
10 mL . 1L/1000 mL = 0.01 L
4.79×10⁻⁴ mol / 0.01 L = 0.048 mol/L
Start with 100.00 mL of 0.10 M acetic acid, CH3COOH. The solution has a pH of 2.87 at 25 oC. a) Calculate the Ka of acetic acid at 25 oC. b) Determine the percent dissociation for the solution.
Answer: a) The [tex]K_a[/tex] of acetic acid at [tex]25^0C[/tex] is [tex]1.82\times 10^{-5}[/tex]
b) The percent dissociation for the solution is [tex]4.27\times 10^{-3}[/tex]
Explanation:
[tex]CH_3COOH\rightarrow CH_3COO^-H^+[/tex]
cM 0 0
[tex]c-c\alpha[/tex] [tex]c\alpha[/tex] [tex]c\alpha[/tex]
So dissociation constant will be:
[tex]K_a=\frac{(c\alpha)^{2}}{c-c\alpha}[/tex]
Give c= 0.10 M and [tex]\alpha[/tex] = ?
Also [tex]pH=-log[H^+][/tex]
[tex]2.87=-log[H^+][/tex]
[tex][H^+]=1.35\times 10^{-3}M[/tex]
[tex][CH_3COO^-]=1.35\times 10^{-3}M[/tex]
[tex][CH_3COOH]=(0.10M-1.35\times 10^{-3}=0.09806M[/tex]
Putting in the values we get:
[tex]K_a=\frac{(1.35\times 10^{-3})^2}{(0.09806)}[/tex]
[tex]K_a=1.82\times 10^{-5}[/tex]
b) [tex]\alpha=\sqrt\frac{K_a}{c}[/tex]
[tex]\alpha=\sqrt\frac{1.82\times 10^{-5}}{0.10}[/tex]
[tex]\alpha=4.27\times 10^{-5}[/tex]
[tex]\% \alpha=4.27\times 10^{-5}\times 100=4.27\times 10^{-3}[/tex]
The Ka of acetic acid is calculated using the pH and the definition of Ka using an ICE table, then percent dissociation is calculated using the initial concentration and the concentration of H+ found. We first calculate [H+] using the pH, then input those values into the Ka expression to find Ka, then use the formula for percent dissociation to find the percent dissociation.
Explanation:To solve this problem, we will first calculate the Ka using the known pH and the equilibrium relationship between the pH, Ka and [H+] concentration. Then we will calculate the percent dissociation of the acetic acid.
Given that pH = 2.87, we can use the pH definition pH=-log[H+] to find that [H+] = 10^-2.87. Knowing that acetic acid dissociates as CH3COOH ⇌ H+ + CH3COO-, and given that the initial concentration of the acetic acid is 0.1 M, the equilibrium concentrations are 0.1–x for CH3COOH and x for both H+ and CH3COO-.From the ICE table, we know that [H+] = x = 10^-2.87. Substituting this into the Ka expression gives Ka = ([H+][CH3COO-])/([CH3COOH]) = x^2 / (0.1 - x). Solving this gives the Ka of acetic acid.To find the percent dissociation, we can use the formula %Dissociation = ([H+]/initial concentration of acid)*100%. Substituting the respective values will give the %Dissociation.Learn more about Acid Dissociation here:https://brainly.com/question/33454852
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Ketones undergo a reduction when treated with sodium borohydride, NaBH4. The product of the above reaction has the following spectroscopic properties; propose a structure. MS: M+ = 86 IR: 3400 cm-1 1H NMR: 1.56 δ (4H, triplet); 1.78 δ (4H, multiplet); 3.24 δ (1H, quintet); 3.58 δ (1H, singlet) 13C NMR: 24.2, 35.5, 73.3 δ
Answer:
The product is cyclohexanol
Explanation:
Firstly,
A ketone undergo a borohydride reduction reaction to form an alcohol as below,
R-CO-R' ⇒ R-CO(OH)-R'
IR Spectrum confirms that alcohol group is existed with the peak at 3400 cm⁻¹From 1H-NMR, the product has 10 hydrogen atoms, the MS suggest that the formula is C₅H₁₀O (M = 86). With this formula, the alcohol is monosaturated. Since, the substance already underwent reduction reaction, the only way to suggest a monosaturated compound is a cyclic alcohol. So the compound is cyclopentanol.Check with other spectroscopic properties,3 signals of 13C NMR confirms the structure is symmetrical, δ 24.2, (-CH₂-CH₂-CH(CH₂-)-OH), δ 35.5 (-CH₂-CH₂-CH(CH₂-)-OH), δ 73.3 (-CH₂-CH₂-CH(CH₂-)-OH).1H NMR confirms,1.56 δ (4H, triplet) - (-CH₂-CH₂-CH-OH) ; triplet as coupling with 2 H,
1.78 δ (4H, multiplet) - (-CH₂-CH₂-CH-OH); multiplet as coupling with 2H of CH₂, 1 H of CH
3.24 δ (1H, quintet); - (-CH₂-CH₂-CH(CH₂-)-OH), coupling with4 H of 2 group of CH₂
3.58 δ (1H, singlet); - (-CH₂-CH₂-CH(CH₂-)-OH), hydrogen of alcohol group, not tend to coupling with other hydrogen
If one wished to obtain 0.15 moles of glacial acetic acid, how many milliliters of glacial acetic acid should be obtained? Enter only the number with two significant figures.
Answer:
8.6 mL
Explanation:
Glacial acetic acid (or anhydrous acetic acid) can be thought of as pure acetic acid, and has a density of 1.05 g/cm³ (or g/mL).
To answer this problem first we convert moles of acetic acid (CH₃CO₂H) to grams, using its molar mass (60 g/mol):
0.15 mol acetic acid * 60 g/mol = 9.0 g acetic acid.Now we convert grams of acetic acid to mL, using its density:
9.0 g ÷ 1.05 g/mL = 8.6 mLYou analyze a sample of unknown metal as you would in this experiment. You measure the volume of H2(g) generated to be 71.85 mL and the water temperature to be 20.0°C. You calculate PH2 to be 0.781 atm. Use the ideal gas law to calculate the number of moles of hydrogen gas generated.
Answer: The number of moles of hydrogen gas generated is [tex]2.33\times 10^{-3}mol[/tex]
Explanation:
To calculate the number of moles of hydrogen gas, we use the equation given by ideal gas which follows:
[tex]PV=nRT[/tex]
where,
P = pressure of the gas = 0.781 atm
V = Volume of the gas = 71.85 mL = 0.07185 L (Conversion factor: 1 L = 1000 mL)
T = Temperature of the gas = [tex]20^oC=[20+273]K=293K[/tex]
R = Gas constant = [tex]0.0821\text{ L. atm }mol^{-1}K^{-1}[/tex]
n = number of moles of hydrogen gas = ?
Putting values in above equation, we get:
[tex]0.781atm\times 0.07185L=n\times 0.0821\text{ L.atm }mol^{-1}K^{-1}\times 293K\\\\n=\frac{0.781\times 0.07185}{0.0821\times 293}=2.33\times 10^{-3}mol[/tex]
Hence, the number of moles of hydrogen gas generated is [tex]2.33\times 10^{-3}mol[/tex]
The energy changes for many unusual reactions can be determined using Hess’s law.
(a) Calculate ΔE for the conversion of F⁻(g) into F⁺(g).
(b) Calculate ΔE for the conversion of Na⁺(g) into Na⁻(g).
Answer:
(a) F⁻(g) ⇒ F⁺(g) ΔE=2009 KJ/mol
(b) Na⁺(g) ⇒ Na⁻(g) ΔE=-548.6 KJ/mol
Explanation:
Hess's Law
"Energy changes for a reaction is independent of steps or route involved."
(a) F⁻(g) ⇒ F⁺(g)Now the conversion F⁻ to F⁺ is a two step reaction. In first step F⁻ loses an electron e⁻ to become neutral atom.
F⁻ ⇒ F + e⁻ ΔH=328 KJ/mol (i)
Kindly note this energy is derived from electron affinity, which is the energy changes while adding an electron to neutral atom and its same while removing an electron from ion.
In second step,
Another electron is removed from F.
F ⇒ F⁺ + e⁻ ΔH= 1681 KJ/mol (ii)
This energy is 1st ionization energy for F, which is the energy required to remove first electron from outer most shell of neutral atom.
Now the total energy change can be found by applying Hess's law and adding equations (i) and (ii);
F⁻ ⇒ F + e⁻ ΔH=328 KJ/mol
F ⇒ F⁺ + e⁻ ΔH= 1681 KJ/mol
F⁻ ⇒ F⁺ + 2e⁻ ΔH= 2009 KJ/mol
(b) Na⁺(g) ⇒ Na⁻(g)This reaction involves two steps in first Na⁺(g) gains electron and become neutral atom while in second step Na(g) accepts another electron to become Na⁻(g).
Na⁺(g) + e⁻⇒ Na(g) ΔH=-495.8 KJ/mol (i)
Na(g) + e⁻ ⇒ Na⁻(g) ΔH=-52.8 KJ/mol (ii) (By Applying Hess's Law)
Na⁺(g) + 2e⁻ ⇒ Na⁻(g) ΔH=-548.6 KJ/mol
Kindly note as there is no molar change in above mentioned reactions therefore enthalpy changes(ΔH) and internal energy changes(ΔE) are same.
A student reacts 20.0 mL of 0.248 M H2SO4 with 15.0 mL of 0.195 M NaOH. Write a balanced chemical equation to show this reaction. Calculate the concentrations of H2SO4 and NaOH that remain in solution after the reaction is complete, and the concentration of the salt that is formed during the reaction.
Answer: The concentration of salt (sodium sulfate), sulfuric acid and NaOH in the solution is 0.0418 M, 0.0999 M and 0 M respectively.
Explanation:
To calculate the number of moles for given molarity, we use the equation:
[tex]\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}[/tex] .....(1)
For NaOH:
Initial molarity of NaOH solution = 0.195 M
Volume of solution = 15.0 mL = 0.015 L (Conversion factor: 1 L = 1000 mL)
Putting values in equation 1, we get:
[tex]0.195M=\frac{\text{Moles of NaOH}}{0.015L}\\\\\text{Moles of NaOH}=(0.195mol/L\times 0.015L)=2.925\times 10^{-3}mol[/tex]
For sulfuric acid:
Initial molarity of sulfuric acid solution = 0.248 M
Volume of solution = 20.0 mL = 0.020 L
Putting values in equation 1, we get:
[tex]0.248M=\frac{\text{Moles of }H_2SO_4}{0.020L}\\\\\text{Moles of }H_2SO_4=(0.248mol/L\times 0.020L)=4.96\times 10^{-3}mol[/tex]
The chemical equation for the reaction of NaOH and sulfuric acid follows:
[tex]2NaOH+H_2SO_4\rightarrow Na_2SO_4+2H_2O[/tex]
By Stoichiometry of the reaction:
2 moles of NaOH reacts with 1 mole of sulfuric acid
So, [tex]2.925\times 10^{-3}[/tex] moles of KOH will react with = [tex]\frac{1}{2}\times 2.925\times 10^{-3}=1.462\times 10^{-3}mol[/tex] of sulfuric acid
As, given amount of sulfuric acid is more than the required amount. So, it is considered as an excess reagent.
Thus, NaOH is considered as a limiting reagent because it limits the formation of product.
Excess moles of sulfuric acid = [tex](4.96-1.462)\times 10^{-3}=3.498\times 10^{-3}mol[/tex]
By Stoichiometry of the reaction:
2 moles of KOH produces 1 mole of sodium sulfate
So, [tex]2.925\times 10^{-3}[/tex] moles of KOH will produce = [tex]\frac{1}{2}\times 2.925\times 10^{-3}=1.462\times 10^{-3}mol[/tex] of sodium sulfate
For sodium sulfate:Moles of sodium sulfate = [tex]1.462\times 10^{-3}moles[/tex]
Volume of solution = [15.0 + 20.0] = 35.0 mL = 0.035 L
Putting values in equation 1, we get:
[tex]\text{Molarity of sodium sulfate}=\frac{1.462\times 10^{-3}}{0.035}=0.0418M[/tex]
For sulfuric acid:Moles of excess sulfuric acid = [tex]3.498\times 10^{-3}mol[/tex]
Volume of solution = [15.0 + 20.0] = 35.0 mL = 0.035 L
Putting values in equation 1, we get:
[tex]\text{Molarity of sulfuric acid}=\frac{3.498\times 10^{-3}}{0.035}=0.0999M[/tex]
For NaOH:Moles of NaOH remained = 0 moles
Volume of solution = [15.0 + 20.0] = 35.0 mL = 0.035 L
Putting values in equation 1, we get:
[tex]\text{Molarity of NaOH}=\frac{0}{0.050}=0M[/tex]
Hence, the concentration of salt (sodium sulfate), sulfuric acid and NaOH in the solution is 0.0418 M, 0.0999 M and 0 M respectively.
After the reaction between solutions of H2SO4 and NaOH is complete, the remaining concentrations are 0.0317 M H2SO4, 0 M NaOH, and 0.0836 M Na2SO4 are formed.
Explanation:The balanced chemical equation for the reaction between sulfuric acid (H2SO4) and sodium hydroxide (NaOH) is: H2SO4 + 2NaOH -> Na2SO4 + 2H2O. This equation shows that one mole of sulfuric acid reacts with two moles of sodium hydroxide to produce one mole of sodium sulfate (Na2SO4) and two moles of water.
Using this stoichiometry, the moles of H2SO4 and NaOH in the solutions can be calculated with moles = concentration x volume (in L). That gives us 0.020 L * 0.248 mol/L = 0.00496 mol for H2SO4 and 0.015 L * 0.195 mol/L = 0.002925 mol for NaOH.
Since one mole of H2SO4 reacts with two moles of NaOH, all of the NaOH will react, and you will have 0.00496 mol - 2* 0.002925 mol = 0.00111 mol of H2SO4 left. The concentration of H2SO4 then is 0.00111 mol / 0.035 L (sum of volumes) = 0.0317 M.
NaOH is completely reacted, so its concentration is 0 M. From the reaction stoichiometry, 0.002925 mol of Na2SO4 is formed, so its concentration is 0.002925 mol / 0.035 L = 0.0836 M.
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A sodium atom has a mass number of 23. Its atomic number is 11. How many electrons does a neutral sodium atom have?
Answer:
it contains 11 electrons
Explanation:
A student measured the length of a piece of paper and determined it to be 21.6cm using a metric measurement and 8 1/2 inches using the English measurement. What is the ratio of cm/inch using the students data.Do not give a fraction answer, calculate it to get an answer with 2 digits past the decimal
Explanation:
As per the measurements that are made by the student are as follows.
21.6 cm is equivalent to 8.5 inch
Therefore, 1 cm for the given situation will be equivalent to inch as follows.
1 cm = [tex]\frac{8.5}{21.6}[/tex] inch
= 0.393 inch
Hence, we will calculate the ratio of cm : inch as follows.
Ratio of cm : inch = [tex]\frac{8.5}{21.6}[/tex]
= 0.39
Thus, we can conclude that the ratio of cm/inch using the students data is 0.39.
Balance the following expression: __ CH3CH2COOH + __ O2 → __ CO2 + __ H2O How many moles of O2 are required for the complete combustion of 6 mol of propanoic acid?
Answer:
We need 21.0 moles of O2
Explanation:
Step 1: Data given
Moles of propanoic acid = 6.0 moles
CH3CH2COOH = propanoic acid
Step 2: The balanced equation
2CH3CH2COOH + 7O2 → 6CO2 + 6H2O
Step 3 :Calculate moles O2
For 2 moles propanoic acid we need 7 moles O2 to produce 6 moles CO2 and 6 moles H2O
For 6.0 moles propanoic acid we need 6.0 * 3.5 = 21 moles O2
We need 21.0 moles of O2
To combust 6 moles of propanoic acid completely, 15 moles of [tex]O_{2}[/tex] are required, based on the balanced chemical equation for the combustion of propanoic acid. 2 [tex]CH_{3} CH_{2} COOH[/tex] + 5 [tex]O_{2}[/tex] → 6 [tex]CO_{2}[/tex] + 4 [tex]H_{2}O[/tex] is the balanced chemical equation.
The question asks for the amount of Oxygen ([tex]O_{2}[/tex]) needed for the complete combustion of propanoic acid ([tex]CH_{3} CH_{2} COOH[/tex]). The combustion of propanoic acid can be represented by a balanced chemical equation:
2 [tex]CH_{3} CH_{2} COOH[/tex] + 5 [tex]O_{2}[/tex] → 6 [tex]CO_{2}[/tex] + 4 [tex]H_{2}O[/tex]
This means that 2 moles of propanoic acid require 5 moles of [tex]O_{2}[/tex] for complete combustion. If we have 6 moles of propanoic acid, a simple stoichiometric calculation would be to multiply the amount of [tex]O_{2}[/tex] required for 2 moles of propanoic acid by 3 (since 6 moles is three times larger than 2 moles), resulting in:
5 moles [tex]O_{2}[/tex]/2 moles [tex]CH_{3} CH_{2} COOH[/tex] × 6 moles [tex]CH_{3} CH_{2} COOH[/tex] = 15 moles [tex]O_{2}[/tex]
Therefore, to combust 6 moles of propanoic acid completely, 15 moles of [tex]O_{2}[/tex] are required.
For an ideal gas, which pairs of variables are inversely proportional to each other (if all other factors remain constant)?
Answer:
Pressure and volume
Explanation:
The ideal gas equation is given below:
PV = nRT
P = nRT /V
If nRT is constant,
P will be inversely proportional to V(P & 1/V)
Therefore, pressure and volume are inversely proportional if all factors remains constant
In an ideal gas, pressure and volume are inversely proportional to each other if all other factors remain constant. This is known as Boyle's Law.
Explanation:In an ideal gas, pressure and volume are inversely proportional to each other if all other factors remain constant.
This is known as Boyle's Law, which states that when the volume of a given amount of gas is decreased, the pressure increases, and vice versa.
For example, if the volume of a gas is reduced by half, the pressure will double, and if the volume is doubled, the pressure will be halved.
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Why do the deprotonated organic acid (RCO2-) and protonated organic base (RNH3+) go into the aqueous layer?
Answer:
The deprotonated organic acid (RCO2-) and protonated organic base (RNH3+) go into the aqueous layer because Organic compounds that are neither acids or bases do not react with either NaOH or HCl and, therefore remain more soluble in the organic solvent and are not extracted.
Explanation: