Answer:The gaseous material primarily extruded from a hydrothermal vent is primarily Hydrogen Sulfide (H2S).
Explanation:
Hydrothermal vents are underwater geothermal systems that occur on the ocean floor. They are formed when seawater seeps into the Earth's crust, gets heated by volcanic activity, and then rises back to the surface. These vents are often found near tectonic plate boundaries, such as mid-ocean ridges.
The primary gaseous material extruded from hydrothermal vents is hydrogen sulfide (H2S). Hydrogen sulfide is a colorless and highly toxic gas with a distinct rotten egg odor. It is produced as a result of chemical reactions that occur within the vent system.
At hydrothermal vents, seawater reacts with hot rocks and minerals in the Earth's crust. This process leads to the formation of various chemical compounds, including hydrogen sulfide. The hot, mineral-rich water released from the vents carries dissolved hydrogen sulfide gas along with other dissolved gases.
The release of hydrogen sulfide gas from hydrothermal vents has significant ecological implications. It serves as an energy source for specialized bacteria that thrive in these extreme conditions. These bacteria, known as chemosynthetic bacteria, use the hydrogen sulfide as an energy source to convert it into organic matter through a process called chemosynthesis. This chemosynthetic activity forms the basis of unique ecosystems around hydrothermal vents, supporting diverse communities of organisms.
While other gases may also be present in lower concentrations, hydrogen sulfide is the primary gaseous material extruded from hydrothermal vents due to its abundance and importance in supporting the unique ecosystems that exist in these extreme environments.
The gaseous material primarily extruded from a hydrothermal vent is hydrogen sulfide (H2S).
High amounts of hydrogen sulphide gas, as well as other gases including carbon dioxide (CO2) and methane (CH4), are known to be released from hydrothermal vents.
The habitats and microbial communities that are found surrounding hydrothermal vents are unique because of the chemical composition and conditions that these gases contribute to. So hydrogen sulphide is the right response.
A seafloor fissure known as a hydrothermal vent is where hot, mineral-rich fluids are released into the surrounding water. Typically at mid-ocean ridges or in regions where tectonic plates are sliding apart, these vents are found in volcanically active regions.
Magma that exists beneath the surface of the Earth heats the fluids that are emitted by hydrothermal vents. When seawater seeps into fissures and fractures, it heats up and reacts with the nearby rocks, leaching away different minerals and metals in the process.
Hot, mineral-rich fluids are released via the vent apertures when the superheated water hits the seafloor.
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which reaction characteristics are changing by the addition of a catalyst to a reaction at constant temperature?
The addition of a catalyst to a reaction at a constant temperature can affect several reaction characteristics:
Reaction Rate: A catalyst can increase the rate of a chemical reaction by providing an alternative reaction pathway with lower activation energy. It provides an alternative mechanism for the reaction to proceed, allowing the reactants to form products more quickly. As a result, the reaction rate is enhanced. Activation Energy: Catalysts lower the activation energy required for the reaction to occur. By providing an alternative pathway with lower energy barriers, a catalyst allows the reactant molecules to overcome the activation energy hurdle more easily, facilitating the reaction. Equilibrium Position: A catalyst does not affect the equilibrium position of a reversible reaction. It can speed up the attainment of equilibrium by accelerating the forward and backward reactions equally. However, the actual concentrations of the reactants and products at equilibrium remain the same. Reaction Selectivity: Catalysts can influence the selectivity of a reaction, promoting the formation of specific products while suppressing undesired side reactions. They can facilitate specific bond-breaking and bond-forming steps, favoring certain reaction pathways over others.
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A 0.148 M solution of a monoprotic acid has a percent ionization of 1.55%. Determine the acid ionization constant (Ka) for the acid. O 2.48 x 10 4 O 1.80 x 105 O 2.61* 10-4 O 3.61 x 105 O 1.32 x
Acid ionization constant is defined as the equilibrium constant for the dissociation reaction of an acid in an aqueous solution. It is represented by the symbol Ka.
To determine the acid ionization constant (Ka) for the monoprotic acid, we will use the following formula: Ka = [H+][A-] / [HA]
Let's solve the problem using the given information: Concentration of the acid (HA) = 0.148 MPercent ionization = 1.55%
Therefore, the concentration of H+ ions will be: H+ concentration = 1.55% of 0.148 M= 0.0155 × 0.148= 0.00229 MThe concentration of the conjugate base (A-) will also be equal to 0.00229 M. The total concentration of the acid (HA) in the solution will be the sum of the ionized and unionized acid: [HA] = [H+] + [HA-]= 0.00229 M + 0.14571 M= 0.148 MNow, we can substitute the values into the formula for Ka:Ka = [H+][A-] / [HA]= (0.00229 M)2 / 0.14571 M= 3.61 × 10-5
Therefore, the acid ionization constant (Ka) for the given monoprotic acid is 3.61 × 10-5.
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carbonic acid can form water and carbon dioxide upon heating. how much carbon dioxide is formed from 6.20 g of carbonic acid? h2co3 → h2o co2
To determine the amount of carbon dioxide formed from 6.20 g of carbonic acid (H2CO3), we need to consider the molar ratios between carbonic acid and carbon dioxide in the balanced chemical equation.
The balanced equation for the decomposition of carbonic acid is H2CO3 → H2O + CO2 From the equation, we can see that for every 1 mole of carbonic acid, 1 mole of carbon dioxide is produced.First, let's calculate the number of moles of carbonic acid using its molar mass Molar mass of H2CO3 = 2(1.00794 g/mol) + 12.0107 g/mol + 3(15.9994 g/mol) ≈ 62.0247 g/mol.
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you have prepared a saturated solution of x at 20 ∘c using 43.0 g of water. how much more solute can be dissolved if the temperature is increased to 30 ∘c ?
The expression mass of solute = (solubility at 30°C / 100) × mass of solvent, where the solubility of x at 30 °C is greater than its solubility at 20 °C.
Using the given data, we can calculate the solubility of the solute, x at 20 °C as follows:
The solubility of a solute at a certain temperature is defined as the amount of solute in grams that dissolves in 100 g of solvent to prepare a saturated solution at that temperature.
This is given by the expression: solubility = (mass of solute / mass of solvent) × 100So, the solubility of x at 20 °C is:solubility at 20°C = (mass of solute / mass of solvent) × 100We can write this as:mass of solute = (solubility at 20°C / 100) × mass of solventmass of solute = (solubility at 20°C / 100) × 43.0gTo find the mass of solute x that can be dissolved at 30 °C, we need to use the expression:solubility at 30°C = (mass of solute / mass of solvent) × 100We can write this as:mass of solute = (solubility at 30°C / 100) × mass of solventSo, we need to find the solubility of x at 30 °C to solve for the mass of solute. The solubility of most solids increases with an increase in temperature.
This means that more solute can be dissolved at a higher temperature than at a lower temperature, provided the initial solution was not saturated.So, we can conclude that the mass of solute that can be dissolved at 30 °C will be greater than the mass of solute that was dissolved at 20 °C.
Summary: To summarize, we can say that to find the mass of solute x that can be dissolved in the solution at 30 °C, we need to use the expression mass of solute = (solubility at 30°C / 100) × mass of solvent, where the solubility of x at 30 °C is greater than its solubility at 20 °C.
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what is the ph of a buffer prepared by adding 0.405 mol of the weak acid ha to 0.305 mol of naa in 2.00 l of solution? the dissociation constant ka of ha is 5.66×10−7 .
A buffer solution is an aqueous solution that resists changes in its pH on the addition of small amounts of an acid or a base. Buffer solutions are made of a weak acid and its conjugate base, or a weak base and its conjugate acid. The pH of a buffer prepared by adding 0.405 mol of the weak acid HA to 0.305 mol of NaA in 2.00 L of solution can be calculated as follows:The initial molar concentration of HA is, \[\left[\ce{HA}\right]=\frac{0.405 \;mol}{2.00 \;L}=0.203 \;M\]The initial molar concentration of A- is,\[\left[\ce{A-}\right]=\frac{0.305 \;mol}{2.00 \;L}=0.1525 \;M\]. The dissociation constant (Ka) of HA is 5.66 × 10⁻⁷. This value is related to the acid dissociation equation for the acid HA,\[\ce{HA + H2O <=> H3O+ + A-}\]From this equation,\[K_a=\frac{\left[\ce{H3O+}\right]\left[\ce{A-}\right]}{\left[\ce{HA}\right]}\]Since we are interested in pH, we rearrange this equation into the form, \[\left[\ce{H3O+}\right]=K_a\frac{\left[\ce{HA}\right]}{\left[\ce{A-}\right]}\]Plugging in the values, \[\left[\ce{H3O+}\right]=5.66 \times 10^{-7}\; \frac{0.203}{0.1525}=7.54 \times 10^{-7}\;M\]. Therefore, pH = -log[H₃O⁺] = -log(7.54 × 10⁻⁷) = 6.12 (rounded to 2 decimal places). Hence, the pH of a buffer prepared by adding 0.405 mol of the weak acid HA to 0.305 mol of NaA in 2.00 L of solution is 6.12.
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The pH of the buffer solution is 6.084. A buffer solution is a chemical substance that resists changes in pH levels when small amounts of acid or base are added to it. The pH of a buffer solution is controlled by its chemical composition and the ratio of its components.
A buffer is a solution that resists pH changes when small amounts of an acid or a base are added to it. Buffers consist of weak acids and their conjugate bases or weak bases and their conjugate acids. They have the property of being able to absorb excess H+ ions or OH- ions, without leading to a significant change in pH.
The dissociation constant of an acid, Ka is the product of the concentration of the hydronium ions and the concentration of the acid in the solution divided by the concentration of the dissociated form of the acid.
Ka= ( [H+][A-] ) / [HA]The acid dissociation constant of the weak acid HA is given as Ka= 5.66 x 10^-7.
We know that the weak acid HA dissociates according to the following equation:HA ⇌ H+ + A-So, [H+] = √Ka[HA]Now, we know that 0.405 moles of the weak acid HA and 0.305 moles of its salt NaA have been added to 2.00 L of solution. Therefore, the molar concentration of HA is0.405 mol/2.00 L = 0.2025 M
The molar concentration of NaA is 0.305 mol/2.00 L = 0.1525 M
To calculate the pH of the buffer, we need to determine the concentration of H+ ions. Thus, we can use the Henderson-Hasselbalch equation. It is given as:pH = pKa + log [A-]/[HA]pKa = -log Ka = -log 5.66 x 10^-7= 6.246log [A-]/[HA] = log [0.1525 M]/[0.2025 M]= -0.162Therefore, pH = 6.246 – 0.162 = 6.084
Thus, the pH of the buffer solution is 6.084.
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what is the value of δgo in kj at 25 oc for the reaction between the pair: pb(s) and sn2 (aq) to give sn(s) and pb2 (aq) ?
The value of ΔG° for the reaction between the pair Pb(s) and Sn2(aq) to give Sn(s) and Pb2(aq) at 25°C is -493.6 kJ/mol. The reaction of the reaction between the pair Pb(s) and Sn2(aq) to give Sn(s) and Pb2(aq) at 25°C can be represented by the following equation: Pb(s) + Sn2(aq) → Sn(s) + Pb2(aq)
The value of δG° (in kJ) at 25°C can be calculated by using the Gibbs free energy equation:ΔG° = ΔH° − TΔS°where ΔH° and ΔS° are the standard enthalpy and standard entropy changes, respectively, and T is the temperature in Kelvin.
To calculate the value of ΔH°, we need to use the standard enthalpy of formation of the reactants and products.
The values are as follows: Reactants: Pb(s) → ΔH°f = 0 kJSn2(aq) → ΔH°f = 0 kJProducts:Sn(s) → ΔH°f = 0 kJPb2(aq) → ΔH°f = -493.8 kJ/mol
The change in enthalpy for the reaction is given by:ΔH° = Σ(ΔH°f of products) − Σ(ΔH°f of reactants)ΔH° = [0 kJ/mol + (-493.8 kJ/mol)] − [0 kJ/mol + 0 kJ/mol]ΔH° = -493.8 kJ/mol. The standard entropy change can be calculated using the molar entropy values of the reactants and products.
The values are as follows:Reactants:Pb(s) → S°m = 22.6 J/mol·KSn2(aq) → S°m = 189.5 J/mol·KProducts:Sn(s) → S°m = 41.5 J/mol·KPb2(aq) → S°m = 163.3 J/mol·K
The change in entropy for the reaction is given by:ΔS° = Σ(S°m of products) − Σ(S°m of reactants)ΔS° = [41.5 J/mol·K + 163.3 J/mol·K] − [22.6 J/mol·K + 189.5 J/mol·K]ΔS° = -6.3 J/mol·K
Now, we can calculate the value of ΔG° using the Gibbs free energy equation:ΔG° = ΔH° − TΔS°ΔG° = [-493.8 kJ/mol] − [(25 + 273.15) K × (-6.3 J/mol·K/1000 J/kJ)]ΔG° = -493.8 kJ/mol + 0.158 kJ/molΔG° = -493.6 kJ/mol
Therefore, the value of ΔG° for the reaction between the pair Pb(s) and Sn2(aq) to give Sn(s) and Pb2(aq) at 25°C is -493.6 kJ/mol.
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the second-order rate constant of hydroxyl radicals for methyl ethyl ketone is
The half-life of methyl ethyl ketone (MEK) in a batch reactor, given an OH concentration of 10⁻¹² mol/L and a second-order rate constant of 9 x 10⁸ L/(mol·s), can be calculated using the integrated rate law for second-order reactions.
The integrated rate law for a second-order reaction is given by the equation:
1/[A]t = kt + 1/[A]0
Where:
[A]t = concentration of MEK at time t
[A]0 = initial concentration of MEK
k = rate constant
In this case, we are interested in the half-life, which is the time it takes for half of the initial concentration to be consumed. When [A]t = [A]0/2, we can substitute these values into the integrated rate law and solve for t.
1/([A]0/2) = k * t + 1/[A]0
Simplifying the equation:
2/[A]0 = k * t + 1/[A]0
Rearranging the equation and solving for t:
t = (2/[A]0 - 1/[A]0) / k
= 1/[A]0k
Given that [A]0 = 10⁻¹² mol/L and k = 9 x 10⁸ L/(mol·s), we can substitute these values into the equation:
t = 1 / (10⁻¹² mol/L * 9 x 10⁸ L/(mol·s))
= 1 / (9 x 10⁻⁴ s⁻¹)
= 1111.11 s
Therefore, the half-life of MEK in a batch reactor, with an OH concentration of 10⁻¹² mol/L and a second-order rate constant of 9 x 10⁸ L/(mol·s), is approximately 1111.11 seconds.
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The complete question is:
Advanced oxidation processes (AOPs). The second-order rate constant of hydroxyl radicals (OH) for methyl ethyl ketone (MEK) is 9 x 10⁹ L/(mols). Calculate the half-life of MEK in a batch reactor for a "OH concentration of 10⁻¹² mol/L.
how many grams of mgo are producedd when 40.0g of o2 reaction completely with mg
The mass of MgO produced is given as; Mass of MgO = 40 g O2 x (2 mol MgO / 1 mol O2) x (40.31 g MgO / 1 mol MgO) Mass of MgO = 3224.8 g / 100 wordsMass of MgO = 32.25 g MgO (to 3 significant figures)Therefore, 32.25 g of MgO are produced when 40.0 g of O2 react completely with Mg.
The balanced chemical equation for the reaction of magnesium with oxygen is;2 Mg + O2 → 2 MgOGiven; the mass of O2 = 40 gTo determine the mass of MgO produced, we need to find the limiting reactant. The limiting reactant is the reactant that is completely used up in a reaction and limits the amount of product formed.The mass of MgO produced is given as; Mass of MgO = 40 g O2 x (2 mol MgO / 1 mol O2) x (40.31 g MgO / 1 mol MgO)Mass of MgO = 3224.8 g / 100 wordsMass of MgO = 32.25 g MgO (to 3 significant figures)Therefore, 32.25 g of MgO are produced when 40.0 g of O2 react completely with Mg.
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A galvanic cell is constructed that carries out the reaction Pb^2+ (aq) + 2 Cr^2+(aq) rightarrow Pb(s) + 2 Cr^3+ (aq) If the initial concentration of Pb^2+(aq) is 0.15 M, that of Cr^2(aq) is 0.20 M, and that of Cr^3+(aq) is 0.0030 M, calculate the initial voltage generated by the cell at 25 Degree C.
The initial voltage generated by the galvanic cell at 25°C is 0.61 V due to the balanced equation of Pb2+ (aq) + 2Cr2+ (aq) Pb (s) + 2Cr3+ (aq).
The initial voltage generated by the galvanic cell can be calculated using the following equation;
E° cell = E° cathode - E° anode The balanced equation for the reaction taking place in the galvanic cell can be written as;
Pb2+ (aq) + 2Cr2+ (aq) → Pb (s) + 2Cr3+ (aq)
At the anode, Cr2+ is oxidized to Cr3+ and loses two electrons as shown below;
Cr2+ → Cr3+ + e- (oxidation)At the cathode, Pb2+ accepts two electrons and is reduced to Pb(s) as shown below;
Pb2+ + 2e- → Pb (s) (reduction)
Therefore, the cell reaction can be written as;Pb2+ (aq) + 2Cr2+ (aq) → Pb (s) + 2Cr3+ (aq)From the reduction table, the reduction potentials for Pb2+/Pb and Cr3+/Cr2+ half-cells are -0.13 V and -0.74 V, respectively. E° cell = E° cathode - E° anode= -0.13 - (-0.74)= + 0.61 V
Therefore, the initial voltage generated by the galvanic cell at 25°C is 0.61 V.
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The initial voltage generated by the cell at 25°C is 1.779 V. The reaction given is a redox reaction. Pb2+ (aq) acts as the reducing agent as it loses two electrons to form Pb(s) while Cr2+ (aq) acts as the oxidizing agent as it gains two electrons to form Cr3+ (aq).
Given: Pb2+ (aq) + 2 Cr2+ (aq) → Pb(s) + 2 Cr3+ (aq) The reaction given is a redox reaction. Pb2+ (aq) acts as the reducing agent as it loses two electrons to form Pb(s) while Cr2+ (aq) acts as the oxidizing agent as it gains two electrons to form Cr3+ (aq).
The initial cell voltage can be calculated using the Nernst equation.E cell = E° cell – (RT/nF) ln QWhere,E° cell = standard cell potentialR = gas constant = 8.314 J mol-1 K-1
T = temperature in Kelvin, F = Faraday’s constant = 96485 C mol-1, n = moles of electrons exchanged, Q = reaction quotient
Initially, the concentrations of Pb2+ (aq), Cr2+ (aq), and Cr3+ (aq) are 0.15 M, 0.20 M, and 0.0030 M respectively.
Thus, the reaction quotient Q will be: Q = [Pb(s)][Cr3+(aq)] / [Pb2+(aq)][Cr2+(aq)]Q = (1)[0.0030] / (0.15)(0.20)
Q = 0.01
E°cell for the reaction given can be calculated by adding the standard reduction potential of Pb2+ (aq) to that of Cr3+ (aq).
E°cell = E°red,Pb2+ (aq) – E°red,Pb(s) + E°red,Cr3+ (aq) – E°red,Cr2+ (aq)
E°cell = (-0.13 V) – 0.00 V + 0.74 V – (-0.91 V)E°cell = 1.72 V
Substituting the given values into the Nernst equation,E cell = E° cell – (RT/nF) ln QE cell = 1.72 V – (8.314 J mol-1 K-1)(298 K)/(2 * 96485 C mol-1) ln 0.01
E cell = 1.72 V – 0.059 V log 0.01E cell = 1.72 V + 0.059 V
E cell = 1.779 V
The initial voltage generated by the cell at 25°C is 1.779 V.
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When 3.0 g of solid ionic compound X is dissolved in 500 g of water at 20.7 °C in a coffee cup calorimeter, the final temperature of the solution that is formed ends up at 14.3 °C a) Did heat transfer into or out of the water? Justify your answer. What do you predict for the sign of puutar here? b) Was there an initial temperature difference between the two samples of matter that were mixed in this scenario that caused heat to transfer into or out of the water (like in the scenario in Question 1?
Regarding the sign of putter, since heat transferred out of the water, we expect the value of the puutar to be negative. This is because the system lost energy in the form of heat, which means the internal energy of the system decreased. This results in a negative value for puutar.
a) Heat transferred out of the water in this scenario. The initial temperature of the water was 20.7 °C, and after dissolving the ionic compound X, the final temperature dropped to 14.3 °C. This decrease in temperature indicates that the water lost heat to the surroundings and the process was endothermic. The sign of "puutar" (possibly referring to heat or energy) would be positive, as the system absorbed heat from the surroundings.
b) There was likely an initial temperature difference between the solid ionic compound X and the water, causing heat to transfer out of the water. The dissolution of the ionic compound is an endothermic process, which means it absorbed heat from the water, resulting in a lower final temperature for the solution. Yes, there was an initial temperature difference between the two samples of matter. The solid ionic compound X had a temperature of 20.7 °C, while the water had a lower temperature. This temperature difference caused heat to transfer from the solid to the water, which led to an increase in the temperature of the water. However, once compound X was completely dissolved, the heat transfer direction was reversed, as explained in part a).
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what volume, in milliliters, of a 0.194 m ba(oh)2 solution is needed to completely react 59.9 ml of a 0.205 m hclo4 solution.
The volume of 0.194 M Ba(OH)₂ solution required to completely react with 59.9 mL of 0.205 M HClO₄ solution is 31.7 mL.
To determine the volume of 0.194 M Ba(OH)₂ solution required to completely react with 59.9 mL of 0.205 M HClO₄ solution, we first need to balance the equation of the reaction that occurs between the two solutions.
The balanced chemical equation for the reaction between Ba(OH)₂ and HClO₄ is: Ba(OH)₂ + 2HClO₄ → Ba(ClO₄)₂ + 2H₂OHere, we can see that 1 mole of Ba(OH)₂ reacts with 2 moles of HClO₄. This means that the moles of Ba(OH)₂ required to react with 59.9 mL of 0.205 M HClO₄ solution are: moles of HClO₄ = Molarity x Volume (in liters) = 0.205 M x 0.0599 L = 0.0123 mol
According to the balanced chemical equation, 1 mole of Ba(OH)₂ reacts with 2 moles of HClO₄. Therefore, the number of moles of Ba(OH)₂ required to react with 0.0123 moles of HClO₄ is: moles of Ba(OH)₂ = 0.0123 mol ÷ 2 = 0.00615 mol
Now, we can calculate the volume of 0.194 M Ba(OH)₂ solution required to contain 0.00615 mol of Ba(OH)₂ :Volume = moles ÷ Molarity = 0.00615 mol ÷ 0.194 M = 0.0317 L = 31.7 mL
Therefore, the volume of 0.194 M Ba(OH)₂ solution required to completely react with 59.9 mL of 0.205 M HClO₄ solution is 31.7 mL.
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four elements are shown. use the periodic table to choose the most stable element. a. chlorine b. neon c. sulfur d. carbon
Among the four elements listed, the most stable element is Neon (Ne). Neon (Ne) is an inert gas belonging to the noble gas group on the periodic table.
Noble gases are known for their high stability due to having a complete outer electron shell. They exist as single atoms and do not readily form compounds with other elements. Neon is particularly stable because it has a full set of eight valence electrons, making it highly unreactive. On the other hand, chlorine (Cl), sulfur (S), and carbon (C) are reactive elements that can form compounds with other elements. While they are essential for various chemical reactions and compounds, they are not as inherently stable as neon. Therefore, the most stable element among the given options is Neon (Ne).
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Which of the following is a Brønsted-Lowry base?
Cl2
HCN
CBr4
NH3
None of the above are Brønsted-Lowry bases.
options (Cl2, HCN, CBr4) are not bases according to the Brønsted-Lowry definition. Cl2 is a diatomic molecule, HCN is a weak acid, and CBr4 is a nonpolar molecule.
The Brønsted -Lowry theory defines an acid as a substance that donates a proton, and a base as a substance that accepts a proton. Ammonia (NH3) is a Brønsted - Lowry base, according to this definition. Therefore, NH3 is a Brønsted -Lowry base. The Brønsted Lowry theory is a model that describes acids and bases in terms of proton donation and acceptance, respectively. Any species that accepts a proton is classified as a Brønsted-Lowry base. In order to be able to identify the Brønsted -Lowry base, it is crucial to understand the concept of proton donation or acceptance.mong the options provided, NH3 (ammonia) is a Brønsted-Lowry base. It can accept a proton (H+) from an acid to form its conjugate acid, NH4+ (ammonium ion).
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Which of the following elements or polyatomic ions become cations when ionized? Select all that apply. Chloride Magnesium Potassium Calcium Carbonate
When ionized, the following elements or polyatomic ions become cations: Magnesium, Potassium, Calcium.
Cations are atoms that have lost one or more electrons. This results in a positively charged ion. On the periodic table, metals like Magnesium, Potassium, Calcium are located on the left side and have low electronegativity. When they lose their valence electrons, they will have a positive charge. Chloride and Carbonate are both polyatomic ions that have a negative charge. Polyatomic ions are groups of atoms that carry a charge. Chloride is a negative ion, while Calcium, Potassium, and Magnesium are positive ions when ionized. These ions, when dissolved in water, create electrolytes, which are critical for many biological processes
Magnesium, Potassium, and Calcium ions become cations when ionized.
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the molar heat of solution of kclo4 is 50.9 kj/mol, the volume of water in which the salt will be siddoolved
To calculate the volume of water in which KClO4 will be dissolved, we need to know the mass of KClO4 and its solubility in water. If the molar heat of the solution is 50.9 KJ/mol
Unfortunately, the information provided is not sufficient to determine the volume of water.
The molar heat of solution of KClO4 is given as 50.9 kJ/mol. This value represents the amount of heat released or absorbed when one mole of KClO4 is dissolved in water.
However, this value alone does not provide enough information to determine the volume of water required for dissolving the salt. To do so, we need to know the mass of KClO4 and its solubility in water (i.e., how many grams of KClO4 can be dissolved in 1 L of water).
To answer your question, please provide additional information such as the mass of KClO4 and its solubility in water. With that information, we can calculate the volume of water required to dissolve the given amount of KClO4.
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a proton is located at a distance of 0.048 repulsive electric force
The charge of each proton is 1.07 × 10^-17 C. A proton is located at a distance of 0.048 m from another proton. If the repulsive electric force between them is 4.3 × 10−25 N,
The repulsive electric force is given by Coulomb’s Law as,F = kq1q2/d²Where,F is the repulsive force k is the Coulomb constant which is equal to 9 × 10^9 N.m²/C²q1 and q2 are the charges of the two protons which are separated by a distance, dd is the distance between the two charges.
Now, we can substitute the given values in the above formula.F = 4.3 × 10^-25 Nk = 9 × 10^9 N.m²/C²d = 0.048 mLet q1 = q2 = q be the charge of each proton.As per Coulomb’s Law,F = kq²/d²4.3 × 10^-25 N = (9 × 10^9 N.m²/C²) q²/(0.048 m)²4.3 × 10^-25 N = 9 × 10^9 N.m²/C² × q²/(0.048 m)²q² = 4.3 × 10^-25 N × (0.048 m)² / (9 × 10^9 N.m²/C²)q² = 1.1408 × 10^-34 C²Taking the square root of both sides of the equation, we get,q = 1.07 × 10^-17 C
Therefore, the charge of each proton is 1.07 × 10^-17 C.
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the permanent electric dipole moment of the water molecule (h2o) is 6.2×10−30c⋅m .
The permanent electric dipole moment of the water molecule (H2O) is 6.2×10^−30 C⋅m.
The electric dipole moment is the distance between two equal but opposite charges.
The electric dipole moment for H2O is 6.2 x 10^-30 C⋅m. In general, the electric dipole moment is defined as the product of charge and distance between the charges.The water molecule is polar because of its bent structure and electronegativity.
A permanent dipole is created as a result of the electronegativity difference between hydrogen and oxygen.
Because of the differences in the electronegativity of the atoms, electrons are drawn toward the oxygen atom, generating a negative charge, whereas the hydrogen atoms develop a positive charge as a result of the electron migration, resulting in a net dipole moment of the H2O molecule.
Summary:The water molecule's permanent electric dipole moment is 6.2×10^-30 C⋅m. The dipole moment is created as a result of the polar nature of the molecule, which is caused by differences in electronegativity between the atoms.
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how will you determine the amount of carbon dioxide that reacts in the grignard reaction?
In order to determine the amount of carbon dioxide that reacts in the Grignard reaction, the method for detecting carbon dioxide can be used.
The Grignard reaction involves the addition of an organomagnesium compound to a carbonyl group which results in the formation of an alcohol. The reaction is exothermic and carbon dioxide is produced in the process. A typical method to detect the carbon dioxide formed in the reaction involves the use of an aqueous solution of barium hydroxide and phenolphthalein indicator. Barium hydroxide reacts with carbon dioxide to form barium carbonate. 2Ba(OH)2 + CO2 → BaCO3 + H2OBarium carbonate is insoluble and hence the presence of carbon dioxide can be detected by observing the formation of a white precipitate. Phenolphthalein is used as an indicator and changes color from pink to colorless upon reaction with the carbon dioxide.The amount of carbon dioxide that reacts in the Grignard reaction can be determined by measuring the mass of the product formed. For example, if the product formed is an alcohol, then its mass can be determined by gravimetric analysis. The amount of carbon dioxide that reacted can be calculated by stoichiometry.
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Which of the following must be true for a spontaneous exothermic process? a. only that ASsys 0 b. only that ASsys>0 c. both ASys <0 and the magnitude of ASsys the magnitude of AS e. either ASyr ASy <0 and the magnitude of ASsys < the magnitude of AS R sum sur suIT
Both ΔSsys < 0 and the magnitude of ΔSsys > the magnitude of ΔSsurr must be true for a spontaneous exothermic process. Thus, the correct option is e. either ΔSsys < 0 and the magnitude of ΔSsys > the magnitude of ΔSsurr.For a spontaneous exothermic process, both ΔSsys < 0 and the magnitude of ΔSsys > the magnitude of ΔSsurr must be true
.A spontaneous process is a process that occurs naturally and does not require external energy or intervention to occur. Exothermic reactions are those that release heat energy as a byproduct. Therefore, when a spontaneous process occurs, energy is released from the system to the surroundings, resulting in a decrease in entropy of the system. The entropy of the surroundings increases since the energy released from the system increases the disorder of the surroundings.
The change in entropy of a system is represented by ΔSsys.ΔSsys = Sfinal - SinitialWhat is ΔSsurr?The change in entropy of the surroundings is represented by ΔSsurr.ΔSsurr = - q / Twhere q is the heat absorbed by the surroundings from the system, and T is the temperature at which the heat transfer occurred.A spontaneous process occurs when ΔSsys + ΔSsurr > 0. However, in exothermic reactions, ΔSsys < 0 since energy is released from the system, resulting in a decrease in entropy of the system. Therefore, to satisfy ΔSsys + ΔSsurr > 0, ΔSsurr > 0. This implies that the entropy of the surroundings should increase as a result of the energy released by the system. Since the surroundings are at a lower temperature than the system, the magnitude of ΔSsurr should be greater than the magnitude of ΔSsys. This is represented as:|ΔSsurr| > |ΔSsys|
Therefore, both ΔSsys < 0 and the magnitude of ΔSsys > the magnitude of ΔSsurr must be true for a spontaneous exothermic process. Thus, the correct option is e. either ΔSsys < 0 and the magnitude of ΔSsys > the magnitude of ΔSsurr.
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what is the net charge of the oligopeptide ala–glu–asn–leu–lys at ph 1
At pH 1, the net charge of the oligopeptide Ala-Glu-Asn-Leu-Lys is +1. Oligopeptides are small peptides that have a certain number of amino acid residues. Oligopeptides are also known as peptides because they are compounds made up of two or more amino acids.
A molecule of water is generated when two amino acids are combined together through a peptide bond. An oligopeptide contains up to 20 amino acid residues. They are utilized for a variety of purposes, including in cosmetics and skincare, sports and physical fitness, and healthcare.
The pH of 1 is extremely acidic, indicating that there is a lot of H+ ions. Acidic pHs have a positive impact on the side chains of amino acids. In an acidic medium, the carboxylic acid group of aspartic acid (aspartate) and glutamic acid (glutamate) will be protonated, resulting in a +1 charge.
The protonated amino group of lysine and the carboxylic acid group of aspartic acid (aspartate) and glutamic acid (glutamate) would be neutral at pH 1 since the amino group and carboxylic acid group will be protonated.The peptide bonds will not have any charge because they are neutral. The carboxylic acid group of asparagine will also be neutral because it lacks the ability to be protonated at pH 1.
The net charge for the oligopeptide Ala-Glu-Asn-Leu-Lys at pH 1 would be +1.
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5. which solvent would you order if you determined that a sample required a more polar solvent than what is available above? explain.
The solvent that would be ordered if a sample required a more polar solvent than what is available above is Ethanol.
When there is a need for a more polar solvent than those that are already available, ethanol is ordered.
Ethanol is a polar solvent, meaning it is a solvent that has a positive and a negative end to its molecule, so it is effective in dissolving polar compounds.
Ethanol is widely used as a solvent in various applications, including the extraction of plant materials and as a preservative in medicinal and personal care products.
The summary of the explanation is that Ethanol is a polar solvent that can be ordered when a more polar solvent is required than those that are already available.
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the solubility of srco3 in water at 25°c is measured to be 0.0045gl. use this information to calculate ksp for srco3.
The Ksp for SrCO₃ is calculated as 1.89 x 10⁻⁹. It is given that the solubility of SrCO₃ in water at 25°c is measured to be 0.0045gl.
Step 1: Write the balanced chemical equation for the dissolution of SrCO₃.
SrCO₃(s) ⇌ Sr²⁺(aq) + CO₃²⁻(aq)
Step 2: Write the expression for the Ksp for SrCO₃.Ksp = [Sr²⁺][CO₃²⁻]
Step 3: Determine the molar solubility of SrCO₃.
Molar mass of SrCO₃ = 103.6 g/mol
The solubility of SrCO₃ in water is given as 0.0045 g/L. Therefore, the molar solubility of SrCO₃ is:
Molar solubility = (0.0045 g/L) / (103.6 g/mol) = 4.35 x 10⁻⁵ M
Step 4: Substitute the molar solubility into the Ksp expression and solve for Ksp.
Ksp = [Sr²⁺][CO₃²⁻] = (4.35 x 10⁻⁵ M)(4.35 x 10⁻⁵ M) = 1.89 x 10⁻⁹
Therefore, the Ksp for SrCO₃ is 1.89 x 10⁻⁹
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Calculate the heat of combustion (kJ) of propane, C3H8 using the listed standard enthalpy of reaction data: C3H8(g)+5O2(g)⟶3CO2(g)+4H2O(g)
The heat of combustion of propane is 2220 kJ/mol. Hence, 2220 kJ of heat is evolved per mole of propane burned completely.
Given DataC3H8(g) + 5O2(g) → 3CO2(g) + 4H2O(g)ΔH° = -2220 kJ/mol of C3H8. We are supposed to calculate the heat of combustion (kJ) of propane, C3H8 using the listed standard enthalpy of reaction data: C3H8(g) + 5O2(g) → 3CO2(g) + 4H2O(g).
Solution: We have the balanced chemical equation of the combustion of C3H8, which shows that 1 mole of propane reacts with 5 moles of O2 to produce 3 moles of CO2 and 4 moles of H2O.C3H8(g) + 5O2(g) → 3CO2(g) + 4H2O(g)The amount of heat evolved when one mole of propane burns completely is equal to the enthalpy change (ΔH°) of the above combustion reaction. Thus,ΔH° = -2220 kJ/mol of C3H8The above value indicates that 2220 kJ of heat is evolved when 1 mole of propane burns completely. Hence, 2220 kJ of heat is evolved per mole of propane burned completely.Thus, the heat of combustion of propane is 2220 kJ/mol.
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A 15.0 mL sample of 0.150 M nitrous acid is titrated with a 0.150 M LIOH solution. What is the pH at the half equivalence point of this titration? A. 10.65 B. 335 C. 5.89 D. 700
C. 5.89, Half-equivalence point is a point in titration when half of the total moles of a base required to react with the total moles of acid in the sample have been added.
At this point, the pH of the solution will be equal to the pKa of the weak acid. Follow these steps to find the pH at half-equivalence point:
Step 1: Write down the balanced chemical equation for the reaction. HNO2(aq) + OH-(aq) ⟶ NO2-(aq) + H2O(l)
Step 2: Calculate the number of moles of nitrous acid (HNO2) in the sample. Number of moles = concentration × volume (in liters)Number of moles of HNO2 = 0.150 mol/L × (15.0/1000) L = 0.00225 mol
Step 3: Calculate the volume of the base (NaOH) required to reach half-equivalence point. Since the acid and base have the same concentration, the volume required would be half of the initial volume. Volume of NaOH = (1/2) × 15.0 mL = 7.5 mL
Step 4: Calculate the number of moles of NaOH required to reach half-equivalence point. Number of moles of NaOH = concentration × volume (in liters)Number of moles of NaOH = 0.150 mol/L × (7.5/1000) L = 0.00113 molStep 5: Calculate the number of moles of HNO2 that have reacted with NaOH. Since the reaction is 1:1, the number of moles of HNO2 that have reacted will be equal to the number of moles of NaOH used. Number of moles of HNO2 reacted = 0.00113 mol
Step 6: Calculate the number of moles of HNO2 remaining. Number of moles of HNO2 remaining = 0.00225 mol - 0.00113 mol = 0.00112 mol
Step 7: Calculate the concentration of HNO2 remaining. Concentration of HNO2 = moles/volume (in liters)Concentration of HNO2 = 0.00112 mol/(15.0 - 7.5) mL = 0.200 M
Step 8: Calculate the pKa of HNO2 using the Henderson-Hasselbalch equation.pKa = pH + log([A-]/[HA])We know that at half-equivalence point, [A-] = [HA]Therefore, pKa = pH + log(1) = pHpKa of nitrous acid (HNO2) is 3.35pH = pKa + log([A-]/[HA])pH = 3.35 + log(1) = 3.35pH at half-equivalence point is 3.35.
Converting pH from negative logarithmic scale to the normal scale:pH = -log[H+]H+ = 10-pH= 10-3.35= 4.466 x 10-4MConverting concentration of HNO2 in moles to that in grams:Mass of HNO2 = moles × molar mass
Mass of HNO2 = 0.00112 mol × 63.01 g/mol = 0.0706 g
Concentration of HNO2 = mass/volume (in liters)Concentration of HNO2 = 0.0706 g/(15.0/1000) L = 4.71 g/LThe answer is C. 5.89.
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draw a structural formula for the intermediate in the following reaction:ch2cl2
The structural formula for the intermediate in the following reaction is: C-Cl-OH-OH-Cl-C. The chemical reaction of CH₂Cl₂ is represented by the following equation CH₂Cl₂ + 2 NaOH → CH₂(OH)₂ + 2 NaCl
The intermediate structure of the following reaction has been illustrated in the figure below.
We know that sodium hydroxide (NaOH) is a strong base. A strong base can react with the hydrogen on the hydrogen chloride (HCl) molecule. NaOH will take away H from HCl and produce NaCl (sodium chloride) and water (H₂O).
The reaction proceeds as follows. CH₂Cl₂ → CCl₂ + CH₂CCl₂ + 2NaOH → CCl₂(OH)₂ + 2NaCl. Thus, the structural formula for the intermediate in the following reaction is: C-Cl-OH-OH-Cl-C.
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how many chiral carbons are present in the open-chain form of an aldohexose? a. six b. four c. three d. none e. five
Aldohexose is a monosaccharide with six carbon atoms and an aldehyde functional group. It contains multiple chiral centers, which are carbon atoms bonded to four different groups. To determine the number of chiral carbons, we must count the number of hydroxyl groups or hydrogen atoms.so, correct answer is b) four
An aldohexose is a monosaccharide with six carbon atoms and an aldehyde functional group. It is an example of a hexose, which is a six-carbon sugar.The open-chain form of an aldohexose contains multiple chiral centers, which are carbon atoms that are bonded to four different groups. These chiral centers can exist in two different configurations, resulting in a total of 2^n stereoisomers (where n is the number of chiral centers).Therefore, to determine the number of chiral carbons in an open-chain form of an aldohexose, we must count the number of carbon atoms that are bonded to four different groups.Each carbon atom in an aldohexose can be bonded to one of two types of groups: a hydroxyl group (-OH) or a hydrogen atom (-H). The first carbon atom in the chain (the aldehyde carbon) is not a chiral center since it is bonded to two identical groups (-H and -CHO).
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Water can react as both an acid and a base, depending on its environment. Because of this characteristic, water is a(n) a. amphoteric molecule. O b. autonomous C. complex O d. reactive e. conjugated QUESTION 53 A weak acid is also a a. weak electrolyte b. strong electrolyte c. nonelectrolyte O d. weak base because it produces a low concentration of ions in solution. e. strong acid QUESTION 54 The following reaction is a reversible reaction. Which of the following statements best describes what it means for this reaction to be reversible? HCOOHH2O HCOO H30+ a. This reaction only occurs in the reverse direction as written above. b. All of the reactant molecules react to make product and then all of the product molecules react to make reactants again. c. Forward and reverse reactions proceed at the same rate. d. Forward and reverse reactions occur simultaneously. e. The rate of the reverse reaction is must faster than the rate of the forward reaction.
Water is an amphoteric molecule, meaning it can act as both an acid and a base depending on its environment. A weak acid is a weak electrolyte because it produces a low concentration of ions in solution.
Lastly, a reversible reaction means that the forward and reverse reactions occur simultaneously and can proceed at different rates, with the rate of the reverse reaction potentially being faster than the rate of the forward reaction. In the given reaction, HCOOH + H2O HCOO- + H3O+, the reaction is reversible and can proceed in both the forward and reverse directions.
Water can react as both an acid and a base depending on its environment, making it an amphoteric molecule. A weak acid is also a weak electrolyte because it produces a low concentration of ions in solution. In a reversible reaction like HCOOH + H2O HCOO- + H3O+, forward and reverse reactions occur simultaneously.
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what are the ion concentrations in a 0.12 m solution of alcl3?
The ion concentrations in a 0.12 M solution of AlCl3 can be determined by using the dissociation equation of AlCl3 as AlCl3 → Al3+ + 3 Cl-.Step-by-step explanation:The dissociation equation of AlCl3 is AlCl3 → Al3+ + 3 Cl-.It shows that one AlCl3 molecule produces one Al3+ ion and three Cl- ions. Therefore, the ion concentrations of Al3+ and Cl- ions in the solution can be determined as follows:Ion concentration of Al3+ ion = 0.12 MIon concentration of Cl- ion = (3 x 0.12) M = 0.36 MThus, the ion concentrations in a 0.12 M solution of AlCl3 are 0.12 M for Al3+ ion and 0.36 M for Cl- ion.
AlCl3, also known as aluminum chloride, is a highly soluble inorganic compound.
When it is added to water, it dissociates into aluminum cations (Al3+) and chloride anions (Cl-), resulting in an increase in the concentration of these ions in solution. So, in a 0.12 M solution of AlCl3, we need to determine the concentration of these ions. Let's start by writing the balanced chemical equation for the dissociation of AlCl3:AlCl3 → Al3+ + 3 Cl-As can be seen, each molecule of AlCl3 dissociates to form one aluminum cation and three chloride anions.
This means that in a 0.12 M solution of AlCl3, the concentration of aluminum cations (Al3+) is 0.12 M, while the concentration of chloride anions (Cl-) is three times that, or 0.36 M. Therefore, the ion concentrations in a 0.12 M solution of AlCl3 are as follows:Al3+: 0.12 MCl-: 0.36 MIn summary, a 0.12 M solution of AlCl3 has an ion concentration of 0.12 M for aluminum cations (Al3+) and 0.36 M for chloride anions (Cl-).
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Map deb pling Identify the true statements regarding a 1,6 linkages in glycogen Exactly 4 residues extend from these linkages. O The number of sites for enzyme action on a glycogen molecule is increased through linkages. New a 1,6 linkages can only form if the branch has a free reducing end The reaction that forms a 1,6 linkages is catalyzed by a branching enzyme. At least four glucose residues separate a 1,6 linkages Previous Give Up & View solution 2
Regarding 1,6-linkages in glycogen, the true statements are: 1. The number of sites for enzyme action on a glycogen molecule is increased through 1,6-linkages. 2. The reaction that forms 1,6-linkages is catalyzed by a branching enzyme. 3. At least four glucose residues separate a 1,6-linkage.Hence the option 1,2,3 are correct.
The true statements regarding a 1,6 linkages in glycogen are:
1. Exactly 4 residues extend from these linkages.
2. The number of sites for enzyme action on a glycogen molecule is increased through linkages.
3. New a 1,6 linkages can only form if the branch has a free reducing end.
4. The reaction that forms a 1,6 linkages is catalyzed by a branching enzyme.
5. At least four glucose residues separate a 1,6 linkages.
Regarding 1,6-linkages in glycogen, the true statements are:
1. The number of sites for enzyme action on a glycogen molecule is increased through 1,6-linkages.
2. The reaction that forms 1,6-linkages is catalyzed by a branching enzyme.
3. At least four glucose residues separate a 1,6-linkage.
These linkages play a significant role in the structure and function of glycogen, enabling rapid glucose release when needed.
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the aka of a weak monoprotic acid is 1.31×10−5.1.31×10−5. what is the ph of a 0.0812 m0.0812 m solution of this acid?
The pH of a 0.0812 M solution of a weak monoprotic acid with an acid dissociation constant (Ka) of 1.31×10⁻⁵ is be calculated as 3.69
Step 1: Write the equation for the dissociation of the weak acid in water. HA(aq) + H₂O(l) ⇌H₃O⁺(aq) + A⁻(aq)
Step 2: Write the expression for the acid dissociation constant (Ka) for the weak acid. Ka = [H₃O⁺][A⁻] / [HA]
Step 3: Substitute the known values into the expression for Ka and solve for [H3O+].Ka = [H₃O⁺][A-] / [HA]1.31 × 10⁻⁵ = [H₃O⁺]2 / 0.0812[H₃O⁺] = 2.04 × 10⁻⁴ M
Step 4: Calculate the pH of the solution using the following equation: pH = -log[H₃O⁺]pH = -log(2.04 × 10⁻⁴)pH = 3.69
Therefore, the pH of a 0.0812 M solution of this weak monoprotic acid is 3.69.
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