6 electrons are transferred during the reaction.
The balanced equation for the given redox reaction is as follows:
[tex]CH(OH)_3[/tex] + 3Cu²+ + 2OH⁻ → CrO4²⁻ + 3Cu+ + 3H2O
In this reaction, electrons are transferred from Cu²+ ions to CrO4²⁻ ions.
Each Cu²+ ion loses one electron to form Cu+ ion. The oxidation state of Cu is reduced from +2 to +1.3
Cu²+ → 3Cu+ + 3e⁻
Electrons are also transferred from CrO4²⁻ ions to [tex]CH(OH)_3[/tex] molecules. The oxidation state of Cr is reduced from +6 to +3. CrO4²⁻ + 3H2O →[tex]Cr(OH)_3[/tex] + 4OH⁻
In this reaction, CrO4²⁻ ions gain 3 electrons to form [tex]Cr(OH)_3[/tex] molecules.
4OH⁻ + CrO4²⁻ + 3e⁻ → [tex]Cr(OH)_3[/tex] + 4OH⁻
Thus, 6 electrons are transferred during the reaction.
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At an altitude of 20 km the temperature is 217 K and the pressure is 0.050 atm. (a) What is the rms speed, the mean free path and collision frequency of N2 molecules at these conditions? (b) What is the probability that a nitrogen molecule at 25 C and 1 atmosphere pressure will travel 100 nm without undergoing a collision?
a) The rms speed is [tex]467\text{ m/s}[/tex], mean free path is [tex]0.44\text{ }\mu \text{m}[/tex], and the collision frequency is [tex]3.5\times {{10}^{8}}\text{ collisions/s}[/tex]
b) At a temperature of 25°C and a pressure of 1 atmosphere, the likelihood of a nitrogen molecule traveling a distance of 100 nm without experiencing a collision is 34%.
a) At a height of 20 km, the temperature measures 217 K while the pressure registers at 0.050 atm. The rms speed of N2 molecules is given by the formula shown below.
[tex]{{v}_{rms}}=\sqrt{\frac{3kT}{m}}[/tex]
Where k is the Boltzmann constant, T is the temperature, m is the mass of N2, and vrms is the root-mean-square speed of the N2 molecules. Substitute the values of the constants and the variables given into the formula and solve.
[tex]{{v}_{rms}}=\sqrt{\frac{3(1.38\times {{10}^{-23}}\text{ J/K})(217\text{ K})}{(28\text{ g/mol})(6.02\times {{10}^{23}}\text{ molecules/mol})}}=467\text{ m/s}[/tex]
The mean free path of N2 molecules at these conditions is given by the formula shown below.
[tex]{{\lambda }_{mfp}}=\frac{1}{\sqrt{2}\pi {{d}^{2}}n}\sqrt{\frac{8kT}{\pi m}}[/tex]
Where d is the diameter of a N2 molecule, n is the number density of N2 molecules, m is the mass of N2, k is the Boltzmann constant, T is the temperature, and λmfp is the mean free path of the N2 molecules. Substitute the values of the constants and the variables given into the formula and solve.
[tex]{{\lambda }_{mfp}}=\frac{1}{\sqrt{2}\pi {{(3.64\times {{10}^{-10}}\text{ m})}^{2}}(2.52\times {{10}^{19}}\text{ molecules/m}^{3})}\sqrt{\frac{8(1.38\times {{10}^{-23}}\text{ J/K})(217\text{ K})}{\pi (28\text{ g/mol})(6.02\times {{10}^{23}}\text{ molecules/mol})}}=0.44\text{ }\mu \text{m}[/tex]
The collision frequency of N2 molecules at these conditions is given by the formula shown below.
[tex]{{Z}_{coll}}=n\sqrt{2}\pi {{d}^{2}}{{v}_{rms}}[/tex]
Where n is the number density of N2 molecules, d is the diameter of a N2 molecule, v is the rms speed of the N2 molecules, and Zcoll is the collision frequency of the N2 molecules. Substitute the values of the constants and the variables given into the formula and solve.
[tex]{{Z}_{coll}}=(2.52\times {{10}^{19}}\text{ molecules/m}^{3})\sqrt{2}\pi {{(3.64\times {{10}^{-10}}\text{ m})}^{2}}(467\text{ m/s})=3.5\times {{10}^{8}}\text{ collisions/s}[/tex]
b) The mean free path of N2 molecules at 25°C and 1 atmosphere pressure is given by the formula shown below.
[tex]{{\lambda }_{mfp}}=\frac{1}{\sqrt{2}\pi {{d}^{2}}n}\sqrt{\frac{8kT}{\pi m}}[/tex]
Where d is the diameter of a N2 molecule, n is the number density of N2 molecules, m is the mass of N2, k is the Boltzmann constant, T is the temperature, and λmfp is the mean free path of the N2 molecules.
Substitute the values of the constants and the variables given into the formula and solve.
[tex]{{\lambda }_{mfp}}=\frac{1}{\sqrt{2}\pi {{(3.64\times {{10}^{-10}}\text{ m})}^{2}}(2.69\times {{10}^{25}}\text{ molecules/m}^{3})}\sqrt{\frac{8(1.38\times {{10}^{-23}}\text{ J/K})(298\text{ K})}{\pi (28\text{ g/mol})(6.02\times {{10}^{23}}\text{ molecules/mol})}}=68\text{ nm}[/tex]
The probability that a nitrogen molecule at 25°C and 1 atmosphere pressure will travel 100 nm without undergoing a collision is calculated using the exponential function, as shown below.
[tex]{{P}_{coll}}={{e}^{-\frac{x}{{{\lambda }_{mfp}}}}}[/tex]
Where x is the distance travelled by a nitrogen molecule, and Pcoll is the probability that a nitrogen molecule at 25°C and 1 atmosphere pressure will travel 100 nm without undergoing a collision. Substitute the values of the constants and the variables given into the formula and solve.
[tex]{{P}_{coll}}={{e}^{-\frac{100\text{ nm}}{68\text{ nm}}}}[/tex]=0.34 or 34%
Therefore, at a temperature of 25°C and a pressure of 1 atmosphere, the likelihood of a nitrogen molecule traveling a distance of 100 nm without experiencing a collision is 34%.
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what is reduced and oxidized in the following reaction? 3 h2s (aq) 2 h (aq) 2 no3−(aq) → 3 s(s) 2 no(g) 4 h2o (ℓ)
In the given chemical reaction, H2S is being oxidized to sulfur and NO3− is being reduced to NO gas.
The given balanced chemical equation is:3 H2S (aq) + 2 H (aq) + 2 NO3−(aq) → 3 S(s) + 2 NO(g) + 4 H2O (ℓ)Oxidation is the process in which a chemical species loses electrons in a chemical reaction. Reduction is the process in which a chemical species gains electrons in a chemical reaction.Here, H2S undergoes oxidation and NO3− undergoes reduction. So, H2S is being oxidized to sulfur and NO3− is being reduced to NO gas. The reducing agent in the reaction is H2S while the oxidizing agent is NO3−.
The oxidation and reduction half-reactions can be written as follows:Oxidation half-reaction:H2S → S + 2 H+ + 2 e-Reduction half-reaction:NO3- + 4 H+ + 3 e- → NO + 2 H2OBy adding the two half-reactions, we get the overall equation as shown below:3 H2S (aq) + 2 H+ (aq) + 2 NO3−(aq) → 3 S(s) + 2 NO(g) + 4 H2O (ℓ).
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Complete and balance each of the following equations for gas evolution reactions. hclo4(aq) k2co3(aq)→hclo4(aq) k2co3(aq)→ express your answer as a chemical equation. identify all of the phases in your answer.
The chemical equation for gas evolution reactions involving HClO4(aq) and K2CO3(aq) can be written as follows:HClO4(aq) + K2CO3(aq) → KClO4(aq) + CO2(g).
The above equation is already balanced. It has H, Cl, K, C, and O atoms on both sides of the equation. The states of matter are also included in the above equation. The chemical equation for gas evolution reactions involving HClO4(aq) and K2CO3(aq) can be written as follows:HClO4(aq) + K2CO3(aq) → KClO4(aq) + CO2(g).
The reactants are aqueous (HClO4(aq) and K2CO3(aq)), while the product KClO4(aq) is also aqueous. Only CO2(g) is gaseous. The above equation is already balanced. It has H, Cl, K, C, and O atoms on both sides of the equation. The states of matter are also included in the above equation.
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determine the number of σ bonds and π bonds in each of the molecules.
The two carbon atoms in the ethylene molecule are connected to one another through a π bond. Therefore, the ethylene molecule has three σ bonds and one π bond.
In chemistry, pi and sigma bonds are types of covalent bonds. Sigma bonds result from the interaction of two atomic orbitals, whereas pi bonds result from the interaction of two atomic orbitals overlapping in a lateral or sideways manner to a significant degree. In a molecule, the number of sigma and pi bonds can be calculated to determine the bonding. In this case, the number of σ bonds and π bonds in each of the molecules needs to be determined.In chemistry, the sigma bond is a type of covalent bond in which two atomic orbitals overlap in such a way that their electron density is concentrated in the region along the line connecting the two atomic nuclei. In a molecule, the sigma bond is represented by the Greek letter σ (sigma).Pi bond is a type of covalent bond that occurs between two atoms that are not directly bonded to one another. Pi bonds result from the overlap of two sets of unhybridized atomic orbitals that are parallel to one another. In a molecule, the pi bond is represented by the Greek letter π (pi).For example, in ethylene molecule, C2H4, each carbon atom is connected to two hydrogen atoms and one carbon atom. Therefore, each carbon atom in the ethylene molecule is connected to three atoms through a total of three σ bonds. The two carbon atoms in the ethylene molecule are connected to one another through a π bond. Therefore, the ethylene molecule has three σ bonds and one π bond.
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If the unknown copper salt you worked with was copper (I) instead of copper (II), would the calculated formula weight of the unknown salt change?
a) Yes, it would increase
b) Yes, it would decrease
c) No, it would remain the same
If the unknown copper salt you worked with was copper (I) instead of copper (II), the calculated formula weight of the unknown salt would decrease.
The formula weight is computed by summing the atomic weights of the atoms in the formula and multiplying by the subscript of each element.
The formula weight is influenced by the valence of the ion and the charge on the ion in the formula. Copper (I) has a charge of +1, whereas copper (II) has a charge of +2.
This means that if we're dealing with copper (I) rather than copper (II), we'll have a lower charge in the formula.
To further illustrate, let's consider the example of CuCl. Copper(I) has a charge of +1, while chlorine has a charge of -1. Copper(II), on the other hand, has a charge of +2, while chlorine has a charge of -1.
As a result, CuCl would have a formula weight of 63.5 + 35.5 = 99, whereas CuCl2 would have a formula weight of 63.5 + 2(35.5) = 134.5.
As a result, if we replaced CuCl2 with CuCl, the formula weight would decrease.Answer: b) Yes, it would decrease
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what is the ph of a 0.125 m solution of barium butyrate at 25 °c?
The pH of a 0.125 M solution of barium butyrate at 25 °C is not readily determined without additional information.
To determine the pH of a solution, we need to know the nature of the compound and its dissociation behavior in water. Barium butyrate is a salt composed of the metal barium and the butyrate anion. Without specific information about the dissociation of barium butyrate in water and the presence of any acid-base reactions, we cannot directly calculate the pH of the solution.
However, we can make some general observations. Barium butyrate is a salt formed by the reaction of barium hydroxide (a strong base) and butyric acid (a weak acid). The barium ion (Ba²⁺) is the conjugate acid of a strong base, and the butyrate ion (C₄H₇O₂⁻) is the conjugate base of a weak acid.
Therefore, the solution of barium butyrate may have a slightly basic pH due to the presence of the barium hydroxide. However, the extent of this basicity will depend on the concentration of the barium hydroxide and the degree of dissociation of butyric acid.
In conclusion, without specific information about the dissociation behavior of barium butyrate and the presence of other acids or bases in the solution, the pH of a 0.125 M solution of barium butyrate at 25 °C cannot be determined accurately.
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The pH of a 0.125 M solution of barium butyrate at [tex]25^0C[/tex] depends on the dissociation of the compound in water, which can be determined using the ionization constant (Ka) and the concentration of the solution.
The pH of a solution is a measure of its acidity or basicity and is determined by the concentration of hydrogen ions ([tex]H^+[/tex]) present in the solution. To calculate the pH of a 0.125 M solution of barium butyrate, we need to consider the dissociation of the compound in water. Barium butyrate is a salt that dissociates into its constituent ions in solution, including the barium ion ([tex]Ba^2^+[/tex]) and the butyrate ion ([tex]C_4H_7O_2^-[/tex]).
To calculate the pH, we need to know the ionization constant (Ka) of butyric acid, the parent acid of butyrate. Assuming that the butyrate ion acts as a weak base, we can use the Ka value to determine the concentration of hydroxide ions ([tex]OH^-[/tex]) in the solution. From there, we can calculate the concentration of [tex]H^+[/tex] ions and convert it into pH.
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calculate the number of atp generated from one saturated 12 ‑carbon fatty acid. assume that each nadh molecule generates 2.5 atp and that each fadh2 molecule generates 1.5 atp .
From one saturated 12-carbon fatty acid, approximately 75 ATP molecules can be generated through the process of β-oxidation and the subsequent citric acid cycle.
To calculate the number of ATP generated from one saturated 12-carbon fatty acid, we need to consider the process of β-oxidation, which breaks down the fatty acid into acetyl-CoA units.
In β-oxidation, each round produces:
- One NADH molecule
- One FADH₂ molecule
- One acetyl-CoA unit
For a 12-carbon fatty acid, there will be six rounds of β-oxidation as each round removes two carbon units (acetyl-CoA). Therefore, we will have six NADH and six FADH₂ molecules generated.
Now, let's calculate the total number of ATP generated:
- Each NADH molecule generates 2.5 ATP.
- Each FADH2 molecule generates 1.5 ATP.
Total ATP from NADH: 6 NADH × 2.5 ATP/NADH = 15 ATP
Total ATP from FADH2: 6 FADH2 × 1.5 ATP/FADH2 = 9 ATP
Additionally, each acetyl-CoA unit enters the citric acid cycle (Krebs cycle) where it undergoes further oxidation, producing three NADH molecules, one FADH₂ molecule, and one GTP molecule (which is equivalent to one ATP molecule).
Total ATP from acetyl-CoA units: 6 acetyl-CoA × 3 NADH × 2.5 ATP/NADH = 45 ATP
6 acetyl-CoA × 1 FADH2 × 1.5 ATP/FADH₂ = 9 ATP
6 acetyl-CoA × 1 GTP = 6 ATP
Adding up all the ATP generated:
15 ATP (from NADH) + 9 ATP (from FADH₂) + 45 ATP (from acetyl-CoA) + 6 ATP (from GTP) = 75 ATP
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what mass of na2so4 is needed to prepare 350. ml of a solution having a sodium ion concentration of 0.125 m? 24.98 12.4g 8.88 g 03.11g 6.218
The mass of Na2SO4 that is needed to prepare a 350 ml solution having a sodium ion concentration of 0.125 M is 6.218 g.
Given: Volume of the solution = 350 mL = 0.350 LConcentration of sodium ion = 0.125 mWe have to find the mass of Na2SO4 required to prepare this solution.Let the mass of Na2SO4 be ‘m’.The molar mass of Na2SO4 = 142.04 g/molNumber of moles of sodium ions in the solution = 0.125 mMolarity = (Number of moles of solute) / (Volume of solution in liters)0.125 = (2 × number of moles of Na ions) / 0.350Lnumber of moles of Na ions = 0.125 × 0.350 / 2 = 0.021875 molSince 1 mol of Na2SO4 contains 2 moles of Na ions.
Number of moles of Na2SO4 = 0.021875 / 2 = 0.010938 molmass of Na2SO4 required = Number of moles of Na2SO4 × Molar mass of Na2SO4= 0.010938 × 142.04= 1.551 g ≈ 6.218 gSo, the mass of Na2SO4 that is needed to prepare a 350 ml solution having a sodium ion concentration of 0.125 M is 6.218 g.
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at a particular temperature, a sample of pure water has a kw of 1.7×10−12. what is the hydroxide concentration of this sample?
The hydroxide concentration of this sample can be calculated using the expression; Kw = [OH-] [H+].Explanation:We know that the value of Kw for water is 1.0 x 10^-14. The expression Kw = [OH-] [H+] can be re-written as [OH-] = Kw / [H+].
The [H+] in pure water is equal to the [OH-] concentration (as pure water is neutral). So, the hydroxide concentration of this sample can be found by solving the equation;Kw = [OH-] [H+]= [OH-]^2 [OH-] = Kw / [H+]= 1.0 x 10^-14 / [H+][OH-] = sqrt (1.0 x 10^-14 / [H+]) = sqrt(1.7 × 10^-12)= 1.30 × 10^-6 M The hydroxide concentration of this sample is 1.30 × 10^-6 M. The value you mentioned, 1.7×10^(-12), refers to the equilibrium constant for the self-ionization of water, which is also known as the ion product of water (Kw).
In pure water, the concentration of hydroxide ions (OH-) is equal to the concentration of hydronium ions (H3O+), and both are denoted as [OH-] = [H3O+].
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During stress or trauma, a person can start to hyperventilate. The person may then be instructed to breathe into a paper bag to avoid fainting. CO2(g) + 2H20(1) -- H30+ (aq) + HCO3 (aq) Part A How does blood pH change during hyperventilation? Hyperventilation will lower the CO2 level in the blood, which decreases the H30and increases the blood pH O Hyperventilation will increase the CO2 level in the blood, which increases the H,0* and decreases the blood pH O Hyperventilation does not impact CO2 levels in the blood, thus H3O+ levels remain the same and pH does not change.
During hyperventilation, the blood pH changes as hyperventilation will lower the CO₂ level in the blood, which decreases the H₃₀⁺ and increases the blood pH.
The condition in which a person breathes more rapidly than usual is known as hyperventilation. During stress or trauma, a person can start to hyperventilate. This leads to a rise in the volume of oxygen (O₂) in the blood and a decrease in the volume of carbon dioxide (CO₂).
This lowers the quantity of CO₂ dissolved in the bloodstream. The acid-base balance of the body is governed by the concentration of hydrogen ions (H⁺) in the blood. Blood pH refers to the balance of acids and bases in the bloodstream, which is critical for maintaining normal bodily function.In order to prevent fainting, a person who is hyperventilating may be instructed to breathe into a paper bag.
When a person breathes into a paper bag, they are inhaling the CO₂ that they exhaled previously, which aids in the restoration of normal CO₂ levels in the bloodstream and aids in the maintenance of a healthy pH balance of the blood.
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During stress or trauma, a person can start to hyperventilate. The person may then be instructed to breathe into a paper bag to avoid fainting. The correct option among the given options is "Hyperventilation will lower the CO2 level in the blood, which decreases the H30+ and increases the blood pH".
During stress or trauma, a person can start to hyperventilate. Hyperventilation is a state of rapid breathing where a person breathes faster than the normal rate. As a result of hyperventilation, the concentration of carbon dioxide (CO2) in the blood decreases, which reduces the level of H+ ions and increases pH. Hyperventilation will lower the CO2 level in the blood, which decreases the H30+ and increases the blood pH. This is because
CO2(g) + 2H2O(l) -- H30+ (aq) + HCO3- (aq)
Hyperventilation decreases the concentration of CO2 in the blood, which leads to the reaction shifting to the left. This results in a decrease in the concentration of H3O+ and an increase in pH. As a result, the blood becomes more alkaline (basic) when a person hyperventilates.
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For a gas obeying the van der Waals equation of state, (a) show that (ac,lav)t = 0. (b) develop an expression for Cp Cu (c) develop expressions for [u(T2, v2) u(T1, v1)] and [s(T2, 02) – s(T1, v)]. (d) complete the Au and As evaluations if c, = a + bT, where a and b are constants.
To show that (αc,lav)t = 0 for a gas obeying the van der Waals equation of state, we need to start with the definition of the compressibility factor (Z): Z = (PV) / (RT) where P is the pressure, V is the molar volume, R is the gas constant, and T is the temperature. For the van der Waals equation of state, we have: P = (RT) / (V - b) - (a / V^2).
Now, let's differentiate the equation with respect to T, assuming constant volume (V): (dP / dT)v = R / (V - b) = 0
Since we have assumed constant volume, the derivative of pressure with respect to temperature is zero. Therefore, (αc,lav)t = 0. To develop an expression for Cp and Cu (specific heat at constant pressure and constant volume), we need to start with the first law of thermodynamics:dU = dq - PdV
where dU is the change in internal energy, dq is the heat transferred, P is the pressure, and dV is the change in volume. At constant volume (V), we have: dU = dqv
And at constant pressure (P), we have: dU = dq - PdV = dq - RdT = dqv + PdV - RdT = dqv + VdP
Since dU = dqv + VdP, we can express Cp and Cu as follows:
Cp = (dU / dT)p = (dqv + VdP) / dT
Cu = (dU / dT)v = (dqv) / dT
To develop expressions for [u(T2, v2) - u(T1, v1)] and [s(T2, v2) - s(T1, v1)], we need to consider the internal energy (u) and entropy (s) of the gas. Using the first law of thermodynamics, we can write:du = dq - Pdv
ds = dq / T
For a reversible process, dq = du + Pdv, which can be expressed as:
dq = Cp dT
Substituting this in the equation for ds, we have: ds = Cp dT / T
Integrating the above equation, we get:
s(T2, v2) - s(T1, v1) = ∫[Cp(T) / T] dT
For the expression [u(T2, v2) - u(T1, v1)], we can integrate the equation du = dq - Pdv using the appropriate equations of state for the gas.
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for the reactionkclo4⟶kcl 2o2 assign oxidation numbers to each element on each side of the equation.k in kclo4: k in kcl: cl in kclo4: cl in kcl: o in kclo4: o in o2:
To determine the oxidation state of a chemical species, we must follow the rules mentioned below:Oxygen atoms have an oxidation state of -2 in almost all compounds.
Potassium has an oxidation state of +1.The total charge of a neutral compound is zero.The total charge of an ion is equal to its net charge.The sum of the oxidation numbers for all atoms in a molecule is equal to the charge on the molecule.
When we apply the above rules to the given chemical equation kcLO4 ⟶ KCl + 2O2, we obtain the following oxidation states: k in KClO4: +1 k in KCl: +1 Cl in KClO4: +7 Cl in KCl: -1 O in KClO4: -2 O in O2: 0
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list all of the intermolecular forces present in the following molecules: a.) ch3ch2oh
A) CH₃CH₂OH: Hydrogen bonding, dipole-dipole interactions, and London dispersion forces.
B) CCl₄: London dispersion forces.
C) CHF₁₀: Dipole-dipole interactions and London dispersion forces.
A) CH₃CH₂OH (ethanol):
- Hydrogen bonding: Ethanol contains a hydrogen atom bonded to an electronegative oxygen atom, allowing for hydrogen bonding between ethanol molecules.
- Dipole-dipole interactions: Ethanol is a polar molecule, with the oxygen atom being more electronegative than carbon and hydrogen. This results in dipole-dipole interactions between ethanol molecules.
- London dispersion forces: Ethanol also experiences London dispersion forces, which arise from temporary fluctuations in electron distribution.
B) CCl₄ (carbon tetrachloride):
- London dispersion forces: Carbon tetrachloride is a nonpolar molecule, so the only intermolecular force present is London dispersion forces. The electron distribution in CCl₄ is symmetrical, resulting in no net dipole moment.
C) CHF₁₀ (tetrafluoromethane):
- Dipole-dipole interactions: Tetrafluoromethane is a polar molecule, with the fluorine atoms being more electronegative than carbon and hydrogen. This leads to dipole-dipole interactions between CHF10 molecules.
- London dispersion forces: CHF₁₀ also experiences London dispersion forces, as all molecules do.
The complete question is:
List all of the intermolecular forces present in the following molecules: A.) CH₃CH₂OH B.) CCl₄ C.) CHF₁₀.
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draw a possible structure for the ir spectrum given below. the molecular formula of the compound is c4h6o2.
Here is the possible structure for the IR spectrum of the given molecular formula C4H6O2. The given molecular formula C4H6O2 suggests that the given compound may contain a functional group called carbonyl group -C=O, which absorbs IR radiation in the range of 1600-1700 cm⁻¹.The given IR spectrum is as follows.
IR spectrum, The given spectrum shows the following peaks:• A strong, broad peak at around 3200-3400 cm⁻¹ that indicates the presence of an OH group, which is a characteristic of carboxylic acids and phenols.• A strong peak at around 1710 cm⁻¹ that indicates the presence of a carbonyl group, which is a characteristic of aldehydes, ketones, and carboxylic acids.• A weak peak at around 910-970 cm⁻¹ that indicates the presence of an alkene group, which is a characteristic of C=C stretching vibrations of alkenes. (It is a weak peak because it is not a prominent functional group of the given molecular formula, C4H6O2.)
From the above IR spectrum and the given molecular formula, C4H6O2, we can infer that the given compound is an unsaturated carboxylic acid. The carbonyl group is attached to one end of the carbon chain and an OH group is attached to the other end. Since it contains four carbon atoms, it must be a butanoic acid. Thus, the possible structure for the given IR spectrum is as follows, The carbonyl group (-C=O) absorbs IR radiation at around 1710 cm⁻¹.• The OH group (-OH) absorbs IR radiation at around 3200-3400 cm⁻¹.
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Calculate the pH of a solution formed by mixing 150.0 mL of 0.10 M HC7H5O2 with 100.0 mL of 0.30 M NaC7H5O2. The Ka for HC7H5O2 is 6.5 x 10-5.
O 4.31
O 4.49
O 10.51
O 4.19
O 9.69
The pH of the solution formed by mixing 150.0 mL of 0.10 M HC₇H₅O₂ with 100.0 mL of 0.30 M NaC₇H₅O₂ and the given Ka value can be calculated using the Henderson-Hasselbalch equation. The correct answer is 4.19.
The Henderson-Hasselbalch equation is given by:
pH = pKa + log([A-]/[HA])
Where:
pH is the logarithmic scale of the hydrogen ion concentration,
pKa is the negative logarithm of the acid dissociation constant,
[A-] is the concentration of the conjugate base (acetate, C₇H₅O₂⁻), and
[HA] is the concentration of the acid (acetic acid, HC₇H₅O₂).
First, we need to determine the concentrations of the acetate ion (C₇H₅O₂⁻) and acetic acid (HC₇H₅O₂) in the final solution after mixing.
The initial moles of HC₇H₅O₂ are:
moles_HC₇H₅O₂ = volume_HC₇H₅O₂ × concentration_HC₇H₅O₂
= 150.0 mL × 0.10 M
= 15.0 mmol
The initial moles of NaC₇H₅O₂ are:
moles_NaC₇H₅O₂ = volume_NaC₇H₅O₂ × concentration_NaC₇H₅O₂
= 100.0 mL × 0.30 M
= 30.0 mmol
Since the reaction between HC₇H₅O₂ and NaC₇H₅O₂ is a 1:1 ratio, the moles of HC₇H₅O₂ remaining after reaction will be equal to the moles of NaC₇H₅O₂ reacted. Therefore, the final concentration of HC₇H₅O₂ will be (15.0 - 30.0) mmol.
Next, we can calculate the concentration of the acetate ion (C₇H₅O₂⁻) by adding the moles of NaC₇H₅O₂ and the remaining moles of HC₇H₅O₂ and dividing by the total volume:
concentration_A- = (moles_NaC₇H₅O₂ + moles_HC₇H₅O₂ remaining) / (volume_NaC₇H₅O₂ + volume_HC₇H₅O₂)
= (30.0 mmol + (15.0 - 30.0) mmol) / (100.0 mL + 150.0 mL)
= 15.0 mmol / 250.0 mL
= 0.06 M
The concentration of the acid (acetic acid, HC₇H₅O₂) can be calculated using the remaining moles and the total volume:
concentration_HA = moles_HC₇H₅O₂ remaining / (volume_NaC₇H₅O₂ + volume_HC₇H₅O₂)
= (15.0 - 30.0) mmol / (100.0 mL + 150.0 mL)
= -15.0 mmol / 250.0 mL
= -0.06 M
(Note: The negative sign indicates that the concentration
is less than 0 M, which is expected since some of the acid has reacted.)
Now, we can substitute the values into the Henderson-Hasselbalch equation:
pH = pKa + log([A-]/[HA])
= [tex]-\log(6.5 \times 10^{-5}) + \log(0.06 / 0.06)[/tex]
= -(-4.19) + \log(1)
= 4.19 + 0
= 4.19
Therefore, the pH of the solution formed by mixing the two solutions is 4.19.
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the volume of hydrogen gas at 38.0 c and 763 torr that can be produced by the reaction of 4.33 gram of zinc excess sulfuric acid is ________L.
A. 1.69
B. 0.592
C. 3.69*10^4
d. 2.71*10^-4
e. 2.84
The volume of hydrogen gas at 38.0 °C and 763 torr that can be produced by the reaction of 4.33 gram of zinc excess sulfuric acid is 1.69 L.
Given; Mass of zinc = 4.33 grams. The balanced chemical equation for the reaction of zinc with excess sulfuric acid is; Zn(s) + H2SO4(aq) ⟶ ZnSO4(aq) + H2(g).
The volume of hydrogen gas produced can be calculated by using the ideal gas law equation; PV = nRT whereP = 763 torr V = ?n = number of moles of hydrogen gasR = gas constant = 0.0821 L atm/mol K (This is the value at 38.0 °C)T = temperature in Kelvin.
To find the number of moles of hydrogen gas, we need to first find the number of moles of zinc used.
Number of moles of zinc used = mass of zinc used/molar mass of zinc Molar mass of zinc (Zn) = 65.38 g/mol.
Number of moles of zinc used = 4.33/65.38 = 0.0662 moles. As per the balanced equation, 1 mole of zinc produces 1 mole of hydrogen gas.
Number of moles of hydrogen gas produced = 0.0662 moles. Volume of hydrogen gas produced; PV = nRTV = nRT/PV = (0.0662)(0.0821)(38 + 273.15)/763V = 1.69 L.
Therefore, the volume of hydrogen gas produced is 1.69 L.
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Which of the following pairs has the stronger acid listed first?
Question 9 options:
a) H2AsO3, H2AsO4
b) H2S, HCl
c) HI, HBr
d) H2SO3, H2SO4
e) HClO, HClO3
D: H[tex]_{2}[/tex]SO[tex]_{4}[/tex] and H[tex]_{2}[/tex]SO[tex]_{3}[/tex], has the stronger acid listed first.
The strength of an acid is determined by its ability to donate protons (H+ ions) in an aqueous solution. Generally, stronger acids have a greater tendency to donate protons.
In option D, H[tex]_{2}[/tex]SO[tex]_{4}[/tex] is sulfuric acid, which is a strong acid. It completely dissociates in water to release two H+ ions. On the other hand, H[tex]_{2}[/tex]SO[tex]_{3}[/tex] is sulfurous acid, which is a weak acid. It only partially dissociates in water to release fewer H+ ions.
Therefore, H[tex]_{2}[/tex]SO[tex]_{4}[/tex] is the stronger acid compared toH[tex]_{2}[/tex]SO[tex]_{3}[/tex].
Option D is the correct answer.
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The correct option is: H2SO4, H2SO3.H2SO4, H2SO3, and is the pair that has the stronger acid listed first.
The strength of an acid is identified by its tendency to donate protons. A stronger acid is one that has a greater tendency to donate protons than a weaker acid. H2SO4 and H2SO3 are pairs, and H2SO4 is a stronger acid than H2SO3 because it has a greater tendency to donate protons. The strength of an acid is determined by its ability to donate hydrogen ions. A stronger acid readily donates hydrogen ions, leading to a higher concentration of H+ in solution. In the case of H2SO4, it has two ionizable hydrogen atoms, making it a diprotic acid, while H2SO3 has only one ionizable hydrogen atom, making it a monoprotic acid. Hence, H2SO4 is the stronger acid of the two. Therefore, the answer to the question is H2SO4, H2SO3 as the stronger acid is listed first.
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.1. A 1.00 L buffer solution is 0.150 M in HC7H5O2 and 0.250 M in LiC7H5O2. Calculate the pH of the solution after the addition of 100.0 mL of 1.00 M HCl.
The Ka for HC7H5O2 is 6.5 x 10^-5
2. A 1.50 L buffer solution is 0.250 M in HF and 0.250 M NaF. Calculate the pH of the solution after the addition of 0.0500 moles of solid NaOH. Assume no volume change upon the addition of base.
The Ka for HF is 3.5 x 10^-4
3. Calculate the pH of a solution formed by mixing 100.0 mL of 0.20 M HClO with 200.0 mL of 0.30 M KClO.
The Ka for HClO is 2.9 x 10^-8
The pH of a buffer solution can be calculated using the Henderson-Hasselbalch equation, which relates the concentrations of the conjugate acid and conjugate base in the buffer to the pH of the solution.
How can the pH of a buffer solution be calculated after the addition of an acid or base?1. For the first question, the addition of HCl to the buffer solution will result in the reaction between HCl and the weak acid HC7H5O2, forming its conjugate base C7H5O2-.
The concentration of HC7H5O2 will decrease while the concentration of C7H5O2- will increase.
To calculate the pH, we need to determine the new concentrations of HC7H5O2 and C7H5O2- after the reaction.
Using the Henderson-Hasselbalch equation, pH = pKa + log([A-]/[HA]), where [A-] is the concentration of the conjugate base and [HA] is the concentration of the weak acid, we can substitute the given values into the equation to calculate the pH.
2. For the second question, the addition of solid NaOH will react with the weak acid HF in the buffer solution, forming its conjugate base F-.
The moles of NaOH added will be consumed by an equal amount of moles of HF, resulting in a decrease in the concentration of HF and an increase in the concentration of F-.
To calculate the pH, we can again use the Henderson-Hasselbalch equation, substituting the new concentrations of HF and F- into the equation.
3. For the third question, the solution is formed by mixing two strong electrolytes, HClO and KClO.
Since both HClO and KClO dissociate completely in water, we can calculate the concentration of H+ ions using the stoichiometry of the balanced equation. The pH can be determined by taking the negative logarithm of the concentration of H+ ions.
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If 0.452 moles of OH1- are reacted with an unlimited supply of Br2, how many grams of H2O can be formed?
Answer:
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For a certain chemical reaction, the standard Gibbs free energy of reaction at 5.00 °C is 105. kJ. Calculate the equilibrium constant K for this reaction. Round your answer to 2 significant digits.
The equilibrium constant (K) for this reaction is approximately 6.3 × 10^19.
To calculate the equilibrium constant (K) for a chemical reaction using the standard Gibbs free energy of reaction (∆G°), we use the equation:
∆G° = -RT ln(K)
where:
∆G° is the standard Gibbs free energy of reaction
R is the gas constant (8.314 J/(mol·K) or 0.008314 kJ/(mol·K))
T is the temperature in Kelvin
K is the equilibrium constant
In this case, we have ∆G° = 105 kJ and the temperature is 5.00 °C, which needs to be converted to Kelvin. The conversion from Celsius to Kelvin is done by adding 273.15 to the Celsius value.
T = 5.00 °C + 273.15 = 278.15 K
Now we can plug these values into the equation and solve for K:
105 kJ = -0.008314 kJ/(mol·K) × 278.15 K × ln(K)
To solve for K, we need to rearrange the equation:
ln(K) = -105 kJ / (-0.008314 kJ/(mol·K) × 278.15 K)
ln(K) = 45.402
Now, we can calculate K by taking the exponent of both sides:
K = e^(ln(K)) = e^(45.402)
K ≈ 6.3 × 10^19
Therefore,
Rounding to 2 significant digits, the equilibrium constant (K) for this reaction is approximately 6.3 × 10^19.
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estimate the molar mass of a gas that effuses at 1.6 times the rate of carbon dioxide (co2)
The estimated molar mass of the gas is 17.25 g/mol.
The rate of effusion for a gas is inversely proportional to the square root of its molecular weight. As a result, if a gas effuses faster than another gas, it has a smaller molecular weight (mass).
The ratio of the effusion rates of the two gases can be used to estimate the ratio of their molecular masses. The molecular mass of carbon dioxide (CO2) is 44 g/mol, for example. Assume that the gas has an effusion rate of 1.6 times that of CO2.
According to the equation,
rate of effusion ∝ 1 / √(molecular weight)
For two different gases 1 and 2, the ratio of the rates of effusion is given as: rate of effusion for gas 1 / rate of effusion for gas 2 = √(molecular weight of gas 2) / √(molecular weight of gas 1)
Solving for the molecular weight of the gas, we get:
Molecular weight of gas = Molecular weight of CO2 × (rate of effusion of CO2 / rate of effusion of gas)²= 44 × (1 / 1.6)²= 44 × 0.390625= 17.25
Therefore, the estimated molar mass of the gas is 17.25 g/mol.
The estimated molar mass of the gas is 17.25 g/mol.
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a rydberg atom is an atom whose valence electrons are in states with a very large principal quantum number n. this means it has a probability cloud with a large amplitude a large distance from the nucleus. evidence of such atoms has been detected by radio astronomers in the form of radiation from diffuse hydrogen gas in intersellar space. in fact, there is no theoritical limit on the size an atom can attain, provided it is free from outside influences. 1)what is the smallest value of n such that the bohr radius of a single such hydrogen atom would be greater than 7 microns, roughly the size of a typical single-celled organism. ee
A Rydberg atom is an atom with valence electrons in states with a very high principal quantum number n. The smallest value of n such that the Bohr radius of a single hydrogen atom would be greater than 7 microns is approximately 1,573.
This implies that it has a probability cloud with a high amplitude and is situated at a significant distance from the nucleus. Such atoms' existence has been discovered by radio astronomers through radiation from diffuse hydrogen gas in interstellar space. The largest size an atom can achieve is unlimited, given it is free of external effects.A single-celled organism's typical size is 7 microns. To determine the smallest value of n where the Bohr radius of a single hydrogen atom would be greater than 7 microns,
we will use the following formula for Bohr radius:
r = n²h²/4π²me²k where h is Planck's constant, me is the electron's mass, and k is Coulomb's constant.
We can see that the Bohr radius increases as n². We can therefore express the equation as:n² = 4π²me²k * r / h²When r = 7 µm, we can plug it into the equation and solve for n as:n² = 4π²(9.10938356 × 10^-31 kg) (8.9875517923 × 10^9 N m²/C²)(7 × 10^-6 m) / (6.62607015 × 10^-34 m² kg/s)²n² = 2,469,471.663n ≈ 1,572.90
Therefore, the smallest value of n such that the Bohr radius of a single hydrogen atom would be greater than 7 microns is approximately 1,573.
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Insoluble sulfide compounds are generally black in color. Which of the following combinations could yield a black precipitate?
Check all that apply.
A. Na2S(aq)+KCl(aq)
B. Li2S(aq)+Pb(NO3)2(aq)
C. Pb(ClO3)2(aq)+NaNO3(aq)
D. AgNO3(aq)+KCl(aq)
E. K2S(aq)+Sn(NO3)4(aq)
The combinations that could potentially yield a black precipitate are (A) Na₂S(aq) + KCl(aq) and (B) Li₂S(aq) + Pb(NO₃)₂(aq).
Insoluble sulfide compounds are known for their black color when formed as precipitates. When sulfide ions (S²⁻) are combined with certain cations, they can form insoluble sulfide compounds.
In option A, the reaction between Na₂S(aq) (sodium sulfide) and KCl(aq) (potassium chloride) can result in the formation of an insoluble black sulfide precipitate. Similarly, in option B, the reaction between Li₂S(aq) (lithium sulfide) and Pb(NO₃)₂(aq) (lead(II) nitrate) can lead to the formation of a black precipitate of lead sulfide.
Options C, D, and E do not involve the combination of sulfide ions with cations that typically form insoluble sulfides. Therefore, they would not yield a black precipitate.
In summary, options A. (Na₂S(aq) + KCl(aq)) and B. (Li₂S(aq) + Pb(NO₃)₂(aq)) have the potential to produce black precipitates due to the formation of insoluble sulfide compounds.
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20.00 mL of a H2SO4 solution with an unknown concentration was titrated to a phenolphthalein endpoint with 30.85 mL of a 0.1095 M NaOH solution. What is the concentration of the H2SO4 solution?
The concentration of a 20.00 mL of a H₂SO₄ solution with an unknown concentration was titrated to a phenolphthalein endpoint with 30.85 mL of a 0.1095 M NaOH solution is 84.39 M.
Phenolphthalein is an indicator which turns colourless in acidic medium and pink in basic medium. Therefore, when all the H₂SO₄ has reacted with NaOH to form H₂SO₄ and H₂O, the phenolphthalein will change colour from colourless to pink. This is the end point. We can write the balanced chemical equation as:
H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O
To find out the number of moles of NaOH used.
Number of moles of NaOH used = Molarity × volume
= 0.1095 M × 30.85 × 10⁻³ L
= 0.003375675 mol
Now, according to the balanced chemical equation, one mole of H₂SO₄ reacts with two moles of NaOH. Therefore,
Number of moles of H₂SO₄ = 0.003375675 ÷ 2 = 0.0016878375 mol
Volume of H₂SO₄ = 20.00 mL = 20.00 × 10⁻³ L
Concentration of H₂SO₄ = Number of moles ÷ volume= 0.0016878375 mol ÷ 20.00 × 10⁻³ L= 84.39 M
Therefore, the concentration of the H₂SO₄ solution is 84.39 M.
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The concentration of the H2SO4 solution is 0.084432 M or 0.084 M
To find out the concentration of H2SO4, the balanced chemical equation of H2SO4 and NaOH will be used as shown below:
H2SO4(aq) + 2NaOH(aq) → Na2SO4(aq) + 2H2O(l)
From the equation above, the stoichiometry of H2SO4 and NaOH is 1:2.
Moles of NaOH= Molarity × Volume in litres
= 0.1095 M × 0.03085 L
= 0.00337728 mol
Moles of H2SO4= 1/2 × Moles of NaOH
= 1/2 × 0.00337728
= 0.00168864 mol
Therefore, the concentration of the H2SO4 solution is given by:
C= Number of moles/Volume in litres
= 0.00168864 mol/0.02000 L
= 0.084432 M
Consequently, the concentration of the H2SO4 solution is 0.084432 M or 0.084 M
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what is the density, in g/l, of propane gas (ch3ch2ch3) at 0°c and 808 mmhg? enter your answer to 3 significant figures. do not include units.
The density of propane gas (CH3CH2CH3) at 0°C and 808 mmHg is 2.16 g/L. Propane is a commonly used fuel gas that is stored in tanks or cylinders. Its density is influenced by temperature and pressure.
To calculate the density of propane gas, we can use the ideal gas law equation: [tex]PV = nRT[/tex], where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant, and T is the temperature in Kelvin. First, we need to convert the given temperature from Celsius to Kelvin by adding 273.15: 0°C + 273.15 = 273.15 K.
Next, we convert the pressure from mmHg to atm by dividing it by 760: 808 mmHg / 760 = 1.06316 atm. Since we want the answer in grams per liter (g/L), we need to solve for the molar volume, which is the volume occupied by one mole of gas at a given temperature and pressure. At standard temperature and pressure (STP), the molar volume of any gas is approximately 22.4 L/mol.
By rearranging the ideal gas law equation and substituting the known values, we can solve for the density: density = (mass of propane gas) / (volume of propane gas). We can assume that the molar mass of propane is approximately 44.1 g/mol.
Finally, using the calculated volume of 22.4 L/mol at STP and the molar mass of propane, we can find the density: density = (44.1 g/mol) / (22.4 L/mol) ≈ 1.964 g/L. Rounding to three significant figures, the density of propane gas at 0°C and 808 mmHg is approximately 2.16 g/L.
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1. Which of the following is in the correct order of standard state entropy? I. Liquid water < gaseous water II. Liquid water < solid water III. NH;
The correct order of standard state entropy is given as below: I. Gaseous water > Liquid water II. Solid water < Liquid water III. NH3 > N2H4
Entropy is an important concept of thermodynamics it is defined as the measure of disorder or randomness in a system. A system is said to be in a state of maximum entropy if its entropy is at a maximum and minimum entropy if its entropy is at a minimum. Standard entropy is defined as the entropy of a substance at its standard state, i.e., the most stable state at 1 atm and 25°C.The entropy of water can be represented in three states as gaseous water, liquid water, and solid water. I. Gaseous water has a higher entropy than liquid water. The reason for this is the gaseous water has more freedom of motion as compared to liquid water. Therefore, the entropy of gaseous water is higher than that of liquid water. II. Solid water has a lower entropy than liquid water. The reason for this is that the molecules in solid water have less freedom of motion as compared to liquid water.
Therefore, the entropy of solid water is lower than that of liquid water. III. NH3 has a higher entropy than N2H4. The reason for this is that the NH3 molecule has a higher number of particles as compared to the N2H4 molecule. Therefore, the entropy of NH3 is higher than that of N2H4.The correct order of standard state entropy is given as below: I. Gaseous water > Liquid water II. Solid water < Liquid water III. NH3 > N2H4
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Calculate [H3O+] in each aqueous solution at 25 °C, and classify each solution as acidic or basic.
a. [OH-] = 1.1 x 10-'M b. OH = 2.9 x 10-2 M c. |ОН | = 6.9 x 10-12 M
For [OH-] = 1.1 x 10-'M: The solution is basic as it has a pH value of 12.96.
For OH = 2.9 x 10-2 M: The solution is basic as it has a pH value of 12.46.
For |ОН | = 6.9 x 10-12 M: The solution is acidic as it has a pH value of 2.84.
To calculate the concentration of [H3O+] and classify each solution as acidic or basic, use the equation of
pH=-log[H3O+] to solve for the concentration of [H3O+].
For [OH-] = 1.1 x 10-'M:
The concentration of [H3O+] can be found by using the following formula:
pH=-log[H3O+]
pH=-log[1.1 x 10-'M]
pH=12.96
The solution is basic as it has a pH value of 12.96.
For OH = 2.9 x 10-2 M:
The concentration of [H3O+] can be found by using the following formula:
Kw = [H3O+][OH-]
=1.0 x 10^-14[H3O+]
= Kw/[OH-][H3O+]
= 1.0 x 10^-14/2.9 x 10^-2[H3O+]
= 3.45 x 10^-13 M
Next, use the following formula to find pH:
pH=-log[H3O+]
pH = -log[3.45 x 10^-13]
pH = 12.46
The solution is basic as it has a pH value of 12.46.
For |ОН | = 6.9 x 10-12 M:
The concentration of [H3O+] can be found by using the following formula:
Kw = [H3O+][OH-]
=1.0 x 10^-14[H3O+]
= Kw/[OH-][H3O+]
= 1.0 x 10^-14/6.9 x 10^-12[H3O+]
= 1.45 x 10^-3 M
Next, use the following formula to find pH:
pH=-log[H3O+]
pH = -log[1.45 x 10^-3]
pH = 2.84
The solution is acidic as it has a pH value of 2.84.
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a. The [H₃O⁺] in [OH-] = 1.1 × 10⁻¹¹ M at 25 °C is 8.23 × 10⁻¹² and the solution is acidic.
b. The [H₃O⁺] in [OH] = 2.9 × 10⁻² M at 25 °C is 2.46 × 10⁻¹³ and the solution is acidic.
c. The [H₃O⁺] in [OH] = 6.9 × 10⁻¹² M at 25 °C is 2.05 × 10⁻¹³ and the solution is acidic.
To find the hydronium ion concentration, we can use the relation between the concentration of hydroxide ions and hydronium ions
[H₃O⁺] × [OH₋] = 1.0 × 10⁻¹⁴
Taking logarithm on both sides,
log [H₃O⁺] + log [OH⁻]
= log 1.0 × 10⁻¹⁴
= -14log [H₃O⁺] = -log [OH⁻] - 14 ... (i)
a. When, [OH⁻] = 1.1 × 10⁻¹¹ M, then log [OH⁻] = -11.04
From equation (i),
log [H₃O⁺] = -log 1.1 × 10⁻¹¹ - 14
= 2.04 - 14
= -11.96
[H₃O⁺] = antilog (-11.96)
= 8.23 × 10⁻¹²
This indicates that the aqueous solution is acidic. Therefore, the answer is [H₃O⁺] = 8.23 × 10⁻¹² acidic.
b. When, [OH] = 2.9 × 10⁻² M, then log [OH] = -1.54
From equation (i),
log [H₃O⁺] = -log 2.9 × 10⁻² - 14
= 1.46 - 14
= -12.54
[H₃O⁺] = antilog (-12.54)
= 2.46 × 10⁻¹³
This indicates that the aqueous solution is acidic. Therefore, the answer is [H₃O⁺] = 2.46 × 10⁻¹³ acidic.
c. When, |OH| = 6.9 × 10⁻¹² M, then log |OH| = -11.16
From equation (i),
log [H₃O⁺] = -log 6.9 × 10⁻¹² - 14
= 1.16 - 14
= -12.84
[H₃O⁺] = antilog (-12.84)
= 2.05 × 10⁻¹³
This indicates that the aqueous solution is acidic. Therefore, the answer is [H₃O⁺] = 2.05 × 10⁻¹³ acidic.
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When 0.481 g of biphenyl C12H10 undergoes combustion in a bomb calorimeter, the temperature rises from 26.2 degrees Celsius to 30.3 degrees Celsius. Find ΔErxn for the combustion of biphenyl in kJ/mol. The heat capacity of the bomb calorimeter, determined in a separate experiment, is 5.86 kJ/degrees Celsius.
The given problem is about calculating the change in enthalpy of the combustion of biphenyl in a bomb calorimeter. The heat capacity of the bomb calorimeter, determined in a separate experiment, is 5.86 kJ/degrees Celsius. Let's find ΔErxn for the combustion of biphenyl in kJ/mol.
Solution: Given, Mass of biphenyl, C12H10 = 0.481 temperature change, ΔT = (30.3 - 26.2)°C = 4.1°CTo find the heat evolved in kJ/g, we use the formula = mCΔTwhereq is the heat evolved in kJ/gm is the mass of the substance burnt is the heat capacity in kJ/°CΔT is the change in temperature in °Substituting the given values = 0.481 g × 5.86 kJ/°C × 4.1°Cq = 11.36 kgotla number of moles of biphenyl present in 0.481 g of biphenyl: First, we need to find the molecular weight of biphenyl.
Molecular weight of biphenyl = (12 × 12) + (10 × 1) = 156 gmol-1One mole of biphenyl weighs 156 g0.481 g of biphenyl contain the following number of moles:n = 0.481 g / 156 gmol-1n = 3.08 × 10-3 molΔErxn = q/n = 11.36 kJ / 3.08 × 10-3 mol= 3687.01 kJ/mol Therefore, ΔErxn for the combustion of biphenyl in kJ/mol is 3687.01 kJ/mol.
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arrange the following ions in order of increasing ionic radius: potassium ion, chloride ion, phosphide ion, calcium ion.
The ions arranged in order of increasing ionic radius are: Phosphide ion (P³⁻) < Chloride ion (Cl⁻) < Potassium ion (K⁺) < Calcium ion (Ca²⁺).
To arrange the ions in order of increasing ionic radius, we need to consider the effective nuclear charge and the number of electron shells surrounding the ions.
As we move across a period in the periodic table, the effective nuclear charge increases, resulting in a smaller ionic radius. As we move down a group, the number of electron shells increases, leading to a larger ionic radius.
The given ions are:
1. Potassium ion (K⁺)
2. Chloride ion (Cl⁻)
3. Phosphide ion (P³⁻)
4. Calcium ion (Ca²⁺)
Arranging them in order of increasing ionic radius:
1. Phosphide ion (P³⁻): The phosphide ion has a larger ionic radius due to the addition of three extra electrons compared to the other ions.
2. Chloride ion (Cl⁻): The chloride ion has a smaller ionic radius compared to the phosphide ion as it has fewer electrons.
3. Potassium ion (K⁺): The potassium ion has a smaller ionic radius compared to chloride ion as it has lost an electron, resulting in a higher effective nuclear charge.
4. Calcium ion (Ca²⁺): The calcium ion has the smallest ionic radius among the given ions due to the higher effective nuclear charge and the loss of two electrons.
Therefore, the ions arranged in order of increasing ionic radius are: Phosphide ion (P³⁻) < Chloride ion (Cl⁻) < Potassium ion (K⁺) < Calcium ion (Ca²⁺).
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check the pair of species where the first member is smaller in diameter than the second member. select one: a. li < be2 b. al < al3 c. f < f- d. be < o
The pair of species where the first member is smaller in diameter than the second member is b. Al < Al3.
The given species are:li < Be2al < Al3F < F-Be < OIt is known that when the size of an atom increases, the diameter of the atom increases.
The atomic radius of Al3+ is less than the atomic radius of Al. This is due to the fact that Al3+ has lost three electrons and the nuclear charge has increased in proportion to the remaining number of electrons. The effective nuclear charge, which is the attractive force exerted by the nucleus on electrons, increases as a result of the increased nuclear charge.As a result, Al3+ has a smaller ionic radius than Al. Therefore, it can be seen that the first member of the species, Al, has a larger diameter than the second member, Al3+. Thus, the correct option is b. Al < Al3.
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