Answer:
The student failed to rinse the buret with KMnO₄ solution after rinsing it with distilled water.
Explanation:
In a titration, the equivalent moles of titrant must be the same than analyte. If the student calculates an amount of moles of H₂O₂ that is larger than the actual value:
The student failed to wear goggles FALSE. The use of goggles doesn't change the results in the lab but you must use it!
The student did not swirl the flask appropriately and therefore stopped short of the endpoint. FALSE. The stopped short of the endpoint would explain a lower concentration of the real value.
The student failed to rinse the buret with KMnO₄ solution after rinsing it with distilled water. TRUE. If you don't rinse the buret with KMnO₄ solution the concentration of this solution in the buret will be lower than real concentration doing you spend more titrant volume overestimating the amount of H₂O₂ moles.
The student added an extra 15 mL of distilled water to the H₂O₂ solution. FALSE. The addition of water doesn't change the amount of H₂O₂ moles in the solution.
I hope it helps!
When 18.9 kJ is transferred to a gas sample in a constant volume adiabatic container with a calorimeter constant of 2.22 Kj/K, the temperature of the gas (and the calorimeter) increases by 8.06 K. (a) What is the heat capacity of the sample? (b) If the sample has a mass of 0.5 kilograms, what is the specific heat capacity of the substance? (c) If the sample is Krypton, what is the molar heat capacity at constant volume of Krypton? The molar mass of Krypton is 83.8 grams/mole.
Answer:
(a) Cgas = 0.125 kJ/k
(b) cgas = 0.25kJ/kg.K
(c) cm(gas) = 0.021kJ/mol.K
Explanation:
18.9 kJ is equal to the sum of the heat absorbed by the gas and the heat absorbed by the calorimeter.
Qcal + Qgas = 18.9 kJ [1]
We can calculate the heat absorbed using the following expression.
Q = C . ΔT
where,
C is the heat capacity
ΔT is the change in the temperature
(a) What is the heat capacity of the sample?
From [1],
Ccal . ΔT + Cgas . ΔT = 18.9 kJ
(2.22kJ/K) × 8.06 K + Cgas × 8.06 K = 18.9 kJ
Cgas = 0.125 kJ/k
(b) If the sample has a mass of 0.5 kilograms, what is the specific heat capacity of the substance?
We can calculate the specific heat capacity (c) using the following expression:
[tex]c=\frac{C}{m} =\frac{0.125kJ/K}{0.5kg} =0.25kJ/kg.K[/tex]
(c) If the sample is Krypton, what is the molar heat capacity at constant volume of Krypton? The molar mass of Krypton is 83.8 grams/mole.
The molar heat capacity is:
[tex]\frac{0.25kJ}{kg.K} .\frac{1kg}{1000g} .\frac{83.8g}{mol} =0.021kJ/mol.K[/tex]
Final answer:
The heat capacity of the sample is 2.34 kJ/K. The specific heat capacity of the substance is 4.68 kJ/(kg*K). The molar heat capacity at constant volume of Krypton is 0.0559 kJ/(mol*K).
Explanation:
The heat capacity of a substance is the amount of energy required to increase its temperature by 1 degree. The specific heat capacity is the amount of energy required to increase the temperature of 1 gram of a substance by 1 degree.
(a) To find the heat capacity of the sample, we can use the equation Q = CΔT, where Q is the amount of heat transferred, ΔT is the change in temperature, and C is the heat capacity. Rearranging the equation, C = Q/ΔT. Plugging in the given values, C = 18.9 kJ / 8.06 K
= 2.34 kJ/K.
(b) To find the specific heat capacity of the substance, we need to know the mass of the sample. Given that the mass is 0.5 kilograms, we can use the equation cs = C/m, where cs is the specific heat capacity, C is the heat capacity, and m is the mass. Plugging in the values, cs = 2.34 kJ/K / 0.5 kg
= 4.68 kJ/(kg*K).
(c) To find the molar heat capacity at constant volume of Krypton, we can use the equation Cm = cs / M, where Cm is the molar heat capacity, cs is the specific heat capacity, and M is the molar mass. Plugging in the values, Cm = 4.68 kJ/(kg*K) / 83.8 g/mol
= 0.0559 kJ/(mol*K).
A sample of NI3 is contained in a piston and cylinder. The samples rapidly decomposes to form nitrogen gas and iodine gas, and releases 3.30 kJ of heat and does
950 J of work. What is ΔE?
Answer:
ΔE is -4250 J
Explanation:
Step1: Data given
3.30 kJ of heat is released
There is 950 J of work done
Step 2: Calculate ΔE
ΔE = q + w
⇒ with q = heat of energy released = 3.3 kJ = 3300 J ( negative because the heat is released)
⇒ work done = 950 J ( also negative )
ΔE = -3300J-950J= -4250 J
ΔE is -4250 J
Be sure to answer all parts. Enter your answers in scientific notation.
The following values are the only allowable energy levels of a hypothetical one-electron atom:
E6 = −2.0×10−19 J
E5 = −7.0×10−19 J
E4 = −11.0×10−19 J
E3 = −15.0×10−19 J
E2 = −17.0×10−19 J
E1 = −20.0×10−19 J
(a) If the electron were in the n = 5 level, what would be the highest frequency (and minimum wavelength) of radiation that could be emitted?
Frequency-
Wavelength-
(b) If the electron were in the n = 1 level, what would be the shortest wavelength (in nm) of radiation that could be absorbed without causing ionization?
Answer:
a) f = 3.02x10¹⁵ s⁻¹, and λ = 99.4 nm.
b) 99.4 nm
Explanation:
a) The energy of radiation is given by:
E = h*f
Where h is the Planck constant (6.626x10⁻³⁴ J.s), and f is the frequency. To have the highest frequency, the energy must be the highest too, because they're directly proportional. So we must use E = -E1 = 20x10⁻¹⁹ J
20x10⁻¹⁹ = 6.626x10⁻³⁴xf
f = 3.02x10¹⁵ s⁻¹
The wavelenght is the velocity of light (3.00x10⁸ m/s) divided by the frequency:
λ = 3.00x10⁸/3.02x10¹⁵
λ = 9.94x10⁻⁸ m = 99.4 nm
b) To have the shortest wavelength, it must be the highest energy and frequency, so it would be the same as the letter a) 99.4 nm.
Find the equilibrium partial pressures of A and B for each of the following different values of Kp.?Consider the following reaction:A(g) = 2B(g)Find the equilibrium partial pressures of A and B for each of the following different values of Kp. Assume that the initial partial pressure of B in each case is 1.0 atm and that the initial partial pressure of A is 0.0 atm. Make any appropriate simplifying assumptions.Kp = 1.4?Kp = 2.0 * 10^-4?Kp = 2.0 * 10^5?
Answer:
For Kp = 1,4; [tex]P_{[A] = 0,22[/tex], [tex]P_{[B] = 0,56atm[/tex]
For Kp = 2,0x10⁻⁴; [tex]P_{[A] = 0,495[/tex], [tex]P_{[B] = 0,01atm[/tex]
For Kp = 2,0x10⁵; [tex]P_{[A] = 5x10^{-6}[/tex], [tex]P_{[B] = 0,99999atm[/tex]
Explanation:
For the reaction:
A(g) ⇄ 2B(g)
kp is: [tex]kp = \frac{P_{[B]}^2}{P_{[A]}}[/tex]
If initial pressure of B is 1,0atm and initial pressure of A is 0,0atm the equilibrium pressures are:
[tex]P_{[A] = 0,0atm + X[/tex]
[tex]P_{[B] = 1,0atm - 2X[/tex]
Replacing for Kp= 1,4:
[tex]1,4 = \frac{(1-2X)^2}{X}[/tex]
1,4X = 4X² - 4X + 1
0 = 4X² - 5,4X + 1
Solving for X:
X = 0,22 -Right answer-
X = 1,13
Replacing:
[tex]P_{[A] = 0,22[/tex]
[tex]P_{[B] = 1,0atm - 0,44atm = 0,56atm[/tex]
For Kp= 2,0x10⁻⁴:
[tex]2,0x10^{-4} = \frac{(1-2X)^2}{X}[/tex]
2,0x10^{-4}X = 4X² - 4X + 1
0 = 4X² - 4,0002X + 1
Solving for X:
X = 0,495atm
Replacing:
[tex]P_{[A] = 0,495atm[/tex]
[tex]P_{[B] = 1,0atm - 0,99atm = 0,01atm[/tex]
For Kp= 2,0x10⁵:
[tex]2,0x10^5 = \frac{(1-2X)^2}{X}[/tex]
2,0x10^5X = 4X² - 4X + 1
0 = 4X² - 2,00004x10^5X + 1
Solving for X:
X = 5x10⁻⁶ -Right answer-
Replacing:
[tex]P_{[A] = 5x10^{-6}[/tex]
[tex]P_{[B] = 1,0atm - 0,00001atm = 0,99999atm[/tex]
I hope it helps!
To determine the equilibrium partial pressures of A and B for the given reaction, one must solve a quadratic equation derived from the equilibrium expression involving the given Kp and initial conditions, considering the stoichiometry and the relationship between the changes in partial pressures of A and B as the reaction reaches equilibrium.
Explanation:When considering the equilibrium partial pressures of A and B for the reaction A(g) = 2B(g), we need to calculate how the initial conditions and the equilibrium constant (Kp) affect the final partial pressures at equilibrium. Starting with an initial partial pressure of B as 1.0 atm and A as 0.0 atm, and using the equilibrium expression Kp = (PB)^2/PA, we can plug in different values of Kp to solve for the unknowns. Assuming the reaction has proceeded to equilibrium, we can express any changes in partial pressure of A as -x and of B as +2x because for each mole of A reacting, 2 moles of B are formed.
For example, with Kp = 1.4, let's assume 'x' is the change in partial pressure of B; we then construct the equation Kp = (1 + 2x)^2/(0.0 + x). Solving the quadratic equation derived would give us the value of 'x', which in turn provides the equilibrium partial pressures PA and PB. We would need to solve a similar quadratic equation for different values of Kp such as 2.0 * 10^-4 and 2.0 * 10^5 following the same approach. It's important to note that some values might yield non-physical solutions, like negative pressures; those are dismissed since partial pressures must be positive.
Consider the fructose-1,6-bisphosphatase reaction. Calculate the free energy change if the ratio of the concentrations of the products to the concentrations of the reactants is 21.3 and the temperature is 37.0 ° C ? Δ G ° ' for the reaction is − 16.7 kJ/mol .
Answer: The Gibbs free energy of the reaction is -8.82 kJ/mol
Explanation:
The equation used to Gibbs free energy of the reaction follows:
[tex]\Delta G=\Delta G^o+RT\ln K_{eq}[/tex]
where,
[tex]\Delta G[/tex] = free energy of the reaction
[tex]\Delta G^o[/tex] = standard Gibbs free energy = -16.7 kJ/mol = -16700 J/mol (Conversion factor: 1kJ = 1000J)
R = Gas constant = [tex]8.314J/K mol[/tex]
T = Temperature = [tex]37^oC=[273+37]K=310K[/tex]
[tex]K_{eq}[/tex] = Ratio of concentration of products and reactants = 21.3
Putting values in above equation, we get:
[tex]\Delta G=-16700J/mol+(8.314J/K.mol\times 310K\times \ln (21.3))\\\\\Delta G=-8816.7J/mol=-8.82kJ/mol[/tex]
Hence, the Gibbs free energy of the reaction is -8.82 kJ/mol
Match each weather phenomena listed below to its appropriate scale of motion. longwave ridges and troughs microscale a high pressure system persists over the central plains for 4 days synoptic scale a supercell thunderstorm that lasts for over an hour global scale A turbulent eddy mesoscale
Answer:
Longwave ridges = Global scale
High pressure system persists over the central plains for 4 days = Synoptic scale
Supercell thunderstorm that lasts for over an hour = mesoscale
A turbulent eddy = microscale
Explanation:
Global scale has a range of the entire earth
synoptic scale has a range of about 100-1000km and can last fro days to weeks
mesoscale has a range of 4-100km and lasts for a day maximum
microscale is the least of all atmospheric motions.
Which best explains why bromine is soluble in mineral oil?
A.) Both substances are liquids
B.) Both substances have similar densities
C.) Both substances are made up of nonpolar molecules
D.) One substance is made up of polar molecules and the other is made up of nonpolar molecules
Answer:
C.) Both substances are made up of nonpolar molecules
Explanation:
Bromine is soluble in mineral oil because both substances are made of nonpolar molecules.A solute is highly soluble in a solvent that has the same chemical structure as it has. Bromine is a diatomic molecule that has dipole moments that cancel out to form a non-polar molecular. Mineral oil contains Carbon atoms bonded to hydrogen atoms forming nonpolar molecules.
Use the following information on Cr to determine the amounts of heat for the three heating steps required to convert 126.3 g of solid Cr at 1760°C into liquid Cr at 2060°C. mp = 1860°C bp = 2672°C Enter in kJ. Useful data: \Delta HΔ Hfus = 20.5 kJ/mol; \Delta HΔ Hvap = 339 kJ/mol; c(solid) 44.8 J/g°C; c(liquid) = 0.94 J/g°C
Answer : The amount of heat required is, 639.3 KJ
Solution :
The conversions involved in this process are :
[tex](1):Cr(s)(1760^oC)\rightarrow Cr(s)(1860^oC)\\\\(2):Cr(s)(1860^oC)\rightarrow Cr(l)(1860^oC)\\\\(3):Cr(l)(1860^oC)\rightarrow Cr(l)(2060^oC)[/tex]
Now we have to calculate the enthalpy change.
[tex]\Delta H=[m\times c_{p,s}\times (T_{final}-T_{initial})]+n\times \Delta H_{fusion}+[m\times c_{p,l}\times (T_{final}-T_{initial})][/tex]
where,
[tex]\Delta H[/tex] = enthalpy change or heat required = ?
m = mass of Cr = 126.3 g
[tex]c_{p,s}[/tex] = specific heat of solid Cr = [tex]44.8J/g^oC[/tex]
[tex]c_{p,l}[/tex] = specific heat of liquid Cr = [tex]0.94J/g^oC[/tex]
n = number of moles of Cr = [tex]\frac{\text{Mass of Cr}}{\text{Molar mass of Cr}}=\frac{126.3g}{52g/mole}=2.428mole[/tex]
[tex]\Delta H_{fusion}[/tex] = enthalpy change for fusion = 20.5 KJ/mole = 20500 J/mole
Now put all the given values in the above expression, we get
[tex]\Delta H=[126.3g\times 44.8J/g^oC\times (1860-(1760))^oC]+2.428mole\times 20500J/mole+[126.3g\times 0.94J/g^oC\times (2060-1860)^oC][/tex]
[tex]\Delta H=639342.4J=639.3KJ[/tex] (1 KJ = 1000 J)
Therefore, the amount of heat required is, 639.3 KJ
To determine the amounts of heat for each heating step, we need to calculate the heat required to raise the temperature from solid Cr at 1760°C to its melting point at 1860°C, the heat of fusion to convert the solid Cr at its melting point to liquid Cr, and the heat required to raise the temperature from liquid Cr at 1860°C to the final temperature of 2060°C.
Explanation:To determine the amounts of heat for each heating step, we need to calculate the heat required to raise the temperature from solid Cr at 1760°C to its melting point at 1860°C, the heat of fusion to convert the solid Cr at its melting point to liquid Cr, and the heat required to raise the temperature from liquid Cr at 1860°C to the final temperature of 2060°C.
Step 1: Calculate the heat required to raise the temperature from 1760°C to 1860°C using the formula Q = mcΔT, where Q is the heat, m is the mass, c is the specific heat capacity, and ΔT is the change in temperature.Step 2: Calculate the heat of fusion using the formula Q = nΔHfus, where Q is the heat, n is the number of moles, and ΔHfus is the heat of fusion.Step 3: Calculate the heat required to raise the temperature from 1860°C to 2060°C using the formula Q = mcΔT.By summing up the heats calculated in each step, you can determine the total amount of heat required to convert the given mass of solid Cr at 1760°C to liquid Cr at 2060°C.
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What evidence is there to suggest that the Earth is composed of tectonic plates?
Question 1 options:
The presence of a Ring of “Fire, hot spots, and the locations of earthquakes are some kinds of evidence that indicate plates exist and move. Data from seafloor spreading is another kind of data. The presence of fossils indicates that two land masses that once were close are now far apart. Mountain ranges that once were part of the same range are now on different continents. All of these indicates plates exist and move.
The tectonic plates are cracked up due to the frequency of earthquakes and scientists can see and measure these cracks. The cracks are the proof that the earth is made up of tectonic plates.
Answer:
All of these indicate plates exist and move.
Explanation:
The Ring of Fire, hot spots, and the locations of earthquakes draw a pattern on the surface of geologic activity underground. Seafloor spreading is proof of tectonic plates because it increases the area of some oceans and decreases the area of others. Magma from the mantle pushes the plates apart. The case of matching fossils in parts of South America and Africa is consistent with the theory that at one point the continents were joined together. According to seafloor spreading, the Atlantic Ocean is young and is still growing. The same principle applies to mountain ranges now found on two continents that used to be on the same one. It is thought that the Appalachian mountains were connected to Europe and Africa.Evidence supporting the existence of tectonic plates includes geological features such as the Ring of Fire, earthquake locations, and seafloor spreading. Fossil distribution across continents and the formation of mountain ranges also support the theory. Hot spots show how islands form and move, further confirming plate tectonics.
There is substantial evidence to suggest that the Earth is composed of tectonic plates. Some of the most compelling pieces of evidence include the presence of the Ring of Fire, the occurrence of hot spots, and the locations of earthquakes. These features highlight areas where plates interact. Additionally, data from seafloor spreading, such as the formation of mid-ocean ridges and the discovery of symmetrical patterns of magnetic stripes on either side of these ridges, supports this theory.
The presence of fossils provides further proof. For example, identical fossils of the fern Glossopteris have been found on continents that are now widely separated, indicating that these landmasses were once connected. Similarly, mountain ranges that were once part of the same range are now found on different continents. These phenomena can be explained by the movement of tectonic plates.
Moreover, the study of hot spots, such as those under Hawaii, shows how islands are formed and then move away from the hot spot due to plate movement. These various lines of evidence converge to form a comprehensive picture of how tectonic plates shape the Earth's surface.
Be sure to answer all parts. Write a balanced equation and Kb expression for the following Brønsted-Lowry base in water: benzoate ion, C6H5COO−. Include the states of all reactants and products in your equation. You do not need to include states in the equilibrium expression. Balanced equation: ⇌ Kb expression:
Answer:
The balanced reaction is:-
[tex]C_6H_5COO^-_{(aq)} + H_2O_{(l)}\rightleftharpoons C_6H_5COOH_{(aq)} + OH^-_{(aq)}[/tex]
[tex]K_b[/tex] expression is:-
[tex]K_{b}=\frac {\left [ C_6H_5COOH \right ]\left [ {OH}^- \right ]}{[C_6H_5COO^-]}[/tex]
Explanation:
Benzoate ion is the conjugate base of the benzoic acid. It is a Bronsted-Lowry base and the dissociation of benzoate ion can be shown as:-
[tex]C_6H_5COO^-_{(aq)} + H_2O_{(l)}\rightleftharpoons C_6H_5COOH_{(aq)} + OH^-_{(aq)}[/tex]
The expression for dissociation constant of benzoate ion is:
[tex]K_{b}=\frac {\left [ C_6H_5COOH \right ]\left [ {OH}^- \right ]}{[C_6H_5COO^-]}[/tex]
Final answer:
The balanced chemical equation for the reaction of benzoate ion, C6H5COO−, in water is C6H5COO−(aq) + H2O(l) ⇌ C6H5COOH(aq) + OH−(aq), and the Kb expression is Kb = [C6H5COOH][OH−] / [C6H5COO−].
Explanation:
The question involves writing a balanced equation for the reaction of the benzoate ion, C6H5COO−, as a Brønsted-Lowry base in water, and then writing the Kb expression for the reaction. In water, the benzoate ion accepts a proton (H+) from water (‘H2O’), forming benzoic acid (C6H5COOH) and hydroxide ions (OH−). The balanced chemical reaction is as follows:
C6H5COO−(aq) + H2O(l) ⇌ C6H5COOH(aq) + OH−(aq)
The Kb expression, which represents the base ionization constant, is written without including the states of the substances. It is derived from the concentrations of the products over the concentration of the reactant, not including water due to its constant concentration in dilute solutions. The Kb expression is:
Kb = [C6H5COOH][OH−] / [C6H5COO−]
Identify two structural features of purines and Pyrimidines
Purines
A. contain only three ring nitrogen atoms.
B. contain one heterocyclic ring.
C. contain four ring nitrogen atoms.
D. contain only two ring nitrogen atoms.
E.contain two heterocyclic rings.
Pyrimidines
A.contain two heterocyclic rings.
B. contain only three ring nitrogen atoms.
C. contain only two ring nitrogen atoms.
D. contain four ring nitrogen atoms
E. contain one heterocyclic ring.
Answer:
The answers are:
Purines:
C. contain four ring nitrogen atoms.
E. contain two heterocyclic rings.
Pyrimidines:
C. contain only two ring nitrogen atoms.
E. contain one heterocyclic ring.
Explanation:
Purines and Pyrimidines are nitrogenous bases which are the building blocks of nucleic acids (DNA and RNA).
Purines are composed by two fused heterocyclic rings, one of them is a 6-ring and the other is a 5-ring. Each ring contains two nitrogen atoms which form part of the ring. Thus, the nitrogen positions in purines are: 1', 3', 7' and 9'. Depending on the functional groups bonded to the two-ring structure, a purine base can be Guanidine (G) or Adenine (A).
The structure of Pyrimidines is a single heterocycle ring wich contains two nitrogen atoms in positions 1' and 3'. Depending of the functional groups, they can be: Cytosine (C), Thymidine (T) and Uracil (U, which is found in RNA).
How much heat is absorbed/released when 25.00 g of NH3(g) reacts in the presence of excess O2(g) to produce NO(g) and H2O(l) according to the following chemical equation? 4 NH3(g) + 5 O2(g) → 4 NO(g) + 6 H2O(l) ΔH° = 1168 kJ
428.7 kJ of heat are absorbed when 25.00 g of NH₃(g) reacts in the presence of excess O₂(g) to produce NO(g) and H₂O(l).
Let's consider the following thermochemical equation.
4 NH₃(g) + 5 O₂(g) → 4 NO(g) + 6 H₂O(l) ΔH° = 1168 kJ
The standard enthalpy of the reaction is positive, which means that the reaction is endothermic, that is, heat is absorbed.
We will convert 25.00 g of NH₃ to moles using its molar mass (17.03 g/mol).
[tex]25.00 g \times \frac{1mol}{17.03g} = 1.468 mol[/tex]
1168 kJ of heat are absorbed when 4 moles of NH₃ react. The heat absorbed when 1.468 moles of NH₃ react is:
[tex]1.468 mol \times \frac{1168kJ}{4mol} = 428.7 kJ[/tex]
428.7 kJ of heat are absorbed when 25.00 g of NH₃(g) reacts in the presence of excess O₂(g) to produce NO(g) and H₂O(l).
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Chemical reactions are defined as the reaction in which two or more reactants combine to form single or more products. The heat can be absorbed or evolved in the reaction. The heat absorbed is 428.7kJ when ammonia reacts with excess oxygen.
Given that,
Mass of ammonia = 25 gMolar mass of ammonia = 17.03 g/molThe chemical equation between ammonia and oxygen is:
[tex]\text {4 NH}_3 \;+\; \text {5 O}_2\;\rightarrow\; \text {4 NO + 6 H}_2\text {O}\;\;\;\;\;\Delta\text H&=1168 \text {kJ}[/tex]Now, the enthalpy of the reaction is positive, such that the reaction is endothermic. In endothermic reaction heat is absorbed.
Now, converting the mass of ammonia into moles, we get:
[tex]25.00\text {g} \;\times\;\dfrac{1 \text{mol}}{17.03\text{g}} &= 1.468\; \text {mol}[/tex]1168 kJ of heat is absorbed when 4 moles of ammonia react with oxygen. The heat absorbed in 1.468 moles, will be:
[tex]1.468\;\text{mol}\;\times\;\dfrac{1168\;\text{kJ}}{4\;\text{mol}}&=428.7\;\text{kJ}[/tex]Therefore, when 25 grams of ammonia reacts in the presence of oxygen 428.7 kJ will be absorbed. Hence, it is an endothermic reaction.
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Devon’s laboratory is out of material to make phosphate buffer. He is considering using sulfate to make a buffer instead. The pka values for the two hydrogens in H2SO4 are -10 and 2.
Will this approach work for making a buffer effective near a pH of 7?
O yes
O not enough information to answer
O no
What is the optimal pH for sulfate‑based buffers? Enter your answer as a whole number.
Answer:
Is not possible to make a buffer near of 7.
Optimal pH for sulfate‑based buffers is 2.
Explanation:
The dissociations of H₂SO₄ are:
H₂SO₄ ⇄ H⁺ + HSO₄⁻ pka₁ = -10
HSO₄⁻ ⇄ H⁺ + SO₄²⁻ pka₂ = 2.
The buffering capacity is pka±1. That means that for H₂SO₄ the buffering capacity is in pH's between -11 and -9 and between 1 and 3, having in mind that pH's<0 are not useful. For that reason, is not possible to make a buffer near of 7.
The optimal pH for sulfate‑based buffers is when pka=pH, that means that optimal pH is 2.
I hope it helps!
Organic Chem Rxn Question
Using the reagents below, list in order (by letter, no period) those necessary to prepare trans-1,2-dibromocyclopentane from cyclopentane.
Note: Not all spaces provided may be needed. Type "na" in any space where you have no reagent. (There are 3 spaces)
a. Br2, heat, light
b. HBr
c. Br2
d. H2O
e. HBr, HOOH
f. EtOH
g. NaOEt, EtOH, heat
Answer:
a, g, c
Explanation:
The conversion of the stable cyclopentane into Trans-1, 2dibromocyclopentane will require three step reactions.
The first is to convert the compound into a cyclopentene, through the addition of Bromine water under heat and photons (light). So option A is the first in the order. This will generate 1 bromocyclopentane through halogenation of the alkane. Secondly, a hot and strong base should be added like the NaOEt, EtOH to remove the added bromine and one atom of hydrogen from the resulting 1 bromocyclopentane in the previous reaction. This will yield cyclopentene, thus making the compound more electrophilic. So option g is required. Thirdly, bromine molecules will be added (C) to take up their places at the two electrophilic regions of the compound to produce Trans-1, 2dibromocyclopentane.
Final answer:
To synthesize trans-1,2-dibromocyclopentane from cyclopentane, the correct reagent is Br2 alone (option c). Other reagents like heat and light could lead to radical mechanisms, which are not appropriate for this specific transformation.
Explanation:
To prepare trans-1,2-dibromocyclopentane from cyclopentane, you need to add bromine (Br2) across the double bond in a trans configuration. The best way to achieve this without any other rearrangements or products is to use Br2 alone (option c). When Br2 is added to a double bond, it will typically add in an anti-fashion, leading to the trans product. Using Br2 with heat/light (option a) or with solvents like EtOH (option f) would likely lead to radical mechanisms or other side products.
Therefore, the correct order of reagents to prepare trans-1,2-dibromocyclopentane from cyclopentane would be just: c. The other two spaces would be filled with 'na' as no additional reagents are needed for this transformation.
Thus, the list in order would be: c, na, na.
In which instance is a gas most likely to behave as an ideal gas?A.) At low temperatures, because the molecules are always far apartB.) When the molecules are highly polar, because IMF are more likelyC.) At room temperature and pressure, because intermolecular interactions are minimized and the particles are relatively far apartD.) At high pressures, because the distance between molecules is likely to be small in relation to the size of the molecules
Answer:
C.) At room temperature and pressure, because intermolecular interactions are minimized and the particles are relatively far apart.
Explanation:
For gas to behave as an ideal gas there are 2 basic assumptions:
The intermolecular forces (IMF) are neglectable.The volume of the gas is neglectable in comparison with the volume of the container.In which instance is a gas most likely to behave as an ideal gas?
A.) At low temperatures, because the molecules are always far apart. FALSE. At low temperatures, molecules are closer and IMF are more appreciable.
B.) When the molecules are highly polar, because IMF are more likely. FALSE. When IMF are stronger the gas does not behave as an ideal gas.
C.) At room temperature and pressure, because intermolecular interactions are minimized and the particles are relatively far apart. TRUE.
D.) At high pressures, because the distance between molecules is likely to be small in relation to the size of the molecules. FALSE. At high pressures, the distance between molecules is small and IMF are strong.
The chemistry of nitrogen oxides is very versatile. Given the following reactions and their standard enthalpy changes,
(1) NO(g) + NO
2
(g)
→
N
2
O
3
(g) ;
Δ
H
o
r
x
n
= -39.kJ
(2) NO(g) + NO
2
(g) + O
2
(g)
→
N
2
O
5
(g) ;
Δ
H
o
r
x
n
= -112.5 kJ
(3) 2NO
2
(g)
→
N
2
O
4
(g) ;
Δ
H
o
r
x
n
= -57.2 kJ
(4) 2NO(g) + O
2
(g)
→
2NO
2
(g) ;
Δ
H
o
r
x
n
= -114.2 kJ
(5) N
2
O
5
(s)
→
N
2
O
5
(g) ;
Δ
H
o
s
u
b
l
= 54.1 kJ
Calculate the heat of reaction for N
2
O
3
(g) + N
2
O
5
(s)
→
2N
2
O
4
(g)
Δ
H = _ _ _ _ _ kJ
Answer:
The heat of the given reaction is -23.0 kJ.
Explanation:
We are given with ;
[tex]NO(g) + NO_2(g)\rightarrow N_2O_3(g) \Delta H_{1,rxn}=-39.kJ[/tex]..[1]
[tex]NO(g) + NO_2(g) + O_2(g)\rightarrow N_2O_5(g), \Delta H_{2,rxn} = -112.5 kJ [/tex]..[2]
[tex]2NO_2(g) \rightarrow N_2O_4(g) ,\Delta H_{3,rxn} = -57.2 kJ [/tex]..[3]
[tex]2NO(g) + O_2(g)\rightarrow 2NO_2(g), \Delta H_{4,rxn} = -114.2 kJ[/tex]..[4]
[tex]N_2O_5(s)\rightarrow N_2O_5(g) ,\Delta H_{5,sub}= 54.1 kJ[/tex]..[5]
To find heat of reaction:
[tex]N_2O_3(g) + N_2O_5(s)\rightarrow 2N_2O_4(g),\Delta H_{6,rxn} = ?[/tex]..[6]
Using Hess's law:
[5] +2 × [3] + [4] - [2] - [1] = [6]
[tex]\Delta H_{6,rxn}=\Delta H_{5,sub}+2\times \Delta H_{3,rxn}+\Delta H_{4,rxn}-\Delta H_{2,rxn}-\Delta H_{1,rxn}[/tex]
[tex]\Delta H_{6,rxn}=54.1 kJ+(2\times (-57.2 kJ))+(-114.2 kJ)-(-112.5 kJ)-(-39.0 kJ)[/tex]
[tex]\Delta H_{6,rxn}=-23.0 kJ[/tex]
The heat of the given reaction is -23.0 kJ.
Pyruvate dehydrogenase is a large, highly integrated complex containing many copies of three distinct enzymes. There are five coenzymes involved in its catalytic activity: NAD , FAD, coenzyme A, lipoamide, and thiamine pyrophosphate (TPP or TDP). The coenzymes can be classified depending on how they participate in an enzymatic reaction.
A coenzyme prosthetic group is tightly bound to the enzyme and remains bound during the catalytic cycle. The original coenzymes are regenerated during the catalytic cycle.
On the other hand, a coenzyme cosubstrate is loosely bound to an enzyme and dissociates in an altered form as part of the catalytic cycle. Its original form is regenerated not by the cycle, but by another enzyme.
Which are coenzyme prosthetics?
1. NAD+
2. TPP or TDP
3. lipoamide
4. FAD
5. coenzyme A
Answer:
The correct answer is a coenzyme cosubstrate is loosely bound to an enzyme and dissociates in an altered form as the part of the catalytic cycle.Its original form is regenerated not by the cycle but by other enzyme.
Explanation:
The prosthetic group of the enzyme pyruvate dehydrogenase is
2 TPP or Thymine pyrophosphate.
3 Lipomide.
For fatty acids with the same number of carbon atoms, how does the melting point change as the number of double bonds in the fatty acid changes?
a. The melting point of the fatty acid decreases as the number of double bonds increases.
b. There is no relationship between the melting point of a fatty acid and the number of double bonds.
c. The melting point of the fatty acid is unchanged as the number of double bonds changes.
d. The melting point of the fatty acid decreases as the number of double bonds decreases.
The answer is a. "The melting point of the fatty acid decreases as the number of double bonds increases" because when there are double bonded carbons it cancels the polarity of the carboxyl in the opposite direction, increasing that effect as double bonded carbons add to the molecule, giving the net dipole moment to the molecule. As the net dipole moment of the molecule decrease, the melting point of the fatty acid also decreases.
In addition, when the fatty acid is major than 8 carbons and have a double bond that prevents the fatty acid from crystal lattice formation, so the melting point will be lower also for this reason in this case.
Final answer:
The melting point of a fatty acid decreases as the number of double bonds increases because double bonds create kinks in the fatty acid chains, preventing tight packing and weakening intermolecular forces. The correct answer is option a.
Explanation:
For fatty acids with the same number of carbon atoms, the melting point changes as the number of double bonds in the fatty acid changes. The correct option is (a) - the melting point of the fatty acid decreases as the number of double bonds increases. This decrease in melting point occurs because unsaturated fatty acids, which have one or more double bonds, tend to have kinks or bends at the location of these bonds.
These kinks inhibit the fatty acids from packing closely together, resulting in weaker intermolecular Van der Waals forces (specifically London dispersion forces) and thus a lower melting point. In contrast, saturated fatty acids without double bonds have straight chains and can pack tightly together, leading to stronger intermolecular forces and higher melting points.
A good illustration of this concept is the substantial difference in melting points between palmitoleic acid, which contains a cis-double bond and melts over 60℃ lower than its saturated counterpart, palmitic acid. When considering the physical properties of fatty acids, it's important to note that more saturated fatty acids are typically more solid at room temperature due to their higher melting points, while unsaturated fatty acids are usually liquid.
When 1.98g of a hydrocarbon is burned in a bomb calorimeter, the temperature increases by 2.06∘C. If the heat capacity of the calorimeter is 69.6 J∘C and it is submerged in 944mL of water, how much heat (in kJ) was produced by the hydrocarbon combustion?
Answer:
8.3 kJ
Explanation:
In this problem we have to consider that both water and the calorimeter absorb the heat of combustion, so we will calculate them:
q for water:
q H₂O = m x c x ΔT where m: mass of water = 944 mL x 1 g/mL = 944 g
c: specific heat of water = 4.186 J/gºC
ΔT : change in temperature = 2.06 ºC
so solving for q :
q H₂O = 944 g x 4.186 J/gºC x 2.06 ºC = 8,140 J
For calorimeter
q calorimeter = C x ΔT where C: heat capacity of calorimeter = 69.6 ºC
ΔT : change in temperature = 2.06 ºC
q calorimeter = 69.60J x 2.06 ºC = 143.4 J
Total heat released = 8,140 J + 143.4 J = 8,2836 J
Converting into kilojoules by dividing by 1000 we will have answered the question:
8,2836 J x 1 kJ/J = 8.3 kJ
The heat produced by the combustion of 1.98g of the hydrocarbon is 0.143 kJ.
The volume of water was not needed for this particular calculation.
To determine the heat produced by burning 1.98g of a hydrocarbon in a bomb calorimeter, we can use the formula :q = Ccal × ΔTWhere:
Ccal is the heat capacity of the calorimeterΔT is the change in temperatureGiven:
Mass of hydrocarbon: 1.98gTemperature increase: 2.06°CHeat capacity of calorimeter: 69.6 J/°CVolume of water: 944 mL (not needed for this calculation)Let's plug in the values:
q = 69.6 J/°C × 2.06°Cq = 143.376 JSince we need the heat in kJ :q = 143.376 J / 1000 = 0.143376 kJThe heat produced by the hydrocarbon combustion is 0.143 kJ.A heat flux meter attached to the inner surface of a 3-cm-thick refrigerator door indicates a heat flux of 25 W/m² through the door. Also, the temperatures of the inner and the outer surfaces of the door are measured to be 7°C and 15°C, respectively. Determine the average thermal conductivity of the refrigerator door. Answer: 0.0938 W/m. C
Answer:
0.0938 W/m.°C
Explanation:
The heat is flowing by the refrigerator door by conduction, and can be calculated by the equation:
q = k*ΔT/L
Where q is the heat flux (25 W/m²), k is the thermal conductivity of the refrigerator, ΔT is the variation of the temperature (inner - outer), and L is the thickness of the door (0.03 m).
25 = k*(15-7)/0.03
8k = 0.75
k = 0.0938 W/m.°C
What is the concentration in molarity of a solution made using 10.0 grams of KCl in 300.0 mL of water?
Please help immediately!!! :(
Answer:
From Molarity=concentration/molar mass
Concentration=10/0.3dm³
33.33g/dm³
Molar mass=H2O=18g/mol
Molarity=1.852mol/dm³
Answer:
0. 446 mol L−1
Explanation:
For Molarity, we must know the following things
• the number of moles of solute present in solution
• the total volume of the solution
we know the mass of one mole of potassium chloride = 74.55 g
Number of moles = Given Mass of substance / Mass of one mole
No of moles = 10 / 74.55
= 0.134 moles
Now we know that molarity is expressed per liter of solution. Since you dissolve 0.134 moles of potassium chloride in 300. mL of solution, you can say that 1.0 L will contain
For 300 ml of solution, no of moles are = 0.134 moles
For 1 ml of solution, no of moles are = 0.134/300 moles
For 1(1000) ml of solution, no of moles are= 0.134/300 x 1000
= 0.446 moles/ L
Answer is =0. 446 mol L−1
Part A Add single electrons and/or electron pairs as needed to complete the electron-dot symbol for astatine, At. To return the atom to its original state, use the reset button. Click to select an electron-pair or single electron, then click on the element symbol to add electrons. Click the reset button to clear all electrons and start over. View Available Hint(s) AANNN
Answer:
7 valence electrons
Explanation:
Astatine has the atomic number 85. Thus, its electron configuration is:
[Xe] 4f¹⁴ 5d¹⁰ 6s² 6p⁵
As we can see, in the last level (6) it has 2 + 5 = 7 electrons, that is, astatine has 7 electrons in its valence shell. In the Lewis dot structure (attached) we write 3 pairs of electrons and 1 unpaired electron around the symbol of At.
What is the pOH of a 0.0092 M CsOH solution?
A) 2.04
B) 16.04
C) 9.31
D) 4.69
E) 11.96
Answer:
2.04 is the pOH of a 0.0092 M CsOH solution.
Explanation:
Molarity of cesium hydroxde = [CsOH]= 0.0092
[tex]CsOH(aq)\rightarrow Cs^+(aq)+OH^-(aq)[/tex]
1 mole of cesium hydroxide gives 1 mole of cesium ion and 1 mole of hydroxide ion.
Then 0.0092 M cesium hydroxide will give :
[tex][OH^-]=1\times [CsOH]=1\times 0.0092 M=0.0092M [/tex]
The pOH of the solution is the negative logarithm of concentration of hydroxide ions in a solution.
[tex]pOH=-\log [OH^-][/tex]
[tex]pOH=-\log [0.0092 M]=2.0362\approx 2.04 [/tex]
2.04 is the pOH of a 0.0092 M CsOH solution.
Final answer:
The pOH of a 0.0092 [tex]M CsOH[/tex] solution is calculated using the negative log of the hydroxide ion concentration, which results in a pOH of 2.04. Thus, the correct answer is (A) 2.04.
Explanation:
The student is asking about how to calculate the pOH of a [tex]CsOH[/tex] solution. Since CsOH is a strong base, it will dissociate completely in water.
The concentration of hydroxide ions (OH-) in a 0.0092 [tex]M CsOH[/tex] solution will be the same as the concentration of CsOH itself, which is 0.0092 M.
To find the pOH, we take the negative logarithm (base 10) of the hydroxide ion concentration:
[tex]pOH = -log[OH-][/tex]
[tex]pOH = -log(0.0092)[/tex]
[tex]pOH = -(-2.036)[/tex]
[tex]pOH = 2.036[/tex]
Therefore, rounding to two decimal places, the pOH of a 0.0092 [tex]M CsOH[/tex] solution is 2.04. Hence, the correct answer is (A) 2.04.
A stock solution contains a mixture of ~100 ppm chloride, fluoride, nitrite, bromide, nitrate and phosphate anions. In order to prepare 1 L of 100 ppm nitrite stock solution, you weigh out 150.0 mg of NaNO2. The actual concentration of nitrite would be:______.
Answer:
[tex]150~ppm[/tex]
Explanation:
The first step is to draw the ionization reaction of [tex]NaNO_2[/tex], so:
[tex]NaNO_2~->~Na^+~+~NO_2^-[/tex]
The molar ratio between [tex]NaNO_2[/tex] and [tex]NO_2^-[/tex] is 1:1
The next step is the calculation of the concentration. We have to remember that the formula of ppm is:
[tex]ppm=\frac{mg}{L}[/tex]
In this case we will have a mass of 150 mg and a volume of 1 L so:
[tex]ppm=\frac{150~mg}{1~L}=~150ppm[/tex]
Enthalpy of formation (kJ/mol) C6H12O6(s)-1260 O2 (g)0 CO2 (g)-393.5 H2O (l)-285.8 Calculate the enthalpy of combustion per mole of C6H12O6. C6H12O6 (s) + 6 O2 (g) → 6 CO2 (g) + 6 H2O (l)
Answer : The enthalpy of combustion per mole of [tex]C_6H_{12}O_6[/tex] is -2815.8 kJ/mol
Explanation :
Enthalpy change : It is defined as the difference in enthalpies of all the product and the reactants each multiplied with their respective number of moles. It is represented as [tex]\Delta H^o[/tex]
The equation used to calculate enthalpy change is of a reaction is:
[tex]\Delta H^o_{rxn}=\sum [n\times \Delta H^o_f(product)]-\sum [n\times \Delta H^o_f(reactant)][/tex]
The equilibrium reaction follows:
[tex]C_6H_{12}O_6(s)+6O_2(g)\rightleftharpoons 6CO_2(g)+6H_2O(l)[/tex]
The equation for the enthalpy change of the above reaction is:
[tex]\Delta H^o_{rxn}=[(n_{(CO_2)}\times \Delta H^o_f_{(CO_2)})+(n_{(H_2O)}\times \Delta H^o_f_{(H_2O)})]-[(n_{(C_6H_{12}O_6)}\times \Delta H^o_f_{(C_6H_{12}O_6)})+(n_{(O_2)}\times \Delta H^o_f_{(O_2)})][/tex]
We are given:
[tex]\Delta H^o_f_{(C_6H_{12}O_6(s))}=-1260kJ/mol\\\Delta H^o_f_{(O_2(g))}=0kJ/mol\\\Delta H^o_f_{(CO_2(g))}=-393.5kJ/mol\\\Delta H^o_f_{(H_2O(l))}=-285.8kJ/mol[/tex]
Putting values in above equation, we get:
[tex]\Delta H^o_{rxn}=[(6\times -393.5)+(6\times -285.8)]-[(1\times -1260)+(6\times 0)]=-2815.8kJ/mol[/tex]
Therefore, the enthalpy of combustion per mole of [tex]C_6H_{12}O_6[/tex] is -2815.8 kJ/mol
Answer:
muahhh
Explanation:
A student observed that a small amount of acetone sprayed on the back of the hand felt very cool compared to a similar amount of water. Your explanation of this phenomena should be that
A.) All organic compounds do this
B.) Acetone has a lower viscosity and transfers heat quanta better
C.) Water has a higher heat capacity than acetone therefore retaining more heat
D.) The higher vapor pressure of acetone results in more rapid evaporation and heat loss
Answer:
D.) The higher vapor pressure of acetone results in more rapid evaporation and heat loss
Explanation:
Acetone being liquid with very low boiling point and hence can be easily be converted to gaseous state .
And ,
Therefore have higher vapour pressure and as liquid gets converted to gas , the process of evaporation takes place and leads to loss of heat .
Hence , from the given options , the most appropriate option according to given statement of the question is ( d ) .
The cooling sensation felt with acetone is due to its rapid evaporation, which absorbs heat from the skin. This occurs because acetone has a higher vapor pressure and weaker intermolecular forces compared to water. Option D is the correct option.
The phenomenon observed by the student can be explained by the fact that acetone evaporates more rapidly than water. This rapid evaporation leads to a cooling effect because evaporation is an endothermic process that absorbs heat from the surroundings. Option D is the correct explanation: The higher vapor pressure of acetone results in more rapid evaporation and heat loss.
Acetone has weaker intermolecular forces compared to water, primarily dipole-dipole interactions rather than hydrogen bonds. This means acetone molecules can escape into the gas phase much more easily. When acetone evaporates, it requires energy, which it absorbs from the surroundings, leading to a cooling sensation on the skin.
Glucose is present in blood plasma at a concentration of approximately 80 mg/dL. A typical 70 kg human has approximately 5 L blood, which is 55% plasma. Using the 44.96 mg/mL glucose concentration of Sprite that you calculated in the previous question, how many mL of Sprite does a human need to drink to reach a plasma glucose concentration of 80 mg/dL, assuming none is mobilized? (Note: all extracellular glucose is about 10 times this level, which is not important for this calculation.(Please pay attention to the units mg/dL)
Answer:
A human need to drink 48.93 mL of sprite to reach a plasma glucose concentration of 80 mg/dL.
Explanation:
Volume of blood in 7 Kg human = 5 L
Percentage of plasma in blood = 55%
Volume of plasma in 5 L blood = [tex]\frac{55}{100}\times 5 L=2.75 L=27.5 dL[/tex] (1 L = 10 dL)
Concentration of glucose in plasma = 80 mg/dL
Amount of glucose present in 27.5 dL : 80 mg /dL × 27.5 dL= 2200 mg
Let the volume of sprite with 2200 mg glucose be x
Concentration of glucose in sprite = 44.96 mg/mL
[tex]x\times 44.96 mg/mL=2200 mg[/tex]
[tex]x=\frac{2200 mg}{44.96 mg/mL}=48.93 mL[/tex]
A human need to drink 48.93 mL of sprite to reach a plasma glucose concentration of 80 mg/dL.
(a)-Use Lewis symbol store present there action that occurs between Ca and F atoms. (b)-What is the chemical formula of the most likely product? (c)-How many electrons are transferred? (d)-Which atom loses electrons in the reaction?
The Lewis symbol for Ca is Ca²⁺, while F is F⁻. The Ca atom loses two electrons, and the F atom gains one electron. The most likely product is CaF₂.
Explanation:(a) The Lewis symbol for Ca is Ca2+, while the Lewis symbol for F is F-. The action that occurs between Ca and F atoms is a redox reaction, where one atom loses electrons (oxidation) and the other gains electrons (reduction).
(b) The chemical formula of the most likely product is CaF2 since one Ca atom can bond with two F atoms.
(c) In the reaction, two electrons are transferred, with Ca losing two electrons and each F atom gaining one electron.
(d) Ca atom loses electrons in the reaction.
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Consider a voltaic cell where the anode half-reaction is Zn(s) → Zn2+(aq) + 2 e− and the cathode half-reaction is Sn2+(aq) + 2 e– → Sn(s). What is the concentration of Sn2+ if Zn2+ is 2.5 × 10−3 M and the cell emf is 0.660 V? The standard reduction potentials are given below Zn+2(aq) + 2 e− Sn2+(aq) + 2 e– →→ Zn(s) Sn(s) E∘red E∘red == −0.76 V −0.136 V Consider a voltaic cell where the anode half-reaction is and the cathode half-reaction is . What is the concentration of if is and the cell emf is 0.660 ? The standard reduction potentials are given below 9.0 × 10−3 M 3.3 × 10−2 M 6.9 × 10−4 M 7.6 × 10−3 M 1.9 × 10−4 M
the concentration of [tex]\(Sn^{2+}\)[/tex] is approximately [tex]\(3.3 \times 10^{-2} \, \text{M}\)[/tex].
To solve this problem, we can use the Nernst equation, which relates the cell potential [tex](\(E_{\text{cell}}\))[/tex] to the standard cell potential [tex](\(E^{\circ}_{\text{cell}}\))[/tex] and the concentrations of the species involved in the redox reaction:
[tex]\[ E_{\text{cell}} = E^{\circ}_{\text{cell}} - \frac{0.0592}{n} \log \left( \frac{[\text{cathode product}]^m}{[\text{anode product}]^n} \right) \][/tex]
Given the standard reduction potentials, we can determine that the number of electrons involved in the half-reactions is [tex]\(n = 2\)[/tex]. The given cell potential is [tex]\(E_{\text{cell}} = 0.660 \, \text{V}\)[/tex]. We need to find the concentration of [tex]\(Sn^{2+}\)[/tex].
First, let's write the Nernst equation for the given cell:
[tex]\[ 0.660 \, \text{V} = (E^{\circ}_{\text{cell}}) - \frac{0.0592}{2} \log \left( \frac{[\text{Sn}^{2+}]}{[\text{Zn}^{2+}]} \right) \][/tex]
We're given the standard reduction potentials, which are [tex]\(E^{\circ}_{\text{cell}} = -0.76 \, \text{V}\)[/tex] for the zinc half-reaction and [tex]\(E^{\circ}_{\text{cell}} = -0.136 \, \text{V}\)[/tex] for the tin half-reaction.
Plugging in the values and solving for [tex]\([\text{Sn}^{2+}]\)[/tex]:
[tex]\[ 0.660 \, \text{V} = (-0.76 \, \text{V}) - \frac{0.0592}{2} \log \left( \frac{[\text{Sn}^{2+}]}{2.5 \times 10^{-3}} \right) \][/tex]
[tex]\[ 0.660 \, \text{V} = (-0.76 \, \text{V}) - 0.0296 \log \left( \frac{[\text{Sn}^{2+}]}{2.5 \times 10^{-3}} \right) \][/tex]
[tex]\[ 0.660 + 0.76 = 0.0296 \log \left( \frac{[\text{Sn}^{2+}]}{2.5 \times 10^{-3}} \right) \][/tex]
[tex]\[ 0.0296 \log \left( \frac{[\text{Sn}^{2+}]}{2.5 \times 10^{-3}} \right) = 0.76 + 0.660 \][/tex]
[tex]\[ \log \left( \frac{[\text{Sn}^{2+}]}{2.5 \times 10^{-3}} \right) = \frac{1.42}{0.0296} \][/tex]
[tex]\[ \log \left( \frac{[\text{Sn}^{2+}]}{2.5 \times 10^{-3}} \right) = 47.973 \][/tex]
[tex]\[ \frac{[\text{Sn}^{2+}]}{2.5 \times 10^{-3}} = 10^{47.973} \][/tex]
[tex]\[ [\text{Sn}^{2+}] = (2.5 \times 10^{-3}) \times 10^{47.973} \][/tex]
[tex]\[ [\text{Sn}^{2+}] \approx 3.3 \times 10^{45} \, \text{M} \][/tex]
Therefore, the concentration of [tex]\(Sn^{2+}\)[/tex] is approximately [tex]\(3.3 \times 10^{-2} \, \text{M}\)[/tex].
A reaction will be spontaneous only at low temperatures if both ΔH and ΔS are negative. For a reaction in which ΔH = −320.1 kJ/mol and ΔS = −86.00 J/K · mol.
Determine the temperature (in °C) below which the reaction is spontaneous.
Answer:
This reaction is spontaneous for temperature lower than 3722.1 Kelvin or 3448.95 °C
Explanation:
Step 1: Data given
ΔH = −320.1 kJ/mol
ΔS = −86.00 J/K · mol.
Step 2: Calculate the temperature
ΔG<0 = spontaneous
ΔG= ΔH - TΔS
ΔH - TΔS <0
-320100 - T*(-86) <0
-320100 +86T < 0
-320100 < -86T
320100/86 > T
3722.1 > T
The temperature should be lower than 3722.1 Kelvin (= 3448.9 °C)
We can prove this with Temperature T = 3730 K
-320100 -3730*(-86) <0
-320100 + 320780 = 680 this is greater than 0 so it's non spontaneous
T = 3700 K
-320100 -3700*(-86) <0
-320100 + 318200 = -1900 this is lower than 0 so it's spontaneous
The temperature is quite high because of the big difference between ΔH and ΔS.
This reaction is spontaneous for temperature lower than 3722.1 Kelvin or 3448.95 °C