Answer:
Heat transfer = 3564 Jolues
The same value
Explanation:
The heat of combustion is the heat released per 1 mole of substunce experimenting the combustion at standard conditions of pressure and temperature ( 1 atm, 298 K):
Qtransfer = - mol x ΔHºc Qtransfer
So look up in appropiate reference table ΔHºc and solve the problem:
ΔHºc = - 891 kJ/mol
Qtransfer = - (4 x 10³ mol x -891 kJ/mol ) = 3564 J
if the combustion were achieved with 100 % excess air, the result will still be the same. As long as the standard conditions are maintained, the heat of combustion remains constant. In fact in many cases the combustion is performed under excess oxygen to ensure complete combustion.
Final answer:
The heat transfer from the combustion chamber when 4 kmol of methane is burned completely with stoichiometric air is 3561.6 kJ. With 100 percent excess air, the heat transfer remains the same as excess air does not change the heat released from the combustion reaction.
Explanation:
Heat Transfer During Combustion of Methane
To determine the heat transfer from the combustion chamber when 4 kmol of methane (CH4) is burned with the stoichiometric amount of air, we must use the thermochemical equation for the combustion of methane. The equation indicates that when 1 mol of methane is combusted, 890.4 kilojoules of heat is released. Therefore, for 4 kmol, the heat release would be 4 kmol × 890.4 kJ/kmol = 3561.6 kJ.
If the combustion is achieved with 100 percent excess air, the amount of heat transferred would not change because the stoichiometry of the reaction remains consistent and the same amount of methane is being burned. However, additional air does not contribute to the heat of the reaction unless it causes incomplete combustion or other side reactions, which in this case is disregarded.
Since the conditions specify that reactants and products are maintained at 25°C and 1 atm, and the water in the products is in the liquid form, we can assume that no extra energy is required for heating up or cooling down the substances and the entire heat of combustion is transferred out from the system.
How long must a constant current of 50.0 A be passed through an electrolytic cell containing aqueous Cu2+ ions to produce 5.00 moles of copper metal?
A) 0.187 hours
B) 0.373 hours
C) 2.68 hours
D) 5.36 hours
Answer:
D) 5.36 hours
Explanation:
According to mole concept:
1 mole of an atom contains [tex]6.022\times 10^{23}[/tex] number of particles.
We know that:
Charge on 1 electron = [tex]1.6\times 10^{-19}C[/tex]
Also, copper will produce 2 electrons. So, out of 5 moles of copper, 10 moles of electrons will be produced.
So,
Charge on 10 mole of electrons = [tex]10\times 1.6\times 10^{-19}\times 6.022\times 10^{23}=9.6352\times 10^5C[/tex]
To calculate the time required, we use the equation:
[tex]I=\frac{q}{t}[/tex]
where,
I = current passed = 50.0 A
q = total charge = [tex]9.6352\times 10^5C[/tex]
t = time required = ?
Putting values in above equation, we get:
[tex]50.0A=\frac{9.6352\times 10^5C}{t}\\\\t=\frac{9.6352\times 10^5C}{50.0A}=19270.4s[/tex]
Converting this into hours, we use the conversion factor:
1 hr = 3600 seconds
So, [tex]19270.4s\times \frac{1hr}{3600s}=5.36hr[/tex]
Hence, the amount of time needed is 5.36 hrs.
Some type of path is necessary to join both half-cells in order for electron flow to occur.
A. True B. False
The given statement, some type of path is necessary to join both half-cells in order for electron flow to occur, is true.
Explanation:
Flow of electrons is possible with the help of a conducting medium like metal wire.
A laboratory device which helps in completion of oxidation and reduction-half reactions of a galvanic or voltaic cell is known as salt bridge. Basically, this salt bridge helps in the flow of electrons from anode to cathode and vice-versa.
If salt bridge is not present in an electrochemical cell, the electron neutrality will not be maintained and hence, flow of electrons will not take place.
Thus, we can conclude that the statement some type of path is necessary to join both half-cells in order for electron flow to occur, is true.
The vapor pressure of Substance X is measured at several temperatures:temperature (C) vapor pressure (atm)34 0.23644 0.29254 0.355Use this information to calculate the enthalpy of vaporization of X. Round your answer to 2 significant digits. Be sure your answer contains a correct unit symbol. Clears your work. Undoes your last action. Provides information about entering answers.
Answer:
ΔHv = 17.04 KJ/mol
Explanation:
T(°C) T(K) Pv(atm) 1/T(K) LnPv
34 307 0.236 0.00325733 - 1.4439
44 317 0.292 0.0031545 - 1.2310
54 327 0.355 0.003058 - 1.0356
Clausius-Clapeyron:
δLnP/δT = ΔH/RT²⇒ δLnP = ΔH/R (δT/T²)
∴ δT/T² = δ/δT(- 1/T )
⇒ δLnP/δT = - ΔH/R
Graphing: LnP vs 1/T
we get an ecuation that corresponds to a straight line:
y = - 2049.6x + 5.2331 ...... R² = 1
where the slope of this line is:
y = mx + b
⇒ m = - 2049.6 = - ΔH/R.....Clausius-Clapeyron
⇒ ΔH = (2049.6)(R)
∴ R = 8.314 E-3 KJ/mol.K
⇒ ΔHv = 17.04 KJ/mol
Label the Following Materials as Polar, Ionic or Nonpolar and Describe Whether They are Soluble or Insoluble in Water HCl O2 CaCl2 N2 C2H6 KNO3 BeCl2 BCl3
Final answer:
HCl is a polar molecule that readily dissolves in water, while O2 and N2 are nonpolar and insoluble. Ionic compounds like CaCl2 and KNO3 are highly soluble in water. BeCl2 is soluble in water, while BCl3 is insoluble.
Explanation:
When classifying materials as polar, ionic, or nonpolar, it's important to consider the types of chemical bonds present and the distribution of charge.
HCl is a polar molecule due to the unequal sharing of electrons between hydrogen and chlorine. It will readily dissolve in water, forming an acidic solution.
O2 is a nonpolar molecule because oxygen atoms share the electrons equally. It is insoluble in water due to the differences in polarity.
CaCl2 is an ionic compound, where calcium ions (positive) and chloride ions (negative) are attracted to each other due to the strong electrostatic forces. It is highly soluble in water.
N2 is a nonpolar molecule with equal sharing of electrons. It is insoluble in water.
C2H6 is a nonpolar molecule, and it is insoluble in water.
KNO3 is an ionic compound, made up of potassium ions and nitrate ions, and it is highly soluble in water.
BeCl2 is a polar molecule due to the unequal sharing of electrons. It is soluble in water, forming acidic solutions.
BCl3 is a nonpolar molecule, and it is insoluble in water.
A voltaic cell is constructed with two silver-silver chloride electrodes, where the halfreaction is AgCl (s) + e- → Ag (s) + Cl- (aq) E° = +0.222 V
The concentrations of chloride ion in the two compartments are 0.0222 M and 2.22 M, respectively. The cell emf is __________ V.
A) 0.212
B) 0.118
C) 0.00222
D) 22.2
E) 0.232
Answer: The cell potential of the cell is +0.118 V
Explanation:
The half reactions for the cell is:
Oxidation half reaction (anode): [tex]Ag(s)+Cl^-(aq.)\rightarrow AgCl(s)+e^-[/tex]
Reduction half reaction (cathode): [tex]AgCl(s)+e^-\rightarrow Ag(s)+Cl^-(aq.)[/tex]
In this case, the cathode and anode both are same. So, [tex]E^o_{cell}[/tex] will be equal to zero.
To calculate cell potential of the cell, we use the equation given by Nernst, which is:
[tex]E_{cell}=E^o_{cell}-\frac{0.0592}{n}\log \frac{[Cl^{-}]_{diluted}}{[Cl^{-}]_{concentrated}}[/tex]
where,
n = number of electrons in oxidation-reduction reaction = 1
[tex]E_{cell}[/tex] = ?
[tex][Cl^{-}]_{diluted}[/tex] = 0.0222 M
[tex][Cl^{-}]_{concentrated}[/tex] = 2.22 M
Putting values in above equation, we get:
[tex]E_{cell}=0-\frac{0.0592}{1}\log \frac{0.0222M}{2.22M}[/tex]
[tex]E_{cell}=0.118V[/tex]
Hence, the cell potential of the cell is +0.118 V
Bobby found that water has a high boiling point and concluded this was because water has weak intermolecular forces. Based on the data, is Bobby’s conclusion correct? A. Yes, because polar covalent molecules have weaker intramolecular forces than nonpolar covalent molecules. B. Yes, because polar covalent molecules have weaker intermolecular forces than nonpolar covalent molecules. C. No, because polar covalent molecules have stronger intramolecular forces than nonpolar covalent molecules. D. No, because polar covalent molecules have stronger intermolecular forces than nonpolar covalent molecules.
Answer:
(D.) No, because polar covalent molecules have stronger intermolecular forces than nonpolar covalent molecules.
Explanation:
In water molecules intermolecular hydrogen bonding occurs between hydrogen and highly electronegative elemnt oxygen that increases the intermolecular forces. Hence water has high boiling point.
Answer:
D. No, because polar covalent molecules have stronger intermolecular forces than nonpolar covalent molecules.
Explanation:
Hello,
Intermolecular forces are those which attract or repel molecules to, or from each other. In this case, since water is the involved molecule, it is known that it is a polar covalent one, as the difference between the electronegativities of oxygen and hydrogen is 1.24, in this manner, it is also known that the more polar the molecule is, the stronger its intermolecular forces are are they are tend to stick the molecules. In such a way, the answer is: D. No, because polar covalent molecules have stronger intermolecular forces than nonpolar covalent molecules.
Best regards.
Indicate what happens to the concentration of Pb2+ in each half cell
a) the left half-cell will increase in concentration; and the right half-cell will decrease in concentration.
b) the left half-cell will increase in concentration; and the right half-cell will increase in concentration.
c) the left half-cell will decrease in concentration; and the right half-cell will decrease in concentration.
d) the left half-cell will decrease in concentration; and the right half-cell wil
Adding sulfuric acid to the half-cell with the lead electrode will decrease the [tex]Pb^{2+}[/tex] ion concentration as [tex]PbSO_{4}[/tex] (s)forms, leading to an increase in cell potential due to a shift in equilibrium to release more [tex]Pb^{2+}[/tex] ions from the electrode. (option C)
The question deals with the behavior of [tex]Pb^{2+}[/tex] ions in a galvanic cell. In general, the addition of sulfuric acid to the half-cell with the lead electrode, causing the formation of [tex]PbSO_{4}[/tex] (s), would lead to a decrease in the concentration of [tex]Pb^{2+}[/tex]ions in the solution.
This is because the [tex]Pb^{2+}[/tex]ions participate in forming the insoluble [tex]PbSO_{4}[/tex] , and the removal of ions from the solution would shift the equilibrium of the half-reaction to replace the ions used in the precipitation.
According to Le Chatelier's principle, this shift in equilibrium would cause more lead ions to be released into the solution from the lead electrode, essentially increasing the cell potential.
Regarding the half-reactions:
The half-reaction occurring at the anode (where oxidation occurs) is typically the loss of electrons by a metal cation.The half-reaction occurring at the cathode (where reduction occurs) is the gain of electrons by ions in the solution.The electrode potentials indicate the tendency of each half-reaction to occur. A negative electrode potential suggests that the substance loses electrons more easily and is therefore prone to oxidation. Conversely, a positive electrode potential indicates a substance that gains electrons readily, suggesting a reduction is favorable.
Consider the following reaction:
CuS(s) + O2(g) ⇌ Cu(s) + SO2(g)
A reaction mixture initially contains 2.9 M O2.
Determine the equilibrium concentration of O2 if Kc for the reaction at this temperature is 1.5.
To determine the equilibrium concentration of O2 in the given reaction, we need to use the equilibrium constant expression and the initial concentration of O2. The equilibrium constant (Kc) for the reaction is 1.5. By substituting the known values into the equilibrium constant expression, we can solve for the equilibrium concentration of O2.
Explanation:In order to determine the equilibrium concentration of O2 in the given reaction, we need to use the equilibrium constant expression and the initial concentration of O2. The equilibrium constant (Kc) for the reaction is 1.5. The reaction is in the form of a reversible reaction, so the concentrations of the reactants and products play a role in determining the equilibrium concentration of O2.
Using the equilibrium constant expression, we have:
Write the expression for Kc using the given reaction equation: Kc = [Cu][SO2] / [CuS][O2]Substitute the known values into the expression: 1.5 = [Cu][SO2] / (x)(2.9)Solve for x to find the equilibrium concentration of O2: x = ([Cu][SO2]) / (1.5 * 2.9)Plug in the values of [Cu] and [SO2] from the given reaction equation to calculate the equilibrium concentration of O2.
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For the given reaction, the equilibrium constant Kc is the ratio of the concentrations of the products to the reactants. Without a change or shift in the equilibrium, the equilibrium concentration of O2 remains the same as its initial concentration, 2.9 M.
Explanation:This problem involves understanding the concept of equilibrium in chemical reactions. From the reaction,
CuS(s) + O2(g) ⇌ Cu(s) + SO2(g)
We know that at equilibrium, the rate of the forward reaction equals to the rate of the reverse reaction. Considering the equilibrium constant, Kc = [Products]/[Reactants], being given as 1.5, we also know that [Cu][SO2]/[CuS][O2] equals 1.5.
Since the concentration of a solid like CuS or Cu in the reaction does not affect the equilibrium, the equilibrium expression is simply [SO2]/[O2] = Kc
In the context of this question, we are provided that the initial concentration of O2 is 2.9 M and are asked to determine the equilibrium concentration of O2. Since there's no shift mentioned in the equilibrium, it indicates that no reaction has taken place. This means the equilibrium concentration of O2 remains the same as its initial concentration, that is, 2.9 M.
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Mr. Auric Goldfinger, criminal mastermind, intends to smuggle several tons of gold across international borders by disguising it as lumps of iron ore. He commands his engineer minions to form the gold into little spheres with a diameter of exactly and paint them black. However, his chief engineer points out that customs officials will surely notice the unusual weight of the "iron ore" if the balls are made of solid gold (density=19.3g/cm^3 ). She suggests forming the gold into hollow balls instead (see sketch at right), so that the fake "iron ore" has the same density as real iron ore (5.15g/cm^3). One of the balls of fake "iron ore," sliced in half. Calculate the required thickness of the walls of each hollow lump of "iron ore." Be sure your answer has a unit symbol, if necessary, and round it to significant digits.
The volume of the hollow gold sphere can be calculated by subtracting the volume of the inner void (hollow space) from the volume of the entire sphere.
What is the thicknessTo get the same density as real iron ore (5.15 g/cm³), the hollow gold spheres must be designed in a way that the total volume of gold is equal to the volume of the iron ore the same mass. The density formula can be rearranged to solve for volume:
Density = Mass / Volume
Volume = Mass / Density
Density of fake "iron ore" = Mass of fake "iron ore" / Volume of fake "iron ore"
The equation:
Density of fake "iron ore" = Mass of fake "iron ore" / Volume of fake "iron ore"Volume of hollow sphere = (4/3) * π * [(radius of outer sphere)³ - (radius of inner sphere)³]
5.15 g/cm³ = Mass of fake "iron ore" / [(4/3) * π * (1³ - (1 - t)³)]
So one then Solve for t:
t = 1 - ( (3 * Mass of fake "iron ore") / (4 * π * 5.15) )^(1/3)
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To achieve the same density as real iron ore, the required thickness of the walls of each hollow lump of 'iron ore' should be approximately 4.02 cm.
To calculate the required thickness of the walls of each hollow lump of 'iron ore,' we can use the formula for density: density = mass/volume.
Since the density of real iron ore is 5.15 g/cm^3 and we want the fake 'iron ore' to have the same density, we can set up the following equation:
5.15 g/cm^3 = (mass of gold)/(volume of hollow sphere)
We know that the mass of gold is the same as the volume of the hollow sphere, since the density of gold is 19.3 g/cm^3. Therefore, we can rewrite the equation as:
5.15 g/cm^3 = 19.3 g/cm^3 / (4/3 * π * (outer radius^3 - inner radius^3))
Solving for the outer radius, we get:
outer radius = (3 * (19.3 g/cm^3) / (5.15 g/cm^3 * 4 * π) )^(1/3) ≈ 4.02 cm
And since the inner radius is 0 (since it's hollow), the required thickness of the walls is the difference between the outer radius and the inner radius:
required thickness = outer radius - inner radius = 4.02 cm - 0 cm = 4.02 cm
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Rose bengal is a chromophore used in biological staining that has an absorption maximum at 559.1 nm and several other shorter wavelength absorption bands in the ultraviolet and visible regions of the spectrum when dissolved in ethanol. What is the energy difference, in kilojoules per mole, between the absorption maximum at 559.1 nm and a band at 263.5nm?
the energy difference is negative, it means that the band at 263.5 nm has a lower energy than the absorption maximum at 559.1 nm.
To calculate the energy difference between two absorption bands, we can use the equation:
[tex]\[ \Delta E = \frac{hc}{\lambda} \][/tex]
Where:
- [tex]\( \Delta E \)[/tex] is the energy difference in joules (J)
- [tex]\( h \)[/tex] is Planck's constant [tex](\(6.62607015 \times 10^{-34} \, \text{J} \cdot \text{s}\))[/tex]
- c is the speed of light in a vacuum [tex](\(3.00 \times 10^8 \, \text{m/s}\))[/tex]
- [tex]\( \lambda \)[/tex] is the wavelength in meters (m)
First, we need to convert the wavelengths from nanometers (nm) to meters (m):
[tex]\[ \lambda_1 = 559.1 \, \text{nm} \times \frac{1 \, \text{m}}{10^9 \, \text{nm}} = 5.591 \times 10^{-7} \, \text{m} \][/tex]
[tex]\[ \lambda_2 = 263.5 \, \text{nm} \times \frac{1 \, \text{m}}{10^9 \, \text{nm}} = 2.635 \times 10^{-7} \, \text{m} \][/tex]
Now, we can calculate the energy difference (\( \Delta E \)):
[tex]\[ \Delta E = \frac{(6.62607015 \times 10^{-34} \, \text{J} \cdot \text{s})(3.00 \times 10^8 \, \text{m/s})}{5.591 \times 10^{-7} \, \text{m}} - \frac{(6.62607015 \times 10^{-34} \, \text{J} \cdot \text{s})(3.00 \times 10^8 \, \text{m/s})}{2.635 \times 10^{-7} \, \text{m}} \][/tex]
[tex]\[ \Delta E = (3.00 \times 10^8 \, \text{m/s})\left(\frac{6.62607015 \times 10^{-34} \, \text{J} \cdot \text{s}}{5.591 \times 10^{-7} \, \text{m}} - \frac{6.62607015 \times 10^{-34} \, \text{J} \cdot \text{s}}{2.635 \times 10^{-7} \, \text{m}}\right) \][/tex]
[tex]\[ \Delta E = (3.00 \times 10^8 \, \text{m/s})(2.3723 \times 10^{-19} \, \text{J} - 2.5161 \times 10^{-19} \, \text{J}) \][/tex]
[tex]\[ \Delta E \approx (3.00 \times 10^8 \, \text{m/s})(-0.1438 \times 10^{-19} \, \text{J}) \][/tex]
[tex]\[ \Delta E \approx -4.314 \times 10^{-12} \, \text{J} \][/tex]
[tex]\[ \Delta E \approx -4314 \, \text{pJ} \][/tex]
Since the energy difference is negative, it means that the band at 263.5 nm has a lower energy than the absorption maximum at 559.1 nm.
The energy difference between the absorption maximum at 559.1 nm and the band at 263.5 nm is 240 kJ/mol.
To calculate the energy difference between the absorption maximum at 559.1 nm and a band at 263.5 nm in kilojoules per mole, follow these steps:
Step 1. Convert Wavelengths to Energy:
Use the equation [tex]\( E = \frac{hc}{\lambda} \)[/tex], where:
(E) is the photon's energy.
Planck's constant [tex](\(6.626 \times 10^{-34}\) J.s)[/tex] is represented as (h).
The speed of light is represented as (c) = [tex](3.00 \times 10^8\)[/tex] m/s.
The wavelength in meters is [tex](\lambda \))[/tex].
Step 2. Calculate Energy for 559.1 nm:
[tex]\[ \lambda_1 = 559.1 \text{ nm} = 559.1 \times 10^{-9} \text{ m} \][/tex]
[tex]\[ E_1 = \frac{6.626 \times 10^{-34} \times 3.00 \times 10^8}{559.1 \times 10^{-9}} = 3.556 \times 10^{-19} \text{ J} \][/tex]
Step 3. Calculate Energy for 263.5 nm:
[tex]\[ \lambda_2 = 263.5 \text{ nm} = 263.5 \times 10^{-9} \text{ m} \][/tex]
[tex]\[ E_2 = \frac{6.626 \times 10^{-34} \times 3.00 \times 10^8}{263.5 \times 10^{-9}} = 7.552 \times 10^{-19} \text{ J} \][/tex]
Step 4. Find the Energy Difference:
[tex]\[ \Delta E = E_2 - E_1 = 7.552 \times 10^{-19} \text{ J} - 3.556 \times 10^{-19} \text{ J} = 3.996 \times 10^{-19} \text{ J} \][/tex]
Step 5. Convert to Kilojoules per Mole:
[tex]\[ 1 \text{ photon} = 3.996 \times 10^{-19} \text{ J} \][/tex]
[tex]\[ 1 \text{ mole of photons} = 3.996 \times 10^{-19} \text{ J} \times 6.022 \times 10^{23} \text{ photons/mol} \][/tex]
[tex]\[ \Delta E_{\text{mol}} = 2.406 \times 10^{5} \text{ J/mol} = 240.6 \text{ kJ/mol} \][/tex]
So, the energy difference between the absorption maximum at 559.1 nm and the band at 263.5 nm is 240.6 kJ/mol.
What is the angle between two of the nitrogen-hydrogen bonds in the ammonium (NH4+) ion?
Answer:
109° 27'
Explanation:
The ammonium ion is tetrahedral in shape, all the HNH bonds are exactly at the tetrahedral bond angle since there are only bond pairs in the structure and no lone pairs. Recall that lone pairs decrease the bond angke from the ideal value in a tetrahedron due to higher repulsion.
The heating curve for a sample of pure ethanol shows the slope of the line for gaseous state is greater than the slope of the line for the liquid state.
Which statement below explains that?a. The specific heat capacity of the gaseous ethanol is less than the specific heat capacity of liquid ethanol.b. The specific heat capacity of the gaseous ethanol is greater than the specific heat capacity of liquid ethanol.c. The heat of vaporization of ethanol is less than the heat of fusion of ethanol.d. The heat of vaporization of ethanol is greater than the heat of fusion of ethanol.
Answer:
a. The specific heat capacity of the gaseous ethanol is less than the specific heat capacity of liquid ethanol.
Explanation:
The heating curve is a curve that represents temperature (T) in the y-axis vs. added heat (Q) in the x-axis. The slope is T/Q = 1/C, where C is the heat capacity. Then, the higher the slope, the lower the heat capacity. For a constant mass, it can also represent the specific heat capacity (c).
Heats of vaporization and fusion cannot be calculated from these sections of the heating curve.
Which statement below explains that?
a. The specific heat capacity of the gaseous ethanol is less than the specific heat capacity of liquid ethanol. YES.
b. The specific heat capacity of the gaseous ethanol is greater than the specific heat capacity of liquid ethanol. NO.
c. The heat of vaporization of ethanol is less than the heat of fusion of ethanol. NO.
d. The heat of vaporization of ethanol is greater than the heat of fusion of ethanol. NO.
The correct option is b. The specific heat capacity of the gaseous ethanol is greater than the specific heat capacity of liquid ethanol.
To understand why option b is correct, let's consider the heating curve for ethanol and what the slope of the line represents. The slope of the line on a heating curve indicates the rate at which the temperature of the substance increases per unit of heat added. A steeper slope means that the temperature rises more quickly for a given amount of heat.
In the context of the heating curve for ethanol:
- The slope of the line for the gaseous state is steeper than the slope for the liquid state. This means that for the same amount of heat added, the temperature of gaseous ethanol increases more than the temperature of liquid ethanol.
- The specific heat capacity (c) of a substance is the amount of heat required to raise the temperature of one gram of the substance by one degree Celsius. A higher specific heat capacity means that more heat is required to achieve the same temperature change.
Now, let's analyze the options:
a. The specific heat capacity of the gaseous ethanol is less than the specific heat capacity of liquid ethanol. - This would mean that gaseous ethanol requires less heat to increase its temperature by a certain amount compared to liquid ethanol. However, the steeper slope for the gaseous state indicates that it takes more heat to increase the temperature of the gas by a certain amount, not less. Therefore, this option is incorrect.
b. The specific heat capacity of the gaseous ethanol is greater than the specific heat capacity of liquid ethanol. - This is consistent with the observation that the temperature of the gas increases more slowly than that of the liquid for the same amount of heat added (as indicated by the steeper slope for the gas). Since the gas requires more heat to increase its temperature, it has a higher specific heat capacity. This option is correct.
c. The heat of vaporization of ethanol is less than the heat of fusion of ethanol. - The heat of vaporization is the amount of heat required to change a substance from the liquid to the gaseous state, while the heat of fusion is the amount of heat required to change a substance from the solid to the liquid state. This option is incorrect because the heat of vaporization is typically much greater than the heat of fusion for most substances, including ethanol.
d. The heat of vaporization of ethanol is greater than the heat of fusion of ethanol. - This statement is generally true for most substances, including ethanol. However, it does not explain the difference in slopes between the gaseous and liquid states on the heating curve. The heat of vaporization is related to the amount of heat required to change the phase of the substance, not the rate at which the temperature changes within a given phase. Therefore, this option is not relevant to the explanation of the slopes on the heating curve.
In conclusion, the correct explanation for the observation that the slope of the line for gaseous ethanol is greater than the slope for liquid ethanol is that the specific heat capacity of gaseous ethanol is greater than that of liquid ethanol. This means that more heat is required to raise the temperature of the gas by a certain amount compared to the liquid, which is consistent with the steeper slope on the heating curve for the gaseous state.
A Mercury(II) chloride dissolves in water to give poorly conducting solutions, indicating that the compound is largely nonionized in solution—it dissolves as HgCl2 molecules. Describe the bonding of the HgCl2 molecule, using valence bond theory. b Phosphorus trichloride, PCl3, is a colorless liquid with a highly irritating vapor. Describe the bonding in the PCl3 molecule, using valence bond theory. Use hybrid orbitals.
Answer:
Using valence bond theory, HgCl2 is sp hybridized while PCl3 is sp3 hybridized.
Explanation:
sp hybridized molecules are largely covalent and cannot conduct electricity hence the nonconducting nature of HgCl2. PCl3 contains a lone pair and three unpaired electrons in sp3 hybridized orbitals.
Nitrogen (N2) enters a well-insulated diffuser operating at steady state at 0.656 bar, 300 K with a velocity of 282 m/s. The inlet area is 4.8 x 10-3 m2. At the diffuser exit, the pressure is 0.9 bar and the velocity is 140 m/s. The nitrogen behaves as an ideal gas with k = 1.4. (a) Determine the exit temperature, in K, and the exit area, in m2. (b) For a control volume enclosing the diffuser, determine the rate of entropy production, in kW/K.
The problem is solved by applying the principles of thermodynamics and fluid dynamics for an ideal gas under isentropic conditions in a diffuser, and involving calculations for temperature, exit area and entropy production.
Explanation:This problem involves the principles of thermodynamics and fluid dynamics, specifically in relation to the behavior of nitrogen gas in a diffuser. To solve it, we need to apply the ideal gas law and the principles of energy and entropy conservation.
For part (a), we can start by using the formula for isentropic relations in an ideal gas, which states that T2/T1 = (P2/P1)^((k-1)/k), where k is the specific heat ratio of the gas. By substituting the given values, and also considering the conservation of mass (continuity equation), we can calculate the exit temperature (T2) and the exit area.
Similarly, for part (b), using the formula for rate of entropy production (ΔS_gen = m_dot*[s2 - s1 + R*ln(P2/P1)]), where m_dot is the mass flow rate and R is the specific gas constant, we can find the rate of entropy production.
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Which of the following provides an instantaneous measure of radioactivity:1 Scintillation counter2. Geiger counter3. Film-badge dosimeterB onlyB and C onlyA and B onlyA, B, and C
Answer:
A and B only
Explanation:
Both Scintillation counter and Geiger counter are used for instantaneous measures of radioactivity.
Both are used to detect and quantify the amount of radiation. GM counter can detect all kind of radiation that alpha, beta and gamma radiation. Whereas Scintillation counter can detect only ionization radiation.
Why do the metals Co, Rh, and Lr form octahedral complexes (rather than tetrahedral or square planar complexes)? Hint: Look at the placement in the periodic table and the respective electron configurations.
Explanation:
Octahedral complexes will be favoured over tetrahedral ones because:
It is more favourable to form six bonds rather than four
The crystal field stabilisation energy is usually greater for octahedral than tetrahedral complexes.
The transition metals Co, Rh, Lr are in group 9 of d block and they have 3d, 4d, and 5d orbitals respectively
Metals like Co, Rh, and Lr form octahedral complexes due to their electron configurations and the stability achieved with six ligands. This coordination results in a geometry that minimizes electron repulsions and maximizes stability.
Transition metals like Co (Cobalt), Rh (Rhodium), and Lr (Lawrencium) typically form octahedral complexes due to their electron configurations and positions in the periodic table. These metals have available d-orbitals that can accommodate six ligands, resulting in a coordination number of six which is most stable in an octahedral geometry.
For instance, Co³⁺ has an electron configuration that favors the formation of octahedral complexes due to the stabilization of its d-electrons in a ligand field that splits the d-orbitals into two energy levels, with four orbitals at lower energy and two at higher energy. Similarly, Rh³⁺ and Lr³+ configurations also favor six-coordinate octahedral structures, providing maximum separation of electron pairs and minimizing electron-electron repulsions.
Examples:
[Co(NH₃)₆]³⁺: An example of an octahedral complex where the cobalt ion is bonded to six ammonia ligands.
[RhCl₆]³⁺: An octahedral complex where the rhodium ion is coordinated by six chloride ligands.
With respect to energy, which is not necessarily a goal of chemistry ?
finding ways to produce energy
finding ways to conserve energy
finding ways to store energy
finding ways to use energy
All listed are common goals about energy in chemistry, including producing, conserving, storing, and using energy. However, these aren't the inherent goals of chemistry, which is more focused on the study of substances and their transformations.
Explanation:In the context of energy within the field of chemistry, all of the options you've listed are typical goals: finding ways to produce energy, conserve energy, store energy, and use energy. Chemistry plays a crucial role in these areas, such as in the development of new energy sources or improving energy efficiency. However, these aren't necessarily inherent goals of chemistry. The inherent goal of chemistry is to study the properties, composition, and structure of substances, and the transformations they undergo.
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When baking soda (sodium bicarbonate or sodium hydrogen carbonate, NaHCO3) is heated, it releases carbon dioxide gas, which is responsible for the rising of cookies, doughnuts, and bread.
a) Write a balanced equation for the decomposition of the compound (one of the products is Na2CO3).
b) Calculate the mass of NaHCO3 required to produce 20.5 g of CO2.
Answer:
a. NaHCO₃ + HCl → NaCl + H₂O + CO₂
b. 39.14 g is the mass of NaHCO₃ required to produce 20.5 moles of CO₂
Explanation:
A possible reaction for NaHCO₃ to make dioxide is this one, when it reacts with hydrochloric to produce the mentioned gas.
NaHCO₃ + HCl → NaCl + H₂O + CO₂
Ratio in this reaction is 1:1
So 1 mol of baking soda, produce 1 mol of CO₂
Let's calculate the moles
20.5 g CO₂ / 44 g/m = 0.466 moles
This moles of gas came from the same moles of salt.
Molar mass baking soda = 84 g/m
Molar mass . moles = mass
84 g/m . 0.466 moles = 39.14 g
What is Δn for the following equation in relating Kc to Kp?SO3(g) + NO(g) ↔ SO2(g) + NO2(g)12-20-1
Answer:
0
Explanation:
The relation between Kp and Kc is given below:
[tex]K_p= K_c\times (RT)^{\Delta n}[/tex]
Where,
Kp is the pressure equilibrium constant
Kc is the molar equilibrium constant
R is gas constant
T is the temperature in Kelvins
Δn = (No. of moles of gaseous products)-(No. of moles of gaseous reactants)
For the first equilibrium reaction:
[tex]SO_3_{(g)}+NO_{(g)}\rightleftharpoons SO_2_{(g)}+NO_2_{(g)}[/tex]
Δn = (2)-(2) = 0
Thus, Kp is:
[tex]K_p= Kc\times \times (RT)^{0}[/tex]
[tex]K_p= Kc[/tex]
Final answer:
The Δn represents the difference in moles of gaseous products and reactants for a reaction when relating Kc to Kp. For the reaction SO3(g) + NO(g) ↔ SO2(g) + NO2(g), Δn would be 0. Kp is related to Kc by the equation Kp = Kc(RT)Δn.
Explanation:
The question relates to the concept of the reaction quotient (Δn) when relating the equilibrium constants Kc (equilibrium constant in terms of concentration) to Kp (equilibrium constant in terms of partial pressure) for a given chemical reaction involving gases. The value of Δn is the difference in the sum of the moles of gaseous products and the sum of the moles of gaseous reactants in a balanced chemical equation. In the example given:
[tex]SO_{3}[/tex] (g) + NO(g) ↔ [tex]SO_{2}[/tex] (g) + [tex]NO_{2}[/tex] (g), Δn would be (1 + 1) - (1 + 1) = 0.
1[tex]N_{2}[/tex](g) + [tex]2H_{2} O[/tex](g) ↔ 2NO(g) + [tex]2H_{2}[/tex] (g), Δn would be (2 + 2) - (1 + 2) = 1.
To relate Kc and Kp, we use the equation Kp = Kc(RT)Δn, where R is the gas constant and T is the temperature in Kelvins. In cases where Δn is zero, as in the first equation, Kp will be equal to Kc because (RT)0 equals 1.
Modern commercial airliners are largely made of aluminum, a light and strong metal. But the fact that aluminum is cheap enough that airplanes can be made out of it is a bit of historical luck. Before the discovery of the Hall-Héroult process in , aluminum was as rare and expensive as gold. What would happen if airplanes had to be made of steel? The fuselage of the Airbus A380, which can carry passengers, is approximately a hollow aluminum cylinder without ends, long, wide, and thick (see sketch at right). The fuselage of an airplane Suppose this fuselage was made of steel (density ) instead of aluminum (density ), and let's say the average passenger has a mass of . We'll also assume the engines can't lift any greater mass than they already do. Calculate the number of passengers that the Airbus A380 could carry if its fuselage was made of steel.
To approach this problem, let's start by understanding what we have and what we need to find. We know:
1. The fuselage is in the shape of a cylindrical shell with approximate dimensions: length of 20 meters, diameter of 4 meters, and thickness of 0.01 meters.
2. The densities of aluminum and steel are 2700 kg/m³ and 7800 kg/m³ respectively.
3. The average mass of a passenger is 70 kg.
Now, we must find: the number of passengers that a steel-made Airbus A380 could carry, given that the total mass that the engines can lift stays the same.
First, we have to calculate the volume of the aluminum fuselage. Since the fuselage is a cylindrical shell, the volume can found using the formula for the volume of a cylinder: `π*r²*h` where r is the radius and h is the height or length of the cylinder in this context. The radius is the width divided by 2. To account for thickness, we subtract the volume of the inner cylinder from the volume of the outer cylinder.
To find the mass of the aluminum fuselage, we multiply the volume just calculated with the given density of aluminum.
Next, we calculate the mass of the equivalent steel fuselage. Given that the volume remains the same, we can multiply the volume of the fuselage with the density of steel to find the mass, as mass is the product of volume and density.
To find the difference in mass between the steel and aluminum fuselages, we subtract the mass of the aluminum fuselage from the steel fuselage.
This gives us 12785.653 kg, which is the additional weight when we switch from an aluminum fuselage to a steel fuselage.
Finally, we calculate how many passengers could be carried with this additional weight. Given the average mass of a passenger, knowing that this mass difference will directly affect how many passengers can be carried, we divide the weight difference by the average passenger mass of 70 kg to get the number of lost seats.
The mass of the aluminum fuselage is approximate 6768.88 kg, the mass of the steel fuselage is approximate 19554.53 kg. The difference in mass is about 12785.65 kg. This extra weight when changing to a steel fuselage would reduce the number of passengers by around 183.
So the aircraft would lose around 183 seats if the fuselage was made from steel instead of aluminum.
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Final answer:
If the Airbus A380's fuselage were made from steel instead of aluminium, it would significantly increase the aircraft's weight, thereby reducing the maximum number of passengers it could carry, given the fixed capacity of the engines.
Explanation:
The question revolves around the impact of using steel instead of aluminium to construct the fuselage of an Airbus A380 and how that would affect the number of passengers it could carry. To address this, one must understand the properties that make aluminum preferable for such applications. Aluminum is a lightweight metal with a density significantly lower than that of steel. This attribute allows airplanes to carry more passengers and cargo without exceeding the maximum lift capacity of the aircraft's engines.
Steel, on the other hand, is much denser than aluminum. If the Airbus A380's fuselage were made from steel, its overall mass would substantially increase, thus reducing the number of passengers it could carry. As the engines are designed with specific thrust capacities that cannot be exceeded, the added weight from using steel would necessitate a reduction in the payload (e.g., passengers). Since the student hasn't provided the exact densities or other needed numerical values, a numeric calculation cannot be performed. However, the general principle is clear: using steel would mean transporting fewer passengers due to the increased weight of the aircraft's structure.
The evolution of aircraft construction materials from wood and fabric to metal alloys and composites has been instrumental in the advancement of modern aviation, leading to more efficient and larger aircraft, as exemplified by the Airbus A380, the world's largest commercial jetliner capable of carrying up to 850 passengers.
Acid chlorides can be prepared from carboxylic acids by treatment with ________.
A) PBr3
B) SOCl_2
C) KCl
D) both A and B
E) both B and C
Answer:D
Explanation:
Both thionyl chloride and phosphorus trichloride are used in the synthesis of acid chlorides from carboxylic acids. The thionyl chloride leads to a chlorosulphite intermediate before the acid chloride is formed
If you lived in Alaska, which of the following natural gases could you keep in an outdoor storage tank in winter: a. Methane (CH4) b. Propane (C3H8) c. Butane (C4HIO)? Explain why.
Answer:
Methane
Explanation:
The gas that you could keep in an outdoor storage tank in winter in Alaska is Methane.
The reason is the extreme low temperature during the winter. The boiling point of butane is 44 ºF ( -1ºC) and that of propane is a higher -43.6 º F but still within the range of average minimum winter temperature in Alaska (-50 ªF). Therefore we will have condensation in the tanks and not enough gas pressure.
Methane having a boling point of -259 ºF will not condense at the low wintertime temperatures in Alaska.
At a particular temperature, K = 1.00×102 for the reaction: H2(g) + F2(g)= 2HF(g) In an experiment, at this temperature, 1.40 mol of H2 and 1.40 mol of F2 are introduced into a 1.14-L flask and allowed to react. At equilibrium, all species remain in the gas phase. What is the equilibrium concentration (in mol/L) of H2?
The equilibrium concentration of H2 in the given reaction is approximately 0.614 mol/L.
Explanation:Given that the equilibrium constant (K) is 1.00×10^2 for the reaction H2(g) + F2(g) ⇌ 2HF(g), we can use the stoichiometry of the reaction to determine the equilibrium concentration of H2. Since 1 mol of H2 reacts with 1 mol of F2 to form 2 mol of HF, the concentration of H2 at equilibrium will be half of its initial concentration. Therefore, the equilibrium concentration of H2 in the 1.14-L flask would be 1.40 mol divided by 2 and then divided by 1.14 L, which gives us approximately 0.614 mol/L.
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The equilibrium concentration of H₂ is calculated to be 0.205 M.
To find the equilibrium concentration of H₂, we first need to set up an ICE (Initial, Change, Equilibrium) table:
Initial concentrations:
Initial concentration of H₂ = 1.40 mol / 1.14 L = 1.23 M Initial concentration of F₂ = 1.40 mol / 1.14 L = 1.23 M Initial concentration of HF = 0 MChanges in concentration:
Change in H₂ = -xChange in F₂ = -xChange in HF = +2xEquilibrium concentrations:
Equilibrium concentration of H₂ = 1.23 - xEquilibrium concentration of F₂ = 1.23 - xEquilibrium concentration of HF = 2xAccording to the equilibrium constant expression (K):
[tex]K = [HF]^2/ [H_2][F_2][/tex]
Substitute the equilibrium concentrations into the expression:
[tex]1.00 \times 10^2 = (2x)^2 / (1.23 - x)(1.23 - x)[/tex]
This simplifies to:
[tex]100 = 4x^2 / (1.23 - x)^2[/tex]
Taking the square root of both sides:
10 = 2x / (1.23 - x)
Solve for x:
10(1.23 - x) = 2x
12.3 - 10x = 2x
12.3 = 12x
x = 1.025
The equilibrium concentration of H₂ is:
[H₂ ] = 1.23 - 1.025 = 0.205 M
The standard enthalpy of certain reaction is approximately constant at +125kJmol-1 from 800K to 1500K. The standard Gibbs energy is +25kJmol-1 at 1150K. Estimate the temperature at witch the equilibrium constant becomes greater than 1.
Answer:
T = 3006.976 K
Explanation:
∴ ΔH° = 125 KJ/mol ( 800K - 1500K)
∴ ΔG° = 25 KJ/mol (1150 K)
⇒ T = ? ∴ K > 1
In the equilibrium:
K = e∧(ΔG°/RT)∴ K > 1
If ΔG°/RT = 1
⇒ e∧(1) = 2.72 > 1
∴ ΔG°/RT = 1
⇒ T = ΔG°/R = (25 KJ/mol)/(8.314 E-3 KJ/mol.K)
⇒ T = 3006.976 K
verifying:
K = e∧(ΔG°/RT)⇒ K = e∧(25/((8.314 E-3)(3006.976)))
⇒ K = 1.000000061 > 1
Final answer:
The student is asking about estimating the temperature at which the equilibrium constant becomes greater than 1 based on the given standard enthalpy and Gibbs energy values.
Explanation:
The given question is related to standard enthalpy and Gibbs energy of a reaction, and the estimation of temperature at which the equilibrium constant becomes greater than 1.
Firstly, it is stated that the standard enthalpy of the reaction is approximately constant at +125 kJmol-1 from 800K to 1500K. This information helps us understand the change in enthalpy with respect to temperature.
Secondly, the standard Gibbs energy is given as +25 kJmol-1 at 1150K. This value allows us to calculate the equilibrium constant using the equation: ΔG = -RT lnK.
To estimate the temperature at which the equilibrium constant becomes greater than 1, we need to solve for temperature in the equation K = e-ΔG/T and find the value of T that results in K > 1.
The use of fertilizers in agriculture has significantly altered several nutrient cycles including:
Answer:
potassium, nitrogen and phosphorous cycle
Explanation:
A fertilizer is a substance which is applied on the plants by farmers to increase the supply of nutrients for the plants. Fertilizers have known to be toxic in many ways such as they alter the potassium, nitrogen and phosphorus cycles. Nitrogen, potassium and phosphorus are present in abundant amounts in the fertilizers. Draining of these fertilizers into rivers and ponds is toxic for the aquatic life. Hence, the use of fertilizers disrupts the natural cycles and is toxic for many aquatic plants and animals.
A meter of polyaramide fiber has a diameter of 710.μm , a mass of 0.059g , and takes 0.13s to produce on an industrial spinneret. How would you calculate the mass of fiber that can be spun in 55.min ?
Answer:
t = 55 min ⇒ m = 1497.692 g
Explanation:
polyaramide fiber:
∴ L = 1 m
∴ d = 710 μm
∴ m = 0.059 g
∴ t = 0.13 s
mass flow:
∴ mf = 0.059 g / 0.13 s = (0.454 g/s)×(60s/min) = 27.23 g/min
If t = 55 min:
⇒ m = mf×t = (27.23 g/min)×(55 min) = 1497.692 g
Answer:
[tex]m_{fiber}=1497.7g[/tex]
Explanation:
Hello,
In this case, as both the diameter and the length of the fiber is the same for the second case, one applies a simple rule of three to compute the mass of fiber that can be spun in 55 min as shown below:
[tex]m_{fiber}=\frac{55min*0.059g}{0.13s*\frac{1min}{60s} }\\m_{fiber}=1497.7g[/tex]
Best regards.
The degradation of CF3CH2F (an HFC) by OH radicals in the troposphere is first order in each reactant and has a rate constant of k = 1.6 x 10^8 M^-1s^-1 at 4°C.
Part A) If the tropospheric concentrations of OH and CF3CH2F are 8.1 x 10^5 and 6.3 x 10^8 molecules/cm^3, respectively, what is the rate of reaction at this temperature in M/s?
Answer:
2.1 × 10⁻¹⁹ M/s
Explanation:
Let's consider the degradation of CF₃CH₂F by OH radicals.
CF₃CH₂F + OH → CF₃CHF + H₂O
Considering the order of reaction for each reactant is 1 and the rate constant is 1.6 × 10⁸ M⁻¹s⁻¹, the rate law is:
r = 1.6 × 10⁸ M⁻¹s⁻¹.[CF₃CH₂F].[OH]
where,
r is the rate of the reaction
If the tropospheric concentrations of OH and CF₃CH₂F are 8.1 × 10⁵ and 6.3 × 10⁸ molecules/cm³, respectively, what is the rate of reaction at this temperature in M/s?
The Avogadro's number is 6.02 × 10²³ molecules/mole.
The molar concentration of OH is:
[tex]\frac{8.1 \times 10^{5}molecules}{cm^{3}}.\frac{1mol}{6.02 \times 10^{23}molecules }.\frac{1000cm^{3} }{1L} =1.3 \times 10^{-15} M[/tex]
The molar concentration of CF₃CH₂F is:
[tex]\frac{6.3 \times 10^{8}molecules}{cm^{3}}.\frac{1mol}{6.02 \times 10^{23}molecules }.\frac{1000cm^{3} }{1L} =1.0 \times 10^{-12} M[/tex]
r = 1.6 × 10⁸ M⁻¹s⁻¹ × 1.0 × 10⁻¹² M × 1.3 × 10⁻¹⁵ M = 2.1 × 10⁻¹⁹ M/s
The composition of a compound used to make polyvinyl chloride (PVC) is 38.4% C, 4.8% H and 56.8% Cl by mass. It took 7.73 min for a given volume of the compound to effuse through a porous plug, but it took only 6.18 min for the same amount of Ar to diffuse at the same temperature and pressure. What is the molecular formula of the compound?
Answer:
C₂H₃Cl
Explanation:
We can calculate the compound's molar mass using the data given by the problem and Graham's law:
Rate₁/Rate₂ = [tex]\sqrt{\frac{M_{2}}{M_{1}} }[/tex]
In this case the subscript 1 refers to the compound and 2 refers to Ar.
Keeping in mind that Rate = volume/time, and that the volume is the same for both compounds, we can rewrite the equation as:
Time₂/Time₁ = [tex]\sqrt{\frac{M_{2}}{M_{1}} }[/tex]
6.18/7.73 = [tex]\sqrt{\frac{39.95}{M_{1}} }[/tex]
M₁ = 62.5 g/mol
Now we determine the molecular formula by using the elemental % analysis:
Assuming we have 1 mol of the compound:
C ⇒ 62.5 g * 38.4/100 = 24 g C = 2 mol C
H ⇒ 62.5 g * 4.8/100 = 3 g H = 3 mol H
C ⇒ 62.5 g * 56.8/100 = 35.5 g Cl = 1 mol Cl
Thus the molecular formula is C₂H₃Cl
The catalytic decomposition of hydrogen peroxide yields oxygen gas and water, according to the reaction given below. In an experiment, the decomposition of hydrogen peroxide yielded 75.3 mL of gas collected over water at 25°C and 0.976 atm. (Water = 0.032 atm at 25°C) 2 H2O2(aq) → 2 H2O(l) + O2(g) If the original H2O2 solution had a volume on 500 mL, what was the molar concentration of hydrogen peroxide prior to decomposition?
Answer:
The molar concentration of hydrogen peroxide is 0.01164 M
Explanation:
Step 1: Data given
Volume of gas yielded = 75.3 mL = 0.0753 L
Temperature = 25.0 °C
Atmosphere = 0.976 atm
(Water = 0.032 atm at 25°C)
The original volume of H2O2 is 500 mL
Molar mass of O2 =32 g/mol
Step 2: The balanced equation
2H2O2(aq) → 2 H2O(l) + O2(g)
Step 3: Calculate pressure of O2
P = P(02) + P(water)
P(O2) = P - P(water)
P(O2) = 0.976 atm - 0.032 atm
P(O2) = 0.944 atm
Step 4: Calculate moles O2
p*V=n*R*T
⇒ with p = the pressure of O2 gas = 0.944 atm
⇒ with V= the volume of O2 = 0.0753 L
⇒ with n = the number of moles of O2
⇒ with R = the gas constant = 0.08206 L*atm/K*mol
⇒ with T = the temperature = 25°C = 298 Kelvin
n = (p*V)/(R*T)
n = (0.944 * 0.0753)/(0.08206*298K)
n = 0.00291 moles O2
Step 5: Calculate moles of H2O2
For 1 mole of O2 produced, we need 2 moles of H2O2
For 0.00291 moles O2 we need 2*0.00291 = 0.00582 moles H2O2
Step 6: Calculate molar concentration of H2O2
Molar concentration = moles / volume
Molar concentration = 0.00582 moles / 0.500L
Molar concentration = 0.01164 M
The molar concentration of hydrogen peroxide is 0.01164 M
A closed vessel system of volume 2.5 L contains a mixture of neon and fluorine. The total pressure is 3.32 atm at 0.0°C. When the mixture is heated to 15°C, the entropy of the mixture increases by 0.345 J/K. What amount (in moles) of each substance (Ne and F2) is present in the mixture? (heat capacity of monoatomic gas = 3/2R and that for a diatomic gas = 5/2R)
Answer: moles of Ne = 0.149 moles
moles of F₂ = 0.221 moles
Explanation:
The process occurs at costant volume.
Neon is a monoatomic gas, Cv =3/2R, Flourine is a diatomic gas, Cv = 5/2R
n = PV/RT ; where n is total = number of moles, P is total pressure = 3.32atm, V is volume = 2.5L, R is molar gas constant = 0.08206 L-atm/mol/K = 8.314 J/mol/K, T is temperature = 0.0°C = 273.15K
n(total) = (3.32 atm)(2.5 L)/(0.08206 L-atm/mol/K)(273.15) = 0.3703 mol
For one mole heated at constant volume,
Change in entropy, ∆S = ∫dq/T = ∫(Cv/T)dT
From T1 to T2, Cvln(T2/T2) = Cvln(288.15/273.15) = 0.05346•Cv
So, for 0.3703 moles,
∆S = (0.3703 mol)(0.05346)Cv = 0.345 J/K
⇒ Cv = 17.43 J/mol/K for the Ne/F₂ mixture.
For pure Ne, Cv = (3/2)R = 1.5 • 8.314 J/mol/K = 12.471 J/mol/K
For pure F₂, Cv = (5/2)R = 2.5 • 8.314 J/mol/K = 20.785 J/mol/K
If Y is the mole fraction of Ne, Y can be determined by setting the observed entropy change (∆S) to the weighted average of the entropy changes expected for the two different gases in the mixture:
17.43 J/mol/K = Y * 12.471 J/mol/K + (1 – Y) * 20.785 J/mol/K
⇒ 20.785 J/mol/K – 8.314 J/mol/K * Y = 17.43J/mol/K
Y = 0.403 ; 1 – Y = 0.597
moles of Ne = (0.403)(0.3703 mol) = 0.149 moles
moles of F₂ = (0.597)(0.3703 mol) = 0.221 moles