Answer:
We need 41.2 L of propane
Explanation:
Step 1: Data given
volume of H2O = 165 L
Step 2: The balanced equation
C3H8(g) + 5O2(g) → 3CO2(g) + 4H2O(g)
Step 3: Calculate moles of H2O
1 mol = 22.4 L
165 L = 7.37 moles
Step 4: Calculate moles of propane
For 1 mol C3H8 we need 5 moles O2 to produce 3 moles CO2 and 4 moles H2O
For 7.37 moles H2O we need 7.37/4 = 1.84 moles propane
Step 5: Calculate volume of propane
1 mol = 22.4 L
1.84 moles = 41.2 L
We need 41.2 L of propane
A lead ball has a mass of 55.0 grams and a density of 11.4 g/cm3. What is the volume of the ball?
A) 0.207 mL
B) 0.207 L
C) 4.82 mL
D) 4.82 L
E) none of the above
Answer:
B
Explanation:
Answer:
The volume of the ball is 4.82 mL (opton C)
Explanation:
Step 1: Data given
Mass of the lead ball = 55.0 grams
Density = 11.4 g/cm³
Step 2: Calculate volume of the balloon
Volume = mass / density
Volume = 55.0 g / 11.4 g/cm³
Volume = 4.82 cm³ = 4.82 mL
The volume of the ball is 4.82 mL (opton C)
A goose is flying south for the winter at a constant speed. Keep in mind that one mile is 1.61 km, and one pound is 454 g.The goose has a mass of 20.1 lb and is flying at 9.60 miles/h. What is the kinetic energy of the goose in joules?
Answer:
kinetic energy = 84.0260 joules
Explanation:
given data
one mile = 1.61 km
one pound = 454 g
mass = 20.1 lb = 9.1172 kg
flying velocity = 9.60 miles/h = 9.60 × 1.61 × [tex]\frac{5}{18}[/tex] = 4.2933 m/s
solution
we get here kinetic energy that is express as
kinetic energy = 0.5 × m × v² .........................1
put here value and we get
kinetic energy = 0.5 × 9.1172 × 4.2933²
kinetic energy = 84.0260 joules
The kinetic energy of the goose, when converted to the same units, is approximately 83.74 joules.
Explanation:The kinetic energy for any object can be found using the equation K.E. = 1/2*m*v^2, where m is mass and v is the velocity. First we need to convert the mass of the goose from pounds to kilograms, and the speed from miles per hour to meters per second. So, the mass should be 20.1 lb * 454 g = 9121.4 g. Converting that to kilograms, we get 9121.4 g * 1 kg/1000 g = 9.1214 kg. The speed of the goose should be 9.6 miles/hour * 1.61 km = 15.44 km/h. Converting that to meters per second, we get 15.44 km/h * 1000 m/1 km * 1 h/3600 s = 4.2889 m/s. Finally, we can substitute these values into the kinetic energy equation
K.E = 1/2 * 9.1214 kg * (4.2889 m/s)^2 = 83.74 joules.
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Draw the structure of the major organic product isolated from the reaction of 1-hexyne with hydrogen chloride (2 mol). Click the "draw structure" button to launch the drawing utility.
Answer:
The structure of the major organic product isolated from the reaction of 1-hexyne with hydrogen chloride (2 mol) is attached below.
Explanation:
Hydracids are added to triple bonds by a mechanism similar to that of the addition to double bonds. The regioselectivity of the addition of H-X to the triple bond follows the rule of Markovnikov, where the Z conformation predominates in the addition of halide to the alkyne, because in the formation of the carbocation it prefers to place the positive charge on the more substituted carbon where the nucleophilic attack of the halide ion will occur.
With respect to the halogenation of the alkene, the same procedure occurs at the time of the formation of the carbocation, joining the nucleophilic ion to the most substituted carbon.
a mammoth skeleton has a carbon-14 content of 12.50 of that found in living organisms
True/False
Correct Question:
A mammoth skeleton has a carbon-14 content of 12.50% of that found in living organisms. When did the mammoth live?
Answer:
17,190 years ago.
Explanation:
The carbon-14 is a carbon isotope that has a half-life of 5730 years. It means that the mass of carbon-14 decreases by half every 5730 years. If we call the initial mass as M and the final mas as m:
m = M/2ⁿ
Where n is the amount of half-live that had past. Thus, in this case, the mass of the skeleton is the final mass, and the mass founded in the living organisms the initial mass, so:
m = 0.1250M, and
0.1250M = M/2ⁿ
2ⁿ = M/0.1250M
2ⁿ = 1/0.1250
2ⁿ = 8
2ⁿ = 2³
n = 3
So it had past 3 half-live or
3*5730 = 17,190 years.
So the mammoth lived 17,190 years ago.
The statement is potentially true, depending on the age of the particular mammoth skeleton. It's expected that a mammoth skeleton, as a remnant of an organism that's been dead for thousands of years, would have a lower carbon-14 content than living organisms. However, the specific amount can vary based on the particular mammoth and its age.
Explanation:The statement is potentially true, but it depends on the specific mammoth and its carbon-14 content. Carbon-14 is a radioactive isotope found in all living organisms. When an organism dies, it stops absorbing carbon-14, and the isotope begins to decay. Over time, the amount of carbon-14 in a dead organism's remains decreases. Thus, it's possible that a mammoth skeleton, which belonged to a species that went extinct thousands of years ago, would have a carbon-14 content lower than that of living organisms today. However, the carbon-14 in a specific mammoth skeleton could also be less than 12.50 percent of that in a present living organism if the skeleton is exceptionally old, or more if the skeleton is relatively recent.
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Free radical chlorination of ethane can produce higher halogenation products (dichlorinated, trichlorinated, etc...) in addition to chloroethane. How could the production of higher halogenated products be minimized? Use an excess of chlorine Use an excess of ethane Use equimolar chlorine and ethane It is not possible to minimize the production of higher halogenated products
Answer:Use an excess of ethane
Explanation:
The halogenation of alkanes is a substitution reaction. All the hydrogen atoms in the alkanes could be potentially substituted. How ever the reaction can be controlled by using an excess of either the alkane or the halogen. If the aim (as it is in this question) is to minimize the yield of halogenated alkanes, an excess of the alkane (in this case, ethane) is used.
IN TEH HYPOTHETIcal case of aliquid that had such strong intermolecular forces it would never evaporate, what temperature change would you expect to see
Answer:
A wide temperature change.
Explanation:
The intermolecular force is the force that puts the molecules together in a substance. If the substance is ionic (formed by ions) than the force is called ion-ion, which is very strong. If it's a covalent compound (the elements share electron pairs), then it can be a polar or a nonpolar substance.
The polar substance has a huge difference in electronegativity of its components, so partial charges are presented. In this case, the intermolecular force is called dipole-dipole, because charged poles are formed. In the other case, of nonpolar, the dipoles are induced, so it's called dipole induced-dipole induced force.
When a polar compound has a hydrogen-bonded to a high electronegative element (F, O, or N), the dipole-dipole is extremely strong and it's called a hydrogen bond.
Thus, as strong is the force (Ionic > hydrogen bond > dipole-dipole > dipole induced - dipole induced), as difficult is to broken these bonds. In a phase change, those bonds must be broken. So if the liquid has so strong forces, it would be necessary a large amount of energy to evaporate it.
The temperature is a measure of the kinetic energy of the molecules, so if the energy applied is too high, the temperature change must be extremely high too!
This situation is not real, because if the forces are so strong than the material would be a solid and not a liquid. Besides, even an extremely strong bond may be broken with the right temperature increase, so the liquid would evaporate!
In a hypothetical liquid with intermolecular forces strong enough to prevent evaporation, we would not expect to see a temperature change due to evaporation. The liquid's temperature would be unaffected by this specific phase change process and would only change due to external thermal influences. This is a theoretical extreme, as in reality, all substances can change phase with sufficient energy.
Explanation:Temperature Change in a Substance with Strong Intermolecular Forces
If we hypothesize a liquid with such strong intermolecular forces that it would never evaporate, we must consider the concept of vapor pressure. Typically, vapor pressure represents the tendency of molecules to escape into the vapor phase; stronger intermolecular forces would mean a lower vapor pressure. In this hypothetical scenario, because evaporation happens when molecules have enough kinetic energy to overcome intermolecular forces, and these forces are too strong to be overcome, one would not expect to see any temperature change due to evaporation.
Moreover, since evaporation is an endothermic process where the liquid absorbs energy to allow molecules to escape, the absence of evaporation implies that the liquid would not absorb energy in this way. Therefore, the temperature of the liquid would remain relatively stable, primarily affected only by external thermal inputs or losses, but not through phase change.
It is important to realize that this is an extreme hypothetical situation, as all real substances have a point at which their intermolecular bonds can be overcome by kinetic energy, leading to a phase transition. Additionally, in real-world scenarios, temperature impacts the kinetic energy of molecules within a substance, potentially affecting state changes at varying temperatures.
What is the concentration of magnesium bromide, in ppm, if 133.4 g MgBr2 dissolved in 1.84 L water. Then solve for the bromine concentration in ppm
Answer: 0.0725ppm
Explanation:
133.4g of MgBr2 dissolves in 1.84L of water.
Therefore Xg of MgBr2 will dissolve in 1L of water. i.e
Xg of MgBr2 = 133.4/1.84 = 72.5g
The concentration of MgBr2 is 72.5g/L = 0.0725mg/L
Recall,
1mg/L = 1ppm
Therefore, 0.0725mg/L = 0.0725ppm
The concentration of magnesium bromide (MgBr2) in the solution is 72467.39 ppm. Considering each molecule of MgBr2 has two atoms of bromine, the concentration of bromine in the solution is 58032.60 ppm.
Explanation:The subject of this question is Chemistry, specifically dealing with the concept of concentration, which is a measure of the amount of solute dissolved in a solvent. The units of ppm (parts per million) imply a measurement of the amount of a particular substance (in this case magnesium bromide and bromine) in a total amount of 1 million parts of the mixture.
First, we find the concentration of magnesium bromide MgBr2 by using the formula ppm = (mass of solute/volume of solution) x 10^6. Given that we have 133.4 g of MgBr2 in 1.84 L of water, we find that the concentration of MgBr2 in ppm is (133.4 g/1.84 L) x 10^6 = 72467.39 ppm.
Next, to solve for the concentration of bromine in ppm, we must consider that each molecule of MgBr2 has two atoms of Br. So, the mass of bromine in 133.4 g of MgBr2 is 133.4 g * (79.9 g/mol Br /159.8 g/mol MgBr2) * 2 mol Br = 106.76 g. The concentration of bromine in ppm is then (106.76 g / 1.84 L) x 10^6 = 58032.60 ppm.
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The N-H bond in ammonia is polar because
A: nitrogen occupies more space than hydrogen.
B: hydrogen is much more electronegative than nitrogen.
C: it is a hydrogen bond.
D: it is an ionic bond.
E: nitrogen is much more electronegative than hydrogen.
Answer: option E. nitrogen is much more electronegative than hydrogen.
Explanation:
Answer: E: nitrogen is much more electronegative than hydrogen.
Explanation:
Each N-H bond is polar because N is more electronegative than H. NH3 is overall asymmetrical in its VSEPR shape, so the dipoles don't cancel out and it is therefore polar.
What volume of 3.00 M HCl will form a solution with an acidic pH when mixed with 100 mL of 3.00 M NaOH
Answer: The volume of acid should be less than 100 mL for a solution to have acidic pH
Explanation:
To calculate the volume of acid needed to neutralize, we use the equation given by neutralization reaction:
[tex]n_1M_1V_1=n_2M_2V_2[/tex]
where,
[tex]n_1,M_1\text{ and }V_1[/tex] are the n-factor, molarity and volume of acid which is HCl
[tex]n_2,M_2\text{ and }V_2[/tex] are the n-factor, molarity and volume of base which is NaOH
We are given:
[tex]n_1=1\\M_1=3.00M\\V_1=?mL\\n_2=1\\M_2=3.00M\\V_2=100mL[/tex]
Putting values in above equation, we get:
[tex]1\times 3.00\times V_1=1\times 3.00\times 100\\\\V_1=\frac{1\times 3.00\times 100}{1\times 3.00}=100mL[/tex]
For a solution to be acidic in nature, the pH should be less than the volume of acid needed to neutralize.
Hence, the volume of acid should be less than 100 mL for a solution to have acidic pH
Calculate the volume in milliliters of a 1.72 mol/L sodium thiosulfate solution that contains 375. mmol of sodium thiosulfate (Na_2S_2O_3).
Answer:
216 mL
Explanation:
Molarity of a substance , is the number of moles present in a liter of solution .
M = n / V
M = molarity ( unit = mol / L or M )
V = volume of solution in liter ( unit = L ),
n = moles of solute ( unit = mol ),
From the question , the respective values are as follows -
M = 1.73 mol / L
n = 375 mmol
Since,
1 mmol = 0.001 mol
n = 375 * 0.001 mol = 0.375 mol
Now , the volume of can be calculated by using the above equation ,
M = n / V
Putting the respective values ,
1.73 mol / L = 0.375 mol / V
V = 0.216 L
Since,
1 L = 1000 mL
V = 0.216 * 1000 mL = 216 mL
Draw the structure of the starting material needed to make 2-methylhept-3-yne using sodium amide in liquid ammonia, followed by 1-bromopropane. The starting hydrocarbon must have no more than five carbons. 1) NaNH2, NH3(l) 2) CH3CH2CH2Br + NaBr
The starting material to synthesize 2-methylhept-3-yne under the given conditions is 4-methyl-1-pentyne, which undergoes alkylation with 1-bromopropane to form the desired product.
The question involves a synthetic route to create 2-methylhept-3-yne using specific reagents and limitations on the carbon count of the starting material. The sodium amide in liquid ammonia indicates a need for a strong base, suggesting an elimination reaction. With the addition of 1-bromopropane, we're looking at an alkylation step which introduces the propyl group.
Given the restrictions, the most fitting starting material would be 4-methyl-1-pentyne. This molecule can undergo alkylation at the terminal methyl group with 1-bromopropane to extend the carbon chain by three carbons, leading to the desired product, 2-methylhept-3-yne. The overall strategy leverages a reaction sequence involving an initial alkyne substrate that can be elongated with an alkyl halide using the action of a strong base.
The starting material for the formation of 2-methylhept-3-yne is 3-methylbutyne. The structure of 3-methylbutyne is attached below
The starting material for the formation of 2-methylhept-3-yne is 3-methylbutyne
Stepwise formation of 2-methylhept-3-yne
Step 1:
The first step is the formation of the Acetylide Ion
3-methylbutyne reacts with sodium amide (NaNH₂) in liquid ammonia to form the acetylide ion:
(CH₃)₂CHC≡CH + NaNH₂ → (CH₃)₂CC≡C⁻Na⁺ + NH₃
Step 2:
The second step is the alkylation of the Acetylide Ion
The acetylide ion reacts with 1-bromopropane (CH₃CH₂CH₂Br) through an SN2 mechanism to form 2-methylhept-3-yne:
(CH₃)₂CC≡C⁻Na⁺ + CH₃CH₂CH₂Br → (CH₃)₂CC≡CCH₂CH₂CH₃ + NaBr
Hence, the final product is 2-methylhept-3-yne:
(CH₃)₂CC≡CCH₂CH₂CH₃
Assume that each atom is a sphere, and that the surface of each atom is in contact with its nearest neighbor. Determine the percentage of unit cell volume that is occupied in (a) a face- centered cubic lattice, (b) a body-centered cubic lattice, and (c) a diamond lattice.
Answer:
The percentage of unit cell volume that is occupied by atoms in a face- centered cubic lattice is 74.05% The percentage of unit cell volume that is occupied by atoms in a body-centered cubic lattice is 68.03% The percentage of unit cell volume that is occupied by atoms in a diamond lattice is 34.01%Explanation:
The percentage of unit cell volume = Volume of atoms/Volume of unit cell
Volume of sphere = [tex]\frac{4 }{3} \pi r^2[/tex]
a) Percentage of unit cell volume occupied by atoms in face- centered cubic lattice:
let the side of each cube = a
Volume of unit cell = Volume of cube = a³
Radius of atoms = [tex]\frac{a\sqrt{2} }{4}[/tex]
Volume of each atom = [tex]\frac{4 }{3} \pi (\frac{a\sqrt{2}}{4})^3[/tex] = [tex]\frac{\pi *a^3\sqrt{2}}{24}[/tex]
Number of atoms/unit cell = 4
Total volume of the atoms = [tex]4 X \frac{\pi *a^3\sqrt{2}}{24} = \frac{\pi *a^3\sqrt{2}}{6}[/tex]
The percentage of unit cell volume = [tex]\frac{\frac{\pi *a^3\sqrt{2}}{6}}{a^3} =\frac{\pi *a^3\sqrt{2}}{6a^3} = \frac{\pi \sqrt{2}}{6}[/tex] = 0.7405
= 0.7405 X 100% = 74.05%
b) Percentage of unit cell volume occupied by atoms in a body-centered cubic lattice
Radius of atoms = [tex]\frac{a\sqrt{3} }{4}[/tex]
Volume of each atom =[tex]\frac{4 }{3} \pi (\frac{a\sqrt{3}}{4})^3[/tex] =[tex]\frac{\pi *a^3\sqrt{3}}{16}[/tex]
Number of atoms/unit cell = 2
Total volume of the atoms = [tex]2X \frac{\pi *a^3\sqrt{3}}{16} = \frac{\pi *a^3\sqrt{3}}{8}[/tex]
The percentage of unit cell volume = [tex]\frac{\frac{\pi *a^3\sqrt{3}}{8}}{a^3} =\frac{\pi *a^3\sqrt{3}}{8a^3} = \frac{\pi \sqrt{3}}{8}[/tex] = 0.6803
= 0.6803 X 100% = 68.03%
c) Percentage of unit cell volume occupied by atoms in a diamond lattice
Radius of atoms = [tex]\frac{a\sqrt{3} }{8}[/tex]
Volume of each atom = [tex]\frac{4 }{3} \pi (\frac{a\sqrt{3}}{8})^3[/tex] = [tex]\frac{\pi *a^3\sqrt{3}}{128}[/tex]
Number of atoms/unit cell = 8
Total volume of the atoms = [tex]8X \frac{\pi *a^3\sqrt{3}}{128} = \frac{\pi *a^3\sqrt{3}}{16}[/tex]
The percentage of unit cell volume = [tex]\frac{\frac{\pi *a^3\sqrt{3}}{16}}{a^3} =\frac{\pi *a^3\sqrt{3}}{16a^3} = \frac{\pi \sqrt{3}}{16}[/tex] = 0.3401
= 0.3401 X 100% = 34.01%
The percentage of unit cell volume occupied is highest in the face-centered cubic lattice, making it the most efficient packing with the least free space. In contrast, the body-centered cubic and diamond lattices have more free space. Cadmium sulfide crystallizes with cadmium ions occupying half of the tetrahedral holes in a closest-packed sulfide ion array, resulting in a 1:1 ratio and a CdS formula.
Explanation:Calculating Unit Cell Volume Occupancy
To calculate the percentage of the unit cell volume that is occupied in different lattice structures, we consider the arrangement and number of atoms within the unit cell and their geometric relationship. In a face-centered cubic (fcc) lattice, each corner atom is shared by eight unit cells and each face atom is shared by two, resulting in 4 atoms per unit cell. In a body-centered cubic (bcc) lattice, there is one atom entirely within the cell and 8 corner atoms shared by 8 unit cells, totaling 2 atoms per unit cell. The diamond lattice is a more complex structure with 8 atoms per unit cell, each with a fractional part inside the unit cell.
The most efficient packing, meaning the least percentage of free space, is found in the face-centered cubic structure. To calculate this, one needs to consider the volume occupied by atoms—which can be delineated through the atomic radius and the volume of the unit cell based on the edge lengths.
Cadmium sulfide crystallizes in a structure where cadmium ions occupy half of the tetrahedral holes of the sulfide ion closest-packed array. Since each sulfide ion allows for two tetrahedral holes, the ratio of cadmium to sulfide ions is 1:1, so the formula of cadmium sulfide is CdS.
Covalent bonds in a molecule absorb radiation in the IR region and vibrate at characteristic frequencies.
A molecular vibration absorbs radiation of wavelength 188 μm. What frequency (in s−1) corresponds to that wavelength? Enter your answer in scientific notation.
Answer:
[tex]Frequency=1.6\times 10^{12}\ Hz[/tex]
Explanation:
The relation between frequency and wavelength is shown below as:
[tex]c=frequency\times Wavelength [/tex]
c is the speed of light having value [tex]3\times 10^8\ m/s[/tex]
Given, Wavelength = 188 μm
Also, 1 μm = [tex]10^{-6}[/tex] nm
So,
Wavelength = [tex]188\times 10^{-6}[/tex] m
Thus, Frequency is:
[tex]Frequency=\frac{c}{Wavelength}[/tex]
[tex]Frequency=\frac{3\times 10^8}{188\times 10^{-6}}\ Hz[/tex]
[tex]Frequency=1.6\times 10^{12}\ Hz[/tex]
The second-order rate constant for the following gas-phase reaction is 0.041 1/MLaTeX: \cdotâs. We start with 0.438 mol C2F4 in a 2.42 liter container, with no C4F8 initially present. C2F4 LaTeX: \longrightarrowⶠ1/2 C4F8 What is the half-life (in seconds) of the reaction for the initial C2F4 concentration? Enter to 1 decimal place.
Answer:
134.8 seconds is the half-life (in seconds) of the reaction for the initial [tex]C_2F_4[/tex] concentration
Explanation:
Half life for second order kinetics is given by:
[tex]t_{\frac{1}{2}=\frac{1}{k\times a_0}[/tex]
Integrated rate law for second order kinetics is given by:
[tex]\frac{1}{a}=kt+\frac{1}{a_0}[/tex]
[tex]t_{\frac{1}{2}[/tex] = half life
k = rate constant
[tex]a_0[/tex] = initial concentration
a = Final concentration of reactant after time t
We have :
[tex]C_2F_4 \longrightarrow \frac{1}{2} C_4F_8[/tex]
Initial concentration of [tex]C_2F_4=[a_o]=\frac{0.438 mol}{2.42 L}=0.1810 mol/L[/tex]
Rate constant = k = [tex]0.041 M^{-1} s^{-1}[/tex]
[tex]t_{\frac{1}{2}=\frac{1}{k\times a_0}[/tex]
[tex]=\frac{1}{0.041 M^{-1} s^{-1}\times 0.1810 mol/L}[/tex]
[tex]t_{1/2}=134.8 s[/tex]
134.8 seconds is the half-life (in seconds) of the reaction for the initial [tex]C_2F_4[/tex] concentration
The second-order rate constant for the following gas-phase reaction
C₂F₄ ⇒ 1/2 C₄F₈
is 0.0411 M⁻¹.s⁻¹. We start with 0.438 mol C₂F₄ in a 2.42-liter container, with no C₄F₈ initially present. What is the half-life (in seconds) of the reaction for the initial C₂F₄ concentration? Enter to 1 decimal place.
The reaction C₂F₄ ⇒ 1/2 C₄F₈, with a second-order rate constant of 0.0411 M⁻¹.s⁻¹, that starts with 0.438 mol C₂F₄ in a 2.42-liter container, has a half-life of 134.4 s.
Let's consider the following reaction following second-order kinetics.
C₂F₄ ⇒ 1/2 C₄F₈
Initially, we have 0.438 mol C₂F₄ in a 2.42-liter container. The initial concentration of C₂F₄ is:
[tex][C_2F_4]_0 = \frac{0.438 mol}{2.42 L} = 0.181 M[/tex]
Given the rate constant (k) is 0.041 M⁻¹.s⁻¹, we can calculate the half-life of a second-order reaction using the following expression.
[tex]t_{1/2}= \frac{1}{k \times [C_2F_4]_0} = \frac{1}{0.0411 M^{-1}s^{-1} \times 0.181 M} = 134.4 s[/tex]
The reaction C₂F₄ ⇒ 1/2 C₄F₈, with a second-order rate constant of 0.0411 M⁻¹.s⁻¹, that starts with 0.438 mol C₂F₄ in a 2.42-liter container, has a half-life of 134.4 s.
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The normal boiling point of liquid methyl acetate is 331 K. Assuming that its molar heat of vaporization is constant at 30.6 kJ/mol, the boiling point of CH3COOCH3 when the external pressure is 1.29 atm is
Answer:
T_2=338.9026K
Boiling point of CH3COOCH3 at external pressure is 338.9026K
Explanation:
We are going to use Clausius-Clapeyron Equation:
[tex]ln\frac{P_2}{P_1} =-\frac{\Delta H}{R}(\frac{1}{T_2}-\frac{1}{T_1})[/tex]
Where:
P_2 is the external pressure
P_1 is the atmospheric Pressure=1 atm
ΔH is the heat of vaporization
T_2 boiling point of CH3COOCH3 at external pressure
T_1 normal boiling point of liquid methyl acetate
Now:
[tex]\frac{1}{T_2}=-ln\frac{P_2}{P_1}*\frac{R}{\Delta H}+\frac{1}{T_1} \\\frac{1}{T_2}=-ln\frac{1.29}{1}*\frac{8.314}{30.6*10^3}+\frac{1}{331} \\\frac{1}{T_2}=2.9507*10^-^3\\T_2=\frac{1}{2.9507*10^-^3} \\T_2=338.9026K[/tex]
Boiling point of CH3COOCH3 at external pressure is 338.9026K
To find the new boiling point of methyl acetate, use the Clausius-Clapeyron equation with the provided values for the initial boiling point, molar heat of vaporization, and new external pressure.
Explanation:This question requires us to use the form of the Clausius-Clapeyron equation:
ln(P2/P1) = -ΔHvap/R * (1/T2 - 1/T1)
Where T1 and P1 are the initial temperature and pressure (331K and 1 atm respectively, since normal boiling point is defined as the temperature at which the vapor pressure of a liquid equals 1 atm), and P2 is the new pressure (1.29 atm). ΔHvap is the molar heat of vaporization (30.6 kJ/mol, or 30.6 * 10^3 J/mol to convert kilojoules to joules), and R is the gas constant (8.314 J/mol*K). Solving the equation for T2 (the new boiling point temperature), we rearrange to get T2 = 1/(1/T1 + (R/ΔHvap)*ln(P2/P1)). Plugging in the values, you should find the answer.
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Where would you expect to find the 1H1H NMR signal of (CH3)2Mg(CH3)2Mg relative to the TMS signal
Answer:
In diethyl ether, (CH₃)₂Mg(CH₃)₂Mg, show peaks at -1.46 and -1.74 ppm, at -100 °CIn tetrahydrofuran (THF), monomer (CH₃)₂Mg show peaks at -1.76 and -1.81 ppm, at -60 °CExplanation:
The 1H NMR signal of (CH₃)₂Mg can be obtained in diethyl ether and tetrahydrofuran. (CH₃)₂Mg tends to polymerize in diethyl ether but mostly remain monomer in tetrahydrofuran. The negative value of chemical shifts shows that the protons of (CH₃)₂Mg are more shielded than the protons of TMS.
In diethyl ether, the structure of (CH₃)₂Mg is predicted in Figure A (attached). The A group in the structure represents the solvent. The chemical shift at -1.46 ppm is associated with protons of the bridging methyl groups while the signal at -1.74 ppm is for the proton of the terminal methyl group.
In THF, the two possible structures of (CH₃)₂Mg are predicted in Figure B (attached). Chemical shift at -1.76 ppm is associated with unsolvated monomer of (CH₃)₂Mg while the shift at -1.81 refers to the protons of methyl group bonded to the solvated (CH₃)₂Mg. The solvent is represented as Y in the Figure B.
This study is reported by J. Heard in "NMR of organomagnessium compounds".
Reaction 1: 2NO(g) + Br2(g) ⇄ 2NOBr(g) ΔH°rxn = −16 kJ Kp = 27 at 25°CReaction 2: CO2(g) ⇄ C(s) + O2(g) ΔH°rxn = +100 kJ Kp = 2.6x10−9 at 25°C8. For which reaction(s) above would an increase in reaction vessel volume at constant temperature result in an increase in product yield? A) Reaction 1 only B) Reaction 2 onlyC) Both reactionsD) Neither reactionFor which reaction(s) above would an increase in temperature at constant volume result in an increase in product yield? A) Reaction 1 only B) Reaction 2 onlyC) Both reactionsD) Neither reaction
Answer:
First question: D)
Second question: B)
Explanation:
A reversible reaction intends to achieve the equilibrium, a state in the velocity of product formation is equal to the velocity of reactants formation. This equilibrium can be disturbed by some alterations in the reaction, and, by Le Chatelier's principle, the reaction will shift in to reestablish the equilibrium.
The equilibrium may be shift for three principal factors: concentration, temperature, and pressure. If a concentration of some of the substance increases, the reaction will shift to consume it; if it decreases, the reaction is shifted to form more this substance.
When the temperature of the system increases, the reaction shifts for the consume of the heat, thus, the endothermic reaction (ΔH°rxn > 0) is favored; if it decreases, the exothermic reaction (ΔH°rxn < 0) is favored. If the direct reaction is endothermic, the inverse is exothermic, and vice versa.
When the pressure of the system increases, the volume decreases, so, the equilibrium shifts for the less gas volume, or the side with fewer moles of gas substance; if the pressure decreases the inverse occurs.
So, if the vessel volume increases, the pressure will decrease, and so, the formation of more gas substances is favored. In reaction 1, in the reactants, there are 3 moles ( 2 of NO + 1 of Br2), and in the products 2 moles, so the reactants are favored, and the product yield decreases. In reaction 2, there is 1 mol on both sides of the reaction, so the pressure doesn't affect the equilibrium.
The increase in temperature favors the endothermic reaction. The reaction 2 has it's direct reaction endothermic, so the product yield increases. In reaction 1, the reactants increases, because the inverse reaction is endothermic.
An increase in temperature at constant volume will result in an increase in product yield in Reaction 1 only.
Explanation:An increase in temperature at constant volume will result in an increase in product yield for Reaction 1 only. This is because the reaction in Reaction 1 is exothermic, meaning it releases heat. When the temperature increases, the reaction will shift in the direction that absorbs heat to counteract the increase in temperature. Since Reaction 1 releases heat, an increase in temperature will favor the formation of products and increase the product yield. For Reaction 2, an increase in temperature will not result in an increase in product yield because the reaction is endothermic, meaning it requires heat as a reactant. Increasing the temperature will shift the reaction in the direction that releases heat, reducing the product yield.
A three level system has energies ε0=0.00J, ε1=5.00x10-21J, and ε2=10.0x10-21J. a. Using the Boltzmann Distribution Law, calcuate the molar internal energy U at T=10.0K. Hint: This means you have to calculate three probabilites at T=10K and use them to calculate the average energy per particle and then calculate the energy of a mole of particles.b. Again using the Bolztmann Distribution Law and the same approach as in part A, calculate the molar internal enerrgy U at T=1000Kc. Using you results from parts A and B, calculate the change in the molar internal energy ΔU when the temperature changes from T=10K to T=1000K.d. Using the Bolzmann Distribution Law probabilities calculated in part A, calculate the molar entropy at T=10K.e. Calculate the change in molar entropy when the temperature of this 3 level system changes from T=10K to T=1000K.
To find the molar internal energy using the Boltzmann Distribution Law, calculate the energy level probabilities at the given temperatures, compute the average energy, then multiply it by Avogadro's number. The change in internal energy and molar entropy is derived from the differences in these quantities between 10K and 1000K.
Explanation:To solve for the molar internal energy U at different temperatures using the Boltzmann Distribution Law, we first need to calculate the probabilities of each energy level at the given temperature T. For a three-level system, these probabilities depend on the energy levels and the temperature through the relation Pi = e(-εi/kBT), where εi is the energy of level i, kB is the Boltzmann constant, and T is the temperature.
For T=10.0K:
Calculate the probability P0, P1, and P2 for each energy level.Find the average energy <E> per particle using the probabilities.Multiply the average energy by Avogadro's number to find the molar internal energy U.Repeat the steps for T=1000K to find U at the higher temperature.
The change in molar internal energy ΔU when the temperature changes from T=10K to T=1000K can be calculated by taking the difference of U values obtained at both temperatures.
To calculate molar entropy S at T=10K, use the probabilities you calculated and the Boltzmann formula S = kB Σ Pi ln(Pi) and sum over all states i.
The change in molar entropy when the temperature changes from T=10K to T=1000K can be calculated by finding the difference in entropy values at both temperatures.
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 ?Set the math up. But don't do any of it. Just leave your answer as a math expression.Also, be sure your answer includes all the correct unit symbols.
In order to calculate the mass of fiber that can be spun in 55 min we have to determine the rate of production.
Generally speaking, all the different types of rates have in common a variable divided over time. In this case the variable we're interested in is the mass of fiber, so we have to divide the mass of fiber over the time it takes to spin, and then if we want to find the mass of fiber for any given time, we just have to multiply the time and the rate together.
Keep in mind that the time that is given to us to find the rate is in seconds, so we have to convert that to minutes
The math expression would be the following:
[tex]mass\ of \ fiber (g)=time(min)*\frac{0,059 g}{0.13s}*\frac{60 s}{1 min} \\\\mass\ of \ fiber (g)=55 min*\frac{0,059 g}{0.13s}*\frac{60 s}{1 min}[/tex]
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A 10.00 mL sample of 8.50 M HNO3 solution is diluted to a new volume of 65.0 mL. What is the concentration of the dilute solution
Answer: 1.31M
Explanation:
V1 = 10mL
C1 = 8.5M
V2 = 65mL
C2 =?
C1V1 = C2V2
10 x 8.5 = C2 x 65
C2 = (10 x 8.5 ) /65
C2 = 1.31M
The concentration of the dilute solution of nitric acid is 1.31 M.
We have 10.00 mL (V₁) of 8.50 M HNO₃ (C₁) and we add water to obtain a dilute solution with a volume (V₂) of 65.0 mL. We can calculate the concentration of the dilute solution (C₂) using the dilution rule.
[tex]C_1 \times V_1 = C_2 \times V_2\\C_2 = \frac{C_1 \times V_1}{V_2} = \frac{8.50 M \times 10.00 mL}{65.0 mL} = 1.31 M[/tex]
The concentration of the dilute solution of nitric acid is 1.31 M.
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In the molecular orbital model of benzene, the six p atomic orbitals combine to form how many molecular orbitals?
Answer:
6 molecular orbital.
Explanation:
According to the molecular orbital theory, number of molecular orbital are obtained is equal to the number of atomic orbital combine.
When six p orbitals of benzene combine, the will form six molecular orbital. These six molecular orbital have 3 molecular orbital with lower energy and three molecular orbital with higher energy.
The orbital with lower energies are called bonding molecular orbital and orbital with higher energies are called antibonding molecular orbital.
In the molecular orbital model of benzene, six p atomic orbitals combine to form six molecular orbitals. These result from the overlapping p orbitals of the six carbon atoms in the benzene ring, forming a 'pi electron cloud'.
Explanation:In the molecular orbital model of benzene, the six p atomic orbitals combine to form six molecular orbitals. This happens because benzene has a cyclic structure where the overlapping p orbitals of six carbon atoms form what is known as a 'pi electron cloud'. This pi electron cloud is a result of side-on overlap of p orbitals encompassing all the six carbon atoms. So, essentially, the six overlapping p orbitals produce six molecular orbitals. These six orbitals then distribute themselves into differing energy levels. Three of these orbitals tend to be lower energy (bonding), and the other three are higher energy (anti-bonding).
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Type of bond Average Bond Enthalpy (kJ/mol) O-O 150O=O 500The conversion of ozone to diatomic oxygen is represented by the equation above. Based on the data in the table above, what is the approximate average bond enthalpy for the oxygen-to-oxygen bonds in ozone? a. 0 kJ/mol b. 150 kJ/mol c. 300 kJ/mol d. 500 kJ/mol
Missing information:
The reaction is 2O₃(g) --> 3O₂ (g) ΔH = -300 kJ/mol
Answer:
c
Explanation:
In the molecule of ozone, 3 oxygens are bonded, and, because each one needs to share two pairs of electrons, these 2 bonds are something between a simple and a double bond.
In the reaction of the transformation of the ozone to oxygen gas, these bonds are broken, and a double bond is formed between two oxygen atoms. The sum of the energy of the broken and the formation of the bond is the enthalpy variation of the reaction.
To break a bond, energy must be added to the system, so it's an endothermic reaction and energy is positive, so the formation is exothermic and the energy is negative. Because there're 2 ozone molecules, 4 bonds will be broken, and because there are 3 oxygen molecules, 3 bonds will be formed:
4*E - 3*500 = -300
4E = -300 + 1500
4E = 1200
E = 300 kJ/mol
So, each O3 bond has 300 kJ/mol as an average energy.
Each ozone molecule has 300 kJ/mol as an average bond energy.
The reaction is
[tex]\bold { 2O_3(g) \rightarrow 3O_2 (g)\ \ \ \ \ \ \ \ \ \ \ \Delta H = -300 kJ/mol}[/tex]
In the molecule of ozone, there 2 double bonds are present with continuous variation.
The enthalpy variation of the reaction is the sum of the energy of the broken and the formation of the bond.
Because of 2 ozone molecules, 4 bonds will be broken, and because there are 3 oxygen molecules, 3 bonds will be formed.
[tex]\bold {4\times E - 3 \times 500 = -300}\\\\\bold {4\times E = -300 + 1500}\\\\\bold {E = 300 kJ/mol}[/tex]
Therefore, Each ozone molecule has 300 kJ/mol as an average bond energy.
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g What mass (in g) of a concentrated solution of nitric acid (69.0% HNO3 by mass) is needed to prepare 364.5 g of a 12.2% solution of HNO3 by mass?
Answer:
64.4 g of concentrated nitric acid solution are needed.
Explanation:
First we calculate the mass of HNO₃ contained in 364.5 g of a 12.2% solution:
364.5 gsolution * 12.2 gHNO₃/100gsolution = 44.47 g HNO₃Now we calculate how many grams of the concentrated solution would contain 44.47 grams of HNO₃:
44.47 g HNO₃ * 100 g solution / 69.0 = 64.45 g concentrated solutionSo 64.4 g of concentrated nitric acid solution are needed.
If a snail crawls at a rate of 0.10 cm/s, what is the speed in miles per hour? (1 mi = 5280 ft)
Answer:
0.0022369 mph
Explanation:
a snail crawls at a rate of 0.10 cm/s
1mile = 5280 ft
Scientifically 1 cm/s = 0.022369 mph
therefore 0.10cm/s = 0.10cm/s x 0.022369 mph = 0.0022369 mph
Predict the products for each of the following Diels–Alder reactions: (a) COOH HOOC + (b) HOOC CN + (c) O O + (d) S O O O + (e) CN NC + (f ) O O O O S MeO OMe + 16.43
Answer:
see explanation below
Explanation:
You are not putting all the structures correctly. Luckily I found the question on another place, so the complete structures you can see them in picture 1.
Now, according to the picture 1, we have all the six reactions. The general mechanism is the same for all and then, if you can have the possibility to rearrange the molecule, you can do that too.
Now, the general mechanism as I stated earlier is the same. The double bond from the diene (The one with one double bond at least) attacks the dienophyle (The first double bond), this bond do resonance to the conjugate bond, and the other double bond attacks the diene, and form a new product.
According to this, the product for each reaction, you can see it in picture 2 and 3:
In Diels-Alder reactions, a diene reacts with a dienophile to form a cyclic compound. Only reactions (b) and (e) will result in Diels-Alder adducts.
Explanation:In order to predict the products of the given Diels-Alder reactions, we need to identify the diene and dienophile components. The diene is usually a compound containing two double bonds, while the dienophile is a compound with a double bond. The reaction between the diene and dienophile will form a cyclic compound known as the Diels-Alder adduct. Let's examine each reaction:
(a) COOH + HOOC: Neither COOH nor HOOC contain diene or dienophile functionality, so no Diels-Alder reaction will occur.
(b) HOOC + CN: This reaction involves a diene (HOOC) and a dienophile (CN). The Diels-Alder adduct will be formed.
(c) O + O: Neither O nor O contain diene or dienophile functionality, so no Diels-Alder reaction will occur.
(d) S + OOO: Both S and OOO contain diene functionality, but no dienophile. Therefore, a Diels-Alder reaction will not occur.
(e) CN + NC: This reaction involves a diene (CN) and a dienophile (NC). The Diels-Alder adduct will be formed.
(f) OOO S MeO OMe: Neither OOO nor S MeO OMe contain diene or dienophile functionality, so no Diels-Alder reaction will occur.
Impure crystalline (solid) substances can be purified by recrystallization from a suitable solvent. Arrange the steps of the recrystallization procedure, from start to finish.
Weigh the crystals into a tared flask.
Dissolve the crude substance in a minimum amount of hot solvent.
Filter the hot solution to remove the solid impurities (if present)
Allow the solution to cool slowly so crystals form.
Filter the crystals.
Dry the crystals
Answer:
1) Dissolve the crude substance in a minimum amount of hot solvent.
2) Filter the hot solution to remove the solid impurities (if present)
3) Allow the solution to cool slowly so crystals form.
4) Filter the crystals.
5) Dry the crystals
6) Weigh the crystals into a tared flask.
Explanation:
Recrystallization is a method used in chemistry to obtain crystals having a high degree of purity for the purpose of analytical or synthetic laboratory work.
The work function of palladium (Pd) is 5.22 eV. What is the minimum frequency of light to observe the photoelectric effect from a Pd surface? If light with a 200 nm wavelength is absorbed by the Pd surface, what is the velocity of the emitted electrons?
Answer:
1)4×10^13Hz
2) 9.95×10-9J
Explanation:
From the image attached, it is clear that the work function of the palladium metal must first be obtained in joules. Then, the frequency is obtained from E=hf.
The kinetic energy if the photoelectrons is obtained as the difference between the energy of the photon and the work function of the metal.
Three of the following statements are wrong. One is correct. Which of the following statements is correct?
Enzymes require optimum pH for maximal activity
Enzymes are denatured at high temperature but in certain exceptional organisms they are effective even at temperatures 80∘−90∘C
Enzymes are highly specific
Most enzymes are proteins but some are lipids
Answer:
The incorrect statement is "Most enzymes are proteins but some are lipids" the others statements are correct regarding to the Enzymes.
Explanation:
Enzymes are complex macromolecules of globular proteins. However, many enzymes contain certain non-protein substances associated with them for their function, which are known as cofactors and can be organic or inorganic compounds. In addition, the enzymes are thermolabile, presenting their best performance at the ideal temperature. Enzyme activity decreases with increasing and increases with increasing temperature and stops at 0 degrees and above 80 degrees. However, the enzymes of the bacteria that inhabit the hot springs have an ideal temperature of 70 degrees or more. Enzymes also show maximum activity at optimal pH. Varying its activity with increasing or decreasing pH. Enzymes are specific in their action. An enzyme can catalyze only a specific type of reaction or even act on a specific substrate. For example, the enzyme lactase catalyzes the hydrolysis of lactose and no other disaccharides.
What kind of intermolecular forces act between a hydrogen iodide molecule and a dichloroethylene
Final answer:
The intermolecular forces between a hydrogen iodide molecule and a dichloroethylene molecule involve both dispersion forces and dipole-dipole forces.
Explanation:
The intermolecular forces between a hydrogen iodide molecule (HI) and a dichloroethylene molecule (C2H2Cl2) involve both dispersion forces and dipole-dipole forces.
Dispersion Forces: Iodine (I) in HI is a larger atom compared to chlorine (Cl) in dichloroethylene. As a result, HI has stronger dispersion forces due to the larger number of electrons, making the attraction between HI molecules stronger than between dichloroethylene molecules.Dipole-Dipole Forces: Dichloroethylene is a polar molecule with a partial positive charge on the hydrogen atom and partial negative charges on the chlorine atoms. HI is also a polar molecule with a partial positive charge on the hydrogen atom and a partial negative charge on the iodine atom. The positive end of the dipole in HI is attracted to the negative end of the dipole in dichloroethylene, creating dipole-dipole interactions between the two molecules.Overall, the intermolecular forces between a hydrogen iodide molecule and a dichloroethylene molecule include both dispersion forces and dipole-dipole forces.
g Tibet (altitude above sea level is 29,028 ft) has an atmospheric pressure of 240. mm Hg. Calculate the boiling point of water in Tibet. The heat of vaporization for water is 40.7 kJ/mol.
Answer: The boiling point of water in Tibet is 69.9°C
Explanation:
To calculate the boiling point of water in Tibet, we use the Clausius-Clayperon equation, which is:
[tex]\ln(\frac{P_2}{P_1})=\frac{\Delta H}{R}[\frac{1}{T_1}-\frac{1}{T_2}][/tex]
where,
[tex]P_1[/tex] = initial pressure which is the pressure at normal boiling point = 1 atm = 760 mmHg (Conversion factor: 1 atm = 760 mmHg)
[tex]P_2[/tex] = final pressure = 240. mmHg
[tex]\Delta H_{vap}[/tex] = Heat of vaporization = 40.7 kJ/mol = 40700 J/mol (Conversion factor: 1 kJ = 1000 J)
R = Gas constant = 8.314 J/mol K
[tex]T_1[/tex] = initial temperature or normal boiling point of water = [tex]100^oC=[100+273]K=373K[/tex]
[tex]T_2[/tex] = final temperature = ?
Putting values in above equation, we get:
[tex]\ln(\frac{240}{760})=\frac{40700J/mol}{8.314J/mol.K}[\frac{1}{373}-\frac{1}{T_2}]\\\\-1.153=4895.36[\frac{T_2-373}{373T_2}]\\\\T_2=342.9K[/tex]
Converting the temperature from kelvins to degree Celsius, by using the conversion factor:
[tex]T(K)=T(^oC)+273[/tex]
[tex]342.9=T(^oC)+273\\T(^oC)=(342.9-273)=69.9^oC[/tex]
Hence, the boiling point of water in Tibet is 69.9°C