Answer:
the minimum [OH−] triggers precipitation = 2.3x10^-6 M
Explanation:
Mg2+ + 2OH- --> Mg(OH)2(s)
Ksp = [Mg2+][OH-]^2
(2.06x10^-13) = (0.039*x^2)
x = [OH-] = 2.3*10^-6 M
the minimum [OH−] triggers precipitation = 2.3x10^-6 M
Final answer:
The minimum [OH-] that triggers the precipitation of Mg2+ ion can be calculated using the Ksp value.
Explanation:
The minimum [OH-] that triggers the precipitation of Mg2+ ion can be calculated using the Ksp value. Given that the concentration of Mg2+ in the solution is 0.039 M, and the Ksp value is 2.06×10⁻¹³, the minimum [OH-] can be determined to initiate precipitation.
Calculate the expected pH of the solution at the equivalence point using YOUR AVERAGE VALUES for the concentrations of NaOH and acetic acid and the volumes of each that you used. (Just like in Prelab Q3, you will only have the conjugate base and spectator ions present at the equivalence point in a volume that is the sum of the volumes of acid, water, and base you combined.) The Ka for acetic acid is 1.8 x 10-5.'
Complete Question
The complete question is shown on the first uploaded image
Answer:
The pH is [tex]pH = 4.94[/tex]
Explanation:
From the question we are told that
The average concentration of NaOH is [tex][NaOH] = 0.101 M[/tex]
The volume of NaOH is [tex]V__{NaOH}} = 15.00 mL[/tex]
The average concentration of Acetic acid is [tex][Acetic \ Acid] =0.497 \ M[/tex]
The volume of Acetic acid is [tex]V__{Acetic \ Acid}} = 5.00 \ mL[/tex]
The chemical equation for this reaction is
[tex]NaOH + CH_3COOH ---> CH_3 COONa + H_2 O[/tex]
The total volume of the solution is
[tex]V__{Total}} = V__{NaOH}} + V__{Acetic \ Acid}}[/tex]
Substituting values
[tex]V__{Total}} = 15 + 5[/tex]
[tex]V__{Total}} = 20mL = 20 *10^{-3} L[/tex]
The number of moles of NaOH is mathematically represented as
[tex]n__{NaOH}} = [NaOH] * V__{NaOH}}[/tex]
substituting values
[tex]n__{NaOH}} = 0.101 * 15*10^{-3}[/tex]
[tex]n__{NaOH}} = 0.001515 \ moles[/tex]
The number of moles of Acetic acid is mathematically represented as
[tex]n__{Acetic acid}} = [Acetic \ acid] * V__{Acetic acid}}[/tex]
substituting values
[tex]n__{Acetic acid}} = 0.497 * 5*10^{-3}[/tex]
[tex]n__{Acetic acid}} = 0.002485\ moles[/tex]
From the chemical equation
1 mole of NaOH reacts with 1 mole of Acetic acid to produce 1 mole of [tex]CH_3 COONa[/tex] salt and 1 mole of [tex]H_2 O[/tex]
So
0.001515 moles of NaOH reacts with 0.001515 moles of Acetic acid to produce 0.001515 moles of [tex]CH_3 COONa[/tex] salt and 0.001515 moles of [tex]H_2 O[/tex]
This implies the number of moles of NaOH remaining after the react would be
[tex]\Delta n__{NaOH}} = 0.001515 - 0.001515[/tex]
[tex]\Delta n__{NaOH}} = 0 \ mole[/tex]
the number of moles of Acetic acid remaining after the react would be
[tex]\Delta n__{Acetic acid}} = 0.002485 - 0.001515[/tex]
[tex]\Delta n__{Acetic acid}} = 0.00097 \ moles[/tex]
the number of moles of [tex]CH_3 COONa \ salt[/tex] remaining after the react would be
[tex]\Delta n__{CH_3 COONa \ salt}} = 0 + 0.001515[/tex]
[tex]\Delta n__{CH_3 COONa \ salt}} = 0.001515 \ moles[/tex]
the number of moles of [tex]H_2 O[/tex] remaining after the react would be
[tex]\Delta n__{H_2O}} = 0 + 0.001515[/tex]
[tex]\Delta n__{H_2O}} = 0.001515 \ moles[/tex]
The expected pH is mathematically evaluated as
[tex]pH = pK_a + log [\frac{[CH_3 COONa]}{[Acetic \ acid]} ][/tex]
Where [tex]pKa[/tex] is mathematically evaluated as
[tex]pK_a = - log (K_a)[/tex]
The concentration of [tex]CH_3 COONa \ salt[/tex] is mathematically evaluated a s
[tex][CH_3 COONa] = \frac{\Delta n_CH_3 COONa \ salt }{V__{Total}}}[/tex]
substituting values
[tex][CH_3 COONa] = \frac{0.001515}{20 *10^{-3}}[/tex]
[tex][CH_3 COONa] = 0.07575M[/tex]
The concentration of Acetic acid is mathematically evaluated as
[tex][Acetic acid] = \frac{\Delta n__Acetic acid}{V__{Total}}}[/tex]
substituting values
[tex][CH_3 COONa] = \frac{0.00097}{20 *10^{-3}}[/tex]
[tex][CH_3 COONa] = 0.0485 M[/tex]
Substituting values into the equation for pH
[tex]pH = - log (1 .8 *10^{-5}) + log [\frac{0.07575}{ 0.0485} ][/tex]
[tex]pH = log [\frac{0.07575}{ 0.0485} ] - log (1 .8 *10^{-5})[/tex]
[tex]pH = log [\frac{1.561856}{1.8*10^{-5}} ][/tex]
[tex]pH = 4.94[/tex]
The isothermal, first-order reaction of gaseous A occurs within the pores of a spherical catalyst pellet. The reactant concentration halfway between the external surface and the center of the pellet is equal to one-fourth the concentration at the external surface. (a) What is the relative concentration of A near the center of the pellet? (b) By what fraction should the pellet diameter be reduced to give an effectiveness factor of 0.77
Answer:
Check the explanation
Explanation:
Kindly check the attached image below to see the step by step explanation to the question above.
Need help setting the problem up
Answer:
4.0 moles
Explanation:
The following data were obtained from the question:
Volume (V) = 12L
Pressure = 5.6 atm
Temperature (T) = 205K
Gas constant (R) = 0.08206 atm.L/Kmol
Number of mole (n) =?
Using the ideal gas equation: PV = nRT, the number of mole of the gas can be obtained as follow
PV = nRT
5.6 x 12 = n x 0.08206 x 205
Divide both side by 0.08206 x 205
n = (5.6 x 12)/(0.08206 x 205)
n = 4.0 moles
Therefore, the number of mole of the gas is 4.0 moles
Push against a door that does not move. Describe the mechanical energy
Answer:
Work done on the door is = 0
Mechanical energy of the person is 100%
Mechanical energy of the door is 50%.
Explanation:
Mechanical energy, M.E., is the energy an object possesses due to its position and motion. When we talk of position, the body is at rest = potential energy. In terms of motion, the body is moving = kinetic energy.
This means that M.E = potential + kinetic
Also, Work done = force × distance
For the person pushing the door, the potential of the energy stored in the person and the kinetics of pushing the door handle are all present. Therefore, M.E of person is 100%.
But despite the doors potential energy present, there is no motion from it, which means the M.E is only half or 50%.
Also, since distance moved by the door is 0, no work is done on the door.
At pH0 the more thermodynamically stable species is ______ and the most strongly oxidizing species is _______ . At pH14 the more thermodynamically stable species is _______ and the most strongly oxidizing species is ______ . At pH0 ______ will tend to disproportionate and pH14 ________will tend to disproportionate. Under acidic conditions mixing Mn3+(aq) and MnO42- (aq) will result in the formation of______.
Answer:
At pH0 the more thermodynamically stable species is lower and the most strongly oxidizing species is higher. At pH14 the more thermodynamically stable species is higher and the most strongly oxidizing species is lower . At pH0 convex curve will tend to disproportionate and pH14 concave curce will tend to disproportionate. Under acidic conditions mixing Mn3+(aq) and MnO42- (aq) will result in the formation of redox oxidation-reduction reaction.
Which choice represents a pair of resonance structures? Note that lone pairs have been omitted for clarity. View Available Hint(s) Which choice represents a pair of resonance structures? Note that lone pairs have been omitted for clarity.
a. N≡C−N≡C− and N≡C−HN≡C−H
b. O=N−FO=N−F and N=O−FN=O−F
c. O−N=OO−N=O and O=N−OO=N−O
d. O=OO=O and F−FF−F
Option b (O=N−F and N=O−F) represents a pair of resonance structures.
Explanation:The correct choice representing a pair of resonance structures is option b. O=N−F and N=O−F.
Resonance structures are different representations of a molecule or ion that show the delocalization of electrons. In option b, the double bond can be moved between the nitrogen and oxygen atoms, resulting in two resonance structures.
Resonance occurs when there are multiple valid Lewis structures that can be drawn for a molecule or ion. It indicates that the actual structure is a combination or hybrid of all the resonance structures.
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Which of the following molecules nonpolar?
○ CHCl3
○ CH2Cl2
○ CH3Cl
○ None, they are polar.
Answer:
CHC13
Explanation:
You need to make 25 microliters of a 3M NaOh solution for a test reagent. Your laboratory routinely stocks 500 milliliters of a 10M NaOh solution. How would you prepare your solution?
Answer:
0.0075 milliliters (7.5 microliters) will be taken from the stock solution and then diluted with 0.0249925 milliliters (24.9925 microliters) of water to make 0.025 milliliters (25 microliters) of NaOH solution.
Explanation:
This is a problem of dilution using the equation:
initial concentration x initial volume = final concentration x final volume.
The final volume to be prepared is 25 microliters.
The final concentration to be prepared is 3 M.
The initial volume to be taken is not known yet.
The initial concentration is 10 M.
Now, let's substitute these parameters into the the equation above.
10 x initial volume = 3 x 25
Initial volume = 3 x 25/10
= 7.5 microliters
Note that: 1 microliter = 0.001 milliliters
Hence,
7.5 microliters = 0.0075 milliliters
This means that an initial volume of 0.0075 milliliters (7.5 microliters) will be taken from the stock solution. This amount will then be diluted with 0.0249925 milliliters (24.9925 microliters) of water to make 0.025 milliliters (25 microliters) of NaOH solution.
On occasion, it has been found that the oxidation of borneol doesn't go to completion (possibly because of poor stirring or insufficient Oxone). This reaction would probably be easy to monitor via TLC, however. Which component should have a lower Rf, and why
Answer:
Check the explanation
Explanation:
Here, Nitrogen (N) undergoes oxidation and Chlorine (Cl) undergoes reduction.
To answer your question:
N is oxidized from an oxidation number of -3 to an oxidation number of -1.
Cl is reduced from oxidation number of +1 to an oxidation number of -1.
Now,
Borneol should have a lower Rf because of boiling point.
a gas occupies 34.5 liters at a pressure of 345 torr what will be the volume of this gas if the pressure becomes 760 torr
Answer:
THE NEW PRESSURE IS 15.5 L.
Explanation:
Using Boyle's law which states that the pressure of a given mass of gas is inversely proportional to the volume of the gas at constant temperature.
Mathematically,
P1V1 =P2V2
P1 = 345 torr = 345/760 atm = 0.45 atm
P2 = 760 torr = 760/760 atm = 1 atm
V1 = 34.5 L
V2 = ?
Rearranging the variables, making V2 the subject of the formula, we obtain:
V2 = P2 V1 / P1
V2 = 1 * 34.5 / 0.45
V2 = 15.525 Litres
The volume of the gas when the pressure becomes 760 torr or 1 atm is 15.5 Litres
what kind of solution would have a Ka value that is much less than 1
Answer:
A weakly acidic solution.
Explanation:
Final answer:
A solution with a Ka value much less than 1 refers to a weak acid, where only a small portion of the acid is ionized, resulting in relatively low concentrations of hydronium ions and the conjugate base.
Explanation:
The type of solution that would have a Ka value much less than 1 is a weak acid solution. This is because the Ka (acid dissociation constant) value indicates the strength of the acid in solution; the lower the Ka, the weaker the acid. For a Ka value significantly less than 1, this means that only a small fraction of the acid molecules donate protons (H+) to the water, resulting in few hydronium ions (H₋₃O⁺) and the conjugate base (A-).
In contrast, strong acids have a Ka value that approaches infinity since they are almost completely ionized in solution. Acetic acid, with a Ka value of 1.75 x 10⁻⁵, is an example of a weak acid, indicating that in equilibrium, the concentration of the undissociated acetic acid is much greater than the concentrations of the acetate and hydronium ions.
True False The colors emitted depends on the number of free electrons passing through the lamp. True False When a free electron hits an atom, the atom is always excited to the highest energy level possible. True False The kinetic energy of the free electron at the point of collision increases as the voltage of the battery increases. True False The kinetic energy of the free electron at the point of collision is higher if the atom is closer to the source of electrons. True False The only way to emit IR photons is if there are empty electronic energy levels really close to the ground state (lowest energy level). True False When atomic electrons are excited to a higher level, they always return to their lowest energy level by jumping down one level at a time.
Answer:
False
False
True
False
False
False
Explanation:
The colour emitted by an excited atom depends on the emission Spectra of the atom rather than on the number of free electrons passing through the lamp.
When a free electrons hits an atom, the atom can be excited to various intermediate states other than the highest energy level depending on the amount of energy it absorbed.
Increase in voltage of the battery also increases the kinetic energy of the electron
The kinetic energy of the electron depends on the voltage difference not on the distance from the atom.
The Paschen, Brackett, and Pfund series of lines are due to transitions from higher-energy orbits to orbits with n = 3, 4, and 5, respectively; these transitions release substantially less energy, corresponding to infrared radiation.
Excited atoms can jump from a higher level to the ground state in a series of steps or directly to the ground state
What is the MOLAR heat of combustion of methane(CH₄) if 64.00g of methane are burned to heat 75.0 ml of water from 25.00°C to 95.00°C?(Water heat capacity-4.184 J/g°C)
Answer:
-5.51 kJ/mol
Explanation:
Step 1: Calculate the heat required to heat the water.
We use the following expression.
[tex]Q = c \times m \times \Delta T[/tex]
where,
c: specific heat capacitym: massΔT: change in the temperatureThe average density of water is 1 g/mL, so 75.0 mL ≅ 75.0 g.
[tex]Q = 4.184J/g.\°C \times 75.0g \times (95.00\°C - 25.00\°C) = 2.20 \times 10^{3} J = 2.20 kJ[/tex]
Step 2: Calculate the heat released by the methane
According to the law of conservation of energy, the sum of the heat released by the combustion of methane (Qc) and the heat absorbed by the water (Qw) is zero
Qc + Qw = 0
Qc = -Qw = -22.0 kJ
Step 3: Calculate the molar heat of combustion of methane.
The molar mass of methane is 16.04 g/mol. We use this data to find the molar heat of combustion of methane, considering that 22.0 kJ are released by the combustion of 64.00 g of methane.
[tex]\frac{-22.0kJ}{64.00g} \times \frac{16.04g}{mol} = -5.51 kJ/mol[/tex]
Given the following reaction:
\ce{2KClO3 -> 2KCl + 3O2}2KClOX
3
2KCl+3OX
2
How many moles of \ce{KCl}KClK, C, l will be produced from 15.0 \text{ g}15.0 g15, point, 0, start text, space, g, end text of \ce{KClO3}KClOX
3
?
moles (round to three significant figures)
Answer:
0.122 moles
Explanation:
Final answer:
0.122 moles of KCl will be produced from the reaction of 15.0 g of KClO3, using stoichiometry and the molar mass of KClO3 which is 122.55 g/mol.
Explanation:
To determine the amount of KCl produced from 15.0 g of KClO3, we first need to know the molar mass of KClO3, which is 122.55 g/mol according to LibreTexts. By using stoichiometry, we can find out how many moles of KClO3 we have and, subsequently, how many moles of KCl will be produced according to the balanced chemical equation.
First, calculate the number of moles of KClO3:
15.0 g KClO3 × (1 mol KClO3 / 122.55 g) = 0.122 mol KClO3
From the balanced equation Î{2KClO3 -> 2KCl + 3O2}, we can see that 2 moles of KClO3 produce 2 moles of KCl. Therefore, the number of moles of KCl produced is equal to the number of moles of KClO3 available:
0.122 mol KClO3 × (2 mol KCl / 2 mol KClO3) = 0.122 mol KCl
Thus, 0.122 mol of KCl are produced from the reaction of 15.0 g of KClO3.
Fill in the blanks with the word that best completes each statement. Scientists develop knowledge by making about the natural world that may lead to a scientific question. A scientific question may lead to a(n) , which can be tested. The results of can lead to changes in scientific knowledge.
A scientific question may lead to a(n)
The answer is may lead to a hypothesis.
Explanation:
Answer: observations
Hypothesis
Experimentation
Explanation:in order and right on edge
The ph of an aqueous solution at 25.0°c is 10.66. what is the molarity of h+ in this solution? the ph of an aqueous solution at 25.0°c is 10.66. what is the molarity of h+ in this solution? 4.6 à 10-4 3.3 2.2 à 10-11 1.1 à 10-13 4.6 à 1010
Answer: The molarity of [tex]H^+[/tex] in this solution is [tex]2.2\times 10^{-11}M[/tex]
Explanation:
pH or pOH is the measure of acidity or alkalinity of a solution.
pH is calculated by taking negative logarithm of hydrogen ion concentration.
[tex]pH=-\log [H^+][/tex]
Given : pH = 10.66
Putting in the values:
[tex]10.66=-\log[H^+][/tex]
[tex][H^+]=10^{-10.66}[/tex]
[tex][H^+]=2.2\times 10^{-11}[/tex]
Thus the molarity of [tex]H^+[/tex] in this solution is [tex]2.2\times 10^{-11}M[/tex]
The molarity of hydrogen ions, H⁺ in this solution is 2.2 * 10⁻¹¹
The molarity of a solution is a measure of the concentration in moles of a substance present in a liter of solution of that substance.
The pH of a solution is the negative logarithm in base 10 of the hydrogen ion concentration of the solution.
pH = -log[H⁺]
-pH = log[H⁺]
pH of solution = 10.66
-10.66 = log[H⁺]
taking antilogarithm of both sides
[H⁺] = 10⁻¹⁰°⁶⁶
[H⁺] = 2.2 * 10⁻¹¹
Therefore, the molarity of hydrogen ions, H⁺ in this solution is 2.2 * 10⁻¹¹
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What is the concentration of hydronium ion ( [H3O+]) in a solution with a PH of _1,3?
Answer: [H3O+]= 0.05 M
Explanation:
Message me for extra explanation.
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Determine the amount of energy required to boil 50 g of
ethanol.
Answer:
42050 J.
Explanation:
Data obtained from the question:
Mass (M) of ethanol = 50g
Heat of vaporisation (ΔHv) of ethanol = 841 J/g
Heat (Q) =.?
The heat required to boil 50g of ethanol can be obtained as follow:
Q = MΔHv
Q = 50 x 841
Q = 42050 J.
Therefore, the heat required to boil 50g of ethanol is 42050 J.
A geochemist in the field takes a sample of water from a rock pool lined with crystals of a certain mineral compound . He notes the temperature of the pool, , and caps the sample carefully. Back in the lab, the geochemist first dilutes the sample with distilled water to . Then he filters it and evaporates all the water under vacuum. Crystals of are left behind. The researcher washes, dries and weighs the crystals. They weigh 3.00 g.
a. Using only the information above, can you calculate the solubility of X in water at 26 degrees Celsius.?
b. If yes than calculate the solubility of X. Round your answer to 3 significant digits
Answer:
Check the explanation
Explanation:
Kindly check the attached image below to see the step by step explanation to the question above.
The U.S. requires automobile fuels to contain a renewable component. The fermentation of glucose from corn produces ethanol, which is added to gasoline to fulfill this requirement: C6H12O6(s) → 2 C2H5OH(l) + 2 CO2(g) Calculate ΔH o , ΔS o and ΔG o for the reaction at 25°C. Is the spontaneity of this reaction dependent on temperature?
Answer:
Explanation:
The chemical equation for the reaction is :
[tex]C_6H_{12}O_6_{(s)} \to 2C_2H_5OH _{(l)} + 2CO_{2(g)}[/tex]
The standard enthalpy of formation [tex]\Delta H ^0_f[/tex] of the above equation is as follows:
[tex]\Delta H^0_{f, C_6H_{12}O_6}[/tex] = -1274.4 kJ/mol
[tex]\Delta H ^0_{f, C_2H_{5}OH[/tex] = -277.7 kJ/mol
[tex]\Delta H ^0_{f, CO_2}[/tex] = -393.5 kJ/mol
[tex]\Delta H ^0_{rxn }= \sum n_p \Delta H ^0_{f,p} - \sum n_r \Delta H ^0_{f,r}[/tex]
where ;
[tex]n_p[/tex] = stochiometric coefficients of products
[tex]n_r=[/tex] stochiometric coefficients of reactants
[tex]\Delta H^0_{f.p}[/tex] = formation standard enthalpy of products
[tex]\Delta H^0_{f.r}[/tex] = formation standard enthalpy of reactants
[tex]\Delta H ^0_{rxn }= (2 \ mol* -2777.7 \ kJ/mol + 2 \ mol * - 393.5 \ kJ/mol) - (1\ mol *(-1274.4 \ kJ/mol)[/tex]
[tex]\Delta H ^0_{rxn }= -68 \ kJ[/tex]
For [tex]\Delta S ^0[/tex] ;
The standard enthalpy of formation of [tex]\Delta S ^0_f[/tex] of the reactant and the products are :
[tex]\Delta S ^0 _{f \ C_6H_{12}O_6} = 212 \ \ J/Kmol[/tex]
[tex]\Delta S ^0 _{f \ C_2H_5O_H} = 160.7 \ \ J/Kmol[/tex]
[tex]\Delta S ^0 _{f \ CO_2} = 213.63 \ \ J/Kmol[/tex]
The [tex]\Delta S ^0_{rxn}[/tex] is as follows:
[tex]\Delta S ^0_{rxn} = \sum n_p \Delta S^0_{f.p} - \sum n_r \Delta S^0_{f.r}[/tex]
[tex]\Delta S ^0_{rxn} = (2 \ mol *160.7 \ \ J/Kmol + 2 \ mol *213.63 \ \ J/Kmol) -(1*212 J/Kmol)[/tex]
[tex]\Delta S ^0_{rxn} =536 . 7 \ J/K[/tex] (to kJ/K)
[tex]\Delta S ^0_{rxn} =536 . 7 \ J/K * \frac{1 \ kJ}{1000 \ J}[/tex]
[tex]\Delta S ^0_{rxn} =0.5367 \frac{kJ}{K}[/tex]
Given that;
at T = 25°C = ( 25 + 273) K = 298 K
[tex]\Delta G^0 _{rxn} = \Delta H^0_{rxn} - T \Delta S^0_{rxn}[/tex]
[tex]\Delta G^0 _{rxn} = -68 \ kJ - 298 * 0.5367 \frac{kJ}{K}[/tex]
[tex]\Delta G^0 _{rxn} = -227. 9 \ \ \ kJ[/tex]
As [tex]\Delta G^0 _{rxn}[/tex] is negative; the reaction is spontaneous
[tex]\Delta H^0 _{rxn}[/tex] = negative
[tex]\Delta S^0 _{rxn}[/tex] = positive; Therefore , the reaction is spontaneous at all temperature , We can then say that the spontaneity of this reaction [tex]\Delta G^0 _{rxn} = \Delta H^0_{rxn} - T \Delta S^0_{rxn}[/tex] is dependent on temperature.
Which of these statements indicates an experiment is systematic?
A. The experiment is repeated many times.
B. The experiment is about a testable question.
C. The experiment is very complicated and uses lots of chemicals.
D. The experiment includes a good plan for what measurements to take.
Answer:
D. The experiment includes a good plan for what measurements to take.
Explanation:
Does a part or slice of a substance have a different density than the whole piece?
Pls help!
A slice or part of a homogeneous substance has the same density as the whole piece because the density is uniform throughout. In a heterogeneous substance, the density can vary and depends on the composition and structure of the part being measured. Density is found by dividing mass by volume and is useful for identifying substances.
The density of a substance is defined as its mass divided by its volume. In a homogeneous material, the density is consistent throughout, which means a slice or part of the substance would have the same density as the whole piece. For example, a solid iron bar is homogeneous, and therefore any part of it would have the same density as the entire bar. However, in a heterogeneous material, the density can vary from one part to another; an instance of this is Swiss cheese, which contains air pockets resulting in variable local densities.
To determine the density of a substance, you would divide the mass by the volume of the substance. This method applies to both whole objects and parts thereof. In the case of a cube, the volume is found by cubing the edge length. In practice, density is often used to help identify substances by comparison with known values, as different substances have characteristic densities.
The molecular formula of butane is C4H10. It is obtained from petroleum and is used commonly in LPG (Liquefied Petroleum Gas) cylinders (a common source of cooking gas). It has two arrangements of carbon atoms: a straight chain and a branched chain. Using this information, draw the structure of the tertiary butyl radical that will form upon removal of a hydrogen atom. Draw the molecule on the canvas by choosing buttons from the Tools (for bonds), Atoms, and Advanced Template toolbars. Include all free radicals by right-clicking on an atom on the canvas and then using the Atom properties to select the monovalent radical.
Answer:
See explaination
Explanation:
Please see the attached file for the structural diagram of the molecular formation.
It is detailed and properly presented at the attachment.
Ethanol has a 38% lower energy density by volume than gasoline. Partially offsetting this disadvantage is the higher octane rating of ethanol, which allows it to be used in engines having a higher compression ratio. In fact, a standard gasoline-powered engine typically runs at a compression ratio r = 10, while an ethanol-powered one can run at r = 16. Internal combustion engines can be approximated by the ideal Otto cycle, for which the efficiency is given by eo = 1 − r −0.4. Assume that a real engine has one-third the ideal Otto efficiency, and calculate how much improvement this would make to the efficiency of the ethanol-fueled engine over a standard gasoline engine. Is it enough to offset the lower fuel energy density of ethanol?
Answer:
Find the given attachment
Show the mechanism for the following reaction: cyclohexene bromine yields a dibromocyclohexane Unlike previous problem, this problem does require illustration of stereochemistry. Draw wedge-and-dash bond stereochemical structures – including H atoms on ALL chirality center (product and intermeidate both)– and include charges, electrons, and curved arrows. Details count. Draw one enantiomer only for any racemates. NOTE: A bromonium ion bromine should have TWO lone pairs, not three.
Answer:
Check the explanation
Explanation:
Kindly check the attached image below for the step by step explanation to the question above.
Write a balanced net-ionic equation: A zinc wire is placed in a solution of FeSO4. (Use the lowest possible coefficients. Use the pull-down boxes to specify states such as (aq) or (s). If a box is not needed, leave it blank.) + → + Calculate E°, ΔG°, and K at 25°C. E° = V ΔG° = kJ K =
Answer:
Zn(s) + Fe^2+(aq) ---------> Zn^2+(aq) + Fe(s)
∆G= 61760J
K= 6.5 ×10^10
Explanation:
Zn(s) + Fe^2+(aq) ---------> Zn^2+(aq) + Fe(s)
E°reaction= E°Fe - E°Zn
E°reaction= (-0.44)-(-0.76)
E°reaction= 0.32V
We know that
∆G= -nFE°
Since n= 2 from the reaction equation and F= 96500C
∆G= -(2×96500×0.32)
∆G= 61760J
From
E°= 0.0592/n log K
0.32= 0.0592/2 log K
log K= 0.32/0.0296
log K= 10.81
K= Antilog(10.81)
K= 6.5 ×10^10
The reaction between sucrose and water to produce fructose and glucose is first order // overall. The data below shows the change in concentration of sucrose over time at 298 K. C12H22011 (aq)sucrose + H2O(l)艹C6H1206(aq)fructose + C6H1206(aq)oucose C12H2201l M Time (minutes) 1.002 0.808 0.630 0.0 60.0 130.0 a. Find the average rate ofdisappearance of C12H22011 from t = 0 min to t-60 min. b. Find the average rate of appearance of fructose from t 0 min to t 60 min. c. Calculate the rate constant, k, for the decomposition of sucrose at 25°C. Include units. d. How long will it take for the concentration of sucrose to drop from 1.002 to 0.212 M? e. What is the half-life for the decomposition of sucrose at 25°C?
Answer:
(a)
Rate of appearance of sucrose = - d[C12H22O11] / dt = - ( 0.808 - 1.002 ) / ( 60.0 - 0.0) = 0.00323 M/s
(b)
Rate of appearance of fructose = d[C6H12O6] / dt = (1.002 - 0.808) / (60.0 - 0.0) = 0.00323 M/s
(c)
k = (1 / t ) * ln[A]/[A]t
k = ( 1 / 60.0 ) * ln[1.002 / 0.808]
k = 0.00359 min-1
(d)
0.00359 = ( 1 / t ) * ln[1.002 / 0.212]
t = 432.6 min
(e)
Half life time = 0.693 / k = 0.693 / 0.00359 = 193 min
Explanation:
First-order reactions are defined as the chemical reactions in which rate of the reaction is linearly dependent on the concentration of only one reactant.
The answers can be explained as:
(a) Rate of appearance of the sucrose from the chemical reaction is:
Rate = [tex]\dfrac{\text d [\text C_{12}\text H_{22}\text O_{11}]}{\text {dt}}[/tex] = [tex]\dfrac{0.808 - 1.002}{60.0 -0.0}[/tex]
Rate = 0.00323 m/s
(b) Rate of appearance of Fructose from the given chemical reaction is:
Rate = [tex]\dfrac{\text d [\text C_{6}\text H_{12}\text O_{6}]}{\text {dt}}[/tex] = [tex]\dfrac{1.002 - 0.808 }{60.0 -0.0}[/tex]
Rate = 0.00323 m/s
(C) Rate constant for the reaction is:
[tex]\text k &= \dfrac{1}{\text t}\times \dfrac {\text{ln [A]}}{\text {[A]} \text t}[/tex]
[tex]\text k &= \dfrac{1}{60}\times \dfrac {\text{ln} (1.002)}{(0.808)}[/tex]
k = 0.00359 minute⁻¹
(d) Time required for the concentration of sucrose to drop from 1.002 to 0.212 M is:
[tex]0.00359 &= \dfrac{1}{t} \times {\text{ln}\dfrac{[1.002]}{[0.212]}[/tex]
t = 432.6 minutes
(e) The half-life of the decomposition of sucrose at 25°C is:
Half-life = [tex]\dfrac{0.693}{\text k} = \dfrac{0.693}{0.00359}[/tex]
Half-life = 193 minutes.
To know more about first-order reaction, refer to the following link:
https://brainly.com/question/12446045
The following balanced equation shows the formation of ethane (C2H6).
C2H2 + 2H2 - C2H6
How many moles of hydrogen are needed to produce 13.78 mol of ethane?
3,445 mol
6.890 mol
27.56 mol
55.12 mol
Answer:
C
Explanation:
Well, we can see from the equation that hydrogen gas and ethane are in a 2:1 ratio based on their mole coefficients.
You need 2 mol H2 to make 1 mol C2H6. So that means you need 2*13.78 mol H2 to make 13.78 mol C2H6!
2*13.78 = 27.56 mol
C is correct.
How many milliliters of 1.50 M magnesium sulfate soulution is required to supply 2.50 mole of this salt?
Answer:
1670 ml
Explanation:
molarity x Volume (Liters) = moles => Volume (Liters) = moles/Molarity
Volume needed = 2.50mol/1.50M = 1.67 Liters = 1670 ml.
Answer:
hes right
Explanation:
ik all
A balloon at 30.0°C has a volume of 222 mL. If the temperature is
increased to 53.1°C and the pressure remains constant, what will the new volume be, in ml?
Answer : The new volume will be, 238.9 mL
Explanation :
Charles' Law : It states that volume of the gas is directly proportional to the temperature of the gas at constant pressure and number of moles of gas.
Mathematically,
[tex]\frac{V_1}{T_1}=\frac{V_2}{T_2}[/tex]
where,
[tex]V_1\text{ and }T_1[/tex] are the initial volume and temperature of the gas.
[tex]V_2\text{ and }T_2[/tex] are the final volume and temperature of the gas.
We are given:
[tex]V_1=222mL\\T_1=30.0^oC=(30.0+273)K=303K\\V_2=?\\T_2=53.1^oC=(53.1+273)K=326.1K[/tex]
Putting values in above equation, we get:
[tex]\frac{222mL}{303K}=\frac{V_2}{326.1K}\\\\V_2=238.9mL[/tex]
Therefore, the new volume will be, 238.9 mL