Convert the Haworth projection of the carbohydrate below into its corresponding Fischer projection (in the standard format with the most oxidized carbon at the top) and chair conformation (correctly drawing equatorial and axial bonds) of the opposite anomer. Then answer the questions in the problem. Upload a photo of your answer as a jpeg file. If we can't read it, we can't grade it!
Answers
Answer:
See explaination
Explanation:
H O / I MOH the OH da Anomer My nequatorial bovel SOH niematerial and CH2OH Yr axial boud - H o H o . H (Auro) 2 flip . note: 1)Fischer projection : L- Sugar ( OH group is in the left side ) at C-5 carbon (bottom most chiral center )
In the Haworth projection given above in the qsn it is Beta anomer( -CH2OH & C-OH are in same side i.e. Cis to each other ). It is a L -sugar .
I omitted C2 -H bond (axial ) in the chair conformation of alpha anomer.
See attachment for further solution
Transitioning from a Haworth to a Fischer projection, every carbon atom becomes a line end or intersection point. Correct position and orientation of the hydroxyl group determines the chair configuration. Carbohydrates, which contain carbon, hydrogen, and oxygen atoms in a 1:2:1 ratio, have monosaccharide, disaccharide, and polysaccharide subtypes.
Explanation:The question is asking you to convert a Haworth projection of a carbohydrate into a Fischer projection and a chair conformation of its opposite anomer. This involves understanding several concepts in organic chemistry. The Haworth projection is a common way of representing cyclic sugars, which are typically found in ring forms in aqueous solutions. The Fischer projection is used to represent the stereochemistry of the molecule. In converting the Haworth projection to a Fischer projection, each carbon atom in the ring becomes the end of a line or a point where lines intersect in the Fischer projection. The most oxidized carbon (typically the one attached to an oxygen atom) is placed at the top.
Next, to draw the chair conformation, you must understand axial and equatorial bonds, which describe the orientation of chemical groups around the carbon atoms in the ring. Axial bonds are vertical and shunted to the side, while equatorial bonds are more horizontal and radiate outwards. For the opposite anomer, if the original has the hydroxyl group below the plane in an alpha position, the opposite anomer would have it above the plane in a beta position.
Carbohydrates bear the stoichiometric formula (CH₂O)n, indicating a carbon to hydrogen to oxygen ratio of 1:2:1. Classified into monosaccharides, disaccharides, and polysaccharides, carbohydrates are a crucial chemical group in biology.
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Related Questions
what kind of solution would have a Ka value that is much less than 1
Answers
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.
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______.
Answers
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.
Need help setting the problem up
Answers
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
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
Answers
Answer:
Check the explanation
Explanation:
Kindly check the attached image below to see the step by step explanation to the question above.
The magnesium and calcium ions present in seawater ([Mg2+] = 0.059 M and [Ca2+] = 0.011 M) can be separated by selective precipitation with KOH.
If the concentration of Mg2+ in the solution were 0.039 M, what minimum [OH−] triggers precipitation of the Mg2+ ion? (Ksp=2.06×10−13.)
Answers
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.
Radiation detectors use which of the following properties to detect radioactivity?
○ charge
○ ionizing
○ mass
○ energy
Answers
Answer:
ionizing
I hope I helped :)
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.
Answers
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
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?
Answers
Answer:
Find the given attachment
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
Answers
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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Determine the amount of energy required to boil 50 g of
ethanol.
Answers
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 0.0200 gram piece of unknown alkaline earth metal, M, is reacted with excess 0.500 M H 2 SO 4 , and the hydrogen gas produced is collected over water. The total gas pressure inside the collecting tube is 1.01 atm., the temperature 24.0 o C. The volume of gas collected is 19.6 mL. The gas in the tube contains water vapor, at a pressure of 0.029 atm. M(s) + H 2 SO 4 (aq) → MSO 4 (aq) + H 2 (g). What is the partial pressure of the dry hydrogen gas collected in the tube?
Answers
Answer:
0.981atm
Explanation:
According ot Dalton's law total pressure of a mixture of non-reactive gas is equal to sum of partial pressures of individual gases.
total pressure= 1.01at
Number of gases=2
Gases: water vapor and hydrogen
partial pressure of water vapor= 0.029atm
1.01= partial pressure of water vapor+ partial pressure of hydrogen
1.01= 0.029 + partial pressure of hydrogen
partial pressure of hydrogen = 0.981atm
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.'
Answers
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]
How many milliliters of 1.50 M magnesium sulfate soulution is required to supply 2.50 mole of this salt?
Answers
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
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?
Answers
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.
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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?
Answers
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.
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?
Answers
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
On January 2, 2020, Nash Company purchases a call option for $290 on Merchant common stock. The call option gives Nash the option to buy 1,050 shares of Merchant at a strike price of $51 per share. The market price of a Merchant share is $51 on January 2, 2020 (the intrinsic value is therefore $0). On March 31, 2020, the market price for Merchant stock is $54 per share, and the time value of the option is $210.
Answers
Final answer:
A call option is a financial contract that gives the buyer the right to purchase shares of a stock at a predetermined price. The net profit from this call option transaction can be calculated by determining the difference between the market price and the strike price, multiplied by the number of shares.
Explanation:
A call option is a type of financial contract that gives the buyer the right, but not the obligation, to purchase a specified number of shares of a stock at a predetermined price within a certain time frame. In this case, Nash Company purchased a call option to buy 1,050 shares of Merchant stock at a strike price of $51 per share. On March 31, 2020, the market price for Merchant stock is $54 per share, and the time value of the option is $210.
The net profit from this stock transaction can be calculated by determining the difference between the market price of the stock and the strike price, multiplied by the number of shares:
Net Profit = (Market Price - Strike Price) x Number of Shares
Using the given information, the net profit from this call option transaction would be:
Net Profit = ($54 - $51) x 1,050 = $3 x 1,050 = $3,150
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?
Answers
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.
Answers
Answer:
D. The experiment includes a good plan for what measurements to take.
Explanation:
The elementary gas-phase reaction takes place isobarically and isothermally in a PFR where 63.2% conversion is achieved. The feed is pure A. It is proposed to put a CSTR of equal volume upstream of the PFR. Based on the entering molar flow rate to A to the first reactor, what will be the intermediate from the CSTR, X 1 , and exit conversion from the PFR, X 2 , based on the feed to first reactor?
Answers
Answer:
Explanation:
check below for explanation in the attached files.
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)
Answers
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.
17
¿Cuál de las siguientes
propiedades distingue una solución
sobresaturada de una diluida?
Answers
Answer:
Which of the following
properties distinguishes a solution
oversaturated with a dilute?
The supersaturated solution is one in which the solvent has dissolved more solute than it can dissolve in the saturation equilibrium. The solute can be a solid, or a gas. The molecules of the solvent surround those of the solute and seek to open space between themselves to be able to harbor more amount of solute.A dilute solution is a solution that has not reached the maximum concentration of solute dissolved in a solvent. The additional solute will dissolve when added in a dilute solution and will not appear in the aqueous phase. It is considered a state of dynamic equilibrium where the speeds in which the solvent dissolves the solute are greater than the recrystallization rate.when 1564 J of heat energy is added to a sample of gold at 25.0°C, the temperature of the gold increases to 449°C. what is the means of the gold sample? (specific heat of gold= 0.129 J/g-°C
Answers
Answer:
The mass of the gold sample is 28.59 g
Explanation:
Calorimetry is the measurement and calculation of the amounts of heat exchanged by a body or a system. The amount of heat received or transferred by a body when it undergoes a temperature variation (Δt) without changing its physical state (solid, liquid or gaseous) is calculated by:
Q = c * m * ΔT
Where Q is the heat exchanged by a body of mass m, made up of a specific heat substance c and where ΔT is the temperature variation.
In this case:
Q= 1564 Jc= 0.129 [tex]\frac{J}{g*C}[/tex]m=?ΔT= Tfinal - Tinitial= 449°C - 25°C= 424 °CReplacing:
[tex]1564 J=0.129\frac{J}{g*C}*m*424 C[/tex]
Solving:
[tex]m=\frac{1564 J}{0.129\frac{J}{g*C}*424 C}[/tex]
m=28.59 g
The mass of the gold sample is 28.59 g
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
Answers
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.
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.
Answers
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.
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 =
Answers
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
Push against a door that does not move. Describe the mechanical energy
Answers
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.
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
Answers
Answer:
Check the explanation
Explanation:
Kindly check the attached image below to see the step by step explanation to the question above.
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
Answers
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
Calculate the percent oxygen in sodium oxide.
Answers
Answer:
% is 51.6% of Oxygen
Explanation:
According to the working in the photo
51.6% is the percent of oxygen in sodium oxide. It represents the amount of solute in a given amount of solution.
What is mass percentage?A concentration is expressed as a mass percent. Furthermore, it defines the component of a particular blend. The solution component can be expressed in terms of mass percentage. It represents the amount of solute in a given amount of solution.
The amount of solute is measured in mass or moles. We shall study the mass percent equation using numerous solved numerical examples in this post. Mass is often represented in grams, although any unit of measurement is acceptable as long as an equivalent unit is used both for the component or solute mass and thus the total or solution mass.
mass percentage= (mass of oxygen / mass of sodium oxide)×100
=(32/62)×100
= 51.6%
Therefore, 51.6% is the percent of oxygen in sodium oxide.
To learn more about mass percentage, here:
https://brainly.com/question/28973847
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