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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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.
A solution of calcium chlorate was poured into a sodium fluoride solution. Would you expect a precipitate to form if 255.0 mL of the calcium chlorate solution (2.0x10-5 M) was mixed with 300.0 mL of a 2.5x10-3 M sodium fluoride solution? A. No, because Qsp < Ksp for calcium fluoride B. Yes, because Qsp < Ksp for calcium fluoride C. Yes, because Qsp = Ksp for sodium chlorate D. Yes, because Qsp = Ksp for calcium fluoride E. No, because Qsp = Ksp for sodium chlorate F. No, because Qsp < Ksp for sodium chlorate G. No, because Qsp > Ksp for calcium fluoride H. No, because Qsp > Ksp for sodium chlorate I. Yes, because Qsp < Ksp for sodium chlorate J. No, because Qsp = Ksp for calcium fluoride K. Yes, because Qsp > Ksp for sodium chlorate L. Yes, because Qsp > Ksp for calcium fluoride
Answer:
A) No , because Qsp<Ksp for calcium fluoride.
Check attachment for calculation
Without specific calculations, to determine if a precipitate forms when mixing calcium chlorate with sodium fluoride, one assesses if the product of the ion concentrations exceeds the Ksp for calcium fluoride; if Qsp > Ksp, precipitation occurs.
Explanation:To determine if a precipitate will form when calcium chlorate is mixed with sodium fluoride, one must consider the ionic products formed and compare the reaction quotient (Qsp) with the solubility product constant (Ksp) of the relevant compound, in this case, calcium fluoride (CaF2). The Ksp for calcium fluoride is 3.45 × 10∑11. In a mixture of calcium chlorate and sodium fluoride, the calcium ions (Ca2+) from calcium chlorate can react with the fluoride ions (F∑) from sodium fluoride to potentially form solid calcium fluoride. However, without the exact concentrations of Ca2+ and F∑ after mixing, one cannot directly calculate Qsp. The principle, however, is that if Qsp > Ksp, a precipitate of calcium fluoride will form, whereas if Qsp < Ksp, no precipitate forms because the solution is not supersaturated with respect to CaF2.
The concentration of each ion in the final solution depends on the dilution that occurs when the two solutions are mixed. Given the initial conditions and the principle that the formation of a precipitate requires Qsp to exceed Ksp, we generally assess the likelihood of precipitation by considering the concentrations of the reactive ions and the solubility product of the insoluble compound they may form. In this scenario, without performing a specific calculation, the correct answer would be based on understanding whether the mixing leads to a situation where the product of the concentrations of calcium and fluoride ions exceeds the Ksp of calcium fluoride.
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.
A student states that this structural formula represents a hydrocarbon.
H H
H-C-0-0-H
I-0-I
Is the student correct?
Yes, a hydrocarbon is any compound that contains a carbon backbone.
Yes, a hydrocarbon is any compound that contains hydrogen and oxygen.
No, a hydrocarbon contains only carbon and hydrogen atoms.
No, a hydrocarbon contains at least one double or triple bond.
Answer:
Yes, a hydrocarbon is any compound that contains hydrogen and oxygen.
Explanation:
Which of the following molecules nonpolar?
○ CHCl3
○ CH2Cl2
○ CH3Cl
○ None, they are polar.
Answer:
CHC13
Explanation:
Absorption of 1.0 rad of which of the following types of radiation will result in the greatest tissue damage?
○ alpha
○ beta
○ gamma
○ They all result in the same amount of damage.
Absorption of 1.0 rad of gamma radiation would likely cause the greatest tissue damage due to its high energy and penetration ability.
The amount of tissue damage caused by radiation depends on various factors, including the type of radiation and its energy level.
Alpha particles are relatively heavy and highly ionizing, but they have low penetration power, so they are generally less damaging to tissues externally but can be more damaging if ingested or inhaled.
Beta particles have less mass and energy compared to alpha particles and can penetrate deeper into tissues, potentially causing more damage.
Gamma rays are electromagnetic radiation with high energy and penetration power. They can penetrate deeply into tissues, causing damage along their path.
Considering these factors, absorption of 1.0 rad of gamma radiation would likely result in the greatest tissue damage due to its high energy and penetration ability.
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.
What kind of reaction does luminol go through?
Answer: This reaction is called as Chemiluminescence. A type of reaction in which light is produced due to chemical reaction. While in case of Phosphorescence and Fluorescence light is emitted due to absorption of photons.
Explanation: i looked the answer up but your welcomee ❤
Final answer:
Luminol undergoes a chemiluminescent reaction that produces light when it reacts with hydrogen peroxide and a catalyst, often used in forensic science to detect blood.
Explanation:
Luminol goes through a chemiluminescent reaction, which refers to a process where a chemical reaction produces a molecule in an excited state. In the presence of hydrogen peroxide and a catalyst, such as the iron found in hemoglobin, luminol reacts to generate 3-aminophthalate in an excited state, emitting a bluish light. This reaction is highly valuable in forensic science for the detection of blood.
Chemiluminescence is a phenomenon where the energy from a chemical reaction is released as light. This reaction is possible because certain chemicals, like peroxide and ozone used in chemiluminescent reactions, contain unstable or energetic chemical bonds. The process is a rare occurrence in chemical reactions and is a direct transduction of chemical energy into radiant energy.
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 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]
Common additives to drinking water include elemental chlorine, chloride ions, and phosphate ions. Recently, reports of elevated lead levels in drinking water have been reported in cities with pipes that contain lead, Pb(s). When Cl2(aq) flows through a metal pipe containing Pb(s), some of the lead atoms oxidize, losing two electrons each, and aqueous chloride ions form. (a) Write a balanced, net-ionic equation for the reaction between Pb(s) , and
Answer:
Check the explanation
Explanation:
Kindly check the attached image below to see the step by step explanation to the question above.
A common addiction to the common water is the elemental chlorine and chlorine ions, the phosphate ions. Recent reports of the higher levels of lead in cities have to lead to the Pb(s) as CI 2 flows in petal pipes some may be oxidized.
They balanced equation of the net ionic reactions.
Pb(s) + Cl2(g) ----- Pb2+ + 2Cl- oxidation reaction :Pb(s) --- Pb2+ + 2e- Reduction reaction:Cl2 + 2e- -----Cl-Learn more about the chloride ions and phosphate ions.
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how many moles of 0.225 m ca oh 2 are present in 0.350 l of solution
Answer: 0.225, x, 0.0788
Explanation:
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.
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
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.
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 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.
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The combustion of titanium with oxygen produces titanium dioxide: Ti(s) + O 2(g) → TiO 2(s) When 2.060 g of titanium is combusted in a bomb calorimeter, the temperature of the calorimeter increases from 25.00°C to 91.60°C. In a separate experiment, the heat capacity of the calorimeter is measured to be 9.84 kJ/K. The heat of reaction for the combustion of a mole of Ti in this calorimeter is ________ kJ/mol.
Answer: The heat of reaction for the combustion of titanium is 15240 kJ/mol
Explanation:
The quantity of heat required to raise the temperature of a substance by one degree Celsius is called the specific heat capacity.
[tex]Q=C\times \Delta T[/tex]
Q = Heat absorbed by calorimeter =?
C = heat capacity of calorimeter = 9.84 kJ/K
Initial temperature of the calorimeter = [tex]T_i[/tex] = [tex]25.00^0C=(25.00+273)=298.00K[/tex]
Final temperature of the calorimeter = [tex]T_f[/tex] = [tex]91.60^0C=(91.60+273)K=364.6K[/tex]
Change in temperature ,[tex]\Delta T=T_f-T_i=(364.6-298.0)K=66.60K[/tex]
Putting in the values, we get:
[tex]Q=9.84kJ/K\times 66.60K=655.3kJ[/tex]
As heat absorbed by calorimeter is equal to heat released by combustion of titanium
[tex]Q=q[/tex]
[tex]\text{Moles of titanium}=\frac{\text{given mass}}{\text{Molar Mass}}=\frac{2.060g}{47.8g/mol}=0.0430mol[/tex]
Heat released by 0.0430 moles of titanium = 655.3 kJ
Heat released by 1 mole of titanium = [tex]\frac{655.3}{0.0430}\times 1=15240kJ[/tex]
The heat of reaction for the combustion of titanium is 15240 kJ/mol
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.
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.
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.
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
An aluminum calorimeter with a mass of 100 g con- tains 250 g of water. The calorimeter and water are in thermal equilibrium at 10.0°C. Two metallic blocks are placed into the water. One is a 50.0-g piece of copper at 80.0°C. The other has a mass of 70.0 g and is originally at a temperature of 100°C. The entire system stabilizes at a final temperature of 20.0°C. (a) Determine the spe- cific heat of the unknown sample.
Answer:1.587J / g degC.
Explanation:SOLUTION
Specific heat of aluminum = 0.897 J / g degC.
Heat gained by calorimeter = 100 * 10 * 0.897 J = 897 J
Specific heat of water = 4.18 J / g degC.
Heat gained by water = 250 * 10 * 4.18 J = 10450 J
Specific heat of copper = 0.385 J / g degC.
Heat loss by copper block = 50 * (80-10) * 0.385 J =1347.5 J
Let specific heat of unknown block be x J / g degC.
Heat loss by unknown block = 70 * (100-10) * x J = 6300x J
Heat gain = Heat loss
897 + 10450 = 1347.5 + 6300x
6300x = 9999.5
x =1.58722= 1.587
Specific heat of unknown substance is 1.587J / g degC.
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
What is the concentration of hydronium ion ( [H3O+]) in a solution with a PH of _1,3?
Answer: [H3O+]= 0.05 M
Explanation:
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snap- parkguy786
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
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.
HELPPP!!! what is the volume (in liters) of a 5.72 gram sample of 02 at STP? (Hint: remember to change grams of oxygen to moles of oxygen first) molar mass of 02 =32g/mol
Answer:
4L
Explanation:
To obtain the volume of O2 at stp, first, we need to determine the number of mole of O2.
From the question given above,
Mass of O2 = 5.72g
Molar Mass of O2 = 32g/mol
Number of mole =Mass/Molar Mass
Number of mole of O2 = 5.72/32
Number of mole of O2 = 0.179 mole
Now, we can calculate the volume of O2 at stp as follow:
1 mole of a gas occupy 22.4L at stp.
Therefore, 0.179 mole of O2 will occupy = 0.179 x 22.4 = 4L
Therefore, the volume occupied by the sample of O2 is 4L
The volume of a 5.72 gram sample of oxygen at STP is approximately 4.01 liters, after converting the mass to moles and using the molar volume of a gas at STP.
To calculate the volume of oxygen at Standard Temperature and Pressure (STP) from the given mass, first convert the mass of oxygen to moles. At STP, 1 mole of gas occupies 22.4 liters. Use the molar mass of Oxygen (32 g/mol) to convert grams to moles and then use the volume of 1 mole of gas at STP to find the volume in liters.
First, convert the given mass of oxygen (5.72 g) to moles:
5.72 g [tex]O_{2}[/tex]
x
1 mol [tex]O_{2}[/tex]/ 32 g [tex]O_{2}[/tex]
= 0.179 moles [tex]O_{2}[/tex]
Then calculate the volume at STP (using the fact that 1 mole of any gas at STP occupies 22.4 liters):
0.179 moles [tex]O_{2}[/tex]
x
22.4 liters/mol
= 4.0096 liters
Therefore, the volume of a 5.72 gram sample of [tex]O_{2}[/tex] at STP is approximately 4.01 liters.
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.
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