Complete combustion of 1 mole of acetone (C3H6O) liberates 1790 kJ of energy as shown in the reaction below: C3HsO (€) + 402 (g) --> 3C02 (g) + 3H20 (€) AHSrxn = -1790 kJlmol Using the information provided (AHSrxn = -1790 kJlmol) together with the data below; Calculate the enthalpy of formation (AH?f) of acetone (C3H6O): AH?f, CO2: -393.5 kJlmol, AH?f, H2O: -285.83 kJlmol -679 kJlmol 2038 kJlmol +1019 kJlmol -248 kJlmol +1790 kJlmol
Answers
When burning one mole of acetone, 1790 kJ is released. The enthalpy of formation of acetone, C₃H₆O is equal to -247.9 kJ/mol.
The combustion reaction of a substance means the burning of a substance in the presence of oxygen to form carbon dioxide and water. For example, the combustion of carbon proceeds as follows: C + O₂ --> CO₂. We have the molar enthalpy of combustion of acetone,
C₃H₆O(l) + 4O₂(g)→ 3CO₂(g) + 3H₂O(l)
Where, ∆H꜀= −1790 kJ. We need to calculate the enthalpy of formation of acetone. Now, the standard enthalpy of formation, *ΔHf(O₂)= 0 kJ/mol, *ΔHf(CO₂)
=−393.4 kJ/mol, *ΔHf(H₂O)= -285.8 kJ/mol. We determine the enthalpy of formation of acetone as follows:
∆H꜀ = 3×ΔHf (CO₂) + 3×ΔHf (H₂O) – 1× ∆Hf(C₃H₆O) – 4× ΔHf (O₂)
=> −1790 kJ = 3 mol × −393.4 kJ/mol + 3 mol × -285.8 kJ/mol - 1 mol × ∆Hf(C₃H₆O) - 4 mol × 0
=> −1790 kJ = - 1180.5 kJ - 857.4 kJ - ∆Hf(C₃H₆O)
=> 1 mol×∆Hf(C₃H₆O)= -1180.5 kJ - 857.4 kJ + 1790 kJ
=> ∆Hf(C₃H₆O) = - 247.9 kJ/mol
Therefore, the required enthalpy value for acetone is -247.9 kJ/mol.
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Related Questions
defintion for concentration
Answers
Answer:
concentration is the abundance of a constituent divided by the total volume of a mixture.
Explanation:
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6. How many moles are in 8.30 x 1023 molecules of CO₂?
a.
b.
C.
d.
1.37
2.8
55.5
100
Answers
Given the potential energy diagram representing a reaction below, which numbered interval represents the heat of reaction?
Answers
Answer:
2
Explanation:
Heat of reaction is usually the difference of heat of products and heat of reactions. The difference here is represented by the letter 2.
The potential energy diagram has been representing the heat of reaction at interval 2.
The heat of reaction has been defined as the difference in the heat of product and reactant. The potential energy has been the stored energy of the compounds.
Potential Energy Diagram:In the given potential energy curve, the heat of reaction has been found to be the increase in potential energy. There has been an increment in the energy at interval 2.
Thus, the potential energy diagram has been representing the heat of reaction at interval 2.
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Help me solve please
Answers
Answer:
..........
Explanation:
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Here is a second order reaction A→ P. If the initial concentration of A 0.0818 M goes down 30.0% in 3.15 minutes, what is the rate constant for the reaction?
Answers
The rate constant of the second-order reaction is 0.111 M^-1 min^-1.
The given data represents a second-order reaction where the rate of the reaction is proportional to the square of the concentration of A.
The integrated form of the second-order reaction is:
1/[A]t = kt + 1/[A]0
where [A]t and [A]0 are the concentrations of reactant A at time t and time zero, respectively, k is the rate constant.
We can use the given information to calculate the rate constant (k) of the reaction for the given half-life (t1/2) of 3.15 minutes:
t1/2 = (1 / k[A]0)
Using the percentage decrease in concentration and the given initial concentration, we can calculate the concentration of A at time t:
[A]t = [A]0 - 0.30[A]0 = 0.57126 M
Substituting the given values, we get:
3.15 min = (1 / k)(0.0818 M) / (0.0818 M - 0.57126 M)
Simplifying the equation above, we can solve for k:
k = 0.111 M^-1 min^-1
Therefore, the rate constant of the second-order reaction is 0.111 M^-1 min^-1.
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Select the correct answer.
Which phrase best describes heat?
O A. the energy that an object has as a result of its temperature
OB. the average translational kinetic energy of the particles in an object
ОС. the energy transferred between objects at different temperatures
OD
the total amount of energy possessed by the particles in an object
Answers
Answer:
C is the answer hope I helped you have a good day by
Why is it dangerous to stand under tall trees and building during thunder?
Answers
A sample of approximately 40.0% iron is dissolved in acid. In order for a good analysis to take place it has to be diluted between .0040 and .006 percent. The method chosen was to dilute in three dilutions of ten mL each. The first dilution is a 1/5, the second dilution is a 1/6 and the third dilution is a 1/4. Determine the amount of acid (the diluent) to add to tube 1 and how much 40.0% solution to add to tube 1. Determine the amount from tube 1 and the amount of acid (the diluent) to put in tube 2. Determine the amount of acid (the diluent) and the amount from tube 2 to put into tube 3. Determine the concentration of each tube. Are we in the proper range?
Answers
The amount of solution to add to:
tube 1 is approximately 1.67 mL, and the amount of diluent (acid) is approximately 8.33 mL.tube 2 is approximately 1.43 mL, and the amount of diluent (acid) is approximately 8.57 mL.tube 3 is 2 mL, and the amount of diluent (acid) is 8 mL.How to determine amount and concentration?To determine the amounts of the 40.0% iron solution and the diluent (acid) to add to each tube, calculate the dilution factors and use them to find the required volumes.
Start with tube 1:
The first dilution is 1/5. This means that 1 part of the 40.0% iron solution will be diluted with 4 parts of diluent (acid).
Let x be the volume (in mL) of the 40.0% iron solution in tube 1.
The volume of diluent (acid) in tube 1 will be 5 times the volume of the iron solution, which is 5x.
The total volume of tube 1 is 10 mL.
Therefore, x + 5x = 10 mL
6x = 10 mL
x = 10 mL / 6
x ≈ 1.67 mL
The amount of the 40.0% iron solution to add to tube 1 is approximately 1.67 mL, and the amount of diluent (acid) is approximately 8.33 mL.
Moving on to tube 2:
The second dilution is 1/6. This means that 1 part of the solution from tube 1 will be diluted with 5 parts of diluent (acid).
Let y be the volume (in mL) of the solution from tube 1 added to tube 2.
The volume of diluent (acid) in tube 2 will be 6 times the volume of the solution from tube 1, which is 6y.
The total volume of tube 2 is 10 mL.
Therefore, y + 6y = 10 mL
7y = 10 mL
y = 10 mL / 7
y ≈ 1.43 mL
The amount of the solution from tube 1 to add to tube 2 is approximately 1.43 mL, and the amount of diluent (acid) is approximately 8.57 mL.
Finally, for tube 3:
The third dilution is 1/4. This means that 1 part of the solution from tube 2 will be diluted with 3 parts of diluent (acid).
Let z be the volume (in mL) of the solution from tube 2 added to tube 3.
The volume of diluent (acid) in tube 3 will be 4 times the volume of the solution from tube 2, which is 4z.
The total volume of tube 3 is 10 mL.
Therefore, z + 4z = 10 mL
5z = 10 mL
z = 10 mL / 5
z = 2 mL
The amount of the solution from tube 2 to add to tube 3 is 2 mL, and the amount of diluent (acid) is 8 mL.
Now, to determine the concentration of each tube:
Tube 1: The volume of the iron solution in tube 1 is 1.67 mL, and the total volume is 10 mL. Therefore, the concentration of tube 1 is (1.67 mL / 10 mL) × 40.0% ≈ 6.67%.
Tube 2: The volume of the solution from tube 1 in tube 2 is 1.43 mL, and the total volume is 10 mL. Therefore, the concentration of tube 2 is (1.43 mL / 10 mL) × 6.67% ≈ 0.96%.
Tube 3: The volume of the solution from tube 2 in tube 3 is 2 mL, and the total volume is 10 mL. Therefore, the concentration of tube 3 is (2 mL / 10 mL) × 0.96% ≈ 0.19%.
Based on the calculated concentrations, it seems that the dilutions are within the desired range of 0.0040 to 0.006%. Tube 1 has a concentration of approximately 6.67%, which is higher than the desired range. However, tubes 2 and 3 have concentrations of approximately 0.96% and 0.19%, respectively, which fall within the desired range.
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. Does a more negative heat of formation (a larger negative number) mean that a compound is more stable or less stable than an isomer with a less negative heat of formation
Answers
Answer:
See explanation
Explanation:
Heat of formation, also called standard heat of formation, enthalpy of formation, or standard enthalpy of formation, the amount of heat absorbed or evolved when one mole of a compound is formed from its constituent elements, each substance being in its normal physical state (gas, liquid, or solid)(Encyclopedia Britannica).
The greater the magnitude of the negative value of the heat of formation(the more negative), the greater stability of the compound formed. Hence, a more negative heat of formation (a larger negative number) means that a compound is more stable than an isomer with a less negative heat of formation.
A more negative heat of formation (a larger negative number) mean a compound is more stable.
What is standard heat of formation?The change in enthalpy during the creation of 1 mole of a material from its constituent elements is known as standard heat of formation of a compound.
The standard heat of formation is the amount of heat absorbed or released when one mole of a compound is formed from its constituent elements.
The larger the negative value of the heat of formation (the more negative), the more stable the compound created will be.
Thus, the more negative heat of formation (a larger negative number) mean a compound is more stable.
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amobarbital sodium react with ethanolic sodium hyrooxide
Answers
Amobarbital (like all barbiturates) works by being incontestible to the GABAA receptor at either the alpha or the beta subunit.
What is the mechanism of amobarbital?Amobarbital (like all barbiturates) works by binding to the GABAA receptor at either the alpha or the beta subunit. These are compulsory sites that are distinct from GABA itself and also distinct from the benzodiazepine binding site.
Amobarbital is a barbiturate classified as having a halfway duration of action, meaning that the effects of the drug can last from 4-6 amobarbital increases the effects of benazepril by apparatus: pharmacodynamic synergism.
So we can conclude that Amobarbital, 5-ethyl-5-isoamyl barbituric acid like all barbiturates.
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PLEASE HELP MEEEE ASAP If a 28.5 g piece of copper is added to a graduated cylinder containing 45.5 mL of water and the water level rose to 49.1 mL, what is the density of the copper?
Answers
Answer:
3.6
Explanation:
subtract both numbers!
Answer:Approx....
8
⋅
g
⋅
m
L
−
1
....
Explanation:
ρ
,
density
=
Mass
Volume
=
28.5
⋅
g
(
49.10
−
45.50
)
⋅
m
L
=
7.92
⋅
g
⋅
m
L
−
1
....
Explanation:
How many atoms are in 80.7 g of pure aluminum?
Answers
Answer:
Like Roman said Avogadro's number is the key to your problem. The value that you'll use on paper is 6.022x10^23. Luckily in our case a mol specifically refers to the amount of atoms. Its a weird concept but 1.97 mol of Aluminum is the exact same amount of atoms as say helium. The only differences lay in the total mass of the sample.
PLZZZZZZZZZZZZZZZ HELP IF RIGHT WILL GIVE BRAINLIEST
Answers
Answer:
I guess it all of these
as life cycle of an organism can be predicted and yeah it begins with birth hatching or germination for mammals, animals and plants respectively. then it's same for the organism of the same species .
Question:-
Which of the following is true of an organism's life cycle?
Answer:-
It is the same for other organisms of the same species.It is predictableIt begins with birth,hatching or germination.All of these \(\checkmark\)9.42 g of lead (ii) nitrate was reacted with 0.655 mol/L potassium iodide to produce 2.64 g of precipitate. What volume of potassium iodide was used in this reaction?
Answers
The volume of potassium iodide used will be 0.087 L or 87 mL
Stoichiometric calculationThe equation of the reaction goes thus: \(Pb(NO_3)_2(aq) + 2KI --- > PbI_2(s) + 2KNO_3(aq)\)
The mole ratio of the 2 reactants is 1:2.
Mole of 9.42 g \(Pb(NO_3)_2\) = 9.42/331.2 = 0.02844 mol
Equivalent mole of KI = 0.02844 x 2 = 0.05688 mol
Volume of 0.05688 mol, 0.655 mol/L KI = 0.05688/0.655 = 0.087 L or 87 mL
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if there is 5 atoms of oxygen in the reactant, how many oxygen atoms must be in the product
Answers
Answer:
5
Explanation:
It requires 20 chracters to fill this answer
Name the structure. Group of answer choices 1-chloro-3-cycloheptene 4-chloro-1-cycloheptene 4-chloro-1-cyclohexene 6-chloro-1-cycloheptene
Answers
The correct name of the compound is 4-chloro-1-cycloheptene according to IUPAC nomenclature.
What is IUPAC nomenclature?The IUPAC nomenclature is a system of naming compounds that was put together by the international union of pure and applied chemistry. The system enables the structure of a compound to be easily written from its name.
For the compound shown in the image attached to this answer, the correct name of the compound is 4-chloro-1-cycloheptene according to IUPAC nomenclature.
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The pOH of a solution is 6.0. Which statement is correct?
Use pOH = -log[OH-] and PH+pOH = 14.
The pH of the solution is 20.0.
O The concentration of OH ions is 1.0 x 108 M.
The concentration of OH ions is 1.0 x 106 M.
O The pH of the solution is 8.0.
A
Answers
At pOH value of 6.0 the pH value of the following solution is 8.0 and the concentration of [\(H^{+}\) ] ion is \(10^{-8}\)
In this question we will apply the formula
pH +pOH = 14 . . . . . . . . . . . . .(1)
where pH = concentration of [\(H^{+}\) ] ion
pOH = concentration of [\(OH^{-}\) ] ion
As per the question
pOH =6.0
Putting the value of pOH in equation (1) we get the value of pH
pH + 6.0 =14
pH = 14 -6.0
pH = 8.0
The value of pH if the pOH value is 6.0 is 8.0
To find the concentration of \(H^{+}\) ion we will use the following formula
This is calculated by the formula
[\(H^{+}\)} = \(10^{-pH}\)
where we will write the values of pH
Hence the concentration of [\(H^{+}\)} ion is \(10^{-8}\)
Therefore at pOH of 6.0 the pH value of the following solution is 8.0 and the concentration of [\(H^{+}\) ] ion is \(10^{-8}\)
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The complete question is -
What is the pH value and concentration of [\(H^{+}\) ] ion of the following if the pOH value of the solution is 6.0 ?
738.90 m has ____ significant figures
Answers
Answer: 4
Explanation: because the zero doesn't count
Which of the following groups of materials would
most likely be used to build an electromagnet?
A bare wire, plastic rod, battery
B. bare wire, iron rod, light bulb
Cinsulated wire, iron rod, battery
D. insulated wire, plastic rod, light bulb
Answers
Answer:
C
Explanation:
There is a structure of CH3CH2CHCHCH2CH3, with a –CH2CH3 group attached to the third and fourth carbons.
Spell out the full name of this compound
Answers
Explanation:
2,3-diethyl hexane
At first we select a long chain.
Then, we number that chain from that side where substituent position is closer.
Then, we write it's IUPAC name
Position of substituent + substituent name + chain name + suffix
Here,
2,3 + -diethyl + hex + -ane
= 2,3-diethyl hexane
1 mole of sulfur atoms has how much mass
Answers
Answer:
One atom of sulfur has a mass of 32.07 AMU; one mole of S atoms has a mass of 32.07 g.
Explanation:
Therefore, the answer should be 32.07 g
What is the experimental mole ratio of baking soda (NaHCO3 ) to sodium chloride (NaCl)?
Answers
Answer: 1:1
Explanation:
Since Na trades place with Na it makes it 1:1
What’s the order can someone help?
Answers
(That’s from lowest to highest don’t get confused) :)
Question 2 of 10
Which chemical equation is balanced?
A. CO₂ + H₂O → H₂CO₂
B. K+ H₂O → K₂O + H₂
C. CaO2 + HCl → CaCl₂ + H₂O
OD. MgO + 2HCl → MgCl₂ + H₂O
Answers
The equation MgO + 2HCl → MgCl₂ + H₂O is a balanced equation.
A balanced chemical equation contains equal number of atoms on both sides of the equation.The equation is MgO + 2HCl → MgCl₂ + H₂O which contains 1 atom of Mg, 2 atoms of chlorine, 2 atoms of chlorine, 1 atom of oxygen on both reactants and products sides.So the equation is balanced.An equation for a chemical reaction is said to be balanced if both the reactants and the products have the same number of atoms and total charge for each component of the reaction. In other words, both sides of the reaction have equal amounts of mass and charge.It follows law of conservation of mass.Mass is neither created nor destroyed rather it can be transferred from one form to another.Learn more about balanced equation at:
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What type of decay is illustrated by the equation below?
A. alpha decay
B. beta decay
C. positron emission
D. electron capture
Answers
B. beta decay
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Answer:
Beta Decay (B)
Which solids are insoluble in water.
Answers
Some types of solids that are insoluble in water are:
Metals. (most of them)Non-Metallic ElementsMetal OxidesSome Non-Metallic ElementsMetal Carbonates (most of them)Metal Sulfides (most of them)Salts (some of them)Which solids are insoluble in water?Many solids are insoluble in water, meaning they do not dissolve in water to a significant extent. Here are some examples of common solids that are generally insoluble in water:
Metals: Most metals, such as gold, silver, platinum, and copper, are insoluble in water.
Non-Metallic Elements: Many non-metallic elements, such as carbon (in the form of graphite or diamond), sulfur, phosphorus, and iodine, are insoluble in water.
Metal Oxides: Some metal oxides, particularly those of less reactive metals, are insoluble in water. Examples include aluminum oxide (Al2O3), iron(III) oxide (Fe2O3), and lead(II) oxide (PbO).
Metal Carbonates: Most metal carbonates are insoluble in water. Examples include calcium carbonate (CaCO3), lead(II) carbonate (PbCO3), and copper(II) carbonate (CuCO3).
Metal Sulfides: Many metal sulfides are insoluble in water. Examples include lead(II) sulfide (PbS), silver sulfide (Ag2S), and mercury(II) sulfide (HgS).
Insoluble Salts: Certain salts have limited solubility in water. Examples include silver chloride (AgCl), lead(II) iodide (PbI2), and calcium sulfate (CaSO4).
It's important to note that while these solids are generally insoluble in water, they may exhibit some solubility to a small extent. The solubility of a solid in water can vary depending on factors such as temperature, pressure, and the presence of other solutes.
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Please help. Thank you so much
Answers
Enthalpy change (H) and entropy change (S) are 11.7 103 Jmol-1 and 105 Jmol-1K-1, respectively, for a reaction at 25 °C.
What is the change in the free basic energy at 25 °C?The absolute entropies of a reactants and their products are S°(N2H4) (= half of this period J/(mol•K), S°(N2) = 191.6 J/(mol•K), or S°(H2) = 130.7 J/(mol•K) at 25°C, where the standard enthalpy change (H°) is 50.6 kJ/mol.
What is the calomel electrode's reduction potential at 25 C?E0 is known as 0.268 V at standard potential at 25°C, despite a slight variation in the computed value above. Similar to a silver-silver chloride electrode, the electrode potential is dependent on the chloride ion concentration.
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I NEED THIS RIGHT NOW!! Daria had some sand from the beach. The mass of the sand was 72 grams. She used the graduated cylinder below to measure the volume.
What is the volume of the sand found in the graduated cylinder? _____ mL
Answers
Daria had some beach sand with her. The sand has a 72 gramme mass. She calculated the volume using the graduated cylinder below. The graduated cylinder contains 15 mL of sand.
The volume of the sand is calculated using the graduated cylinder below. The sand's bulk is specified as 72 grammes.
We can use the water displacement method to calculate the volume of the sand. Following is a description of how to estimate the amount of sand using the water displacement method:
The graduated cylinder of water should first be measured for volume.
The graduated cylinder's water volume should then be measured after adding the sand to it. The volume of water increases by the same amount.
Let's use the provided problem to implement this approach.
In the beginning, there is 10 mL of water in the graduated cylinder. The graduated cylinder contains 25 mL of water once the sand has been added.
The amount of sand is therefore equal to the difference between the two volumes, which is: Sand volume equals final water volume minus initial water volume (25 - 10 = 15 mL).
As a result, there are 15 mL of sand in the graduated cylinder.
Answer : 15
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What is the average atomic mass of 10 hydrogen -1 molecules?
Answers
Answer:
1.674 x 10^-23 grams
Explanation:
Hydrogen-1 is called Protium
wikipedia
atomic mass of Protium is 1.00794 amu
sciencedirectcom
atomic mass of 10 Protiums is 10.0794 amu
10.0794 amu in grams is
1.6737236x10^-23 grams
barium hydroxide is insoluble in water. do you do any reaction that produces this compound in this experiment. If so, write down the equation for this reaction.