Given question is incomplete. The complete question is as follows.
[tex]AB_{2}[/tex] is a molecule that reacts readily with water. Calculate the bond energy of the A–B bond using the standard enthalpy of reaction and the bond energy data provided. Enter a number in kJ to 0 decimal places.
[tex]2AB_{2}(g) + 2H_{2}O(g) \rightarrow O=O(g) + 4HB(g) + A_{2}[/tex] [tex]\Delta H^{o}[/tex] = –142 kJ
Bond: O–H O=O H–B [tex]A \rightarrow A^{+}[/tex]
Bond energy (kJ/mol): 467 498 450 321
Explanation:
The given reaction is as follows.
[tex]2AB_{2} + 2H_{2}O \rightarrow O_{2} + 4HB + A_{2}[/tex]
Now, we will calculate the enthalpy of reaction as follows.
[tex]\Delta H^{o}_{R}[/tex] = -142 kJ
Also, we know that
[tex]\Delta H^{o}_{R} = \Delta H_{reactants} - \Delta H_{products}[/tex]
= [tex][(2 \times 2 (A-B) + 2 \times 2 (O-H)] - [(O=O) + 4(H-B) + (A-A)][/tex]
-142 = [tex]4(A-B) + 4 \times 467 - 498 - 4(450) - 321[/tex]
[tex]4(A-B)[/tex] = -142 - 1868 + 498 + 1800 + 321
= 609
(A-B) = 152.25 kJ/mol
Thus, we can conclude that the bond energy of the A–B bond is 152.25 kJ/mol.
Which of the following properties of matter can be explained on a submicroscopic level?
Question 7 options:
Structure
Composition
Behavior
All the above
The properties of matter including structure, composition, and behavior can all be explained on a submicroscopic level. The atoms or molecules' arrangement, what they are composed of, and their response to different conditions, helps in understanding the properties of the matter.
Explanation:The properties of matter such as structure, composition, and behavior can all be explained on a submicroscopic level. In fact, observing matter at this level allows us to understand why material behaves in the way that it does.
Structure refers to the arrangement of atoms in a material. For example, diamonds and graphite are both made of carbon atoms, but their different structures make them drastically different in hardness.
Composition refers to what a material is made of at the elemental level. For example, water is composed of two hydrogen atoms and one oxygen atom, leading to its unique properties.
Behavior can be understood as the way that material reacts under different conditions, this can be due to the interaction of its molecules. For example, water boils at 100 degrees Celsius because that's the temperature at which its molecules have enough energy to change from a liquid to a gas.
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Thea dissolves 0.33 mole of glucose into 450 mL of water. What is the concentration of the solution?
a
0.73M
b
1.5M
c
1.0M
d
14M
Answer:
0.73M
Explanation:
Select each of the reactions and observe the reactants taking part in the reactions. Suppose you are carrying out each of these reactions starting with five moles of each reactant. Observe that some reactants are in excess whereas some reactants limit the amount of products formed.
Classify each of the reactants as a limiting reactant or an excess reactant for a reaction starting with five moles of each reactant.
A. H2 in formation of water
B. O2 in formation of water
C. CH4 in combustion of methane
D. N2 in formation of ammonia
E. H2 in formation of ammonia
F. O2 in combustion of methane
Answer:
A. In the formation of water, H2 is the limiting reactant and O2 is the excess reactant.
B. In the Combustion of methane (CH4), O2 is the limiting reactant and CH4 is the excess reactant.
C. In the formation of ammonia (NH3), H2 is the limiting reactant and N2 is the excess reactant.
Explanation:
The question suggests that each reactant has 5 moles.
A. Formation of water. Water is produced when H2 and O2 combine together according to the balanced equation below:
2H2 + O2 —> 2H2O
From the balanced equation above,
2 moles of H2 reacted with 1 mole of O2.
Therefore, 5 moles of H2 will react with = 5/2 = 2.5 moles of O2.
From the above calculations, we can see that there are left over for O2 as only 2.5 moles reacted out of the 5 moles that was given.
Therefore, H2 is the limiting reactant and O2 is the excess reactant.
B. Combustion of methane. Combustion is simply a reaction in the presence of oxygen. The balance equation for the Combustion of methane (CH4) is given below:
CH4 + 2O2 —> CO2 + 2H2O
From the balanced equation above,
1 mole of CH4 reacted with 2 moles of O2.
Therefore, 5 moles of CH4 will react with = 5 x 2 = 10 moles of O2.
From the calculations made above, we can see that it requires higher amount of O2 to react with 5 moles CH4. Therefore, O2 is the limiting reactant and CH4 is the excess reactant.
C. Formation of ammonia.
Ammonia is obtained when N2 and H2 combine according to the balanced equation below:
N2 + 3H2 —> 2NH3
From the balanced equation above,
1 mole of N2 reacted with 3 moles of H2.
Therefore, 5 moles of N2 will react with = 5 x 3 = 15 moles of H2.
From the calculations made above, a higher amount of H2 is required to react with 5 moles of N2. Therefore, H2 is the limiting reactant and N2 is the excess reactant.
Final answer:
To determine limiting or excess reactants, compare the stoichiometry of the balanced equations to the amounts provided. For reactions involving the formation of water or ammonia, and the combustion of methane, the limiting reactant is based on the stoichiometric ratio required for each reactant. Hence, O2 typically ends up as the limiting reactant for the formation of water and combustion of methane, while N2 is the limiting reactant in the formation of ammonia.
Explanation:
When determining whether a reactant is a limiting reactant or an excess reactant, you must look at the balanced chemical equations for the reactions. Consider the stoichiometric coefficients, which tell you the proportions in which reactants combine to form products. If starting with five moles of each reactant:
A. H2 in formation of water: The balanced equation is 2H2 + O2To identify the limiting reactant, compare the molar ratios of the reactants to those in the balanced equation. The reactant that is in a smaller proportion relative to its stoichiometric coefficient in the equation is the limiting reactant. The one in the larger proportion is the excess reactant.
pure potassium hydrogen phthalate is used for the standardization of the sodium hydroxide solution. suppose that the potassium hydrogen phthalate is not completely dry. will the reported molar concentration of the sodium hydroxide solution be too high too low or unaffected because of the moistness of the potassium phthalate ? explain
Answer:
the reported molar concentration of the sodium hydroxide solution will be too high.
Explanation:
The term "standardization" in science or in analytical Chemistry simply means the process involved in the determination of a standard or the concentration of a particular substance.
One of the advantages and the most important of the advantages of Standardization is that it Helps in making sure that we get a result with the least error.
So, let me explain the answer. The reported molar concentration of the sodium hydroxide solution will be TOO HIGH because the potassium hydrogen phthalate is NOT completely dry. The weight of the moisture will create an additional weight which will increase the weight of potassium hydrogen phthalate in the solution
If potassium hydrogen phthalate is not completely dry, it will make the reported molar concentration of the sodium hydroxide solution too low. This is due to the added water from the moisture which causes a dilution effect.
Explanation:In the situation where the potassium hydrogen phthalate is not completely dry, the reported molar concentration of the sodium hydroxide solution will be too low. This is because the water from the phthalate's moisture can dilute the sodium hydroxide solution, which leads to a lower than actual molar concentration. It's essential to have a completely dry potassium hydrogen phthalate when standardizing sodium hydroxide solutions to get an accurate molar concentration.
For instance, when potassium hydroxide, a highly soluble ionic compound, is dissolved in a dilute solution, it completely dissociates, giving a certain concentration of [OH-] ions. Suppose this reaction occurs in an environment where excess water is introduced due to the moisture in potassium hydrogen phthalate. In that case, this equilibrium will be affected, leading to a lower concentration of hydroxide ions and thus, a lower reported molar concentration of the sodium hydroxide solution.
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What is the fate of glucose 6‑phosphate, glycolytic intermediates, and pentose phosphate pathway intermediates in this cell? Glycolytic intermediates can only enter the pentose phosphate pathway through conversion to pyruvate and subsequent gluconeogenesis. The oxidative pentose phosphate pathway reaction catalyzed by glucose 6‑phosphate dehydrogenase is slowed down. Most of the glucose 6‑phosphate enters the pentose phosphate pathway. One molecule of glyceraldehyde 3‑phosphate and two molecules of fructose 6‑phosphate are used to generate three molecules of ribose 5‑phosphate. Most of the glucose 6‑phosphate enters the glycolytic pathway and is converted to fructose 6‑phosphate and glyceraldehyde 3‑phosphate. Under the given conditions, all triose phosphates are converted to pyruvate by the glycolytic pathway.
Answer:
The Phosphorylated glucose(glucose +inorganic phosphate), with the energy supplied from ATP hydrolysis formed glucose 6- phosphate, which is later converted to 2 molecules of fructose 6-phosphate- this is phosphorylation.And represented the fate of glucose -6-phosphate.
The fructose 6-phosphate are converted to triose phosphate- which is a 2-molecules of 3C compound. The latter is oxidized by NAD→ NADH+ to form intermediates in the glycolytic pathways .
These intermediates are converted to ribose 5-phosphates in the presence of transketolase and transaldolase enzymes.And they are finally converted to pyruvate in the glycolytic pathway with the production of 2ATPs per molecule of glucose.
Basically the phosphate pathway reaction is very slow due to enzyme catalysis.
How many L of carbon dioxide at 1.00 atm and 298.15 K are released from a car's engine upon consumption of a 60.0 L LIQUID tank gasoline? (Gasoline density: 0.77 kg/L, Molar mass of C₈H₁₈: 114.2 g/mol) C₈H₁₈ (l) + ²⁵/₂ O₂ (g) → 8 CO₂ (g) + 9 H₂O (g)
Answer:
79,000 litersExplanation:
1. Number of moles of gasoline
a) Convert 60.0 liters to grams
density = 0.77kg/literdensity = mass / volumemass = density × volume mass = 0.77kg/liter × 60.0 liter = 46.2 kg46.2kg × 1,000g/kg = 46,200gb) Convert 46,200 grams to moles
molar mass of C₈H₁₈ = 114.2 g/molnumber of moles = mass in grams / molar massnumber of moles = 46,200g / (114.2 gmol) = 404.55 mol2. Number of moles of carbon dioxide, CO₂ produced
a) Balanced chemical equation (given):
C₈H₁₈ (l) + ²⁵/₂ O₂ (g) → 8 CO₂ (g) + 9 H₂O (g)b) mole ratio:
1 mol C₈H₁₈ / 8 mol CO₂ = 404.55 mol C₈H₁₈ / xSolve for x:
x = 404.55mol C₈H₁₈ × 8 mol CO₂ / 1mol C₈H₁₈ = 3,236.4 mol CO₂3. Convert the number of moles of carbon dioxide to volume
Use the ideal gas equation:
pV = nRTV = nRT/pp = 1 atmT = 298.15Kn = 3,236.4 molR = 0.08206 (mol . liter)/ (K . mol)Substitute and compute:
V =3,236.4 mol × 0.08206 (mol . liter) / (K . mol) 298.15K / 1 atmV = 79,183 literRound to two significant figures (because the density has two significant figures): 79,000 liters ← answer
The liters L of carbon dioxide at 1.00 atm and 298.15 K released from the car's engine upon consumption of a 60.0 L LIQUID tank gasoline is 79230.76 L
From the given information;
The equation for the reaction can be expressed as:
[tex]\mathbf{C_8H_8 _{(l)} + \frac{25}{2}O_2_{(g)} \to 8 CO_2_{(g)} + 9H_2O_{(g)}}[/tex]
Given that:
Pressure = 1.00 atmTemperature = 298.15 Kamount of gasoline used = 60.0 Lgasoline density = 0.77 kg/LMolar mass of C₈H₁₈: 114.2 g/molThe first thing to do is to determine the mass amount of gasoline used by using the relation:
[tex]\mathbf{Density = \dfrac{mass}{volume}}[/tex]
[tex]\mathbf{0.77 \ kg/L = \dfrac{mass}{60 \ L}}}[/tex]
mass amount of gasoline used = 0.77 kg/L × 60 L
mass amount of gasoline used = 46.2 kg
mass amount of gasoline used = 46.2 × 1000g
mass amount of gasoline used = 46200 g
Now, since we know the mass amount of the gasoline used, we can determine the number of moles of the gasoline by using the formula:
[tex]\mathbf{number of moles = \dfrac{mass}{molar \ mass}}[/tex]
[tex]\mathbf{number of moles = \dfrac{46200 \ g}{114.2 \ g/mol}}[/tex]
[tex]\mathbf{number \ of \ moles = 404.6 \ moles }[/tex]
From the given reaction, 1 mole of gasoline react to produce 8 moles of CO₂
As such, 404.6 moles of gasoline will produce = (404.6 moles × 8)/ 1
= 3236.8 moles of CO₂ is produced.
Now, using the ideal gas equation to determine the volume of CO₂ that are released;
PV = nRT
[tex]\mathbf{1.00 atm \times V = 3236.8 moles \times 0.0821 atm L /K/mol \times 298.15 \ K}[/tex]
[tex]\mathbf{V = \dfrac{3236.8 moles \times 0.0821 atm L /K/mol \times 298.15 \ K}{1.00 atm}}[/tex]
V = 79230.76 L
Therefore, we can conclude that the liters L of carbon dioxide at 1.00 atm and 298.15 K released from the car's engine upon consumption of a 60.0 L LIQUID tank gasoline is 79230.76 L
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A rapid reaction is distinguished by A rapid reaction is distinguished by having a small value of activation energy. having a large value of activation energy. being unaffected by catalysts. having a large heat of reaction. having a small heat of reaction.
Answer:
A rapid reaction is distinguished by having a relatively small value of activation energy.
Explanation:
Chemical kinetics is involved in determining the rate of a reaction, how fast or slow a reaction will occur in a particular condition. The factors affecting the rate of reaction determining whether it will be a rapid reaction includes nature of the reactants, temperature, pressure, surface area of solid state, catalysts, concentration and so on. Based on temperature, temperature affects the collision frequency of a reaction and this contributes to a portion of the increased rate of reaction. At a given temperature, the rate of a reaction depends on the magnitude of the activation energy, pre-exponential factor A, molar gas constant, R, and temperature. This is true based on the Arrhenius equation K = Ae^-Ea/(RT). So therefore, from the equation, it is revealed that at small activation energies, reaction rate is rapid and slow at high activation energies.
Final answer:
A rapid reaction is characterized by a low activation energy, which allows for a faster reaction rate. Catalysts can lower this energy barrier, further speeding up the reaction without altering the overall energy change of the reaction.
Explanation:
A rapid reaction is distinguished by having a small value of activation energy. Activation energy (Ea) is the barrier that must be overcome for reactants to transform into products. A low activation energy indicates that the reactants can more easily reach the transition state and react to form products, leading to a faster reaction rate.
Catalysts are substances that lower the activation energy and provide a new pathway for the reaction to occur, thus speeding up the reaction without being consumed in the process. They do not affect the overall energy change of the reaction (∆H), but they make it easier for the reaction to occur.
This is especially valuable in biological systems where catalysis allows for important cellular reactions to occur at appreciable rates without the need for high temperatures that could harm the cell.
It is the heat of the reaction (∆H) that determines whether a reaction is exothermic or endothermic, not the activation energy.
Therefore, whether a reaction has a large heat of reaction or a small heat of reaction does not directly relate to its speed.
For the process O2(g)→2O(g), ΔH° = 498 kJ·mol–1. What would be the predicted sign of ΔS°rxn and the conditions under which this reaction would be spontaneous? Question 13 options: 1) ΔS°rxn is positive, and the process is spontaneous at low temperatures only. 2) ΔS°rxn is positive, and the process is spontaneous at high temperatures only. 3) ΔS°rxn is negative, and the process is spontaneous at high temperatures only. 4) ΔS°rxn is negative, and the process is spontaneous at low temperatures only.
Answer:
2) ΔS°rxn is positive, and the process is spontaneous at high temperatures only.
Explanation:
ΔS = ΔH / T , ΔS is change in entropy , ΔH is change in enthalpy
Since ΔH is positive , ΔS is positive .
ΔG = ΔH - TΔS
For spontaneous reaction . ΔG should be negative .
As ΔS is positive , at high temperature the value of TΔS will be more and hence the value of TΔS will be higher than Δ H . Hence ΔG will be negative.
Hence at higher temperature , the reaction will be spontaneous.
The predicted sign of ΔS°rxn and the conditions under which this reaction would be spontaneous should be option 2) ΔS°rxn is positive, and the process is spontaneous at high temperatures only.
Predicted sign of ΔS°rxn and its conditons:We know that
ΔS = ΔH / T
Here ΔS means a change in entropy
And, ΔH means a change in enthalpy
Since ΔH is positive so ΔS is positive.
Now
ΔG = ΔH - TΔS
However, For spontaneous reaction. ΔG should be negative .
Since ΔS is positive, at high temperature the value of TΔS should be more and due to this the value of TΔS will be more than Δ H . Thus, ΔG will be negative.
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The molar mass of an unknown organic liquid (M ~ 100) is determined by placing 5 mL of the liquid in a weighed 125-mL conical flask fitted with a piece of Al foil with a pin hole in it. The flask is heated in a boiling water bath until the liquid evaporates to expel the air and fill the flask with the unknown vapor at atmospheric pressure. After cooling to room temperature the flask and its contents are reweighed.
The uncertainty in which piece of apparatus causes the largest percentage error in the molar mass?
A) balance (+/-0.01 g)
B) barometer (+/-0.2 mm Hg)
C) flask (+/-1.0 mL)
D) thermometer (+/-0.2 ˚C)
Write the balanced chemical equations for the reactions in which sodium phosphate is added to copper(II) sulfate and iron(III) chloride, respectively. Include states of matter.
Explanation:
1. When sodium phosphate is added to copper (II) sulfate it gives solid precipitate of copper(II) phosphate and solution of sodium sulfate.
The balanced equation is given as:
[tex]2Na_3PO_4(aq)+3CuSO_4(aq)\rightarrow Cu_3(PO_4)_2(s)+3Na_2SO_4(aq)[/tex]
2. When sodium phosphate is added to iron(III) chloride it gives solid precipitate of iron (III) phosphate and solution of sodium chloride.
The balanced equation is given as:
[tex]Na_3PO_4(aq)+Fe(Cl)_3(aq)\rightarrow FePO_4(s)+3NaCl(aq)[/tex]
2) A professor wanted to set up a similar experiment as the one you performed in lab. He wanted to use Al(OH)3 in place of Ca(OH)2. Calculate how many mL of saturated Al(OH)3 solution it would take to titrate against 12.00 mL of 0.0542 M HCl solution. The Ksp of Al(OH)3 is 3.0x10-34. Show your work to receive credit. Finally, do you think this would be reasonable experiment for a general chemistry lab
Explanation:
The only flaw I can find is you squared 3 instead of cubing it and it will be 27X^4 instead of 9x^4.
This reduces the amount slightly, but the number is still incredibly high (about 10 ^ 5 L is what I've calculated). Your professor might want to point out that this will not be a effective experiment due to the large volume of saturated
The Ksp value of Ca(OH)2 on the site (I used 5.5E-6 [a far more soluble compound than Al(OH)3]) and estimated how much of it will be needed. My calculation was approximately 30 ml. If you were using that much in the experiment, it implies so our estimates for Al(OH)3 are right, that the high amount is unreasonably big and that Al(OH)3 will not be a suitable replacement unless the procedure was modified slightly.
What are the two main differences between the 2 ecosystem in terms of organism population
Answer:
The answer to the question above is explained below
Explanation:
There are so many differences between the two ecosystem (terrestrial ecosystems and aquatic ecosystems). The most important ones will be highlighted:
1. Aquatic environments are so rich in nutrients they support more live than equivalent terrestrial ecosystems as aquatic ecosystems have availability of water more than the terrestrial ecosystems. Presenting the consequent importance of water as a limiting factor in the terrestrial ecosystems.
2. Aquatic ecosystems are much more stable than terrestrial ecosystems, with smaller fluctuations in temperature and other variables.
3. In terrestrial ecosystems, there is hardly ever a shortage of light, while it can be a limiting factor in some aquatic ecosystems.
4. Gravity has so much influence on terrestrial animals. That is not the case for their aquatic counterparts where water supports aquatic organisms.
An ecosystem is a large community of living organisms (plants, animals and microbes) in a particular area in conjunction with the non-living components of their environment, interacting as a system.
Terrestrial Ecosystem includes: Grasslands, Forests, Desert.
Aquatic Ecosystem includes: Freshwater ecosystem, Marine Ecosystem.
At the equivalence point for a weak acid-strong base titration an equal number of moles of OH- and H have reacted, producing a solution of water and salt. What affects the pH at the equivalence point for a weak-acid/strong-base titration
Answer:
The pH shifts less with small additions of titrant near the equivalence point.
Explanation:
Explanation:
In an acid-base titration, the titration curve reflects the strengths of the corresponding acid and base.
If one reagent is a weak acid or base and the other is a strong acid or base, the titration curve is irregular, and the pH shifts less with small additions of titrant near the equivalence point.
Acid-base titrations depend on the neutralization between an acid and a base when mixed in solution.
The endpoint and the equivalence point are not exactly the same: the equivalence point is determined by the stoichiometry of the reaction, while the endpoint is just the color change from the indicator.
Provide the IUPAC name for the following compound: a. 2,4,5-Trimethyl-4-propylheptane b. 4-Isobutyl-4,5-dimethylheptane c. 4-sec-Butyl-2,4-dimethylheptane d. 3,4,6-Trimethyl-4-propylheptane e. 4-Isobutyl-3,4-dimethylheptane
Final answer:
The IUPAC name for the compound is 4-sec-Butyl-2,4-dimethylheptane, which is determined by identifying the longest continuous carbon chain and naming the substituents.
Explanation:
The IUPAC name for the compound is 4-sec-Butyl-2,4-dimethylheptane.
The name is determined by identifying the longest continuous carbon chain, which in this case is heptane. The substituents are then named using prefixes such as sec-, iso-, or tert-, which indicate the position of the substituent on the main chain. The substituents are listed alphabetically, followed by the position number and the main chain name.
In this case, the compound has a butyl group attached at the 4th position on the main chain, and two methyl groups attached at the 2nd and 4th positions. Therefore, the IUPAC name is 4-sec-Butyl-2,4-dimethylheptane.
Mass is defined as anything that can be observed.
Question 4 options:
True
False
In the reaction, 2 Al (s) + 6HCL (aq) --> 2ALCL3 (aq) + 3 H2 (g), 2.00 g of Al will react with how many milliliters of 0.500 M HCl?
Answer:
The volume of HCl required is [tex]V = 420 mL[/tex]
Explanation:
From the question we are told that
The chemical equation for this reaction is
[tex]2 Al _{(s)} + 6 HCl _{(aq)} -----> 2 Al Cl_{3} _{(aq)} + 3 H_2 _{(g)}[/tex]
The mass of Al is [tex]m__{Al}} = 2.00g[/tex]
The concentration of HCl is [tex]C__{HCl}} = 0.500M[/tex]
The number of moles of [tex]Al[/tex] given is [tex]n__{Al}} = \frac{mass \ of Al}{Molar \ mass \ of \ Al}[/tex]
The molar mass of Al is [tex]M = 27 g/mol[/tex]
So
[tex]n__{Al} = \frac{2}{27}[/tex]
[tex]n__{Al} = \ 0.07 \ moles[/tex]
From the balanced equation
2 moles of Al reacts with 6 moles of HCl
So 0.07 moles of Al will react with x
Therefore
[tex]x = \frac{0.07 *6}{2}[/tex]
[tex]x = 0.21 \ moles[/tex]
Now the volume of HCl can be obtained as
[tex]Volume(V) = \frac{moles}{concentration }[/tex]
So [tex]V = \frac{0.21}{0.500}[/tex]
[tex]V = 420 mL[/tex]
the internal energy of an ideal gas depends only on its temperature. Do a first-law analysis of this process. A sample of an ideal gas is allowed to expand at constant temperature against atmospheric pressure.
Answer:
Check Explanation
Explanation:
The Mathematical expression of the first law is given as
ΔU = Q - W
Note that depending on convention, the mathematical expression for the first law can be written as
ΔU = Q + W
a) Yes, the gas does work on its surroundings. Since it expands; indicating that there is a change in volume.
The work done is given as -PΔV if we are using the convention of (ΔU = Q + W). And this work is negative work done, since the system expands and does work on the surroundings.
But if we are using the convention of (ΔU = Q - W), the work done is given as PΔV and the workdone by the system on the environment is positive work during expansion.
b) Since this all takes place at constant temperature and against a constant atmospheric pressure, the change in internal energy for this system is 0. Change in internal energy depends on a change in temperature.
So, from the mathematical expression,
ΔU = Q + W or ΔU = Q - W
If ΔU = 0,
Q = - W or Q = W
But either ways,
Q = PΔV
(because, W = -PΔV for ΔU=Q+W and W = PΔV for ΔU=Q-W)
So, either ways, the heat transfer is the same and it is positive. This indicates heat is transferred from the surroundings to the system.
(c) What is ΔU for the gas for this process?
Since this all takes place at constant temperature and against a constant atmospheric pressure, the change in internal energy for this system is 0. Change in internal energy depends on a change in temperature.
Hope this Helps!!!
Based on the given question, we can state that Yes, the gas does work on its surroundings because it expands, this shows that there is a change in volume.
Based on the second question, we can state that because this takes place at constant temperature and against a constant atmospheric pressure, then the change in internal energy for this system is zero.
Based on the third question, we can see that the ΔU for the gas for this process is zero because of the constant temperature and atmospheric pressure.
The Mathematical expression of the first lawThis is given as
ΔU = Q - W
Hence, from the mathematical expression,
ΔU = Q + W or ΔU = Q - WIf ΔU = 0,Q = - W or Q = WRegardless,
Q = PΔV
This means that the heat transfer remains the same and is positive
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Feldspars producing clay is an example of ___
Answer: A
Explanation:
Hydrolysis reactions break down not only feldspars but many other silicate minerals as well, amphiboles, pyroxenes, micas, and olivines.
The fuel used in many disposable lighters is liquid butane, C4H10
C
4
H
10
. Butane has a molecular weight of 58.1 grams in one mole. How many carbon atoms are in 3.50 g
g
of butane?
Answer:
1.45 *10^23 atoms C
Explanation:
3.50 g butane * 1 mol butane/58.1 g butane =0.06024 mol butane
in 1 mol C4H10 -------- 4 mol C
in 0.06024 mol C4H10 -------- 4*0.6024 = 0.241 mol C
0.241 mol C * 6.02*10^23 atoms C/1 mol C = 1.45 *10^23 atoms C
How many hydrogen atoms are present in a 2.65 gram sample of hydrazine (N2H4)?
Answer:
4.95×10^22 atoms of hydrogen.
Explanation:
Now, we must remember that the number of elementary entities in one mole of a substance (atoms, molecules, ions,etc) is given by the Avogadro's number. Avogadro's number is numerically equal to 6.02×10^23.
We must first find the number of moles of hydrazine corresponding to 2.65 g
We find that from;
Number of moles= mass of hydrazine given/ molar mass of hydrazine
Molar mass of hydrazine= 32.0452 g/mol
Number of moles of hydrazine= 2.65/32.0452 g/mol= 0.0823 moles
If 1 mole of hydrazine contains 6.02×10^23 hydrogen atoms
0.0823 moles of hydrazine will contain 0.0823 moles × 6.02×10^23 = 4.95×10^22 atoms of hydrogen.
Which element is an actinoid?
Answer:Neptunium; Protactinium; Thorium; Uranium
Explanation:
Write the balanced chemical reaction for the following nitrogen cycle reactions. These reactions are mediated by bacteria and plants in nature and in wastewater treatment. (a) nitrification of ammonium to nitrite using molecular oxygen. (b) nitrification of nitrite to nitrate using molecular oxygen. (c) denitrification of nitrate to N2 using succinic acid as the carbon
Answer:
Nitrification is a biological process by which ammonia is oxidized to nitrites ([tex]NO_{2} ^{-}[/tex]) and then to nitrates ([tex]NO_{3} ^{-}[/tex]) by action of specialized bacteria in nature (such as nitrosomonas and nitrobacter).
Explanation:
This two-step process utilizes molecular atmospheric oxygen and ammonia and ammonium components of soil and nature.
(a.) Ammonia and ammonium compounds to nitrite
[tex]NH_{4} ^{+} + O_{2}[/tex] → [tex]NO_{2} ^{-} + 4H^{+}[/tex]
[tex]2NH_{3} ^{+} + 3O_{2}[/tex] → [tex]2NO_{2} ^{-} + 2H^{+} + 2H_{2} O[/tex]
(b.) Nitrite to nitrate
[tex]2NO_{2} ^{-} + O_{2}[/tex] → [tex]2NO_{3} ^{-} + energy[/tex]
(c.) The reverse process involves reduction of nitrates to molecular nitrogen, called denitrification (with succinic acid in the equation below).
[tex]4NO_{3} ^{-} + 10H^{+} + (CH_{2})_{2}(COOH)_{2}[/tex] → [tex]2N_{2} +4CO_{2} + 8 H_{2} O[/tex]
([tex]NO[/tex] and [tex]N_{2}O[/tex] is an intermediate product in this process.)
Draw the structure of the organic product(s) of the Grignard reaction between dimethyl oxalate and excess methylmagnesium bromide, followed by aqueous workup. You do not have to consider stereochemistry. If a compound is formed more than once, add another sketcher and draw it again. Alternatively, you may use the square brackets tool to add stoichiometries greater than one. Draw one structure per sketcher. Add additional sketchers using the drop-down menu in the bottom right corner. Separate multiple products using the + sign from the drop-down menu.
Answer:
Check the explanation
Explanation:
The Grignard Reaction refers to the organomagnesium halide (also known as the Grignard reagent) addition to a ketone or aldehyde, just to create a secondary or tertiary alcohol, respectively. When there’s a reaction with the formaldehyde, it leads to a primary alcohol. It is a multipurpose material that can be utilized in creating new carbon–carbon bonds.
Kindly check the image below to see the structure of the organic product(s) of the Grignard reaction between dimethyl oxalate and excess methylmagnesium bromide.
The product of the reaction is a diol obtained by nucleophilic addition of methylmagnesium bromide to each ester group in dimethyl oxalate, followed by aqueous acid workup.
Explanation:The organic product of the Grignard reaction between dimethyl oxalate and excess methylmagnesium bromide, followed by aqueous workup, is the diol resulting from the nucleophilic addition of the Grignard reagent to the carbonyl group of the ester. Each ester group in the dimethyl oxalate is replaced by a methyl group from methylmagnesium bromide. After acidic workup, the final product is a diol with two primary alcohol groups. The mechanism involves the formation of a magnesium alkoxide intermediate, which upon acid treatment, delivers the alcohol. To start, the Grignard reagent methylmagnesium bromide acts as a strong nucleophile, attacking the electrophilic carbonyl carbon of the first ester group, then similar addition occurs to the second ester group after another equivalent of the Grignard reagent. Finally, hydrolysis during the aqueous workup completes the reaction.
Gaseous methane will react with gaseous oxygen to produce gaseous carbon dioxide and gaseous water . Suppose 9.3 g of methane is mixed with 52.3 g of oxygen. Calculate the maximum mass of carbon dioxide that could be produced by the chemical reaction. Round your answer to significant digits.
Answer:
26 g
Explanation:
Write the balanced reaction first
CH4 + 2 O2 --> CO2 + 2 H2O
9.3g + 52.3g --> ? CO2
You must determine how much carbon dioxide can be made from each of the reactants. The maximum mass that can be made is the lower of the two.
From CH4:
9.3g CH4 (1molCH4/16.05gCH4) (1molCO2 / 1molCH4) (44.01g CH4 / 1molCO2) = 26 g
From O2:
52.3g O2 (1molO2/32gO2) (1molCO2/2molO2)(44.01g/1molCO2) = 36 g
Calcium carbonate can be broken
down by heating what . What is the scientific
term for this? (Hint: your answer
should be two words.)
Answer:
Thermal decomposition
Explanation:
Calcium carbonate can be broken down by heating. The following equation represents this chemical change.
CaCO₃(s) ⇒ CaO(s) + CO₂(g)
The scientific term for this process is thermal decomposition.
Thermal refers to the fact that heat is required.Decomposition refers to the kind of reaction in which a big substance breaks down into smaller substances.The breakdown of calcium carbonate by heating is called thermal decomposition. This process involves a single substance breaking down into simpler compounds upon heating. The process is demonstrated in the reaction - CaCO3 (s) -> CaO(s) + CO2 (g), where calcium carbonate decomposes into calcium oxide and carbon dioxide.
The process of breaking down Calcium Carbonate (CaCO3) by heating is scientifically known as Thermal Decomposition. This is a chemical process where a single substance breaks down into two or more simpler substances when heated. To demonstrate this, let's use the following reaction:
CaCO3(s) → CaO(s) + CO2(g)
In this instance, calcium carbonate (which can come from limestone or chalk) decomposes into calcium oxide and carbon dioxide when heated. It's important to note that the reaction is reversible and to ensure a 100% yield of CaO, the CO2 should be allowed to escape.
Learn more about Thermal Decomposition here:
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Which two factors are most likely to cause a plants guard cell to open its stomata
Explanation:
Water Content of Epidermal Cells
Temperature: Increase in the temperature causes stomata to open
Answer:
Temperature: Increase in the temperature causes stomata to open
Explanation:
Propose a mechanism for the acid-catalyzed hydrolysis of cyclohexanone dimethyl acetal. (Remember that this reaction is reversible, so each step in the hydrolysis is the same as the acetal synthesis, only in reverse.)
Answer:
The proposed mechanism is shown on the first uploaded image
Explanation:
The noble gases were, until relatively recently, thought to be entirely unreactive. Experiments in the early 1960s showed that Xe could, in fact, form compounds with fluorine. The formation of compounds containing Xe is made possible by ________.
Final answer:
The formation of xenon compounds such as xenon difluoride is made possible due to the displacement of xenon's outer electrons, which can occur under certain conditions like high pressure and temperature.
Explanation:
The formation of compounds containing Xe is made possible by the fact that the outer electrons of the larger noble gases like Xe (xenon) are far enough away from the nucleus that they can be displaced under certain conditions. The noble gases were long thought to be entirely unreactive due to their filled outer electron shells, which provide a stable electronic configuration. Despite this, experiments in the early 1960s confirmed that noble gas compounds can indeed be synthesized, with xenon forming stable compounds with fluorine. For instance, xenon difluoride (XeF2), xenon tetrafluoride (XeF4), and xenon hexafluoride (XeF6) are all compounds that form when xenon reacts with varying amounts of fluorine, producing stable crystals that are inert at room temperature. These reactions typically require the noble gas to be exposed to high pressure and temperature conditions.
can decomposition determine the property of a substance
Answer:The general properties of matter such as color, density, hardness, are examples of physical properties. Properties that describe how a substance changes into a completely different substance are called chemical properties. Flammability and corrosion/oxidation resistance are examples of chemical properties.
Explanation:
Suppose you are titrating vinegar, which is an acetic acid solution of unknown strength, with a sodium hydroxide solution according to the equation
H
C
2
H
3
O
2
+
N
a
O
H
⟶
H
2
O
+
N
a
C
2
H
3
O
2
If you require 32.22 mL of 0.1943 M
N
a
O
H
solution to titrate 10.0 mL of
H
C
2
H
3
O
2
solution, what is the concentration of acetic acid in the vinegar?
Answer:
0.0626 M
Explanation:
Equation of the reaction
CH3COOH(aq) + NaOH(aq) ---------> CH3COONa(aq) + H2O
Concentration of acid CA= ???
Concentration of base CB= 0.1943 M
Volume of acid VA= 10.0ml
Volume of base VB= 32.22 ml
Number of moles of acid NA= 1
Number of moles of base NB= 1
From
CA= CB VB NA/VA NB
Hence ;
CA= 0.1943 M × 32.22 ml × 1/10.0ml ×1
CA= 0.0626 M