Answer:
The correct answer would be stoichiomtery!
Answer:
This is known as stoichiometry. Stoichiometry, by definition, is the calculation of the quantities of reactants or products in a chemical reaction using the relationships found in the balanced chemical equation.
A 32.5 g piece of aluminum (which has a specific heat capacity of 0.921 J/g°C) is heated to 82.4°C and dropped into a calorimeter containing water initially at 22.3°C. The final temperature of the water is 24.2°C. Ignoring significant figures, calculate the mass of water in the calorimeter.
Answer:
The mass of water = 219.1 grams
Explanation:
Step 1: Data given
Mass of aluminium = 32.5 grams
specific heat capacity aluminium = 0.921 J/g°C
Temperature = 82.4 °C
Temperature of water = 22.3 °C
The final temperature = 24.2 °C
Step 2: Calculate the mass of water
Heat lost = heat gained
Qlost = -Qgained
Qaluminium = -Qwater
Q = m*c*ΔT
m(aluminium)*c(aluminium)*ΔT(aluminium) = -m(water)*c(water)*ΔT(water)
⇒with m(aluminium) = the mass of aluminium = 32.5 grams
⇒with c(aluminium) = the specific heat of aluminium = 0.921 J/g°C
⇒with ΔT(aluminium) = the change of temperature of aluminium = 24.2 °C - 82.4 °C = -58.2 °C
⇒with m(water) = the mass of water = TO BE DETERMINED
⇒with c(water) = 4.184 J/g°C
⇒with ΔT(water) = the change of temperature of water = 24.2 °C - 22.3 °C = 1.9 °C
32.5 * 0.921 * -58.2 = -m * 4.184 * 1.9
-1742.1 = -7.95m
m = 219.1 grams
The mass of water = 219.1 grams
To calculate the mass of water in the calorimeter, you equalize the heat gained by the water to the heat lost by the heated aluminum (since energy is conserved). You then rearrange the resulting equation to solve for water mass, substituting the known values into the equation.
Explanation:This question is related to the concept of heat transfer in Physics. Here, we are dealing with a piece of aluminium that was heated and then dropped into a calorimeter with water. We want to find the mass of the water contained in the calorimeter.
The heat lost by aluminum will be the heat gained by the water, so we should equalize the heat gained and lost: (mass of Aluminum)*aluminum heat capacity*(Initial temperature-Final temperature) = (water mass)*water heat capacity*(final temperature-initial temperature).
To find the mass of the water, rearrange the equation to solve for it: Mass_water = (mass of Aluminum * aluminum heat capacity * (Initial temperature - Final temperature)) / (water heat capacity * (final temperature - initial temperature)). Plugging in the known values and working through the math should yield the mass of the water in the calorimeter.
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Consider benzene in the gas phase, C6H6(g)Use the heat of formation, ΔH∘f , values below to answer the questions.
Substance H(g) C(g) C6H6(g) ΔH∘f (kJ/mol ) 217.94 718.4 82.9
Part A
What is the standard enthalpy change for the reaction that represents breaking all the bonds in gaseous benzene, C6H6(g) ?
Express your answer to one decimal place and include the appropriate units.
SubmitHintsMy AnswersGive UpReview Part
Correct
Part B
What is the chemical equation for the reaction that corresponds to breaking just the carbon-carbon bonds in gaseous benzene, C6H6(g) ? Indicate the phase of each species in your answer.
Express your answer as a chemical equation including phases.
Answer: 5535.1KJ/mol
Explanation:The equation for breaking of all the bonds in the benzene is written as follows:
C_6 H_6(g) --> 6C(g) +6H (g)
∆H_rxn= ∑moles of product X∆H_PRODUCT -∑moles of reactants x ∆H_REACTANT
={6mols X ∆H_°f(C) + 6mols X ∆H_°f(H)} –{1 mol ∆H_°fC_6 H_6 (g)
={6 X 217.74+ 6X 718.4} – {82.9}KJ/mol
= 5618.04 – 82.9 KK/mol
=5535.1KJ/mol
b) The equation for the breaking of the carbon-carbon bonds in benzene is illustrated below as
C_6 H_6(g) --> 6C---H (g)
A) The standard enthalpy change for the reaction is ; 5535.1 KJ/mol
B) The chemical equation for the reaction that corresponds to breaking just the carbon-carbon bonds is : C₆H₆(g) --> 6C---H(g)
A) The standard enthalpy change for the reaction which represents breaking of all bonds
Reaction equation ; C₆H₆(g) --> 6C(g) + 6H (g)
∆H = ∑ ( moles of product * ∆Hpro ) - ( moles of reactant * ∆H reactant )
= ( 6 * 217.94 + 6 * 718.4 ) – ( 82.9) KJ/mol
= ( 5618.04 ) - ( 82.9 ) = 5535.1 KJ/mol
B) The chemical equation for the reaction that corresponds to breaking the carbon-carbon bonds
chemical equation = C₆H₆(g) --> 6C---H(g)
Hence we can conclude that The answers to your questions are 5535.1 KJ/mol and C₆H₆(g) --> 6C---H(g)
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How many molecules are there in 0.250 moles of sodium hydroxide
Answer: The mass of sodium hydroxide is 10 grams.
Explanation: the atomic weight: 39.997 g/mol
Complete the reaction, which is part of the electron transport chain. The abbreviation FMN represents flavin mononucleotide. Use the appropriate abbreviation for the product. NADH + H + + FMN ⟶ NAD +FMNH_{2} NAD + FMNH 2 The reactant that is reduced is FMN . In complex III, electrons are transferred from coenzyme Q to cytochrome c, which contains iron. QH 2 + 2 cyt c ( Fe 3 + ) ⟶ Q + 2 cyt c ( Fe x ) + 2 H + Determine the oxidation number for iron on the right side of the reaction arrow.
Answer:
Oxidation number of Fe(iron) on right side of reaction arrow = (+2)
Explanation:
NADH+H+ + FMN FMNH2 + NAD+
FMNH2 - Reduced FMN (1,5-Dihydroriboflavin 5'-(dihydrogen phosphate)
The reactant that is reduced is Flavin mononucleotide(FMN)
QH2 + 2cyt c(fe³+) -----------------------> Q + 2Cytc ( fe²+) + 2H+
Oxidation number of Fe(iron) on right side of reaction arrow = (+2)
The equation of the transfer of two electrons from NADH to FMN is given below:
NADH + H⁺ + FMN ⟶ NAD⁺ + FMNH₂
The oxidation number of iron on the right side of the reaction arrow is +2 (Fe²⁺)
In the electron transport chain, electrons are passed from electron carriers such as NADH through various carriers and eventually to oxygen. Water and energy in the form of ATP is produced.
The electron carriers are organized into complexes; Complex I, II, III, and IV
In complex I, also known as NADH Dehydrogenase, electrons are passed from NADH to ubiquinone through an FMN-containing flavoprotein and several iron-sulfur centers.
The equation below shows how two electrons are passed from NADH to FMN:
NADH + H⁺ + FMN ⟶ NAD⁺ + FMNH₂
In complex III, electrons are transferred from coenzyme Q to cytochrome c, a single electron-carrier which contains iron. The equation for the electron transfer is given below:
QH₂ + 2 cyt c ( Fe³⁺) ⟶ Q + 2 cyt c ( Fe²⁺ ) + 2H⁺
In the reaction above, the two electron-carrier ubiquinone transfers its two electrons to two molecules of the one electron-carrier cytochrome c containing iron in the oxidized iron (iii) state, Fe³⁺. The electrons accepted reduces the Fe³⁺ in cytochrome c to Fe²⁺.
Therefore, the oxidation number of iron on the right side of the reaction arrow (reduced cytochrome c) is +2.
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The substance fluorine has the following properties: normal melting point: 53.5 K normal boiling point: 85.0 K triple point: 1.6×10-4 atm, 53.4 K critical point: 55 atm, 144.1 K A sample of fluorine at a pressure of 1.00 atm and a temperature of 90.3 K is cooled at constant pressure to a temperature of 49.3 K. Which of the following are true? Choose all that apply The sample is initially a gas. The liquid initially present will vaporize. The final state of the substance is a solid. One or more phase changes will occur. The final state of the substance is a liquid. Submit Answer
Answer:
The true statements include;
- The sample is initially a gas.
- The final state of the substance is a solid.
- One or more phase changes will occur.
The untrue/false statements include;
- The liquid initially present will vaporize.
- The final state of the substance is a liquid.
Explanation:
A couple pieces of informatton on Fluorine is imitially provided.
The substance fluorine has the following properties: normal melting point: 53.5 K normal boiling point: 85.0 K triple point: 1.6×10-4 atm, 53.4 K critical point: 55 atm, 144.1 K
So, a question is now attached about a sample of Fluorine. A sample of fluorine at a pressure of 1.00 atm and a temperature of 90.3 K is cooled at constant pressure to a temperature of 49.3 K.
We are then told to examine a group of options to find the ones that are correct/apply.
Taking the options one at a time
- The sample is initially a gas.
The initial state of the Fluorine sample has its temperature at 90.3 K, which is above the gas' boiling point. Hence, the sample can be concluded to initially be a gas.
- The liquid initially present will vaporize.
The sample doesn't initially contain liquid. And even of it did, the temperature is cooled, not heated , Hence, this statement is wrong.
- The final state of the substance is a solid.
The sample of Fluorine moves from a temperature higher than boiling point (85.0 K), with the sample in gaseous form, to one that is at a lower temperature (49.3 K) than the gas' normal melting point (53.5 K).
At temperatures lower than melting point, a substance exists in the solid form. Hence, this statement is true. The final state of the substance is solid.
- One or more phase changes will occur.
In moving from 90.3 K to 49.3 K for the sample and passing through the substance's boiling and melting points (85.0 K and 53.5 K respectively) along the way, it is logical to conclude that there would be one or more phase changes will occur. This statement is true.
- The final state of the substance is a liquid.
This is false as we already established that the final state of the substance is a solid. Hence, this statement is false.
Hope this Helps!!!
Final answer:
At 1.00 atm, fluorine starts as a gas at 90.3 K, then condenses to a liquid as it cools, and finally becomes a solid at 49.3 K, indicating that both condensation and freezing phase changes occur.
Explanation:
The substance fluorine has different states at various temperatures and pressures. To determine the state changes of fluorine when cooling from a temperature of 90.3 K to 49.3 K at constant pressure of 1.00 atm, we refer to the given melting and boiling points of fluorine. According to the information:
Normal melting point: 53.5 K
Normal boiling point: 85.0 K
At the starting temperature of 90.3 K and 1.00 atm, fluorine is above the boiling point, so the sample is initially a gas. As the temperature cools to below the boiling point but still above the melting point, any liquid that may be present will not vaporize; instead, the gas will condense to form a liquid. Since the final temperature of 49.3 K is below the melting point of 53.5 K, the final state of the substance is a solid. Throughout this process, one or more phase changes will occur; specifically, the gas will condense to a liquid and then freeze into a solid. Therefore, the final state of the substance will not remain a liquid; this statement is false.
Sulfur dioxide and oxygen react to form sulfur trioxide during one of the key steps in sulfuric acid synthesis. An industrial chemist studying this reaction fills a 500. mL flask with 3.7 atm of sulfur dioxide gas and 2.3 atm of oxygen gas, and when the mixture has come to equilibrium measures the partial pressure of sulfur trioxide gas to be 2.2 atm. Calculate the pressure equilibrium constant for the reaction of sulfur dioxide and oxygen at the final temperature of the mixture. Round your answer to 2 significant digits.
Answer:
The pressure equilibrium constant is [tex]K_p = 323[/tex]
Explanation:
From the question we are told that
The volume of the flask is [tex]V = 50 mL = 50 *10^{-3} L[/tex]
The pressure of sulfur dioxide is [tex]P_s = 3.7 \ atm[/tex]
The pressure of oxygen gas [tex]P_o = 2.3 \ atm[/tex]
The pressure of sulfur trioxide at equilibrium is [tex]P_t = 2.2 \ atm[/tex]
The chemical equation for this reaction is
[tex]2 SO_2_{(g)} + O_2_{(g)}[/tex] ⇄ [tex]2SO_3_{(g)}[/tex]
The partial pressure of oxygen at equilibrium is mathematically evaluated as
[tex]P_p__{O}} = P_o - P_t[/tex]
Substituting values
[tex]P_p__{O}} = 2.3 -2.2[/tex]
[tex]P_p__{O}} = 0.1 \ atm[/tex]
The partial pressure of sulfur dioxide at equilibrium is mathematically evaluated as
[tex]P_p__{s}} = P_s - P_t[/tex]
Substituting values
[tex]P_p__{S}} = 3.7 -2.2[/tex]
[tex]P_p__{O}} = 1.5 \ atm[/tex]
From the chemical equation pressure constant is mathematically represented as
[tex]K_p = \frac{[P_t]^2}{[P_p__{o}} ]^2 [P_p__{s}}]}[/tex]
Substituting values
[tex]K_p = \frac{[2.2]^2}{[ 0.1 ]^2 [{ 1.5}]}[/tex]
[tex]K_p = 323[/tex]
Ice melts into water. The correct equation to calculate the heat of this reaction is
Answer:
The correct equation to calculate the heat of this reaction is:
ΔH = m*s*∆T
Explanation:
During any chemical reaction, heat can either be absorbed from the environment or released to the environment through the reaction. The heat exchange between a chemical reaction and its environment is known as the reaction enthalpy, or H. However, H cannot be measured directly; the change in temperature of a reaction over time is used to find the enthalpy change over time (denoted as ΔH).
In general ΔH = m*s*∆T, where m is the mass of the reactants, s is the specific heat of the product, and ΔT is the change in the reaction temperature.
As a medication, Chang's doctor prescribed him a drug with serious restrictions. However, Chang started overdosing on it. This act had severe
Implications on his physical and mental state. To which category could this drug belong?
Answer: Schedule (2) two drug
Explanation: Although it’s virtually difficult or impossible to design a set of defining drug classification standards because even experts have struggled on which drugs should be and not be on a particular schedule but drugs are generally categorised based on their abuse rate (misuse and physical dependency) and their medical use. Drugs with no medical use and higher abuse rate which has physical and mental implications are placed under schedule one, drugs with medical use but also higher abuse rate with physical and mental implications are placed under schedule two, drugs with lower or moderate abuse rate are placed under schedule three, drugs with low potential for dependency are place under schedule four and drugs which are mostly use for antidiarrheal, analgesic and antitussive are placed under schedule five.
Chang's drug has medical use because it was prescribed by his doctor, it also has a high abuse rate because he started overdosing on them which led to a severe physical and mental implications.
Therefore Cheng's drugs will mostly likely be categorised as a schedule (2) two drug.
g Solid sulfur hexafluoride evaporates when heated at atmospheric pressure rather than liquefying. What is the correct name of the type of phase transition that is represented by this process? a) This process is an example of vaporization. b) This process is an example of deposition. c) This process is an example of sublimation. d) This process is an example of freezing. e) This process is an example of condensation.
Answer: c) This process is an example of sublimation.
Explanation:
When liquid is further heated, the molecules gain more kinetic energy and molecules move farther and convert to gaseous state and the process is called vaporization.
Sublimation is a process of converting a substance from solid state to gaseous state without the formation of liquid at constant temperature.
Deposition is a process of converting a substance from gaseous state to solid state without the formation of liquid at constant temperature.
In liquid the particles are loosely packed as the inter molecular forces of attraction are not that strong. As they are cooled, the kinetic energy of the molecules decreases and thus the molecules start getting closer and convert to solid state. The process is called freezing.
2. A 47.7 g sample of SrCl2 is dissolved in 112,500.0 g of solvent. Calculate the molality of the
solution.
A. 0.301 m
B. 2.67 m
C. 0.0339 m
D. 3.99 m
Answer:
I have the same question
Explanation:
Which of the following statements concerning hydrocarbons is/are correct?
1.
Saturated hydrocarbons may be cyclic or acyclic molecules.
2.
An unsaturated hydrocarbon molecule contains at least one double bond.
3.
Ethylenediamine, H2NCH2CH2NH2, is an example of a saturated hydrocarbon.
Answer:
1. Saturated hydrocarbons may be cyclic or acyclic molecules.
2. An unsaturated hydrocarbon molecule contains at least one double bond.
Explanation:
Hello,
In this case, hydrocarbons are defined as the simplest organic compounds containing both carbon and hydrogen only, for that reason we can immediately discard the third statement as ethylenediamine is classified as an amine (organic chain containing NH groups).
Next, as saturated hydrocarbons only show single carbon-to-carbon bonds and carbon-to-hydrogen bonds, they may be cyclic (ring-like-shaped) or acyclic (not forming rings), so first statement is true
Finally, since we can find saturated hydrocarbons which have single carbon-to-carbon and carbon-to-hydrogen bonds only and unsaturated hydrocarbons which could have double or triple bonds between carbons and carbon-to-hydrogen bonds, the presence of at least one double bond makes the hydrocarbon unsaturated.
Therefore, first and second statements are correct.
Best regards.
Final answer:
Statements 1 and 2 are correct regarding saturated hydrocarbons being cyclic or acyclic and unsaturated hydrocarbons containing at least one double bond. Statement 3 is incorrect because ethylenediamine is not a hydrocarbon.
Explanation:
To address which statements concerning hydrocarbons are correct:
Saturated hydrocarbons may indeed be cyclic or acyclic molecules. When cyclic, they have the general formula CnH₂n, such as cycloalkanes, which are saturated compounds. Acyclic saturated hydrocarbons, also known as alkanes, have single bonds only and follow the general formula CHn₂+2.
An unsaturated hydrocarbon molecule contains at least one double or triple bond. Molecules with one or more double bonds are alkenes, with the simplest being ethene (or ethylene), C₂H₄.
Ethylenediamine, H₂NCH₂CH₂NH₂, is not an example of a saturated hydrocarbon. It contains amine groups (NH₂) and therefore is not a hydrocarbon.
Hence, statements 1 and 2 are correct, while statement 3 is incorrect.
Which of the following is a component of cell theory that is supported by the diagram
Option 4. All living organisms are composed of cells. The component of cell theory that is supported by the diagram is option 4.
The cell theory, which is a fundamental concept in biology, states the following:
All living organisms are composed of one or more cells: This means that cells are the basic structural and functional units of all living things.
The cell is the basic unit of life: Cells are the smallest units that can carry out all the processes necessary for life, including metabolism and reproduction.
All cells arise from pre-existing cells: New cells are produced by the division of existing cells, and no new cells are spontaneously generated.
From the diagram, the first image is a simple unicellular organism and the others are multicellular organism tissues. These shows that all living things are made up of cell/cells depending on the complexity of the organism.
Complete question
A sample of hexane (C6H14) has a mass of 0.580 g. The sample is burned in a bomb calorimeter that has a mass of 1.900 kg and a specific heat of 3.21 J/giK. What amount of heat is produced during the combustion of hexane if the temperature of the calorimeter increases by 4.542 K?
A. 8.46 kJ
B. 16.1 kJ
C. 27.7 kJ
D. 47.8 kJ
Answer:
27.7 KJ
Explanation:
Q= mC dT
m= 1900 g+0.580 g= 1900.58 g
Q= (1900.58 g * 3.21 KJ/ gK* 4.542 K)
Q=27710 J= 27.7 KJ
Answer:
C) 27.7 kJ
Explanation:
which molecule is a product of photosynthesis?
Carbon dioxide
Sugar
Water
Nitrogen
A galvanic cell is powered by the following redox reaction: br2 (l) 2NO(g) (l)(aq) (aq) (aq) Answer the following questions about this cell. If you need any electrochemical data, be sure you get it from the ALEKS Data tab. Write a balanced equation for the half-reaction that takes place at the cathode. Write a balanced equation for the half-reaction that takes place at the anode. Calculate the cell voltage under standard conditions. Round your answer to decimal places.
Final answer:
In the galvanic cell, the reduction of Br₂ to 2Br⁻ occurs at the cathode, while the oxidation of 2NO to N₂O₄ occurs at the anode. The cathode is the silver electrode and the anode is the copper electrode. The silver electrode is the positive electrode and the copper electrode is the negative electrode.
Explanation:
In the galvanic cell powered by the redox reaction between Br₂ and 2NO, the half-reaction that takes place at the cathode is the reduction of Br₂ to 2Br⁻. The half-reaction that takes place at the anode is the oxidation of 2NO to N₂O₄.
For the electrode assignment, the cathode is the electrode where reduction occurs, so it is the silver electrode. The anode is the electrode where oxidation occurs, so it is the copper electrode.
The positive electrode in this galvanic cell is the cathode, which is the silver electrode, and the negative electrode is the anode, which is the copper electrode.
4) What is the difference between tetrahedral bent and trigonal planar bent?
Answer:
A tetrahedral bent molecular geometry involves four electron pairs including two lone pairs and two bonding groups while a trigonal planar bent molecular geometry involves three electron pairs two bonding groups and one lone pair.
Explanation:
According to valence shell electron pair repulsion theory, the number of electron pairs on the central atom of a molecule influences its shape. It follows that, the shape of a molecule is a consequence of arrangement of valence shell electron pairs. Electron pairs must be positioned as far apart in space as possible.
Electron pairs may be bonding pairs or lone pairs. The repulsion between two lone pairs is greater than the repulsion between a lone pair and a bond pair which in turn is greater than the repulsion between two bond pairs. Hence the presence of lone pairs causes much distortion of the expected molecular geometry of the molecule as predicted by VSEPR theory. This distortion usually leads to the assumption of a bent geometry, depending on the expected geometry of the molecule based on VSEPR theory.
A tetrahedral bent molecular geometry involves four electron pairs including two lone pairs and two bonding groups while a trigonal planar bent molecular geometry involves three electron pairs; two bonding groups and one lone pair. For a bent tetrahedral geometry, the bond angle is much less than 109° while for a bent trigonal planar geometry, the bond angle is much less than 120°.
The difference between tetrahedral bent and trigonal planar bent molecular geometries lies in the number of electron pairs around the central atom and the resulting shape.
In a tetrahedral bent geometry, the central atom is surrounded by four electron pairs, which are arranged around the atom in a tetrahedral shape. This arrangement leads to bond angles of approximately 109.5 degrees.
An example of a molecule with a tetrahedral bent shape is water (H2O), where the central oxygen atom has two bonded pairs and two lone pairs of electrons. The lone pairs exert greater repulsion than the bonded pairs, causing the bond angles to be slightly less than the ideal tetrahedral angle, resulting in a bent shape.
On the other hand, in a trigonal planar bent geometry, the central atom is surrounded by three electron pairs, which are arranged in a plane around the atom in a trigonal planar shape.
This arrangement leads to bond angles of approximately 120 degrees. However, when one or more of these electron pairs are lone pairs, the geometry can become bent due to the lone pair-bond pair repulsion being greater than bond pair-bond pair repulsion.
An example of a molecule with a trigonal planar bent shape is sulfur dioxide (SO2), where the central sulfur atom has two bonded pairs and one lone pair of electrons. The presence of the lone pair causes the molecule to have a bent shape with bond angles less than 120 degrees.
In summary, the key differences are:
- Tetrahedral bent involves four electron pairs with a tetrahedral arrangement, while trigonal planar bent involves three electron pairs with a trigonal planar arrangement.
- Tetrahedral bent typically results from two bonded pairs and two lone pairs, leading to a bond angle of slightly less than 109.5 degrees.
- Trigonal planar bent typically results from two bonded pairs and one lone pair, leading to a bond angle of less than 120 degrees.
Therefore, the main difference is the number of electron pairs and the resulting molecular shape and bond angles.
Consider the following reaction.
CO2(g) + H2(9)
-
CO(g) + H2O(1)
What is being oxidized?
O carbon
carbon dioxide
oxygen
o hydrogen
Answer:
Hydrogen
Explanation:
It gains oxygen from carbon dioxide to form water
Oxidation is the addition of oxygen to an element in a chemical reaction
Answer:
Its D
Explanation:
Trust me just get ronas and cheat
Electrophilic addition to an alkene proceeds via Markovnikov regiochemistry due to the formation of the more stable carbocation intermediate. In the case of conjugated dienes, that is dienes that are separated by one sigma bond, the carbocation that is formed is stabilized additionally by resonance. Addition of the nucleophile to the carbocation intermediate can therefore give two types of products: direct addition to the double bond, also called 1,2-addition, and conjugate addition to the resonance stabilized carbocation, also called 1,4-addition. Allylic carbocation stability is affected by both the nature of the carbocation (primary allylic, secondary allylic, or tertiary allylic) and by the degree of substitution of the double bond. The latter is typically the dominant effect and so a primary allylic carbocation with a trisubstituted double bond is more stable than a tertiary allylic carbocation with a monosubstituted double bond. Electrophilic addition to a conjugated diene is temperature dependent where reaction at or below room temperature typically leads to a mixture of products in which the 1,2 adduct (or direct addition product) predominates, this is termed kinetic control. At elevated temperatures the reactions have time to come to equilibrium and typically the 1,4 adduct (or conjugate addition product) will predominate, this is termed thermodynamic control. Draw curved arrows to show the movement of electrons in this step of the mechanism.
Answer:
Check the explanation
Explanation:
Kindly check the attached image below for the step by step explanation to the question above.
EDTA EDTA is a hexaprotic system with the p K a pKa values: p K a1 = 0.00 pKa1=0.00 , p K a2 = 1.50 pKa2=1.50 , p K a3 = 2.00 pKa3=2.00 , p K a4 = 2.69 pKa4=2.69 , p K a5 = 6.13 pKa5=6.13 , and p K a6 = 10.37 pKa6=10.37 . The distribution of the various protonated forms of EDTA will therefore vary with pH. For equilibrium calculations involving metal complexes with EDTA EDTA , it is convenient to calculate the fraction of EDTA EDTA that is in the completely unprotonated form, Y 4 − Y4− . This fraction is designated α Y 4 − αY4− . Calculate α Y 4 − αY4− at two pH values.
To calculate αY4-, we need to find the fraction of EDTA in its protonated forms at the given pH and divide the concentration of the completely unprotonated form by the total concentration of EDTA.
Explanation:To calculate the fraction of EDTA that is in the completely unprotonated form, Y4- (designated as αY4-), at a given pH value, we need to find the values of αY4- at two different pH values.
To calculate αY4-, we first need to find the fraction of EDTA in its protonated forms at the given pH.
For example, at pH = 2, we can calculate the fractional composition of each protonated form by dividing the concentration of the protonated form by the total concentration of all protonation states of EDTA. We can repeat this calculation for another pH value to find αY4- at that pH.
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The fraction of EDTA in the completely unprotonated form at pH3 is approximately [tex]\( 9.28 \times 10^{-4} \)[/tex], and at pH8 is approximately [tex]\( 8.83 \times 10^{-7} \)[/tex].
To calculate the fraction of EDTA that is in the completely unprotonated form, [tex]\( \alpha_{Y^{4-}} \)[/tex], at a given pH, we use the following equation:
[tex]\[ \alpha_{Y^{4-}} = \frac{1}{1 + \sum_{i=1}^{6} 10^{(pH - pKa_i)}} \][/tex]
Here, [tex]\( pKa_i \)[/tex] represents the [tex]\( i \)-th[/tex] acid dissociation constant for EDTA. Since EDTA is a hexaprotic acid, it has six [tex]\( pKa \)[/tex] values, which are given as:
[tex]\( pKa_1 = 0.00 \), \( pKa_2 = 1.50 \), \( pKa_3 = 2.00 \), \( pKa_4 = 2.69 \), \( pKa_5 = 6.13 \), and \( pKa_6 = 10.37 \).[/tex]
Calculate [tex]\( \alpha_{Y^{4-}} \)[/tex] for two pH values, say pH = 3 and pH = 8.
For pH = 3:
[tex]\[ \alpha_{Y^{4-}} = \frac{1}{1 + 10^{(3 - 0.00)} + 10^{(3 - 1.50)} + 10^{(3 - 2.00)} + 10^{(3 - 2.69)} + 10^{(3 - 6.13)} + 10^{(3 - 10.37)}} \][/tex]
[tex]\[ \alpha_{Y^{4-}} = \frac{1}{1 + 10^{3} + 10^{1.5} + 10^{1} + 10^{0.31} + 10^{-3.13} + 10^{-7.37}} \][/tex]
[tex]\[ \alpha_{Y^{4-}} = \frac{1}{1 + 1000 + 31.62 + 10 + 2.08 + 0.00074 + 4.17 \times 10^{-8}} \][/tex]
[tex]\[ \alpha_{Y^{4-}} \approx \frac{1}{1 + 1000 + 31.62 + 10 + 2.08} \][/tex]
[tex]\[ \alpha_{Y^{4-}} \approx \frac{1}{1033.62 + 31.62 + 10 + 2.08} \][/tex]
[tex]\[ \alpha_{Y^{4-}} \approx \frac{1}{1077.32} \][/tex]
[tex]\[ \alpha_{Y^{4-}} \approx 9.28 \times 10^{-4} \][/tex]
For pH = 8:
[tex]\[ \alpha_{Y^{4-}} = \frac{1}{1 + 10^{(8 - 0.00)} + 10^{(8 - 1.50)} + 10^{(8 - 2.00)} + 10^{(8 - 2.69)} + 10^{(8 - 6.13)} + 10^{(8 - 10.37)}} \][/tex]
[tex]\[ \alpha_{Y^{4-}} = \frac{1}{1 + 10^{8} + 10^{6.5} + 10^{6} + 10^{5.31} + 10^{1.87} + 10^{-2.37}} \][/tex]
[tex]\[ \alpha_{Y^{4-}} = \frac{1}{1 + 10000000 + 316227.76 + 1000000 + 20794.42 + 74.12 + 0.00417} \][/tex]
[tex]\[ \alpha_{Y^{4-}} \approx \frac{1}{10000000 + 316227.76 + 1000000 + 20794.42 + 74.12} \][/tex]
[tex]\[ \alpha_{Y^{4-}} \approx \frac{1}{1131992.3} \][/tex]
[tex]\[ \alpha_{Y^{4-}} \approx 8.83 \times 10^{-7} \][/tex]
Therefore, the fraction of EDTA in the completely unprotonated form at pH 3 is approximately [tex]\( 9.28 \times 10^{-4} \)[/tex], and at pH 8 is approximately [tex]\( 8.83 \times 10^{-7} \)[/tex].
The complete question is:
EDTA is a hexaprotic system with the PK, values:
PKa1 = 0.00, pKa2 = 1.50, pKa3 2.00, pKa4 = 2.69,
pKa5 6.13, and pKa6 10.37.
The distribution of the various protonated forms of EDTA will therefore vary with pH. For equilibrium calculations involving metal complexes with EDTA, it is convenient to calculate the fraction of EDTA that is in the completely unprotonated form, Y4-. This fraction is designated αY4−. Calculate αY4− at two pH values.
pH = 3.10
pH = 10.55
A 2.5-liter sample of gas is at STP. When the temperature is raised to 546 K and the pressure remains constant, what will be the new volume of the gas?
Answer:
5L
Explanation:
Please see the step-by-step solution in the picture attached below.
Hope this answer can help you. Have a nice day!
A 2.5 liter sample of gas is STP. When the temperature is raised to 546 K and the pressure remains constant the new volume of the gas will be 5 L.
What is Ideal Gas Law ?The ideal gas law states that the pressure of gas is directly proportional to the volume and temperature of the gas.
PV = nRT
where,
P = Pressure
V = Volume
n = number of moles
R = Ideal gas constant
T = Temperature
What is STP ?STP is Standard Temperature and Pressure. At STP the temperature is 273 K or 0°C and the standard pressure is 1 atm.
Now, according to question the pressure is constant, then
At constant pressure
V ∝ T
[tex]\frac{V_1}{V_2} = \frac{T_1}{T_2}[/tex]
[tex]\frac{2.5\ L}{V_2} = \frac{273\ K}{546\ K}[/tex]
[tex]V_{2} = \frac{546 \times 2.5\ L}{273}[/tex]
V₂ = 5 L
Thus, from the above conclusion we can say that A 2.5 liter sample of gas is STP. When the temperature is raised to 546 K and the pressure remains constant the new volume of the gas will be 5 L.
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Determine the pH of a 0.188 M NH 3 solution at 25°C. The K b of NH 3 is 1.76 × 10 -5.
Answer:
pH is 8.52
Explanation:
According to the working in the photo
To how much water should 50mL of 12M hydrochloric acid be added to produce a 4M solution?
Answer:
50 mL of 12M HCl should be added into 100 mL of water to produce a 4M solution.
Explanation:
According to law of dilution-
[tex]C_{1}V_{1}=C_{2}V_{2}[/tex]
where [tex]C_{1},C_{2},V_{1}[/tex] and [tex]V_{2}[/tex] are initial concentration, final concentration, initial volume and final volume respectively.
Here, [tex]C_{1}=12M[/tex] , [tex]C_{2}=4M[/tex] , [tex]V_{1}=50mL[/tex]
So, [tex]V_{2}=\frac{C_{1}V_{1}}{C_{2}}[/tex] = [tex]\frac{(12M)\times (50mL)}{(4M)}[/tex] = 150 mL
Hence, final volume of HCl solution should be 150 mL.
So, volume of water needed = (150-50) mL = 100 mL
Therefore 50 mL of 12M HCl should be added into 100 mL of water to produce a 4M solution.
You need to add [tex]\( {150} \)[/tex] mL of water to 50 mL of 12 M hydrochloric acid to produce a 4 M solution.
To determine how much water should be added to 50 mL of 12 M hydrochloric acid (HCl) to produce a 4 M solution, we can use the dilution formula which states:
[tex]\[ C_1V_1 = C_2V_2 \][/tex]
where:
- [tex]\( C_1 \)[/tex] and [tex]\( V_1 \)[/tex] are the concentration and volume of the initial solution (12 M and 50 mL, respectively),
- [tex]\( C_2 \)[/tex] and [tex]\( V_2 \)[/tex] are the concentration and volume of the final solution (4 M and the unknown volume of water, respectively).
Let's solve for [tex]\( V_2 \)[/tex]:
[tex]\[ C_1V_1 = C_2V_2 \][/tex]
Substitute the given values:
[tex]\[ 12 \times 50 = 4 \times V_2 \][/tex]
Solve for [tex]\( V_2 \)[/tex]:
[tex]\[ 600 = 4 \times V_2 \][/tex]
[tex]\[ V_2 = \frac{600}{4} \][/tex]
[tex]\[ V_2 = 150 \text{ mL} \][/tex]
Calculate the number of moles of O2 gas held in a sealed 2.00 L tank at 3.50 atm and 25 ℃.
Answer:
[tex]n=0.286mol[/tex]
Explanation:
Hello,
In this case, we consider oxygen as an ideal gas, for that reason, we use yhe ideal gas equation to compute the moles based on:
[tex]PV=nRT\\\\n=\frac{PV}{RT}[/tex]
Hence, at 3.50 atm and 25 °C for a volume of 2.00 L we compute the moles considering absolute temperature in Kelvins:
[tex]n=\frac{3.50atm*2.00L}{0.082\frac{atm*L}{mol*K}(25+273)K} \\\\n=0.286mol[/tex]
Best regards.
Two radioactive nuclides X and Y both decay to stable products. The half-life of X is about a day, while that of Y is about a week. Suppose a radioactive sample consists of a mixture of these two nuclides. If the mixture is such that the activities arising from X and Y are initially equal, then a few days later the activity of the sample will be due ______________.
Answer:Then a few days later the activity of the sample will be due to have more of Nuclides Y than X
Explanation:
This is because half life of nuclide X is about a day which is less than Y having half life of about a week, After a few days, we would observe that X would have disintegrated more while Y will still be predominant since it disintegrate slower than X. The time it takes for X to disintegrate will always be faster than Y.
What is the total number of distinct 13C NMR signals that may be observed for the product, methyl-3-nitrobenzoate, and for the reactant, methylbenzoate?What is the total number of distinct 13C NMR signals that may be observed for the product, methyl-3-nitrobenzoate, and for the reactant, methylbenzoate?
Answer:
See explaination
Explanation:
methyl-3-nitrobenzoate = 3-NO2-C6H4-COOCH3
Singlet at 4 ppm = CH3 from ester (COOCH3)
Triplet at 7.6 ppm = Aromatic, H5 proton
Doublet at 8.2 ppm - Aromatic, H4 and H6 protons
Singlet at 8.8 ppm - Aromatic, H2 proton
See attached file for diagrammatic representation and further solution.
Methylbenzoate has 8 distinct carbon environments, therefore it would show 8 distinct 13C NMR signals. Methyl-3-nitrobenzoate, due to its symmetrical structure, has 7 distinct carbon environments, therefore it would show 7 distinct 13C NMR signals.
Explanation:
The product, methyl-3-nitrobenzoate, and the reactant, methylbenzoate, are both organic compounds, and their 13C NMR signals can be determined by the number of carbon environments they have.
Methylbenzoate has 8 distinct carbon environments: benzoate carbon framework (7) + methyl group (1). Hence, in the 13C NMR spectroscopy, methylbenzoate would show 8 distinct signals.
Methyl-3-nitrobenzoate on the other hand, due to the introduction of the nitro group, creates symmetry which changes the carbon environments, thus, it has 7 distinct carbon environments: nitrobenzoate carbon framework (6) + methyl group (1). So, methyl-3-nitrobenzoate would show 7 distinct signals in its 13C NMR spectroscopy.
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Using the appropriate Ksp values, find the concentration of K+ ions in the solution at equilibrium after 600 mL of 0.45 M aqueous Cu(NO3)2 solution has been mixed with 450 mL of 0.25 M aqueous KOH solution. (Enter in M.) (Ksp for Cu(OH)2 is 2.6x10-19).
Now find the concentration of OH? ions in this solution at equilibrium.
Answer:
[K⁺] = 0.107 M
[OH⁻] = 1.13 × 10⁻⁹ M
Explanation:
600 mL of 0.45 M Cu(NO3)2 gives equal mole of Cu²⁺ and (NO₃)²⁻
⇒ 0.45 × 600 × 10⁻³
= 0.27 moles of Cu²⁺ and (NO₃)²⁻
450 mL of 0.25 M KOH gives equal moles of K⁺ and OH⁻
⇒ 0.25 × 450 × 10⁻³
= 0.1125 moles of K⁺ and OH⁻
Now after mixing 0.1125 moles of OH⁻ precipitates 0.05625 moles of Cu²⁺ (because 1 Cu²⁺ needs 2 OH⁻)
Therefore , moles of remaining Cu²⁺ = 0.27 - 0.05625
=0.21375 moles which is equal to :
⇒ 0.21375/(( 600+450))× 10⁻³
= 0.21375/1050 × 10⁻³
= 0.20357 M
Given that :
(Ksp for Cu(OH)2 is 2.6 × 10⁻¹⁹)
We know that , Ksp = [Cu²⁺][OH⁻]²
2.6 × 10⁻¹⁹ = 0.20357 × [OH⁻]²
[OH⁻]² = 2.6 × 10⁻¹⁹/0.20357
[OH⁻] = 1.13 × 10⁻⁹ M
[K⁺] = moles of K⁺ /total volume
[K⁺] = 0.1125 / 1050 × 10⁻³
[K⁺] = 0.107 M
Acid indigestion is sometimes neutralized with an antacid such as magnesium hydroxide (Mg(OH)2). What products will be released when the antacid is mixed with the hydrochloric acid in the stomach
Answer:
Magnesium chloride and water
Explanation:
Mg(OH)₂ + 2HCl ⟶ MgCl₂ + 2H₂O
magnesium chloride water
On average, each person in the industrialized world is responsible for the emission of 10,000 kg of carbon dioxide (CO2) every year. This includes CO2 that you generate directly, by burning fossil fuels to operate your car or furnace, as well as CO2 generated on your behalf by electric generating stations and manufacturing plants. CO2 is a greenhouse gas that contributes to global warming.
If you were to store your yearly CO2 emissions in a cube at STP, how long would each edge of the cube be?
Final answer:
To calculate the length of each edge of the cube that would store your yearly CO2 emissions, convert the emissions from kg to metric tons and use the density of carbon dioxide at STP to find the volume. The length of each edge would be approximately 17.9 meters.
Explanation:
To calculate the length of each edge of the cube that would store your yearly CO2 emissions, we first need to convert the emissions from kg to metric tons. Since the average CO2 emissions per person per year is 10,000 kg, this is equal to 10 metric tons (1 metric ton = 1000 kg).
Next, we need to find the volume of the cube. The formula for the volume of a cube is V = s^3, where s represents the length of each edge.
Let's use the given data to solve for s:
CO2 emissions per person per year: 10 metric tons
Density of CO2 at STP: 1.98 kg/m³ (source: https://pubchem.ncbi.nlm.nih.gov/compound/carbon_dioxide)
Using the density, we can convert the metric tons of CO2 to the corresponding volume in cubic meters:
10 metric tons * 1000 kg/metric ton = 10,000 kg
10,000 kg / 1.98 kg/m³ ≈ 5,051.51 m³
Now, let's solve for s:
s^3 = 5,051.51 m³
s ≈ 17.9 meters (rounded to one decimal place)
Therefore, each edge of the cube that would store your yearly CO2 emissions at STP would be approximately 17.9 meters long.
Consider this chemical reaction, where moving from left to right represents moving forward in time. A five panel comic strip. In the first panel, there are ten large red spheres. In the second panel, there are 8 large red spheres and two small blue spheres. In the third panel, there are six large red spheres and four small blue spheres. In the fourth panel, there are four large red spheres and six small blue spheres. In the fifth panel, there are four large red spheres and six small blue spheres. At what point does the reaction first reach equilibrium
Answer:
The reaction reaches equilibrium at the fourth panel.
Explanation:
Chemical Equilibrium is achieved when the overall properties the system seem to be constant, that is, stop changing.
Although, for chemical equilibrium, the right term for this equilibrium is dynamic equilibrium; the rate of forward reaction balances the rate of backward reaction, but concentrations can keep changing.
The point where equilibrium is achieved is when exactly when we reach the panel where the spheres that make up this panel is the same as the next panel and the next, that is, the specific colour and number of spheres start to become unchanged.
And from the description given in the question,
- In the first panel, there are ten large red spheres.
- In the second panel, there are 8 large red spheres and two small blue spheres.
- In the third panel, there are six large red spheres and four small blue spheres.
- In the fourth panel, there are four large red spheres and six small blue spheres.
- In the fifth panel, there are four large red spheres and six small blue spheres.
It is evident that the make-up of the spheres have become the same as at the fourth and fifth panel. This means that the first point where this final configuration of spheres first appeared is the fourth panel.
Hence, equilibrium is first reached at the fourth panel.
Hope this Helps!!!
The reaction reaches equilibrium at the fourth panel. A reaction is at equilibrium when the amounts of reactants or products no longer change. Chemical equilibrium is a dynamic process
What is Chemical Equilibrium?It is the rate of formation of products by the forward reaction is equal to the rate at which the products re-form reactants by the reverse reaction.
The point where equilibrium is achieved is when exactly when we reach the panel where the spheres that make up this panel is the same as the next panel and the next, that is, the specific color and number of spheres start to become unchanged.
As per the descriptions of panel given in question:
It is evident that the make-up of the spheres have become the same as at the fourth and fifth panel. This means that the first point where this final configuration of spheres first appeared is the fourth panel.
Hence, equilibrium is first reached at the fourth panel.
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HURRY PLS HELP PLEASE: A 40.0-L sample of fluorine is heated from 363 Kelvin to 459 K. What volume will the sample occupy at the higher temperature?
19.3 L
31.6 L
50.5 L
82.6 L
Final answer:
The volume of a 40.0-L sample of fluorine gas heated from 363 K to 459 K can be found using Charles's Law. After setting up the equation from Charles's Law and solving for the new volume (V2), we find that the volume at the higher temperature is 50.5 L.
Explanation:
To determine the new volume of fluorine gas when heated from 363 K to 459 K, we can use Charles's Law, which states that for a given mass of an ideal gas at constant pressure, the volume is directly proportional to its absolute temperature. Specifically, the formula is V1/T1 = V2/T2, where V1 is the initial volume, T1 is the initial temperature, V2 is the final volume, and T2 is the final temperature.
In this case, the initial volume (V1) is 40.0 L and the initial temperature (T1) is 363 K. The final temperature (T2) is 459 K. Plugging these values into the formula, we have:
40.0 L / 363 K = V2 / 459 K
Multiplying both sides by 459 K to solve for V2 gives us:
V2 = (40.0 L x 459 K) / 363 K
Upon calculation, V2 = 50.5 L. Therefore, the new volume occupied by the sample at 459 K is 50.5 liters.