The most likely substrates for the oligo-(α1→4)-(α1→4) glucanotransferase catalytic center of glycogen debranching enzyme are maltosyl and maltotriosyl residues.
Explanation:The glycogen debranching enzyme has two activities; it acts as a glucosidase and a glucanotransferase. The (α‑6) glucosidase activity hydrolyzes α-1,6 glycosidic bonds, whereas the oligo-(α1→4)-(α1→4) glucanotransferase activity shifts α-1,4-linked glucose chains from one branch to another, often creating a substrate that the glucosidase can act upon. Based on the mechanism of glycogenolysis, the enzyme glycogen phosphorylase releases glucose units from the linear chain until a few are left near the branching point, and it is here that the glucan transferase action is relevant. The glucan transferase shifts the remaining α-1,4 linked glucose units, leaving a single α-1,6 linked glucose that the glucosidase can then release. Hence, the most likely substrates for the oligo-(α1→4)-(α1→4) glucanotransferase catalytic center are maltosyl and maltotriosyl residues, as these are the short α-1,4 linked glucose chains that are left after phosphorylase action.
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You have been asked to prove that the Oil Spill Eater speeds up the reaction as you found above. Describe the basic set up of a laboratory experiment to measure the effect of enzyme on an oil spill in the ocean. Select all necessary test reactions. Group of answer choices Run a test reaction of crude oil with ocean water over time without Oil Spill Eater present. Run a test reaction of crude oil at a cool temperature with Oil Spill Eater present Run a test reaction of crude oil with ocean water over time with Oil Spill Eater present. Run a test reaction of crude oil at a warm temperature with Oil Spill Eater present Run a test reaction of crude oil at a warm temperature without Oil Spill Eater present Run a test reaction of crude oil at a cool temperature without Oil Spill Eater present
Answer:
The correct answer is option 3. Run a test reaction of crude oil with ocean water over time with Oil Spill Eater present
Explanation:
In any laboratory experiment, all the apparatus needed to carry out a particular experiment must be provided. In this case, our apparatus will be crude oil with ocean water and oil spill eater which is the enzyme used.
We can then run a test reaction of crude oil with ocean water over time with Oil Spill Eater present.
Fatty acid oxidation occurs in the mitochondrial matrix. However, long-chain fatty acyl-CoA molecules cannot cross the inner membrane to enter the matrix. The carnitine shuttle system transfers the acyl group from CoA to carnitine, which can enter the mitochondrial matrix. Label the enzymes and compounds of the carnitine shuttle system. These abbreviations are used: intermembrane space, IMS; carnitine acyltransferase I, CAT1; mitochondrial carnitine acyltransferase II, CAT2; and carnitine/acylcarnitine translocase (carnitine carrier protein), CAT.
Answer:
Explanation:
see answer below in the attached file.
The carnitine shuttle system is an important system that helps in the fatty acid oxidation.
The functions of the enzymes and compounds of the carnitine shuttle system include the following:
Intermembrane space (IMS): The activities of the enzymes of the carnitine system takes place in this space within the cells.Carnitine acyltransferase I, (CAT1): This is also called Carnitine palmitoyltransferase 1. It converts long-chain acyl-CoA species to their corresponding long-chain acyl-carnitines for transport into the mitochondria.Mitochondrial Carnitine acyltransferase II, CAT2: This is an enzyme found inside the mitochondria that oxidizes long-chain fatty acids in the mitochondria.carnitine/acylcarnitine translocase (carnitine carrier protein), CAT: This is a membrane transporter that exchanges cytoplasmic acylcarnitine for mitochondrial carnitine.Learn more here:
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Write a balanced net ionic equation to show why the solubility of Cu(OH)2(s) increases in the presence of a strong acid and calculate the equilibrium constant for the reaction of this sparingly soluble salt with acid. Use the pull-down boxes to specify states such as (aq) or (s).
The solubility of Cu(OH)2(s) increases in the presence of a strong acid due to the dissolution of the salt and formation of Cu2+ ions and water. The balanced net ionic equation for this process is Cu(OH)2(s) + 2H+(aq) → Cu2+(aq) + 2H2O(l). The equilibrium constant (K) for the reaction can be calculated using the concentrations of the reactants and products at equilibrium.
Explanation:The balanced net ionic equation to show why the solubility of Cu(OH)2(s) increases in the presence of a strong acid is:
Cu(OH)2(s) + 2H+(aq) → Cu2+(aq) + 2H2O(l)
Cu(OH)2 is a sparingly soluble salt, meaning it has low solubility in water. However, in the presence of a strong acid, the concentration of H+ ions increases, leading to the dissolution of Cu(OH)2 and the formation of Cu2+ ions and water.
The equilibrium constant for this reaction can be calculated by using the concentrations of the reactants and products at equilibrium. The equation for calculating the equilibrium constant (K) is:
K = [Cu2+][H2O2]/[Cu(OH)2][H+]2
Note: The square brackets indicate the concentrations of the species in the solution.
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Transport of aspirin is expected to be faster in the ____. The speed of nonmediated absorption depends strongly on the polarity and charge of molecules. When p H is low (as it is in the _______), the aspirin molecule will be _______ and thus _______, due to _____ of H+ in the solution. The higher pH is, the _______ charged and polar will become the molecule, leading to the ____--- in the speed of absorption. One more evidence to this conclusion is that the pH in the _______ is much lower than pKa of aspirin comparing to the pH of the _______.
The missing words in the blanks are-
Stomach Stomach highly protonated uncharged high concentration more decrease stomach intestine.The transport of aspirin is more in the bloodstream from the stomach as absorption speed depends on the polarity and charge of molecules of the aspirin.
The pH of the stomach is low then molecules of aspirin become highly protonated due to the high concentration of H+ ion they are uncharged.So it is clear that an increase in pH charge and polarity of the molecules also increases which causes less absorption speed of molecules.the pKa of aspirin in the intestine is much more than the pH in the stomach.Thus,
The missing words in the blanks are-
Stomach Stomach highly protonated uncharged high concentration more decrease stomach intestine.Learn more:
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Which of the following is the net ionic equation for the reaction that occurs when a few drops of HCl are added to a buffer containing a weak base (B) and its conjugate acid (BH+)? B(aq) + OH−(aq) → BOH−(aq) BH+(aq) + OH−(aq) → H2O(l) + B(aq) H+(aq) + OH−(aq) → H2O(l) H+(aq) + B(aq) → BH+(aq) B(aq) + H2O(l) begin mathsize 12px style rightwards harpoon over leftwards harpoon end style BH+(aq) + OH−(aq)
Answer:
H+(aq) + B(aq) → BH+(aq)
Explanation:
When an acid is added onto a buffer, it is neutralized by the base.
So we pretty much have;
HCl + Weak base
Since HCl completely dissociates in water, it is represented as;
HCl(aq) --> H+(aq) + Cl-(aq)
The weak base reacts with the H+
So our Net ionic reaction is given as;
H+(aq) + B(aq) → BH+(aq)
In the absence of sodium methoxide, the same alkyl bromide gives a different product. Draw an arrowpushing mechanism to account for its formation. 6. (a) In the reaction in part 5(a), two additional products, which contain only carbon and hydrogen, are also formed. Draw their structures and propose mechanisms for their formation. Predict which of these two products would be formed in greater quantities. (b) In the reaction in part 5(b), two additional products, which contain only carbon
Answer:
See explanation below
Explanation:
The question is incomplete, cause you are not providing the structure. However, I found the question and it's attached in picture 1.
Now, according to this reaction and the product given, we can see that we have sustitution reaction. In the absence of sodium methoxide, the reaction it's no longer in basic medium, so the sustitution reaction that it's promoted here it's not an Sn2 reaction as part a), but instead a Sn1 reaction, and in this we can have the presence of carbocation. What happen here then?, well, the bromine leaves the molecule leaving a secondary carbocation there, but the neighbour carbon (The one in the cycle) has a more stable carbocation, so one atom of hydrogen from that carbon migrates to the carbon with the carbocation to stabilize that carbon, and the result is a tertiary carbocation. When this happens, the methanol can easily go there and form the product.
For question 6a, as it was stated before, the mechanism in that reaction is a Sn2, however, we can have conditions for an E2 reaction and form an alkene. This can be done, cause the extoxide can substract the atoms of hydrogens from either the carbon of the cycle or the terminal methyl of the molecule and will form two different products of elimination. The product formed in greater quantities will be the one where the negative charge is more stable, in this case, in the primary carbon of the methyl it's more stable there, so product 1 will be formed more (See picture 2)
For question 6b, same principle of 6a, when the hydrogen migrates to the 2nd carbocation to form a tertiary carbocation the methanol will promove an E1 reaction with the vecinal carbons and form two eliminations products. See picture 2 for mechanism of reaction.
Balance the redox reaction by inserting the appropriate coefficients.
H2O + Br- + Al3- = Al + BrO3- + H+
The balanced equation is given as,
2Al³⁺ + 3H₂O + Br⁻ → 2 Al + BrO³⁻ + 6H⁺
Explanation:
H₂O + Br⁻ + Al³⁺ → Al + BrO₃⁻ + H⁺
Here the half reactions are:
Al³⁺→ Al [reduction]
Br⁻ → BrO₃⁻ [oxidation]
Now we have to balance the half reactions as,
Al³⁺ + 3e⁻ → Al
To balance H atoms we have to add water and electrons.
3H₂O + Br⁻→ BrO³⁻ + 6H⁺ + 6e⁻
Now we have to balance the electrons as,
2Al³⁺ + 6e⁻ → 2 Al
3H₂O + Br⁻→ BrO³⁻ + 6H⁺ + 6e⁻
Now we have to add both the equations as,
2Al³⁺ + 6e⁻ + 3H₂O + Br⁻ → 2 Al + BrO³⁻ + 6H⁺ + 6e⁻
6 electrons on both sides of the equation gets cancelled and so the balanced reaction can be written as,
2Al³⁺ + 3H₂O + Br⁻ → 2 Al + BrO³⁻ + 6H⁺
Please help! The Michael reaction is a conjugate addition reaction between a stable nucleophilic enolate ion (the donor) and an α,β-unsaturated carbonyl compound (the acceptor). 1.) Draw the structure of the product of the Michael reaction between ethyl propenoate and 3-oxobutanenitrile.2.)The Michael reaction is a conjugate addition reaction between a stable nucleophilic enolate ion (the donor) and an α,β-unsaturated carbonyl compound (the acceptor). Draw the structure of the product of the Michael reaction between propenamide and 2,4-pentanedione.
Answer:
See explaination andd attachment
Explanation:
Treatment of aldehydes and ketones with a suitable base can lead to the formation of a nucleophilic species called an enolate that reacts with electrophiles. These C nucleophiles are useful for making new carbon-carbon bonds.
Please kindly see attachment for the drawings.
In the laboratory you are given the task of separating Ag+ and Cu2+ ions in aqueous solution.
For each reagent listed below indicate if it can be used to separate the ions. Type "Y" for yes or "N" for no. If the reagent CAN be used to separate the ions, give the formula of the precipitate. If it cannot, type "No"
Y or N Reagent Formula of Precipitate if YES
1. NaI
2. K2S
3. K2CO3
Answer:
1. NaI Y AgI
2. K2S Y CuS
3. K2CO3 N
Explanation:
1. When we add NaI to the mixture, the reaction that takes place is:
NaI(aq) + Ag⁺(aq) → AgI(s) + Na⁺(aq)Such a reaction does not happen with Cu⁺².
2. When we add K₂S to the mixture, the reaction that takes place is:
K₂S(aq) + Cu⁺²(aq) → CuS(s) + 2K⁺(aq)Such a reaction does not happen with Ag⁺.
3. When we add K₂CO₃ to the mixture, the reactions that take place are:
K₂CO₃(aq) + 2Ag⁺(aq) → Ag₂CO₃(s) + 2K⁺(aq)K₂CO₃(aq) + Cu⁺²(aq) → CuCO₃(s) + 2K⁺(aq)This means both Cu⁺² and Ag⁺ would precipitate, thus they would not be separated.
1. Compare and contrast the rate of solution formation between the three physical forms of salt that were placed in the vial and not agitated with the three forms of salt that were placed in the vial and were agitated.
Answer:
Explanation:
The salt that was in pellet form took the longest time before it could be dissolved during all the trials. The salt with fine texture dissolves first during the trials, and the salt with coarse texture took the middle position during all the trials.
Salt with agitation dissolves faster than salts without any agitation.
To prepare a buffer you weigh out 7.20 grams of NaHCO3 and place it into a 400.00 mL volumetric flask. To this flask you add 56.0 mL of 5.60 M H2CO3 and then fill it about halfway with distilled water, swirling to dissolve the contents. Finally, the flask is filled the rest of the way to the mark with distilled water.What is the pH of the buffer that you have created?Acid KaH2CO3 4.3 X 10⁻⁷ HCN 4.9 X 10⁻¹⁰HNO2 4.6 X 10⁻⁴C6H5COOH 6.5 X 10⁻⁵
Answer:
pH = 5.80
Explanation:
The buffer solution is:
H₂CO₃(aq) + H₂O(l) ⇄ HCO₃⁻Na⁺(aq) + H₃O⁺(aq)
To find the pH of the buffer solution we will use the Henderson-Hasselbalch equation:
[tex]pH = pKa + log(\frac{[NaHCO_{3}]}{[H_{2}CO_{3}]})[/tex] (1)
First, we need to find the concentration of the buffer solution. For the NaHCO₃ we have:
[tex][NaHCO_{3}] = \frac{mol}{V} = \frac{m}{M*V}[/tex]
Where:
m: is the mass of the NaHCO₃ = 7.20 g
M: is the molar mass of the NaHCO₃ = 84.007 g/mol
V: is the volume of the solution = 400.0 mL
Hence, the concentration of NaHCO₃ is:
[tex][NaHCO_{3}] = \frac{7.20 g}{84.007 g/mol*400.0 \cdot 10^{-3} L} = 0.214 M[/tex]
Now, the concentration of H₂CO₃ is:
[tex] V_{i}C_{i} = V_{f}C_{f} [/tex]
Where:
Vi: is the initial volume of H₂CO₃ = 56.0 mL
Ci: is the initial concentration of H₂CO₃ = 5.60 M
Vf: is the final volume of H₂CO₃ = 400.0 mL
Cf: is the final concentration of H₂CO₃ (to find)
[tex] C_{f} = \frac{V_{i}C_{i}}{V_{f}} = \frac{56.0 mL*5.60 M}{400.0 mL} = 0.784 M [/tex]
Finally, we can use the equation (1) to find the pH of the buffer solution:
[tex] pH = -log(4.3 \cdot 10^{-7}) + log(\frac{0.214 M}{0.784 M}) = 5.80 [/tex]
I hope it helps you!
Bromobenzene is converted to a compound with the molecular formula C7H7Br in the reaction scheme below. Draw the structures of the product and the two intermediates, and identify the reagents in each of the three steps.
Complete Question:
The first file attached contains the complete question
Answer:
The reagents play a vital role in a reaction to undergo a transformation. Each reagent plays a different role in the chemical reaction.
Grignard reagent: Grignard reagent R - Mg - X . R represents an alkyl group or aryl group. X represents halides (I,Br,Cl) . The main purpose of a Grignard reagent is the formation of new C−C bond. Grignard reagent undergoes reaction with carbonyl groups (like ketone ( {\rm{ - C = O}}−C=O ), Aldehyde ( - C( = O)H}}−C(=O)H ), Ester ( - C( = O)OR}}−C(=O)OR )) to form an addition product.
Nucleophilic addition: the addition of nucleophile (electron rich) with electrophiles (electron deficient). In this reaction, the double bond is converted to a single bond.
Nucleophilic substitution: In this, the Nucleophile (electron rich) forms a bond with an electrophile (electron deficient) and replaces the leaving group.
The attached file contained detailed solution
The quantity of heat required to change the temperature of 1 g of a substance by 1°C is defined as ____.
a joule
a calorie
density
specific heat
Answer:
specific heat
Explanation:
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What is the mass of 0.75 moles of (NH4)3PO4?
The mass of (NH4)3PO4 is 111.75grams.
HOW TO CALCULATE THE MASS:
The mass of a substance can be calculated by multiplying the number of moles by its molecular mass. That is;mass (g) = moles (mol) × molar mass (g/mol)
Molar mass of (NH4)3PO4 is calculated as follows:{14 + 1(4)}3 + 31 + 16(4)
= (14+4)3 + 31 + 64
= 54 + 31 + 64
= 149g/mol
mass in grams = 149 × 0.75
mass in grams = 111.75g
Therefore, the mass of (NH4)3PO4 is 111.75grams.
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The Michael reaction is a conjugate addition process wherein a nucleophilic enolate anion (the donor) reacts with an α,β-unsaturated carbonyl compound (the acceptor). The best Michael reactions are those that take place when a particularly stable enolate anion is formed via treatment of the donor with a strong base. Alternatively, milder conditions can be used if an enamine is chosen as the donor, this variant is termed the Stork reaction. In the second step, the donor adds to the β-carbon of the acceptor in a conjugate addition, generating a new enolate. The enolate abstracts a proton from solvent or from a new donor molecule to give the conjugate addition product.
Draw curved arrows to show the movement of electrons in this step of the mechanism.
Answer:
See the attached file for the structure
Explanation:
See the attached file
A chemist titrates 130.0mL of a 0.4248 M lidocaine (C14H21NONH) solution with 0.4429 M HBr solution at 25 degree C . Calculate the pH at equivalence. The pKb of lidocaine is 7.94 . Round your answer to 2 decimal places. Note for advanced students: you may assume the total volume of the solution equals the initial volume plus the volume of HBr solution added.
Answer:
pH = 3.36
Explanation:
Lidocaine is a weak base to be titrated with the strong acid HBr, therefore at equivalence point we wil have the protonated lidocaine weak conjugate acid of lidocaine which will drive the pH.
Thus to solve the question we will need to calculate the concentration of this weak acid at equivalence point.
Molarity = mol /V ∴ mol = V x M
mol lidocaine = (130 mL/1000 mL/L) x 0.4248 mol/L = 0.0552 mol
The volume of 0.4429 M HBr required to neutralize this 0.0552 mol is
0.0552 mol x (1L / 0.4429mol) = 0.125 L
Total volume at equivalence is initial volume lidocaine + volume HBr added
0 .130 L +0.125 L = 0.255L
and the concentration of protonated lidocaine at the end of the titration will be
0.0552 mol / 0.255 L = 0.22M
Now to calculate the pH we setup our customary ICE table for weak acids for the equilibria:
protonated lidocaine + H₂O ⇆ lidocaine + H₃O⁺
protonated lidocaine lidocaine H₃O⁺
Initial(M) 0.22 0 0
Change -x +x +x
Equilibrium 0.22 - x x x
We know for this equilibrium
Ka = [Lidocaine] [H₃O⁺] / [protonaded Lidocaine] = x² / ( 0.22 - x )
The Ka can be calculated from the given pKb for lidocaine
Kb = antilog( - 7.94 ) = 1.15 x 10⁻⁸
Ka = Kw / Kb = 10⁻¹⁴ / 1.15 x 10⁻⁸ = 8.71 x 10⁻⁷
Since Ka is very small we can make the approximation 0.22 - x ≈ 0.22
and solve for x. The pH will then be the negative log of this value.
8.71 x 10⁻⁷ = x² / 0.22 ⇒ x = √(/ 8.71 X 10⁻⁷ x 0.22) = 4.38 x 10⁻⁴
( Indeed our approximation checks since 4.38 x 10⁻⁴ is just 0.2 % of 0.22 )
pH = - log ( 4.4x 10⁻⁴) = 3.36
The pH at the equivalence point of a titration of lidocaine with HBr is calculated using the pKb of lidocaine to find its pKa. The pH at equivalence is equal to the pKa of lidocaine, which is 6.06.
Explanation:To calculate the pH at the equivalence point in the titration of lidocaine with HBr, we use the pKb of lidocaine to find its pKa and then apply the Henderson-Hasselbalch equation. Since lidocaine is a weak base, at equivalence point it is completely neutralized by HBr, forming its conjugate acid. At this point, the concentration of the conjugate acid equals the original concentration of the base due to stoichiometry.
The pKa of lidocaine is calculated from its pKb using the formula pKa + pKb = pKw, where pKw is 14.00 at 25°C. Therefore, the pKa of lidocaine is 14.00 - 7.94 = 6.06.
At the equivalence point, the pH is equal to the pKa since the concentration of the conjugate acid (lidocaine) equals the concentration of the conjugate base (HBr). Consequently, the pH at the equivalence point is 6.06.
The student realizes that the precipitate was not completely dried and claims that as a result, the calculated Na2CO3 molarity is too low. Do you agree with the student’s claim? Justify your answer
Answer:I disagree, the molarity is meant to be higher if all other factors remain constant.
Explanation:
Molarity=amount of substance/volume of the liquid
So,the lesser,the volume, the higher the molarity and vice versa.
The molarity calculation of Na2CO3 could indeed be lower if the precipitate was not adequately dried. Excess moisture adds to the mass, misrepresenting the actual quantity of Na2CO3, which may lead to a lower calculated molarity.
Explanation:In the field of chemistry, the student's claim about the molarity of Na2CO3 (sodium carbonate) being inaccurately low due to incomplete drying of the precipitate can be valid. As the drying process helps in removing water molecules from the precipitate, any remaining moisture can add extra mass, causing the quantity of pure Na2CO3 to be undervalued. While calculating the molarity, this would imply more volume per mole, hence a lower molarity. This is a common issue in stoichiometry where proper handling and processing of chemical substances significantly affect the outcome of calculations.
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Calcium hydroxide, Ca(OH)2, is used as a calcium nutritional supplement in some foods and beverages, such as orange juice. What is the pH of a solution of 0.0012M calcium hydroxide at 25.0∘C?
Answer:
pH = 11.38
Explanation:
The pH of a solution is given as:
pH = pKw − pOH = 14.00 + log[OH-]
Since the concentration of Ca(OH)2 is 0.0012 M, the [OH−] is twice that, or 0.0024 M, resulting in pOH = 2.62 and pH = 11.38.
Final answer:
The pH of a solution of 0.0012M calcium hydroxide at 25.0°C is 11.38.
Explanation:
The pH of a solution of 0.0012M calcium hydroxide at 25.0°C can be calculated using the concentration of hydroxide ions ([OH-]).
Calcium hydroxide is a strong base that dissociates completely in water to form two hydroxide ions for every formula unit dissolved.
The concentration of the solute is 0.0012M, but because Ca(OH)2 is a strong base, the actual concentration of hydroxide ions ([OH-]) is two times this, or 2 × 0.0012M = 0.0024M.
Since [OH-] = 0.0024M, the pOH can be calculated as pOH = -log([OH-]) = -log(0.0024) = 2.62. The pH can then be calculated using the expression pH = 14 - pOH = 14 - 2.62 = 11.38.
The following reaction was followed by the method of initial rates: 5 Br-(aq) + BrO3-(aq) + H+(aq) → 3 Br2(aq) + 3 H2O(l) with the following results. Exp [Br− ]0 [BrO3− ] H+ (Δ[BrO3− ]/Δt)0 M·s−1 I 0.10 0.10 0.10 6.8 10-4 II 0.15 0.10 0.10 1.0 10-3 III 0.10 0.20 0.10 1.4 10-3 IV 0.10 0.10 0.25 4.3 10-3 Please see Determine Rate Laws for assistance. What is the rate law for the reaction? (Example input 'rate = k . [A]^2 . [B]'.)
Answer:
rate = k [Br⁻][H⁺]²[BrO₃⁻]
Explanation:
Here we are going to determine the rate law for the reaction
5 Br-(aq) + BrO3-(aq) + H+(aq) → 3 Br2(aq) + 3 H2O(l)
by performing experiments in which we vary concentration of the reactants and determining the effect this has on the initial rate.
Exp [ Br− ] [BrO3− ] [ H+] (Δ[BrO3− ]/Δt)0 M·s−1
I 0.10 0.10 0.10 6.8 x 10-4
II 0.15 0.10 1.0 10-3
III 0.10 0.20 0.10 1.4 x 10-3
IV 0.10 0.10 0.25 4.3 10-3
The way to do the comparison is by taking experiments in which we keep constant the concentration of two of the reactants and vary the third and study the effect this change has on the initial rate of the reaction.
Comparing experiments I and III we see that the initial reaction doubled when we doubled the [BrO3− ] while keeping the other two the same. Thus the reaction rate is of order 1 respect to [BrO3− ].
In experiments I and IV we increased the concentration of H⁺ by 2.5 times, and the rate of the reaction increased by a factor of 6.3 which is 2.5 squared . If you do not see it, lets try using logarithms
(0.25/ 0.10)^ x = 4.3 x 10⁻³ / 6.8 x 10⁻⁴
2.5 ^ x = 6.3235
x log 2.5 = log 6.3235
0.40 x = 0.80 ∴ x = 2
Therefore the rate is second order respect to [H⁺].
Comparing experiments I and II we see that increasing the concentration of Br⁻ by a factor of 1.5, the initial rate also went up by a factor of 1.5 ( 1.0 x 10⁻³ / 6.8 x 10⁻⁴ =1.5. Thus the rate is first order respect to [ Br⁻ ].
Then our rate law is
rate = k [Br⁻][H⁺]²[BrO₃⁻]
The rate law for the reaction 5 Br-(aq) + BrO3-(aq) + H+(aq) → 3 Br2(aq) + 3 H2O(l) is determined to be rate = k[Br-]^1[BrO3-]^1[H+]^0 by analyzing how the rate changes with varying initial concentrations of the reactants.
Explanation:To determine the rate law for the reaction: 5 Br-(aq) + BrO3-(aq) + H+(aq) → 3 Br2(aq) + 3 H2O(l), we need to examine how the rate changes with varying initial concentrations of the reactants. The change in rates as concentrations are altered provides the order of reaction with respect to each reactant. Our data from experiments I, II, and III indicate that when the Br- concentration increases by a factor of 1.5 (from 0.10 to 0.15), the rate also increases by a similar factor (from 6.8 x 10-4 to 1.0 x 10-3), suggesting a first-order dependence: [Br-]^1. However, when we double the initial concentration of BrO3 (experiment III vs I), the rate also doubles indicating first-order dependence on BrO3 as well: [BrO3]^1. The experiment IV suggests a zero-order dependence on [H+]. Hence, the rate law for this reaction is rate = k[Br-]^1[BrO3-]^1[H+]^0, where k is the rate constant.
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17. Of the four different substances, which of the following has the highest alkalinity?
Substance A pH = 1.0
Substance B pH = 3.0
Substance C pH = 7.0
Substance D pH = 12.0 *
Answer:
Explanation:
Substance A
When 102 g of water at a temperature of 22.6 °C is mixed with 66.9 g of water at an unknown temperature, the final temperature of the resulting mixture is 48.9 °C. What was the initial temperature of the second sample of water? (The specific heat capacity of liquid water is 4.184 J/g ⋅ K.) Initial temperature = °C
Answer:
[tex]T_{2,H_2O}=89^oC[/tex]
Explanation:
Hello,
In this case, we notice that the energy gained by the first sample of water is lost by the second sample of water as they are heated and cooled respectively, therefore, in terms of heats:
[tex]Q_{1,H_2O}=-Q_{2,H_2O}[/tex]
Which in terms of masses, heat capacities and temperatures is:
[tex]m_{1,H_2O}*Cp_{1,H_2O}*(T_{EQ}-T_{1,H_2O})=-m_{2,H_2O}*Cp_{2,H_2O}*(T_{EQ}-T_{2,H_2O})[/tex]
In such a way, as the heat capacity is the same and the initial temperature of the cold water is required, we solve for it via:
[tex]m_{1,H_2O}*(T_{EQ}-T_{1,H_2O})=-m_{2,H_2O}*(T_{EQ}-T_{2,H_2O})\\T_{EQ}-T_{2,H_2O}=\frac{m_{1,H_2O}*(T_{EQ}-T_{1,H_2O})}{-m_{2,H_2O}} \\\\T_{2,H_2O}=T_{EQ}+\frac{m_{1,H_2O}*(T_{EQ}-T_{1,H_2O})}{m_{2,H_2O}}[/tex]
With the given data we obtain:
[tex]T_{2,H_2O}=T_{EQ}+\frac{m_{1,H_2O}*(T_{EQ}-T_{1,H_2O})}{m_{2,H_2O}}\\\\T_{2,H_2O}=48.9^oC+\frac{102g*(48.9^oC-22.6^oC)}{66.9g} =48.9^oC+40.1^oC\\T_{2,H_2O}=89^oC[/tex]
Which means the second sample was hot.
Regards.
Answer:
The initial temperature of the second sample of water is 8.8 °C
Explanation:
Step 1 :Data given
Mass of water = 102 grams
Initial temperature of water = 22.6 °C =
Mass of other water = 66.9 grams
The final temperature is 48.9 °C
The specific heat capacity of liquid water is 4.184 J/g *K = 4.184 J/g°C
Step 2: Calculate the initial temperature of the second sample water
Heat gained = heat lost
Qgained = -Q lost
Q = m* C* ΔT
m(water1)*C(water)*ΔT(water1) = -m(water2) * C(water) *ΔT(water2)
⇒with m(water1) = the mass of the water at 22.6 °C = 102 grams
⇒with C(water) = the specific heat of water = 4.184 J/g°C
⇒with ΔT = the change of water of the first sample = 48.9 - 22.6 = 26.3 °C
⇒with m(water2) = the mass of the other unknown sample water = 66.9 grams
⇒with with C(water) = the specific heat of water = 4.184 J/g°C
⇒with ΔT = the change of water of the second sample = 48.9 - T1
102 * 4.184 * 26.3 °C = 66.9 * 4.184 * (48.9 - T1)
102 * 26.3 = 66.9 * (48.9 - T1)
2682.6 = 3271.4 - 66.9 T1
-588.8 = -66.9 T1
T1 = 8.8 °C
The initial temperature of the second sample of water is 8.8 °C
Which of the following statements is true? Only the first principle energy level can have s orbitals. Every principle energy level can have only one s orbital. Every principle energy level can have one s orbital, 3 p orbitals, 5 d orbitals, and 7 f orbitals. The fourth principle energy level is the first energy level that has d orbitals. Every principle energy level can have an s orbital and 3 p orbitals.
Answer:
Every principle energy level can have only one s orbital.
Explanation:
For many-electron atoms we use the Pauli exclusion principle to determine electron configurations. This principle states that no two electrons in an atom can have the same set of four quantum numbers. If two electrons in an atom should have the same Principal, Angular Momentum and Magnetic quantum numbers' values (that is, these two electrons are in the same atomic orbital), then they must have different values of Electron Spin Quantum Number. In other words, only two electrons may occupy the same atomic orbital, and these electrons must have opposite spins.
Answer:
B
Explanation:
Every principle energy level can have only one s orbital.
Aufbau principle works in the following formula:
1s2, 2s2 2p6, 3s2 3p6, 4s2 3d10 4p6, and so on.
Standard reduction potentials for zinc(II) and copper(II) The standard reduction potential for a substance indicates how readily that substance gains electrons relative to other substances at standard conditions. The more positive the reduction potential, the more easily the substance gains electrons. Consider the following: Zn2+(aq)+2e−→Zn(s),Cu2+(aq)+2e−→Cu(s), E∘red=−0.763 V E∘red=+0.337 V
The question is incomplete, the complete question is:
Standard reduction potentials for zinc(II) and copper(II)
The standard reduction potential for a substance indicates how readily that substance gains electrons relative to other substances at standard conditions. The more positive the reduction potential, the more easily the substance gains electrons. Consider the following:
Zn2+(aq)+2e−→Zn(s),Cu2+(aq)+2e−→Cu(s), E∘red=−0.763 V E∘red=+0.337 V
Part B
What is the standard potential, E∘cell, for this galvanic cell? Use the given standard reduction potentials in your calculation as appropriate.
Express your answer to three decimal places and include the appropriate units.
Answer:
1.100 V
Explanation:
E∘cell= E∘cathode - E∘anode
E∘cathode= +0.337 V
E∘anode= −0.763 V
E∘cell= 0.337-(-0.763)
E∘cell= 1.1V
Provide a stepwise synthesis of 1-cyclopentylethanamine using the Gabriel synthesis. Collapse question part Testbank, Question 065a Correct answer. Your answer is correct. Using the reagents below, list in order (by letter, no period) those necessary to provide this synthesis. a. Mg(OH)2 b. KOH c. 1-bromo-1-cyclohexylethane d. 1-bromo-1-cyclopentylethane e. NH3 f. xs LiAlH4 g. H3O , then OH- Step 1 Step 2 Step 3 Entry field with correct answer Entry field with correct answer Entry field with correct answer
Answer:
Azide synthesis is the first method on the table of synthesis of primary amines. The Lewis structure of the azide ion, N3−, is as shown below.
an azide ion
An “imide” is a compound in which an N−−H group is attached to two carbonyl groups; that is,
imide linkage
You should note the commonly used trivial names of the following compounds.
phthalic acid, phthalic anhydride, and phthalimide
The phthalimide alkylation mentioned in the reading is also known as the Gabriel synthesis.
If necessary, review the reduction of nitriles (Section 20.7) and the reduction of amides (Section 21.7).
Before you read the section on reductive amination you may wish to remind yourself of the structure of an imine (see Section 19.8).
The Hofmann rearrangement is usually called the Hofmann degradation. In a true rearrangement reaction, no atoms are lost or gained; however, in this particular reaction one atom of carbon and one atom of oxygen are lost from the amide starting material, thus the term “rearrangement” is not really appropriate. There is a rearrangement step in the overall degradation process, however: this is the step in which the alkyl group of the acyl nitrene migrates from carbon to nitrogen to produce an isocyanate.
Explanation:
A salt bridge: Question 3 options: provide a pathway through which the electrons travel from the cathode to the anode. provide a pathway through which the electrons travel from the anode to the cathode. provide a source of counterions to prevent the build-up of charge at both the cathode to the anode. provide a physical connection between the two cells, allowing the solutions in both cells to slowly mix.
Answer:
The correct option is: provide a source of counterions to prevent the build-up of charge at both the cathode to the anode.
Explanation:
A salt bridge is a U-shaped glass tube that is used in a voltaic cell or galvanic cell to connect the oxidation and reduction half-cells and complete the electric circuit.
It allows the ions to pass through it, thus preventing the accumulation of charge on the anode and cathode as the chemical reaction proceeds.
Therefore, the correct option is: provide a source of counterions to prevent the build-up of charge at both the cathode to the anode.
Write the names of the following ionic compounds: 1) NaOH _____________________________________________________ 2) Co 3 P 2 _____________________________________________________ 3) Pb(CO 3 ) 2 _____________________________________________________ 4) MgF 2 _____________________________________________________ 5) Li 2 SO 3 _____________________________________________________ 6) (NH 4 ) 3 PO 4 _____________________________________________________ 7) FeO _____________________________________________________ 8) CaSO 4 _____________________________________________________ 9) Ag 3 N _____________________________________________________ 10) Na 2 S
Answer:
1. NaOH is sodium hydroxide
2. Co3P2 is cobalt (II) phosphide
3. Pb(CO3)2 is lead(iv) carbonate or lead(iv) trioxocarbonate(iv)
4. MgF2 is magnesium fluoride
5. Li2SO3 is lithium sulphite
6. (NH4)3PO4 is ammonium phosphate.
7. FeO is iron(II) oxide
8. CaSO4 is calsium sulphate
9. Ag3N is called silver nitride
10. Na2S is sodium sulfide
Explanation:
Step 1:
Data obtained from the question. This includes:
NaOH
Co3P2
Pb(CO3)2
MgF2
Li2SO3
(NH4)3PO4
FeO
CaSO4
Ag3N
Na2S
Step 2:
Namig the compound
1. NaOH.
NaOH is a binary ionic compound containing sodium Na and hydroxyl OH. It therefore a binary compound (i.e it contains two elements). Binary compounds end with - ide.
NaOH is called sodium hydroxide
2. Co3P2. In this case we must determine the oxidation state of Co. This is illustrated below:
Co3P2 = 0
3Co + 2P = 0
3Co + (2x-3) = 0
3Co - 6 = 0
Collect like terms
3Co = 6
Divide both side by 3
Co = 6/3
Co = +2
Co3P2 contains cobalt Co and phosphorus P. It is binary compound and the oxidation state of Co is +2 in the compound. Therefore, the name of Co3P2 is cobalt (II) phosphide
3. Pb(CO3)2. In this case, we must determine the oxidation state of Pb. This is illustrated below:
Pb(CO3)2 = 0
Pb + 2 [ 4 + (-2x3)] = 0
Pb + 2[ 4 - 6] = 0
Pb + 2[-2] = 0
Pb - 4 = 0
Pb = +4
The name of the compound is lead(iv) carbonate or lead(iv) trioxocarbonate (iv)
4. MgF2 is a binary compound containing magnesium Mg and fluorine F. The name therefore is magnesium fluoride
Note: oxidation number of the group 1, 2 and 3 metals are not indicated in their names since their oxidation number is constant.
5. Li2SO3 is a binary ionic compound containing lithium and sulphite. Therefore, the name is lithium sulphite
6. (NH4)3PO4 contain ammonium NH4 and the phosphate. Therefore the name is ammonium phosphate
7. FeO. This is a binary compound, but we must determine the oxidation state of Fe. This is illustrated below:
FeO = 0
Fe + (-2) = 0
Fe - 2 = 0
Fe = +2
Therefore, the name of FeO is iron(II) oxide
8. CaSO4 is binary ionic compound containing calsium and sulphate. It is name as calsium sulphate
9. Ag3N is a binary compound containing silver and nitrogen. It is therefore called silver nitride
10. Na2S is a binary compound containing sodium and sulphur. It is named as sodium sulfide
The names of the ionic compounds are Sodium hydroxide (NaOH), Cobalt (III) phosphide (Co3P2), Lead (II) carbonate (Pb (CO3)2), Magnesium fluoride (MgF2), Lithium sulfite (Li2SO3), Ammonium phosphate ((NH4)3PO4), Iron (II) oxide (FeO), Calcium sulfate (CaSO4), Silver nitride (Ag3N), Sodium sulfide (Na2S)
Explanation:The names of the given ionic compounds are:
Sodium hydroxide (NaOH)Cobalt(III) phosphide (Co3P2)Lead(II) carbonate (Pb(CO3)2)Magnesium fluoride (MgF2)Lithium sulfite (Li2SO3)Ammonium phosphate ((NH4)3PO4)Iron(II) oxide (FeO)Calcium sulfate (CaSO4)Silver nitride (Ag3N)Sodium sulfide (Na2S)Learn more about Naming ionic compounds here:https://brainly.com/question/32037662
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Phosphorous acid, H 3 PO 3 ( aq ) , is a diprotic oxyacid that is an important compound in industry and agriculture. p K a1 p K a2 1.30 6.70 Calculate the pH for each of the points in the titration of 50.0 mL of 1.5 M H 3 PO 3 ( aq ) with 1.5 M KOH ( aq ) . A molecule of phosphorous acid. A central phosphorus atom is single bonded to a hydrogen atom and two O H groups. An oxygen atom is also double bonded to the phosphorus atom.
a. before addition of any KOH :
b. after addition of 25.0 mL KOH :
c. after addition of 50.0 mL KOH :
d. after addition of 75.0 mL KOH :
e. after addition of 100.0 mL KOH :
Answer:
a. 0.60
b. 1.30
c. 4.00
d. 6.70
e. 10.20
Explanation:
The 3 attached files shows a comprehensive solution
to the problems with answers highlighted as above
8. Compare the rates of effusion for hydrogen and oxygen gases.
When compared to oxygen gas, hydrogen gas emits approximately 2.82 times faster.
Rate of effusion refers to the speed at which a gas escapes or diffuses through a small opening or porous membrane into a vacuum or another gas. It is a measure of how quickly gas molecules can move and pass through a barrier.
The rate of effusion for a gas is determined by its molar mass. According to Graham's law of effusion, the rate of effusion of a gas is inversely proportional to the square root of its molar mass.
The molar mass of hydrogen is 2.02 g/mol, while the molar mass of oxygen is 32.00 g/mol. Therefore, the molar mass of oxygen is significantly larger than that of hydrogen.
Using Graham's law, we can calculate the ratio of the rates of effusion for hydrogen and oxygen:
Rate of effusion of hydrogen / Rate of effusion of oxygen = √ (Molar mass of oxygen / Molar mass of hydrogen)
Rate of effusion of hydrogen / Rate of effusion of oxygen = √ (32.00 g/mol / 2.02 g/mol)
= 2.82
Therefore, on comparing hydrogen gas effuses approximately 2.82 times faster than oxygen gas.
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Final answer:
Hydrogen gas effuses approximately 4 times faster than oxygen gas according to Graham's law of effusion.
Explanation:
According to Graham's law of effusion, the rate of effusion of a gas is inversely proportional to the square root of its molar mass. Since hydrogen has a smaller molar mass than oxygen, it will effuse at a faster rate.
For example, if we consider the rate of effusion of hydrogen to be 1, then the rate of effusion of oxygen would be √((molar mass of hydrogen)/(molar mass of oxygen)).
Therefore, hydrogen gas effuses at a rate that is approximately 4 times faster than oxygen gas.
Which of the following solutions is a good buffer system?Which of the following solutions is a good buffer system?A solution that is 0.10 M Li OH and 0.10 MHNO3A solution that is 0.10 MHCN and 0.10 MLiCNA solution that is 0.10 MNaCl and 0.10 MHClA solution that is 0.10 M Li and 0.10 MHNO3A solution that is 0.10 MHCN and 0.10 M Li Cl
Answer:
A solution that is 0.10 M HCN and 0.10 M LiCN.
Explanation:
A buffer solution is a solution of a weak acid and its conjugate base or a solution of a weak base and its conjugate acid, so the pH of the solution changes in a very little range when an acid or base is added to the solution.
Let's evaluate each statement:
(a) A solution that is 0.10 M LiOH and 0.10 M HNO₃
This is not a buffer solution since the HNO₃ is not the acid conjugate of the LiOH, and also, the LiOH is a strong base and the HNO₃ is a strong acid.
(b) A solution that is 0.10 M HCN and 0.10 M LiCN
This is a buffer solution, with the following reaction:
HCN + H₂O ⇄ CN⁻Li⁺ + H₃O⁺
The acid HCN is a weak acid and the LiCN is its conjugate base. The fact that the concentrations are equal for both is appropriate for the buffer solution.
(c) A solution that is 0.10 M NaCl and 0.10 M HCl
This is not a buffer solution since the HCl is a strong acid. It dissociates in water to form Cl⁻ and H⁺.
(d) A solution that is 0.10 M Li and 0.10 M HNO₃
This is not a buffer solution since the HNO₃ is not the conjugate base of Li, the HNO₃ is a strong acid that dissociates in water to form H⁺ and NO₃⁻.
(e) A solution that is 0.10 M HCN and 0.10 M LiCl
This is not a buffer solution since the LiCl is not the conjugate base of the acid HCN.
Therefore, the correct option is: A solution that is 0.10 M HCN and 0.10 M LiCN.
I hope it helps you!
You have studied the gas-phase oxidation of HBr by O2: 4 HBr(g) + O2(g) → 2 H2O(g) + 2 Br2(g) You find the reaction to be first order with respect to HBr and first order with respect to O2. You propose the following mechanism: HBr(g) + O2(g) → HOOBr(g) HOOBr(g) + HBr(g) → 2 HOBr(g) HOBr(g) + HBr(g) → H2O(g) + Br2(g) a. Confirm that the elementary reactions add to give the overall reaction. (Hint: Use Hess Law) b. Based on the experimentally determined rate law, which step is rate determining? c. What are the intermediates in this mechanism? d. If you are unable to detect HOBr or HOOBr among the products, does this disprove your mechanism?
In the oxidation of HBr by O2, the elementary reactions add up to give the overall reaction and thus confirm Hess's Law. The rate determining step is the first one, and the intermediates in the reaction are HOOBr and HOBr. The inability to detect these among the final products does not disprove this mechanism.
Explanation:The overall reaction for the gas-phase oxidation of HBr by O2 is written as: 4 HBr(g) + O2(g) → 2 H2O(g) + 2 Br2(g)
First, we need to confirm if the elementary reactions add up to the overall reaction. Based on Hess's Law, we can see that if we add the elementary reactions: HBr(g) + O2(g) → HOOBr(g), HOOBr(g) + HBr(g) → 2 HOBr(g), and HOBr(g) + HBr(g) → H2O(g) + Br2(g), they give the same overall reaction, thus confirming Hess's Law.
Second, the experimentally determined rate law is first order with respect to both HBr and O2, which suggests the rate determining step is the first one: HBr(g) + O2(g) → HOOBr(g). This is because the rate-determining step usually determines the order of the reaction.
The intermediates in this reaction are the species that are produced in one step and consumed in another, in this case HOOBr(g) and HOBr(g).
Lastly, the inability to detect HOBr or HOOBr among the final products does not necessarily disprove the mechanism. This is because these are intermediates and are typically used up in subsequent reaction steps, leading them to not appear in the final product of the reaction.
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The given elementary steps add up to the overall reaction. The rate-determining step involves HBr and O₂ as indicated by the rate law. HOBr and HOOBr are intermediates, and their absence among products does not disprove the mechanism.
let's analyze and confirm each part.
a. Confirming the Elementary Reactions:
The given elementary steps are:
HBr(g) + O₂(g) → HOOBr(g)HOOBr(g) + HBr(g) → 2 HOBr(g)HOBr(g) + HBr(g) → H₂O(g) + Br₂(g)Adding these reactions together:
1st Step: HBr(g) + O₂(g) → HOOBr(g)2nd Step: HOOBr(g) + HBr(g) → 2 HOBr(g)3rd Step: HOBr(g) + HBr(g) → H₂O(g) + Br₂(g)Combining the species on both sides gives us:
4 HBr(g) + O₂(g) → 2 H₂O(g) + 2 Br₂(g)
This confirms that the elementary steps add up to the overall reaction.
b. Rate-Determining Step:
The given rate law is first order with respect to both HBr and O₂, indicating that the rate-determining step involves one molecule each of HBr and O₂. Hence, the first step, HBr(g) + O₂(g) → HOOBr(g), is the rate-determining step.
c. Intermediates:
Intermediates are species that appear in the reaction mechanism but not in the overall reaction. Here, HOOBr and HOBr are intermediates.
d. Detecting Intermediates:
Not detecting HOBr or HOOBr among the products does not disprove the mechanism. Intermediates typically have short lifespans and do not accumulate significantly; hence they might not be detected in significant amounts in the product mixture.