Answer: New concentration of [tex]HC_{6}H_{5}O^{2-}_{7}[/tex] is 0.23 M.
Explanation:
The given data is as follows.
Moles of [tex]HC_{6}H_{5}O^{2-}_{7}[/tex] = 0.06 mol
Moles of [tex]H_{2}C_{6}H_{5}O_{7}[/tex] = 0.08 mol
Therefore, moles of [tex]OH^{-}[/tex] added are as follows.
Moles of [tex]OH^{-}[/tex] = [tex]0.125 \times \frac{25}{1000}[/tex]
= 0.003125 mol
Now, new moles of [tex]HC_{6}H_{5}O^{2-}_{7}[/tex] = 0.06 + 0.003125
= 0.063125
Therefore, new concentration of [tex]HC_{6}H_{5}O^{2-}_{7}[/tex] will be calculated as follows.
Concentration = [tex]\frac{0.063125}{0.275}[/tex]
= 0.23 M
Thus, we can conclude that new concentration of [tex]HC_{6}H_{5}O^{2-}_{7}[/tex] is 0.23 M.
This reaction involves a conjugate addition reaction followed by an intramolecular Claisen condensation. The steps involved are as follows: 1. Conjugate addition of methyl carbanion (from the Gilman reagent) to form enolate ion 1; 2. Cyclization to form tetrahedral intermediate 2; 3. Collapse of the tetrahedral intermediate and expulsion of methoxide ion completes the reaction to form the final product. Write out the reaction on a separate sheet of paper, and then draw the structure of tetrahedral intermediate 2.
Answer:
Explanation:
Check below for the answer in the attachment.
Two friends, Pamela and Elaine, compared snack items they planned to enjoy during a work break. Pamela's snack was a small bag of shortbread cookies composed of 2.6 grams fat, 21 grams carbohydrates, and 2.1 grams protein. Elaine had a bag of popcorn composed of 2.9 grams fat, 31 grams carbohydrates, and 3.5 grams protein.
Nutrient Fuel value (kJ/g):Carbohydrate 17, Fat 38, Carbs 17
(a) Using the fuel values provided, calculate the number of calories in each person's snack. Pamela's snack Cal Elaine's snack Cal
(b) Suppose the friends took a 15-minute walk after snacking. If Pamela burned 43 calories and Elaine expended 48 calories, who had the smaller net gain of calories after snacking and walking?
Answer:
a) Pamela's shortbread cookies has 117,471.3 calories = 117.5 kcal
Elaine's popcorn has 166,515.3 calories = 166.5 kcal
b) Pamela had the smaller net gain of calories after snacking and walking.
Explanation:
- Pamela's snack was a small bag of shortbread cookies composed of 2.6 grams fat, 21 grams carbohydrates, and 2.1 grams protein.
- Elaine had a bag of popcorn composed of 2.9 grams fat, 31 grams carbohydrates, and 3.5 grams protein.
Nutrient | Fuel value (kJ/g):
Carbohydrate | 17
Fat | 38
Protein | 17
Noting that instead of Carbs, Protein was intended to be written.
a) For Pamela
Carbohydrates = (21 g) × (17 KJ/g) = 357 KJ
Fats = (2.6 g) × (38 KJ/g) = 98.8 KJ
Protein = (2.1 g) × (17 KJ/g) = 35.7 KJ
Total fuel value = 357 + 98.8 + 35.7 = 491.5 KJ = 491,500 J
Note that 1 cal = 4.184 J
491,500 J = (491,500/4.184) = 117,471.3 calories = 117.5 kcal
Carbohydrates = (31 g) × (17 KJ/g) = 527 KJ
Fats = (2.9 g) × (38 KJ/g) = 110.2 KJ
Protein = (3.5 g) × (17 KJ/g) = 59.5 KJ
Total fuel value = 527 + 110.2 + 59.5 = 696.7 KJ
Note that 1 cal = 4.184 J
696,700 J = (696,700/4.184) = 166,515.3 calories = 166.5 kcal
b) Suppose the friends took a 15-minute walk after snacking. If Pamela burned 43 kcal and Elaine expended 48 kcal, who had the smaller net gain of calories after snacking and walking?
Pamela gains 117.5 kcal
And loses 43 kcal
Net gain = 117.5 - 43 = 74.5 kcal
Elaine gains 166.5 kcal
And loses 48 kcal
Net gain = 166.5 - 48 = 118.5 kcal
Pamela had the smaller net gain of calories after snacking and walking.
Hope this Helps!!!
affects of cholera toxin on adenylyl cyclase the gram negative bacterium vibrio cholerae produces a protein cholera toxin that is respondible for th e characteristic symptoms of cholera. if body fluids and Na are not replaced, severe dehydration results. what is the effect of cholera toxin on cAMP in the intestinal cells
Answer: the effects of cholera toxin on cAMP in the intestinal cells is that it INCREASES cAMP production.
Explanation:
Vibrio cholerae is a gram negative bacterium which produces a protein cholera toxin that is responsible for the characteristic symptoms of cholera such as
watery diarrhea, vomiting, rapid heart rate, loss of skin elasticity, low blood pressure, thirst, and muscle cramps. There is need for body fluids and Na replacement to avoid severe dehydration results which may lead to death.
Cyclic adenosine monophosphate( cAMP) is a derivative used for intracellular signal transduction in organisms. The cholera toxin produced by the bacteria INCREASES the production of cAMP through its polypeptides( which consist of active protomer and binding protomer). The cholera toxin first binds to cell surface receptors, the protomer then enters the cell and bind with and activate the adenylate cyclase. Increasing adenylate cyclase activity will INCREASE cellular levels of cAMP, increasing the activity of ion pumps that remove ions from the cell. Due to osmotic pressure changes, water also must flow with the ions into the lumen of the intestinal mucosa, dehydrating the tissue. I hope this helps, thanks.
CaO(s) + H2O(l) - Ca(OH)2(s)
enthalpy of rxn= -63.7 kJ/molrxn
Calcium oxide, CaO(s), has been proposed as a substance that can be used to heat water quickly for portable heating packs or for cooking. When placed in water, Cao(s) reacts as shown by the equation above.
A student wants to design a heating pad that could heat a 150.0 g sample of water from 25.0°C to 60.0°C.
Calculate the amount of heat, in joules, that the water must absorb for its
temperature to change by this amount. (Assume that the specific heat capacity
of the water is 4.18 J/gK).
Answer:
21,976 J
Explanation:
In order to increase the temperature of a certain amount of a substance by [tex]\Delta T[/tex], the amount of heat that must be supplied to the substance must be:
[tex]Q=mC\Delta T[/tex]
where
m is the mass of the substance
C is the specific heat capacity of the substance
[tex]\Delta T[/tex] is the increase in temperature
For the sample of water in this problem we have:
[tex]m=150.0 g[/tex] is the mass
[tex]C=4.186 J/g^{\circ}C[/tex] is the specific heat capacity of water
[tex]\Delta T=60.0-25.0=35.0^{\circ}C[/tex] is the increase in temperature
Therefore, the amount of heat that must be supplied is
[tex]Q=(150.0)(4.186)(60-0-25.0)=21,976 J[/tex]
To heat 150.0 g of water from 25.0°C to 60.0°C, approximately 21885 joules of heat must be absorbed, calculated using the formula q = mcΔT with the values m = 150.0 g, c = 4.18 J/gK, and ΔT = 35.0°C.
Explanation:The amount of heat required to raise the temperature of 150.0 g of water from 25.0°C to 60.0°C can be calculated using the formula q = mcΔT, where q is the heat absorbed, m is the mass of the water, c is the specific heat capacity of water, and ΔT is the change in temperature.
Using the given values, the mass m = 150.0 g, specific heat capacity c = 4.18 J/gK, and the temperature change ΔT = (60.0 - 25.0)°C = 35.0°C, we get:
q = (150.0 g)(4.18 J/gK)(35.0 K)
q = 150.0 g * 4.18 J/g°C * 35.0°C
q = 21885 J
Therefore, the water must absorb approximately 21885 joules of heat for the temperature to increase from 25.0°C to 60.0°C.
During your reaction, you added 0.3 mL concentrated H2SO4 (18.4 M) as the catalyst. At the end of the reaction, you need to add base to neutralize it. How much volume (in mL) of 10% Na2CO3 to neutralize all the acid present
Answer:
58.72 mL
Explanation:
The chemical equation for the neutralization reaction is :
H₂SO₄(aq) + Na₂CO₃(s) --------------> Na₂SO₄(aq) + H₂O(l) + CO₂(g)
where;
M₁ = Molarity of H₂SO₄
M₂= Molarity of Na₂CO₃
V₁= Volume of H₂SO₄
V₂ = Volume of Na₂CO₃
Given that :
M₁ = 18.4 M
V₁= 0.3 mL
10% Na₂CO₃ means 100 g of solution contain 10 g of Na₂CO₃
i.e. 10 g Na₂CO₃ dissolved and diluted to 100 mL water.
Molar mass of Na₂CO₃ = 106 g/mol
106 g Na₂CO₃ dissolved in 100 mL will give 0.1 M Na₂CO₃ solution.
However;
If, 106 g Na₂CO₃ ≡ 0.1 M Na₂CO₃
Then, 10 g Na₂CO₃ ≡ 'A' M of Na₂CO₃
By cross multiplying; we have:
106 × A = 10 × 0.1
106 × A = 1
A = (1/106) M/100 mL
A = 10 x (1/106)) M/L
A = (10/106) M
A = 0.094 M
Therefore,the molarity of 10% Na₂CO₃ solution is 0.094 M.
For the Neutralization equation, we have:
M₁V₁ = M₂V₂
18.4×0.3 = 0.094×V₂
Making V₂ the subject of the formula;we have:
[tex]V_2 = \frac{18.4*0.3}{0.094}[/tex]
V₂ = 58.72 mL
When electrodes are used to record the electrocardiogram, an electrolyte gel is usually put between them and the surface of the skin. This makes it possible for the metal of the electrode to form metallic ions that move into the electrolyte gel. Often, after prolonged use, this electrolyte gel begins to dry out and change the characteristic of the electrodes. Draw an equivalent circuit for the electrode while the electrolyte gel is fresh. Then discuss and illustrate the way you expect this equivalent circuit to change as the electrolyte gel dries out. In the extreme case where there is no electrolyte gel left, what does the equivalent circuit of the electrode look like
Answer:
The equivalent circuit for the electrode while the electrolyte gel is fresh
From the uploaded diagram the part A is the electrolyte, the part part B is the electrolyte gel when is fresh and the part C is the surface of the skin
Now as the electrolyte gel start to dry out the resistance [tex]R_s[/tex] of the gel begins to increase and this starts to limit the flow of current . Now when the gel is then completely dried out the resistance of the gel [tex]R_s[/tex] then increases to infinity and this in turn cut off flow of current.
The diagram illustrating this is shown on the second uploaded image
Explanation:
Ethylenediamine is a bidentate ligand. The oxalate group, used in this experiment was also a didentate ligand. The structure of ethylenediamine is NH2CH2CH2NH2. The oxalate group formed a coordinated compound using the negative charges on the oxygen. Explain how the ethylenediamine compound will bond.
Answer:
Ethylene diamine will bond to the Central metal via a lone pairs of electrons on nitrogen
Explanation:
Complexes are formed by coordinate bond formation. Before a coordinate bond is formed, one of the species must have a lone lair of electrons available for donation into empty orbitals on the central metal.
Ethylene diammine contains nitrogen which has a lone pair of electrons. The two lone pairs on the two nitrogen atoms can bond with the central metal. This makes ethylene diammine a bidentate ligand (two bonding atoms).
Ethylenediamine will bond to a metal ion through both of its nitrogen atoms, forming a chelate complex.
Ethylenediamine (en) is a bidentate ligand, which means it has two donor atoms that can simultaneously bond to a central metal ion. In the case of ethylenediamine, the donor atoms are the nitrogen atoms, each of which has a lone pair of electrons available for donation. When ethylenediamine bonds to a metal ion, it does so through both nitrogen atoms, creating a five-membered ring structure known as a chelate. The term chelate comes from the Greek word chele, which means claw. This chelation results in a more stable complex because the metal ion is bound to two sites on the ligand, reducing the likelihood of the ligand dissociating from the metal center.The formation of a chelate complex with ethylenediamine can be represented as follows: [tex]\[ \text{M}^{n+} + \text{en} \rightarrow \text{M(en)}^{n+} \][/tex] Here, M represents the metal ion, n+ is its oxidation state, and en is the ethylenediamine ligand. The resulting complex M(en)n+ has the metal ion bound within the chelate ring formed by the ethylenediamine ligand. In contrast, the oxalate group is also a bidentate ligand but with two oxygen atoms as the donor atoms. The oxalate group typically bonds to a metal ion through its negatively charged oxygen atoms, forming a similar chelate complex but with a different donor atom and potentially different coordination geometry. The stability of chelate complexes is often discussed in terms of the chelate effect, which states that metal complexes with chelating ligands are more stable than those with similar non-chelating (monodentate) ligands. This increased stability is due to the entropy effect, where the formation of a chelate ring results in the release of fewer solvent molecules compared to the formation of separate bonds with monodentate ligands.
In the following reaction HF(aq) + HPO42-(aq) = F-(aq) + H2PO4-(aq)
a. HPO42- is an acid and H2PO4- is its conjugate base.
b. H2PO4- is an acid and F- is its conjugate base.
c. HPO42- is an acid and HF is its conjugate base.
d. HF is an acid and F- is its conjugate base.
e. HF is an acid and HPO42- is its conjugate base.
d. HF is an acid and [tex]F^-[/tex] is its conjugate base.
What is acid/ base and its conjugate acid/base?
Whenever an acid donates a proton, the acid changes into a base, and whenever a base accepts a proton, an acid is formed. An acid and a base which differ only by the presence or absence of a proton are called a conjugate acid-base pair.For example: HCl is an acid and its conjugate base is [tex]Cl^-[/tex].Given chemical reaction:
[tex]HF(aq) + HPO_4^{2-}(aq)----> F^-(aq) + H_2PO_4^-(aq)[/tex]
In this reaction, HF is an acid and its conjugate base is [tex]F^-[/tex].
Thus out of all the options; the correct option is d.
Find more information about "Acid" here:
/brainly.com/question/22514615
In the reaction HF(aq) + HPO4^2-(aq) = F^-(aq) + H2PO4^-(aq), HF is the acid and F^- is the conjugate base, making the correct answer (d).
Explanation:In the given reaction HF(aq) + HPO42-(aq) = F-(aq) + H2PO4-(aq), we need to identify the correct acid-base pairs. When looking at this reaction, HF donates a proton (H+) to form its conjugate base F-, which makes HF the acid. On the other hand, since HPO42- receives a proton (H+) to become H2PO4-, HPO42- clearly acts as the base in this reaction and H2PO4- is its conjugate acid, which shows that it is amphoteric.
Based on this explanation, the correct answer is (d) HF is an acid and F- is its conjugate base.
Learn more about Acid-Base Reactions here:https://brainly.com/question/15209937
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You have a 25.0 L cylinder of helium at a pressure of 132 atm and a temperature of 19 [infinity]C. The He is used to fill balloons to a volume of 2.50 L at 732 mm Hg and 27 [infinity]C. How many balloons can be filled with He? Assume that the cylinder can provide He until its internal pressure reaches 1.00 atm (i.e., there are 131 atmospheres of usable He in the cylinder).
Answer: Number of balloons that can be filled with He are 1397.
Explanation:
The given data is as follows.
V = 25 L He , P = 131 atm
T = [tex]19^{o}C[/tex] = (19 + 273) K
= 292 K
According to ideal gas equation,
PV = nRT
where, n = [tex]\frac{PV}{RT}[/tex]
= [tex]\frac{131 \times 25}{0.082 \times 292}[/tex]
= 136.76 mol of He
The data for small balloons is given as follows.
T = [tex]27^{o}C[/tex] = (27 + 273) K
= 300 K
P = 732 = 0.963 atm
V = 2.50 L
Now, we will calculate the number of moles as follows.
n = [tex]\frac{PV}{RT}[/tex]
= [tex]\frac{0.963 \times 2.50}{0.082 \times 300}[/tex]
= 0.098 mol
So, number of balloons that can be filled with He are calculated as follows.
n = [tex]\frac{N_{1}}{N_{2}}[/tex]
= [tex]\frac{136.76}{0.098}[/tex]
= 1397.60 balloons
or, = 1397 balloons (approx)
Thus, we can conclude that number of balloons that can be filled with He are 1397.
Microwave ovens work by irradiating food with microwaves, which are absorbed by the water molecules in the food and converted to heat. Assuming that microwave ovens emit radiation with a wavelength of 29.7 cm, calculate how many photons are emitted per second in a 600-Watt microwave oven. (1 W = 1 J/s)
Answer:
8.969×10²⁶ photons
Explanation:
From the question,
Using,
E = hc/λ........................... Equation 1
Where E = Energy of the photon, h = Planck's constant, c = speed of light, λ = wave length of the photon.
Given: h = 6.626×10⁻³⁴ J.s, c = 3×10⁸ m/s, λ = 0.297 m
Substitute into equation 1
E = 6.626×10⁻³⁴(3×10⁸)/0.297
E = 6.69×10⁻²⁵ J.
The energy of the photon in one seconds = 6.69×10⁻²⁵ J/s
If the power of the microwave oven = 600 J/s
Then,
Number of photons emitted per seconds = 600/(6.69×10⁻²⁵)
Number of photons emitted per seconds = 8.969×10²⁶ photons
In a redox reaction, oxidation is defined by the:
1.gain of electrons, resulting in an increased oxidation number.
2.loss of electrons, resulting in a decreased oxidation number.
3.gain of electrons, resulting in a decreased oxidation number.
4.loss of electrons, resulting in an increased oxidation number.
Answer:
Option 4. loss of electrons, resulting in an increased oxidation number.
Explanation:
Oxidation is a process involving loss of electron(s). When this happens the oxidation number of the atom being oxidised increases. This can be seen when calcium (Ca) reacts with chlorine (Cl2) to form calcium chloride (CaCl2) according to the equation given below:
Ca + Cl2 —> CaCl2
The oxidation number of calcium increases from 0 to +2. This implies that calcium is being oxidised as it loses its electrons. The oxidation number of chlorine decreases from 0 to - 1 as it gains electron.
Now, we can see that the oxidation of calcium i.e lose of electrons increased its oxidation number from 0 to +2.
From the simple illustrations above, we can see clearly that oxidation involves loss of electrons, resulting in an increased oxidation number.
18.67 Consider the reaction Given that DG8 for the reaction at 258C is 173.4 kJ/mol, (a) calculate the standard free energy of formation of NO, and (b) calculate KP of the reaction. (c) One of the starting substances in smog formation is NO. Assuming that the temperature in a running automobile engine is 11008C, estimate KP for the above reaction. (d) As farmers know, lightning helps to produce a better crop. Why
Answer:
a. 86.7KJ/mol
b. 2.5 * 10^30
c. 1.42 * 10^55
d. Lightning produces more amount of NO from nitrogen and oxygen thus giving a better crop as NO is essential for crop growth
Explanation:
Complete question is as follows;
Consider the reaction N2(g)+ O2(g) —-> 2NO(g)
Given that ΔG0 for the reaction at 25 degrees celsius is 173.4 kJ/mol a) calculate the standard freeenergy of formation of NO and B) calculate KP ofthe reaction. C) One of the starting substances in smog formationis NO. Assuming that the temperature in a running automobile engine is 1100 degreesC, estimate KP of the above reaction.D) As farmers know, lighting helps to produce a better crop.Why?
solution
Please check attachment for complete solution and step by step explanation
A chemistry student weighs out 0.120 g of acetic acid into a 250. mL volumetric flask and dilutes to the mark with distilled water. He plans to titrate the acid with 0.0700 M NaOH solution. Calculate the volume of solution the student will need to add to reach the equivalence point. Be sure your answer has the correct number of significant digits.
Answer:
28.6 mL
Explanation:
Step 1: Write the balanced neutralization reaction between acetic acid and sodium hydroxide
CH₃COOH(aq) + NaOH(aq) = CH₃COONa(aq) + H₂O(l)
Step 2: Calculate the moles of acetic acid
The molar mass of acetic acid is 60.05 g/mol. The moles corresponding to 0.120 g are:
[tex]0.120g \times \frac{1 mol}{60.05g} =2.00 \times 10^{-3} mol[/tex]
Step 3: Calculate the moles of sodium hydroxide
The molar ratio of CH₃COOH to NaOH is 1:1. The reacting moles of NaOH are 2.00 × 10⁻³ mol.
Step 4: Calculate the volume of the 0.0700 M NaOH solution
[tex]2.00 \times 10^{-3} mol \times \frac{1L}{0.0700mol} =0.0286 L = 28.6 mL[/tex]
Answer:
We have to add 286 mL of NaOH
Explanation:
Step 1: Data given
Mass of acetic acid (CH3COOH)= 0.120 grams
Volume of acetic acid = 250 mL = 0.250 L
Molarity of NaOH = 0.0700 M
Step 2: The balanced equation
CH3COOH + NaOH → CH3COONa + H2O
Step 3: Calculate moles acetic acid
Moles acetic acid = mass / molar mass
Moles acetic acid = 0.120 grams / 60.05 g/mol
Moles acetic acid = 0.00200 moles
Step 4: Calculate molarity acetic acid
Molarity acetic acid = moles / volume
Molarity acetic acid = 0.00200 moles /0.250 L
Molarity acetic acid = 0.008 M
Step 5: Calculate the volume of solution the student will need to add to reach the equivalence point
C1*V1 = C2*V2
⇒with C1 = the molarity of acetic acid = 0.008 M
⇒with V1 = the volume of acetic acid = 0.250 L
⇒with C2 = the molarity of NaOH = 0.0700 M
⇒with V2 = the volume of NaOH neede = TO BE DETERMINED
0.008 M * 0.250 L = 0.0700 M * V2
V2 = (0.008M * 0.250 L) / 0.0700 M
V2 = 0.286 L = 286 mL
We have to add 286 mL of NaOH
Consider the three ligand field spectra corresponding to octahedral complexes A, B, and C, all formed from the same metal ion.
From the following list, Ti3 , Ni2 , Pt4 , Cu2 , to which metal ions could the spectra correspond and to which would it be very unlikely
Answer:
Ni^2+ is most likely
Ti^3+ is very unlikely
Explanation:
The Crystal Field Stabilization Energy almost always favors octahedral over tetrahedral in very many cases, but the degree of this favorability varies with the electronic configuration. In other words, for d1 there is only a small gap between the octahedral and tetrahedral lines, whereas at d3 and d8 is a very big gap. However, for d0, d5 high spin and d10, there is no crystal field stabilization energy difference between octahedral and tetrahedral. The ordering of favorability of octahedral over tetrahedral is:
d3, d8 > d4, d9> d2, d7 > d1, d6 > d0, d5, d10. This explains the answer choices above.
Ti^3+ being a d1 specie is least likely to exist in octahedral shape while Ni2+ a d8 specie is more likely to exist in octahedral shape.
Which of the following is the balanced overall reaction and standard cell potential of an electrochemical cell constructed from half-cells with the given half reactions? Mn 2+(aq) + 2 e−→Mn(s); E° = –1.180 V Pb2+(aq) + 2 e−→ Pb(s); E° = –0.130 V Group of answer choices Pb 2+(aq) + Mn(s) →Pb(s) + Mn2+(aq); = 1.050 V Pb(s) + Mn2+(aq) →Pb2+(aq) + Mn(s); = −1.050 V Pb 2+(aq) + Mn2+(aq) →Pb(s) + Mn(s); = –1.310 V Pb 2+(aq) + Mn(s) →Pb(s) + Mn2+(aq); =0.525 V Pb(s) + Mn2+(aq) →Pb2+(aq) + Mn(s); = −0.525 V
Answer:
Explanation:
the question has been solve below
The balanced overall reaction is Mn(s) + Pb²⁺ (aq) → Mn²⁺ (aq) + Pb(s) with a standard cell potential of 1.050 V, making option a) the correct choice.
To answer this, let's first identify the standard cell potential (E°) for the given half-reactions:
Mn²⁺ (aq) + 2 e⁻ → Mn(s); E° = –1.180 V
Pb²⁺ (aq) + 2 e⁻ → Pb(s); E° = –0.130 V
In an electrochemical cell, the half-reaction with the more positive reduction potential acts as the cathode (reduction), and the one with the less positive potential acts as the anode (oxidation). Here, E° for Pb²⁺/Pb is more positive (-0.130 V) than E° for Mn²⁺/Mn (-1.180 V).
Therefore, the cell setup will be:
Manganese will be oxidized: Mn(s) → Mn²⁺ (aq) + 2 e⁻ (oxidation at the anode)
Lead will be reduced: Pb²⁺ (aq) + 2 e⁻ → Pb(s) (reduction at the cathode)
The balanced overall reaction:
Mn(s) + Pb²⁺ (aq) → Mn²⁺ (aq) + Pb(s)
Calculating the Cell Potential:
The standard cell potential (E°cell) is calculated as follows:
E°cell = E°cathode - E°anode
E°cell = (-0.130 V) - (-1.180 V) = 1.050 V
So, the correct answer is:
a) Pb²⁺(aq) + Mn(s) → Pb(s) + Mn²⁺(aq); E° = 1.050 V
Complete question:
Which of the following is the balanced overall reaction and standard cell potential of an electrochemical cell constructed from half-cells with the given half reactions? Mn 2+(aq) + 2 e−→Mn(s); E° = –1.180 V Pb2+(aq) + 2 e−→ Pb(s); E° = –0.130 V Group of answer choices
a) Pb 2+(aq) + Mn(s) →Pb(s) + Mn2+(aq); = 1.050 V
b) Pb(s) + Mn2+(aq) →Pb2+(aq) + Mn(s); = −1.050 V
c) Pb 2+(aq) + Mn2+(aq) →Pb(s) + Mn(s); = –1.310 V
d) Pb 2+(aq) + Mn(s) →Pb(s) + Mn2+(aq); =0.525 V
e) Pb(s) + Mn2+(aq) →Pb2+(aq) + Mn(s); = −0.525 V
A chemist must prepare 900.0mL of sodium hydroxide solution with a pH of 13.90 at 25°C. She will do this in three steps: Fill a 900.0mL volumetric flask about halfway with distilled water. Weigh out a small amount of solid sodium hydroxide and add it to the flask. Fill the flask to the mark with distilled water.
Answer:
28.58 g of NaOH
Explanation:
The question is incomplete. The missing part is:
"Calculate the mass of sodium hydroxide that the chemist must weigh out in the second step"
To do this, we need to know how much of the base we have to weight to prepare this solution.
First we know that is a sodium hydroxide aqueous solution so, this will dissociate in the ions:
NaOH -------> Na⁺ + OH⁻
As NaOH is a strong base, it will dissociate completely in solution, so, starting with the pH we need to calculate the concentration of OH⁻.
This can be done with the following expression:
14 = pH + pOH
and pOH = -log[OH⁻]
So all we have to do is solve for pOH and then, [OH⁻]. To get the pOH:
pOH = 14 - 13.9 = 0.10
[OH⁻] = 10⁽⁻⁰°¹⁰⁾
[OH⁻] = 0.794 M
Now that we have the concentration, let's calculate the moles that needs to be in the 900 mL:
n = M * V
n = 0.794 * 0.9
n = 0.7146 moles
Finally, to get the mass that need to be weighted, we need to molecular mass of NaOH which is 39.997 g/mol so the mass:
m = 39.997 * 0.7146
m = 28.58 gAmpicillin is a bacteriostatic antibiotic. (Instead of killing bacteria, it inhibits their growth.) Ramon plated bacteria transformed with a plasmid that confers ampicillin resistance. He plated the bacteria on Thursday and left them in a 37 oC incubator overnight. On Friday, he observed moderately sized bacterial colonies. He decide to leave them in the incubator a little longer before picking the colonies that he wanted to work with further. Unfortunately, he forgot about the plate and left it in the incubator over the weekend. On Monday, his plate had large colonies that were each surrounded by very small colonies
A) Interpret and explain Ramon’s observations.
B) Predict what he will find when he picks some large and some small colonies and follows the plasmid isolation protocol for each of the colonies.
Answer:
A) That resistance in bacteria is produced due to inactivation of ampicillin by the beta lactamase enzyme. This enzyme is expressed by the bla gene found in the plasmid. This enzyme is secreted into the culture medium, thereby inactivating ampicillin. Thanks to this inactivation, the bacteria colonies will be able to develop. After a day of incubation, only those bacteria that took the plasmid that gives them resistance to ampicillin will grow after transformation. After prolonged incubation, two types of colonies can be observed in the culture medium. One large colony with ampicillin resistance, and another small colony, both of which are sensitive to ampicillin.
B) Large colonies are characterized by being resistant to ampicillin. When Ramón isolates the plasmid, he will have the gene that provides resistance to antibiotics. Said plasmid can be used again on those bacteria that are sensitive to ampicillin.
On the other hand, satellite colonies are sensitive to ampicillin. These types of colonies do not have the plasmid that contains the gene that gives ampicillin resistance. It is not possible to isolate any plasmids from these satellite colonies. These satellite bacteria will not be able to grow if they are transferred to a plate containing fresh ampicillin, while large colonies, which possess the plasmid that gives them resistance to ampicillin, will be able to grow on that plate.
Explanation:
An accident happens in the lab of Professor Utonium, and a radioactive element X is released in the form of a gas at around 4:00 am. Element X has a short half-life (25 min), and the lab would be considered safe when the concentration of X drops by a factor of 10. Considering the decomposition of element X is of first-order, what is the earliest time Professor Utonium can come back to do experiments in the lab
Answer:
5:22 am
Explanation:
The gas X decays following a first-order reaction.
The half-life ([tex]t_{1/2}[/tex]) is 25 min. We can find the rate constant (k) using the following expression.
[tex]k = \frac{ln2}{t_{1/2}} =\frac{ln2}{25min} = 0.028 min^{-1}[/tex]
We can find the concentration of X at a certain time ([tex][X][/tex]) using the following expression.
[tex][X] = [X]_0 \times e^{-k \times t}[/tex]
where,
[tex][X]_0[/tex]: initial concentration of X
t: time elapsed
[tex]\frac{[X]}{[X]_0}= e^{-k \times t}\\\frac{1/10[X]_0}{[X]_0}= e^{-0.028min^{-1} \times t}\\t=82min[/tex]
The earliest time Professor Utonium can come back to do experiments in the lab is:
4:00 + 82 = 5:22 am
Final answer:
The earliest Professor Utonium can return to the lab after a radioactive release is approximately 5:15 am, based on the half-life of 25 minutes and the requirement for the concentration of the gas to drop by a factor of 10, corresponding to just over 3 half-lives.
Explanation:
The question asks for the earliest time Professor Utonium can return to the lab after a release of a radioactive gas, X, which has a half-life of 25 minutes, and the lab is considered safe when its concentration drops by a factor of 10. Understanding that the decay of the radioactive element follows first-order kinetics, we can calculate the time required for the concentration to drop by this factor.
First-order decay implies that the time it takes for a substance to decay to half its initial amount is constant, known as the half-life. To reduce the concentration of a substance by a factor of 10, we need to go through a certain number of half-lives. The formula for calculating the amount of substance remaining after a given time is N = N0,[tex](1/2)^{(t/t1/2)}[/tex] where N is the remaining amount, N0 is the initial amount, t is time, and t1/2 is the half-life.
To reduce the concentration by a factor of 10, we effectively need the substance to go through just over 3 half-lives (since (1/2)³ = 1/8, which is just a bit more than one-tenth). Therefore, the calculation is 3 * 25 = 75 minutes after the initial release. Since the accident happened at around 4:00 am, adding 75 minutes means the earliest Professor Utonium can return to the lab is approximately 5:15 am.
Esters can be synthesized by an acid-catalyzed nucleophilic acyl substitution between an alcohol and a carboxylic acid; this process is called the Fischer esterification reaction. Because the alcohol oxygen is a poor nucleophile, the carbonyl carbon is made a better electrophile by protonation of the carbonyl oxygen. The steps of the synthesis are all reversible. The reaction is generally driven to completion by using an excess of the liquid alcohol as a solvent, or by distilling off the product as it forms. Draw curved arrows to show the movement of electrons in this step of the mechanism.
Answer:
See attachment for mechanism.
Explanation:
The Fischer esterification reaction is a nucleophilic substitution in the acyl group of a carboxylic acid, catalyzed by an acid.
1) The protonation of the oxygen of the carbonyl group activates the carboxylic acid...
2) ... against a nucleophilic attack by the alcohol, and produces a tetrahedral intermediate.
3) The transference of a proton from an oxygen atom to another produces a second tetrahedral intermediate and converts the -OH into a good leaving group.
4) The loss of a proton and the expulsion of H₂O regenerates the acid catalyzer and gives an ester as a product.
The Fischer esterification reaction involves reacting a carboxylic acid with an alcohol to form an ester. The reaction is driven to completion by using an excess of alcohol or by distilling off the product.
Explanation:
In the
Fischer esterification reaction
, an alcohol (for example, ethanol) reacts with a carboxylic acid (like acetic acid) to form an ester. The carbonyl carbon of the acid is protonated to make it a better electrophile that can react with the alcohol's oxygen. Here's a simplified version of what happens:
R-C-0-H (a carboxylic acid) + :O: (the oxygen of an alcohol) -> R-C(=O)-O-R' (an ester)
, where R and R' represent any organic groups. The reaction is balanced by distilling off the ester product or using an excess of the alcohol solvent. In a more detailed mechanism, curved arrows would be used to show the movement of electrons, but those can't be represented in text form.
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2.088 g sample of a compound containing only carbon, hydrogen, and oxygen is burned in an excess of dioxygen, producing 4.746 g CO 2 and 1.943 g H 2O. What is the empirical formula of the compound?
Answer:
The empirical formula is C3H6O
Explanation:
Step 1: Data given
Mass of the sample =2.088 grams
The mass contains carbon, hydrogen, and oxygen
Mass of CO2 produced = 4.746 grams
Mass of H2O produced = 1.943 grams
Molar mass of CO2 = 44.01 g/mol
Molar mass of H2O = 18.02 g/mol
Atomic mass of C = 12.01 g/mol
Atomic mass of H = 1.01 g/mol
Atomic mass of O = 16.0 g/mol
Step 2: Calculate moles CO2
Moles CO2 = mass CO2 / molar mass CO2
Moles CO2 = 4.746 grams/ 44.01 g/mol
Moles CO2 = 0.1078 moles
Step 3: Calculate moles C
For 1 mol CO2 we have 1 mol C
For 0.1078 moles CO2 we'll have 0.1078 moles C
Step 4: Calculate mass C
Mass C: moles C * atomic mass C
Mass C: 0.1078 moles * 12.01 g/mol
Mass C= 1.295 grams
Step 5: Calculate moles H2O
Moles H2O = 1.943 grams / 18.02 g/mol
Moles H2O = 0.1078 moles
Step 6: Calculate moles H
For 1 mol H2O we'll have 2 moles H
For 0.1023 moles H2O we'll have 2*0.1078 = 0.2156 moles H
Step 7: Calculate mass H
Mass H = 0.2046 moles * 1.01 g/mol
Mass H = 0.218 grams
Step 8: Calculate mass O
Mass O = 2.088 grams - 1.295 grams - 0.218 grams
Mass O = 0.575 grams
Step 9: Calculate moles O
Moles O = 0.575 grams / 16.0 g/mol
Moles O = 0.0359 moles
Step 10: Calculate the mol ratio
We divide by the smallest amount of moles
C: 0.1078 moles / 0.0359 moles = 3
H: 0.2156 moles / 0.0359 moles = 6
O: 0.0359 moles / 0.0359 moles =1
The empirical formula is C3H6O
The empirical formula of the compound is C₃H₆O
We'll begin by calculating the mass of C, H and O in the compound. This can be obtained as follow:
For C:Mass of CO₂ = 4.746 g
Molar mass of CO₂ = 44 g/mol
Molar mass of C = 12 g/mol
Mass of C =?Mass of C = 12/44 × 4.746
Mass of C = 1.294 gFor H:Mass of H₂O = 1.943 g
Molar mass of H₂O = 18 g/mol
Molar mass of H₂ = 1 × 2 = 2 g/mol
Mass of H =?Mass of H = 2/18 × 1.943
Mass of H = 0.216 gFor O:Mass of C = 1.294 g
Mass of H = 0.216 g
Mass of compound = 2.088 g
Mass of O =?Mass of O = (Mass of compound ) – (mass of C + mass of H)
Mass of O = 2.088 – (1.294 + 0.216)
Mass of O = 0.578 gFinally, we shall determine the empirical formula of the compound. This can be obtained as follow:C = 1.294 g
H = 0.216 g
O = 0.578 g
Empirical formula =?Divide by their molar mass
C = 1.294 / 12 = 0.108
H = 0.216 / 1 = 0.216
O = 0.578 / 16 = 0.036
Divide by the smallest
C = 0.108 / 0.036 = 3
H = 0.216 / 0.036 = 6
O = 0.036 / 0.036 = 1
Therefore, the empirical formula of the compound is C₃H₆O
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What is the solubility in moles/liter for magnesium hydroxide at 25 oC given a Ksp value of 1.1 x 10-11. Write using scientific notation and use 1 or 2 decimal places (even though this is strictly incorrect!)
Answer:
S = 0.00014 moles /L = 1.4 * 10^-4 moles/L
Explanation:
Step 1: Data given
Temperature = 25.0 °C
Ksp = 1.1 * 10^-11
Step 2: The balanced equation
Mg(OH)2(s) ⇆ Mg^2+(aq) + 2OH-(aq)
Step 3: Define Ksp
[Mg(OH)2 = 1.11 * 10^-11 = S
[Mg^2+] = S
[OH-] = 2S
Ksp = [Mg^2+]*[OH-]²
Ksp = S * (2S)²
1.1 * 10^-11 = 4S³
S³ = 2.75 * 10^-12
S = 0.00014 moles /L
Which of the following statements applies to the E2 mechanism? Which of the following statements applies to the E2 mechanism? It occurs with inversion of stereochemistry. It occurs with racemization of stereochemistry. The C-H and C-X bonds that break must be anti. Use of a bulky base gives the more highly substituted alkene product. It proceeds through the more stable carbocation intermediate.
Final answer:
The E2 reaction is a type of elimination mechanism where a base removes a proton from a carbon adjacent to one with a leaving group, leading to the creation of a double bond. It requires the bonds to be anti-aligned and uses a bulky base to favor the formation of more highly substituted alkenes, with no intermediates formed.
Explanation:
The E2 elimination reaction is a concerted process where a base removes a proton (H) adjacent to a carbon with a leaving group (often denoted as 'X'), resulting in the formation of a double bond. The key characteristics of the E2 mechanism include the requirement for the C-H and C-X bonds to be in an anti-periplanar arrangement, ensuring they are aligned properly for the elimination to occur. Additionally, the use of a bulky base typically leads to the formation of the more highly substituted alkene product, favoring the more stable Zaitsev product. There is also no carbocation intermediate in the E2 mechanism; instead, the reaction proceeds through a single concerted step without intermediates. The E2 mechanism is typically observed with secondary and tertiary substrates where steric hindrance inhibits SN2 reactions.
A gas is placed in a storage tank at a pressure of 30.0 atm at 20.3 C. As a safety device, a small metal plug in the tank is made of a metal alloy that melts at 130 C. If the tank is heated, what is the maximum pressure (in atm) that will be attained in the tank before the plug melts and releases gas?
Answer: 41.2 atm
Explanation
To calculate the final pressure of the system, we use the equation given by Gay-Lussac Law. This law states that pressure of the gas is directly proportional to the temperature of the gas at constant pressure.
Mathematically,
[tex]\frac{P_1}{T_1}=\frac{P_2}{T_2}[/tex]
where,
[tex]P_1\text{ and }T_1[/tex] are the initial pressure and temperature of the gas.
[tex]P_2\text{ and }T_2[/tex] are the final pressure and temperature of the gas.
We are given:
[tex]P_1=30.0atm\\T_1=20.3^0C=(20.3+273)=293.3K\\P_2=?\\T_2=130^0C=(130+273)K=403K[/tex]
Putting values in above equation, we get:
[tex]\frac{30.0}{293.3K}=\frac{P_2}{403}\\\\P_2=41.2atm[/tex]
The maximum pressure (in atm) that will be attained in the tank before the plug melts and releases gas is 41.2
The maximum pressure (in atm) that will be attained in the tank before the plug melts and releases gas is 41.22 atm
From the question given above, the following data were obtained:
Initial pressure (P₁) = 30 atm
Initial temperature (T₁) = 20.3 °C = 20.3 + 273 = 293.3 K
Final temperature (T₂) = 130 °C = 130 + 273 = 403 K
Final pressure (P₂) =?The final pressure can be obtained as illustrated below:
[tex]\frac{P_{1}}{T_{1}} = \frac{P_{2}}{T_{2}}\\\\\frac{30}{293.3} = \frac{P_{2}}{403}\\\\[/tex]
Cross multiply
293.3 × P₂ = 30 × 403
293.3 × P₂ = 12090
Divide both side by 293.3
[tex]P_{2} = \frac{12090}{293.3} \\\\[/tex]
P₂ = 41.22 atmTherefore, the maximum pressure (in atm) that will be attained in the tank before the plug melts and releases gas is 41.22 atm
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Isocyanates are good electrophiles that have been used for protein modification. However, they have limited stability in water. As a result isothiocyanates have been developed at less hydrolytically sensitive variants. Molecules like fluorescein isothiocyanate (FITC) have been used extensively to fluorescently modify proteins. Draw the product of a lysine side chain with FITC. (for the sake of simplicity we are just using a model for lysine's side chain butyl amine)
Find figure in the attachment
Answer:
Explanation:
solution to the question is found below
50.g of NaNO3 was dissolved in 1250 mL of water. what is the molality of the solution? [ Molar mass of NaNO3 = 85 g/mol
Answer:
Approximately [tex]0.47\; \rm mol \cdot L^{-1}[/tex] (note that [tex]1\; \rm M = 1 \; \rm mol \cdot L^{-1}[/tex].)
Explanation:
The molarity of a solution gives the number of moles of solute in each unit volume of the solution. In this [tex]\rm NaNO_3[/tex] solution in water,
Let [tex]n[/tex] be the number of moles of the solute in the whole solution. Let [tex]V[/tex] represent the volume of that solution. The formula for the molarity [tex]c[/tex] of that solution is:
[tex]\displaystyle c = \frac{n}{V}[/tex].
In this question, the volume of the solution is known to be [tex]1250\; \rm mL[/tex]. That's [tex]1.250\; \rm L[/tex] in standard units. What needs to be found is [tex]n[/tex], the number of moles of [tex]\rm NaNO_3[/tex] in that solution.
The molar mass (formula mass) of a compound gives the mass of each mole of units of this compound. For example, the molar mass of [tex]\rm NaNO_3[/tex] is [tex]85\; \rm g \cdot mol^{-1}[/tex] means that the mass of one mole of
[tex]\displaystyle n = \frac{m}{M}[/tex].
For this question,
[tex]\begin{aligned}&n\left(\mathrm{NaNO_3}\right) \\ &= \frac{m\left(\mathrm{NaNO_3}\right)}{M\left(\mathrm{NaNO_3}\right)}\\&= \frac{50\; \rm g}{85\; \rm g \cdot mol^{-1}} \\& \approx 0.588235\; \rm mol\end{aligned}[/tex].
Calculate the molarity of this solution:
[tex]\begin{aligned}c &= \frac{n}{V} \\&= \frac{0.588235\; \rm mol}{1.250\; \rm L} \\&\approx 0.47\;\rm mol \cdot L^{-1}\end{aligned}[/tex].
Note that [tex]1\; \rm mol \cdot L^{-1}[/tex] (one mole per liter solution) is the same as [tex]1\; \rm M[/tex].
Answer:
.47
Explanation:
just did it on CK-12
Within each ______ in the periodic table, elements have similar properties because they have the same number of valence electrons.
Period
Row
Group
Metals
Plz this is really really really urgent
Answer:
group
Explanation:
elements in the same group have the same number of valence electrons
There are various kind of elements that are present in periodic table. Some elements are harmful, some are radioactive, some are noble gases. Therefore, the correct option is option C.
What is periodic table?Periodic table is a table in which we find elements with properties like metals, non metals, metalloids and radioactive element arranges in increasing atomic number.
Periodic table help a scientist to know what are the different types of elements are present in periodic table so that they can discover the new elements that are not being discovered yet.
Within each group in the periodic table, elements have similar properties because they have the same number of valence electrons.
Therefore, within each group in the periodic table, elements have similar properties because they have the same number of valence electrons. The correct option is option C.
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A 3.00 L flask contains 2.33 g of argon gas at 312 mm Hg What is the temperature of the gas
Answer:
T= 257.36 k
Explanation:
using the ideal gas law
pv=nRt
first, convert pressure from mmhg to kpa
312 x (101.3\760)= 41.58 kpa
R is constant= 8.31
to get n(number of moles)
n=m\M (m is mass, M is molar mass)
molar mass of argon is 39.948
n= 2.33\39.948
n=0.0583
substitute;
41.58 x 3 = 0.0583 x 8.31 x T
T= (41.58 x 3)\ (0.0583 x 8.31)
T= 257.36 k
The temperature of the gas under ideal conditions is 257.36K
In order to get the temperature of the gas, we will use the ideal gas equation expressed according to the formula:
[tex]PV = nRT[/tex]
P is the pressure of the gas = 312mmHg = 41.58KPa
V is the volume of the gas = 3.00L
n is the moles of the gas = 0.1165moles
R is boltzmann constant = 8.31
T is the required temperature
Mole = mass/molar mass
Mole = 2.33/40
Mole of argon = 0.05825moles
Substitute the given parameters into the formula
[tex]T=\frac{PV}{nR}\\ T=\frac{41.58 \times 3}{0.05825\times8.31}\\T=257.36K[/tex]
Hence the temperature of the gas under ideal conditions is 257.36K
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Enter your answer in the provided box.A mixture of helium and neon gases is collected over water at 28°C and 791 mmHg. If the partial pressure of helium is 381 mmHg, what is the partial pressure of neon? (Vapor pressure of water at 28°C is 28.3 mmHg.)
Answer:
Explanation:
Using Dalton's law of partial pressure
P total pressure = Pressure of helium + Pressure of neon + Vapor pressure of water
P = 28.3 mmHg, Pressure of helium = 381 mmHg, Vapor pressure of water at 28°C
791 mmHg - 381 mmHg - 28.3 mmHg = Pressure of neon
Pressure of neon = 381.7 mmHg
A certain substance X has a normal freezing point of 5.6 °C and a molal freezing point depression constant Kf-7.78 °C-kg·mol-1. A solution is prepared by dissolving some urea ((NH2)2CO) in 550. g of Χ. This solution freezes at-0.9 °C. Calculate the mass of urea that was dissolved. Be sure your answer has the correct number of significant digits.
Answer:
27.60 g urea
Explanation:
The freezing-point depression is expressed by the formula:
ΔT= Kf * mIn this case,
ΔT = 5.6 - (-0.9) = 6.5 °CKf = 7.78 °C kg·mol⁻¹m is the molality of the urea solution in X (mol urea/kg of X)
First we calculate the molality:
6.5 °C = 7.78 °C kg·mol⁻¹ * mm = 0.84 mNow we calculate the moles of urea that were dissolved:
550 g X ⇒ 550 / 1000 = 0.550 kg X
0.84 m = mol Urea / 0.550 kg Xmol Urea = 0.46 molFinally we calculate the mass of urea, using its molecular weight:
0.46 mol * 60.06 g/mol = 27.60 g ureaThe Mannich reaction is one of the few three-component reactions in organic chemistry. In this reaction, a ketone, an aldehyde and an amine react together under acid catalyzed conditions to form the final product. The mechanism involves the following steps: 1. Following initial protonation of the carbonyl oxygen, nucleophilic attack by the amine forms a protonated carbinolamine 1; 2. Proton transfer and elimination of water forms iminium ion 2; 3. The enol form of the ketone attacks the iminium ion to form adduct 3; 4. Deprotonation of adduct 3 leads to the final product. Write out the mechanism on a separate sheet of paper and then draw the structure of iminium ion 2.
Answer:
Step 1, formation of the iminium ion
2) enol reaction with iminium ion to form beta-amino enone
Explanation:
Please find attached the detailed reaction mechanism and the structure of iminium ion separately
The Mannich reaction is a three-component reaction that forms a ß-amino ketone or aldehyde. It involves protonation, nucleophilic attack, formation of an iminium ion, enol attack, and deprotonation.
The Mannich reaction is a three-component organic reaction involving a ketone, an aldehyde, and an amine under acid-catalyzed conditions to form a ß-amino ketone or ß-amino aldehyde. The mechanism of the reaction can be broken down into several steps:
Protonation of the carbonyl oxygen: Initially, the carbonyl oxygen of the ketone or aldehyde is protonated by an acid catalyst.Nucleophilic attack by the amine: The nucleophilic amine attacks the protonated carbonyl, forming a protonated carbinolamine.Formation of iminium ion: Following a proton transfer, the carbinolamine dehydrates to form an iminium ion.Enol attack: The enol form of the ketone then attacks the iminium ion, leading to the formation of adduct 3.Deprotonation: Finally, adduct 3 undergoes deprotonation to yield the final ß-amino product.Given these steps, the structure of iminium ion 2 involves a nitrogen atom double-bonded to a carbon atom derived from the original carbonyl.