"Phenylacetic acid, C7H7COOH, builds up in the blood of people with phenylketonuria. The pKa of phenylacetic acid is 4.32 at 25 ºC. What is the pH of a 0.15 M aqueous solution of calcium phenylacetate, Ca(C7H7COO)2, at 25 ºC?"

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.

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 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

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.

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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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.

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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Answer:

specific heat

Explanation:

please like,rate and tag as brainliest

Final answer:

The sudden rise in cytosolic Ca²+ concentration in skeletal muscle leads to activation of phosphorylase kinase, not inactivation. This process is essential for muscle contraction, where Ca²+ plays a critical role by binding to proteins to expose actin for myosin attachment.

Explanation:

In response to a sudden elevation of cytosolic Ca²+ concentration in a contracting skeletal muscle, the activation of phosphorylase kinase occurs. This elevation is primarily due to excitation-contraction coupling, where an action potential triggers the release of Ca²+ from the sarcoplasmic reticulum into the cytosol. Immediately following this, calcium ions interact with troponin, altering its configuration and moving tropomyosin off the actin-binding sites. This allows the myosin head to bind to actin, resulting in muscle contraction.

Separately, in the context of ß-adrenergic receptor activation by adrenaline, an increase in cyclic AMP (cAMP) inside the muscle cell activates PKA (protein kinase A), leading to the phosphorylation of enzymes involved in glycogen degradation and inhibition of glycogen synthesis. Specifically, cAMP-dependent protein kinase is activated, causing a ready pool of glucose to be available for muscular activity.

Therefore, the correct answer is that a sudden increase in Ca²+ results in the activation of phosphorylase kinase, which is critical for initiating muscle contraction and metabolic responses.

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                                                          

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

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)

Answer:

Explanation:

Substance A

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:

Such a reaction does not happen with Cu⁺².

2. When we add K₂S to the mixture, the reaction that takes place is:

Such a reaction does not happen with Ag⁺.

3. When we add K₂CO₃ to the mixture, the reactions that take place are:

This means both Cu⁺² and Ag⁺ would precipitate, thus they would not be separated.

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.

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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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.

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.

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.

Answer:

See explanation

Explanation:

Enamines are nucleophiles, and will react with alkyl halides to give alkylation products. Subsequent treatment with aqueous acid will give ketones.

The mechanism of an enamine reaction is shown in the image attached.

The structure of the carbanion is also shown in the image attached.

Answer:

12.00

Explanation:

Final answer:

To find the pH of the diluted Ca(OH)₂ solution, calculate the new concentration after dilution, determine the [OH-], find pOH, and then subtract from 14 to get pH. The final pH of the solution after dilution to 1 L will be 12.

Explanation:

The question is about calculating the pH of a diluted solution of Ca(OH)₂ after it has been diluted to 1.00 L. We begin by finding the new concentration of Ca(OH)₂ after dilution, which would be:

Determine the initial number of moles of Ca(OH)₂ : moles = 200 mL times 0.025 M = 0.005 moles.

Calculate the new concentration after dilution: concentration = 0.005 moles / 1.00 L = 0.005 M.

As Ca(OH)₂ dissociates to give 2 OH - per molecule, the [OH-] = 2 times 0.005 M = 0.01 M.

Calculate the pOH: pOH = -log(0.01) = 2.

Finally, calculate the pH: pH = 14 - pOH = 14 - 2 = 12.

The pH of the solution after dilution to 1 L will be 12.

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.

The mass of (NH4)3PO4 is 111.75grams.

HOW TO CALCULATE THE MASS:

mass (g) = moles (mol) × molar mass (g/mol)

{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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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.

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:

Adding these reactions together:

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.

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.

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!

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.

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:

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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.

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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)

The names of the given ionic compounds are:

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Answer:

See the attached file for the structure

Explanation:

See the attached file

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:

Answer: 205.70 mL of acid solution is needed

Explanation: Please see the attachments below

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