You make dilutions of curcumin stock solution and measure the absorbance of each dilution to obtain the following data: Conc. (M) Abs. 1.60E-05 1.93 1.21E-05 1.62 8.09E-06 1.08 4.04E-06 0.44 2.02E-06 0.13 Which data points should you keep when making your calibration curve? Choose all that apply. (Hint: Plot the data.)

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.

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:

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:

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!

Answer:

See the attached file for the structure

Explanation:

See the attached file

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.

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:

The first stage of this reaction involves the 2-propanol reacting with benzophenone which involves the formation of the most stable carbocation, the reaction is then heated under high temperature and pressure to form benzopinacol, on cooling the benzopinacololone crystallizes out

Melting point of benzopinacolol= 184-186°C

Explanation:

Please find attached the detailed step by step mechanism of synthesis of benzopinacololone

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:

[tex]\Delta _fS=0.2724\frac{J}{K}[/tex]

Explanation:

Hello,

In this case, we define the entropy change for such freezing process as:

[tex]\Delta _fS=\frac{n_{Hg}\Delta _fH}{T_f}[/tex]

Thus, we compute the moles that are in 5.590 g of liquid mercury:

[tex]n_{Hg}=5.590 gHg*\frac{1molHg}{200.59gHg} =0.02787molHg[/tex]

Hence, we compute the required entropy change, considering the temperature to be in kelvins:

[tex]\Delta _fS=\frac{0.02787mol*2.29\frac{kJ}{mol} }{(-38.9+273.15)K}\\\\\Delta _fS=2.724x10^{-4}\frac{kJ}{K} *\frac{1000J}{kJ} \\\\\Delta _fS=0.2724\frac{J}{K}[/tex]

Best regards.

Answer:

ΔS = -0.272 J/K

Explanation:

Step 1: Data given

Its normal freezing point is –38.9 °C

molar enthalpy of fusion is ∆Hfusion = 2.29 kJ/mol

Mass of Hg = 5.590 grams

Step 2:

ΔG = ΔH - TΔS

At the normal freezing point, or any phase change in general,  

ΔG =0

0  =  Δ

Hfus

−

Tfus  Δ

S

fus

Δ

S

fus = Δ

Hfus

/Tfus

Δ

S

fus = 2290 J/mol / 234.25 K

Δ

S

fus = 9.776 J/mol*K

Since fusion is  from solid to liquid. Freezing is the opposite process, so the entropy change of freezing is -9.776 J/mol*K

Step 3: Calculate moles Hg

Moles Hg = 5.590 grams / 200.59 g/mol

Moles Hg = 0.02787 moles

Step 4: Calculate the entropy change of the system:

Δ

S = -9.776 J/mol*K * 0.02787 moles

ΔS = -0.272 J/K

Final answer:

The balanced chemical equation for the overall reaction is 2NO(g) + O2(g) → 2NO2(g). The experimentally-observable rate law, assuming the second step is rate-determining, is rate = k[NO]^2[O2], where k is the overall rate constant.

Explanation:

The balanced chemical equation for the overall reaction combining the two steps 2NO(g) → N2O2(g) and NO2(g) + O2(g) → 2NO2(g) is:

2NO(g) + O2(g) → 2NO2(g)

To write the experimentally-observable rate law for the overall reaction, we must identify the rate-determining step. Assuming the second step is the rate-determining step, and given that the first step is a fast equilibrium, the overall rate can be expressed as:

rate = k2[N2O2][O2]

Since [N2O2] is the intermediate formed in the first step and we know that the rate of formation of N2O2 is proportional to the square of [NO] concentration, we can express [N2O2] in terms of [NO]. Substituting into the rate law, we get:

rate = k2k1[NO]2[O2]

Here, k = k1k2 represents the overall rate constant for the reaction. Therefore, the rate law for the overall chemical reaction is:

rate = k[NO]2[O2]

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.

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:

Check the explanation

Explanation:

Answer – Given, [tex]H_3PO_4[/tex] acid and there are three Ka values

[tex]K_{a1}=6.9x10^8, K_{a2} = 6.2X10^8, and K_{a3}=4.8X10^{13}[/tex]

The transformation of [tex]H_2PO_4- (aq) to HPO_4^2-(aq)[/tex]is the second dissociation, so we need to use the Ka2 = 6.2x10-8 in the Henderson-Hasselbalch equation.

Mass of KH2PO4 = 22.0 g , mass of Na2HPO4 = 32.0 g , volume = 1.00 L

First we need to calculate moles of each

Moles of KH2PO4 = 22.0 g / 136.08 g.mol-1

                             = 0.162 moles

Moles of Na2HPO4 = 32.0 g /141.96 g.mol-1

                             = 0.225 moles

[H2PO4-] = 0.162 moles / 1.00 L = 0.162 M

[HPO42-] = 0.225 moles / 1.00 L = 0.225 M

Now we need to calculate the pKa2

pKa2 = -log Ka

       = -log 6.2x10-8

       = 7.21

We know Henderson-Hasselbalch equation

pH = pKa + log [conjugate base] / [acid]

pH = 7.21 + log 0.225 / 0.162

     = 7.35

The pH of a buffer solution obtained by dissolving 22.0 g of KH2PO4 and 32.0 g of Na2HPO4 in water and then diluting to 1.00 L is 7.35

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:

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:

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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Answer:

It produces water.

Explanation:

H+   +    OH-    produces    H2O.

It is a type of Neutralization reaction.

Answer: 205.70 mL of acid solution is needed

Explanation: Please see the attachments below

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:

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:

specific heat

Explanation:

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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:

Explanation:

Larger percentage of the cells would be in mitosis, although the differences in percentages within the different stages of mitosis would still look alike, and interphase might likely still maintain the largest percentage, although it might sometimes be 50% and not  88%.

The slides you viewed in this lab are different from the slides, A larger percentage of the meristematic cells would be in mitosis in treated onion tips

The differences in percentages within the different stages of mitosis look the same, and interphase might likely still maintain the largest percentage.

Thus, The slides you viewed in this lab are different from the slides, A larger percentage of the meristematic cells would be in mitosis in treated onion tips

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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:

Option B is the correct option- Ethanol increases apparent Km without affecting Vmax.

Explanation:

Vmax remains the same, and Km increases in competitive inhibition. There is an increment in Km because competitive inhibitors interfere with substrate binding to the enzyme. Vmax is not affected because the competitive inhibitor cannot bind to ES and therefore does not alter the catalysis.

Option B is the correct option- Ethanol increases apparent Km without affecting Vmax.

In the attached image, the first picture is the Michaelis Menten Plot for competitive inhibition in which an increase in km but constant Vmax is observed.

In the images,  ............................. represents the case with competitive inhibitor, while _______________ represents the case without competitive inhibitor.

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.

The missing words in the blanks are-  

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.

Thus,

The missing words in the blanks are-  

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Answer:

The final pH is 3.80

Explanation:

Step 1: Data given

Volume of acetic acid = 200.0 mL = 0.200 L

Number of moles acetic acid = 0.5000 moles

Volume of NaOH = 100.0 mL = 0.100 L

Molarity of NaOH = 0.500 M

Ka of acetic acid = 1.770 * 10^-5

Step 2: The balanced equation

CH3COOH + NaOH → CH3COONa + H2O

Step 3: Calculate moles

moles = molarity * volume

Moles NaOH = 0.500 M * 0.100 L

Moles NaOH = 0.0500 moles

Step 4: Calculate the limiting reactant

For 1 mol CH3COOH we need 1 mol NaOH to produce 1 mol CH3COONa and 2 moles H2O

NaOH is the limiting reactant. It will completely be consumed (0.0500 moles). CH3COOH is in excess. There will react 0.0500 moles . There will remain 0.500 - 0.0500 = 0.450 moles

There will be produced 0.0500 moles CH3COONa

Step 5: Calculate the total volume

Total volume = 200.0 mL + 100.0 mL = 300.0 mL

Total volume = 0.300 L

Step 6: Calculate molarity

Molarity = moles / volume

[CH3COOH] = 0.450 moles / 0.300 L

[CH3COOH] = 1.5 M

[CH3COONa] = 0.0500 moles / 0.300 L

[CH3COONa]= 0.167 M

Step 7: Calculate pH

pH = pKa + log[A-]/ [HA]

pH = -log(1.77*10^-5) + log (0.167/ 1.5)

pH = 4.75 + log (0.167/1.5)

pH = 3.80

The final pH is 3.80

To calculate the final pH of the solution, you need to consider the reaction between acetic acid (CH3COOH) and NaOH. The concentration of OH- ions in the final solution determines the pH, which can be calculated using the formula pH = -log10(H+ concentration). Using the given information, we can calculate the concentration of OH- ions, and then convert that to the concentration of H+ ions to find the pH of the solution.

To calculate the final pH of the solution, we need to consider the reaction between acetic acid (CH3COOH) and NaOH. The reaction between the two compounds forms sodium acetate (CH3COONa) and water.

The balanced equation for the reaction is:

CH3COOH + NaOH → CH3COONa + H2O

Since NaOH is a strong base, it completely dissociates in water to produce OH- ions. These OH- ions react with the acetic acid to form water. Therefore, the concentration of OH- ions will determine the pH of the final solution.

Using the given information, we can calculate the moles of acetic acid and OH- ions:

Moles of acetic acid = volume of acetic acid solution (L) x concentration of acetic acid (M)

= 0.200 L x 0.5000 M = 0.100 moles

Moles of OH- ions = volume of NaOH solution (L) x concentration of NaOH (M)

= 0.100 L x 0.5000 M = 0.050 moles

Since the mole ratio between acetic acid and OH- ions is 1:1, there will be an equal number of moles of OH- ions and acetic acid in the solution after the reaction is complete.

Therefore, the concentration of OH- ions in the final solution is:

Concentration of OH- ions = Moles of OH- ions / volume of solution (L)

= 0.050 moles / 0.300 L = 0.167 M

To calculate the final pH, we can use the formula: pH = -log10(H+ concentration)

Since in water, OH- ions and H+ ions are inversely proportional, we can calculate the concentration of H+ ions using:

H+ concentration = Kw / OH- concentration

Where Kw is the ion product of water and is equal to 1.0 x 10^-14 at 25°C.

Therefore, H+ concentration = 1.0 x 10^-14 / 0.167 M = 5.99 x 10^-14 M

Finally, calculating the pH:

pH = -log10(5.99 x 10^-14) ≈ 13.22

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