Give the major organic products from the oxidation with KMnO4 for the following compounds. Assume an excess of KMnO4.
a) ethylbenzene
b) m-Xylene (1,3- dimethylbenzene)
c) 4-Propyl-3-t-butyltoluene

Answer:

[I⁻] = 0.0352M

Explanation:

Based on the equilibrium:

I₃⁻(aq) ⇄ I₂(aq) + I⁻(aq)

Kc is defined as:

Kc = 0.25 = [I₂] [I⁻] / [I₃⁻]

The system reaches the equilbrium when the ratio [I₂] [I⁻] / [I₃⁻] is equal to 0.25

In the beginning, you add 0.0401M of both [I₂] [I⁻].  When the reaction reach the equilibrium, xM of both [I₂] [I⁻] is consumed producing xM of  [I₃⁻]. That is written as:

[I₃⁻] = X

[I₂] = 0.0401M - X

[I⁻] = 0.0401M - X

X is known as reaction coordinate.

Replacing in Kc:

0.25 = [I₂] [I⁻] / [I₃⁻]

0.25 = [0.0401M - X] [0.0401M - X] / [X]

0.25X = 0.00160801 - 0.0802X + X²

0 = 0.00160801 - 0.3302X + X²

Solving for X:

X = 0.0049M → Right solution

X = 0.3252M → False solution. Produce negative concentrations

Replacing, equilibrium concentrations will be:

[I₃⁻] = X

[I₂] = 0.0401M - X

[I⁻] = 0.0401M - X

[I₃⁻] = 0.0049M

[I₂] = 0.0352M

The equilibrium concentration (in molarity) of  [I⁻] should be considered as the 0.0352M.

Since

I₃⁻(aq) ⇄ I₂(aq) + I⁻(aq)

Here Kc should be defined

Kc = 0.25 = [I₂] [I⁻] / [I₃⁻]

Also, The system finished the equilibrium at the time when the ratio [I₂] [I⁻] / [I₃⁻] is equivalent to 0.25.

Also,

[I₃⁻] = X

[I₂] = 0.0401M - X

[I⁻] = 0.0401M - X

Also,

0.25 = [I₂] [I⁻] / [I₃⁻]

0.25 = [0.0401M - X] [0.0401M - X] / [X]

0.25X = 0.00160801 - 0.0802X + X²

0 = 0.00160801 - 0.3302X + X²

Now

X = 0.0049M → Right solution

X = 0.3252M → False solution

Now equilibrium concentrations will be:

[I₃⁻] = X

[I₂] = 0.0401M - X

[I⁻] = 0.0401M - X

[I₃⁻] = 0.0049M

[I₂] = 0.0352M

[I⁻] = 0.0352M

Hence, The equilibrium concentration (in molarity) of  [I⁻] should be considered as the 0.0352M.

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Answer:

(1) Before the addition of any HBr, the pH is 12.02

(2) After adding 12.0 mL of HBr, the pH is 10.86

(3) At the titration midpoint, the pH is 10.73

(4) At the equivalence point, the pH is 5.79

(5) After adding 45.1 mL of HBr, the pH is 1.18

Explanation:

First of all, we have a weak base:

(CH₃)₂NH  + H₂O  ⇄  (CH₃)₂NH₂⁺  +  OH⁻            Kb = 5.4×10⁻⁴

0.289 - x                             x                x

Kb = x² / 0.289-x

Kb . 0.289 - Kbx - x²

1.56×10⁻⁴ - 5.4×10⁻⁴x - x²

After the quadratic equation is solved x = 0.01222 → [OH⁻]

- log  [OH⁻] = pOH → 1.91

pH = 12.02   (14 - pOH)

We determine the mmoles of H⁺, we add:

0.286 M . 12 mL = 3.432 mmol

We determine the mmoles of base⁻, we have

27.9 mL . 0.289 M = 8.0631 mmol

When the base, react to the protons, we have the protonated base plus water (neutralization reaction)

(CH₃)₂NH     +      H₃O⁺        ⇄  (CH₃)₂NH₂⁺  +  H₂O

8.0631 mm       3.432 mm                 -

4.6311 mm                                  3.432 mm

We substract to the dimethylamine mmoles, the protons which are the same amount of protonated base.

[(CH₃)₂NH] → 4.6311 mm / Total volume (27.9 mL + 12 mL) = 0.116 M

[(CH₃)₂NH₂⁺] → 3.432 mm / 39.9 mL = 0.0860 M

We have just made a buffer.

pH = pKa + log (CH₃)₂NH  / (CH₃)₂NH₂⁺

pH = 10.73 + log (0.116/0.0860) = 10.86

mmoles of base = mmoles of acid

Let's find out the volume

0.289 M . 27.9 mL = 0.286 M . volume

volume in Eq. point = 28.2 mL

(CH₃)₂NH     +      H₃O⁺        ⇄  (CH₃)₂NH₂⁺  +  H₂O

8.0631 mm       8.0631mm               -

                                                8.0631 mm

We do not have base and protons, we only have the conjugate acid

We calculate the new concentration:

mmoles of conjugated acid / Total volume (initial + eq. point)

[(CH₃)₂NH₂⁺] = 8.0631 mm /(27.9 mL + 28.2 mL)  = 0.144 M

(CH₃)₂NH₂⁺   +  H₂O   ⇄   (CH₃)₂NH  +  H₃O⁻       Ka = 1.85×10⁻¹¹

 0.144 - x                                  x               x

[H₃O⁺] = √ (Ka . 0.144) →  1.63×10⁻⁶ M  

pH = - log [H₃O⁺] = 5.79

This is the point where we add, the half of acid. (14.1 mL)

This is still a buffer area.

mmoles of H₃O⁺ = 4.0326 mmol (0.286M . 14.1mL)

mmoles of base = 8.0631 mmol - 4.0326 mmol

[(CH₃)₂NH] = 4.0305 mm / (27.9 mL + 14.1 mL) = 0.096 M

[(CH₃)₂NH₂⁺] = 4.0326 mm (27.9 mL + 14.1 mL) = 0.096 M

pH = pKa + log (0.096M / 0.096 M)

pH = 10.73 + log 1 =  10.73

Both concentrations are the same, so pH = pKa. This is the  maximum buffering capacity.

mmoles of acid = 45.1 mL . 0.286 M = 12.8986 mmol

mmoles of base = 8.0631 mmoles

This is an excess of H⁺, so, the new [H⁺] = 12.8986 - 8.0631 / Total vol.

(CH₃)₂NH     +      H₃O⁺        ⇄  (CH₃)₂NH₂⁺  +  H₂O

8.0631 mm     12.8986 mm             -

       -               4.8355 mm                        

[H₃O⁺] = 4.8355 mm / (27.9 ml + 45.1 ml)

[H₃O⁺] = 4.8355 mm / 73 mL → 0.0662 M

- log [H₃O⁺] = pH

- log 0.0662 = 1.18 → pH

Answer:

0.250 mol Mg²⁺

0.500 mol Cl⁻

Explanation:

Magnesium chloride (MgCl₂) dissociates into ions according to the following equilibrium:

MgCl₂  ⇒  Mg²⁺ + 2 Cl⁻

1 mol      1 mol   2 mol

1 mol of Mg²⁺ and 2 moles of Cl⁻ are formed per mole of MgCl₂.  If we have 0.250 mol of MgCl₂, the following amounts of ions will be formed:

0.250 mol MgCl₂ x 1 mol Mg²⁺/mol MgCl₂= 0.250 mol Mg²⁺

0.250 mol MgCl₂ x 2 mol Cl⁻/mol MgCl₂= 0.500 mol Cl⁻

Answer:

HEY THE ANSWER ABOVE ME IS RIGHT!! i defientely misclicked my rating :/

5/5 all the way.

Explanation:

Answer:

Explanation:

In weight/volume (w/v) terms,

   1 ppm = 1g m-3 = 1 mg L-1 = 1 μg mL-1

200 mL = 0.2 L

15 / 0.2 mg L-1 =75 ppm

The concentration in ppm of a solution which is prepared by dissolving in 15mg of NaCl in 200ml water is 75 mg/.,

ppm stand for 'part per million' and it is used to define the concentration of any substance as mass of any substance present in per liter of volume of solution, its unit for measurement is mg/L.

Given that, mass of NaCl = 15mg

Volume of solution = 200mL = 0.2L

Concentration in ppm will be calculated as:
ppm = 15/0.2 = 75mg/L

Hence ppm concentration of NaCl is 75 mg/L.

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Answer:

Amount of HCL = 0.00318 L  of 3.18 ml

Explanation:

Given:

HCL = 2.5 M

NaOH = 0.53 M

Amount of NaOH  = 15 ml = 0.015 L

Find:

Amount of HCL

Computation:

HCL react with NaOH

HCl + NaOH ⇒ NaCl + H₂O

So,

Number of moles = Molarity × volume

Number of moles of NaOH  = 0.53 × 0.015

Number of moles of NaOH = 0.00795 moles

So,

Number of moles of HCl needed =  0.00795 mol es

So,

Volume = No. of moles / Molarity

Amount of HCL = 0.00795  / 2.5

Amount of HCL = 0.00318 L  of 3.18 ml

Answer:

1. 0.98 atm

Explanation:

The following data were obtained from the question:

Mass of unknown liquid (m) = 0.3175 g

Temperature (T) = 99 °C

Pressure (P) = 748.2 mmHg

Volume (V) = 145.0 mL

Gas constant (R) = 0.08206 atm.L/Kmol

1. Determination of the pressure in atm.

760 mmHg = 1 atm

Therefore,

748.2 mmHg = 748.2/760 = 0.98 atm

Therefore, the pressure in atm is 0.98 atm.

Answer:

8.00L of ammonia can be produced

Explanation:

The reaction is:

N₂(g) + 3H₂(g) → 2NH₃(g)

Where 1 mole of nitrogen reacts with 3 moles of hydrogen to produce 2 moles of ammonia.

Avogadro's law states that, under constant pressure and temperature, equal volumes of gases contains equal number of moles.

As in the reaction conditions are mantained at STP (Pressure and temperature are constant) you can say of the reaction that:

1 liter of nitrogen reacts with 3 liters of hydrogen to produce 2 liters of ammonia

Thus, if 12.0L of hydrogen reacts and 3L of hydrogen produce 2L of ammonia, liters of ammonia produced are:

12L H₂(g) ₓ (2L NH₃(g)  /  3L H₂(g)) =

Answer:

C

Explanation:

I had this question and C is the right answer

One idea that Dalton taught about atoms was that all atoms of one type were identical in mass and properties.

An atom is defined as the smallest unit of matter which forms an element. Every form of matter whether solid,liquid , gas consists of atoms . Each atom has a nucleus which is composed of protons and neutrons and shells in which the electrons revolve.

The protons are positively charged and neutrons are neutral and hence the nucleus is positively charged. The electrons which revolve around the nucleus are negatively charged and hence the atom as a whole is neutral and stable due to presence of oppositely charged particles.

Atoms of the same element are similar as they have number of sub- atomic particles which on combination do not alter the chemical properties of the substances.

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Answer:

Approximately [tex]24\; \rm g[/tex] (at most.)

Explanation:

Aluminum [tex]\rm Al[/tex] reacts with bromine [tex]\rm Br_2[/tex] at a [tex]2:3[/tex] ratio:

[tex]\rm 2\; Al\, (s) + 3\; Br_2\, (g) \to 2\; AlBr_3\, (s)[/tex].

Look up the relative atomic mass of [tex]\rm Al[/tex] and [tex]\rm Br[/tex]. From a modern periodic table:

Calculate the formula mass of the reactants and of the product:

Calculate the quantity (in number of moles of formula units) of each reactant:

Assume that [tex]\rm Al\, (s)[/tex] is the limiting reactant. From the coefficients:

[tex]\displaystyle \frac{n(\mathrm{AlBr_3})}{n(\mathrm{Al})} = 1[/tex].

Based on the assumption that [tex]\rm Al\, (s)[/tex] is the limiting reactant:

[tex]\begin{aligned}n(\mathrm{AlBr_3}) &= \frac{n(\mathrm{AlBr_3})}{n(\mathrm{Al})} \cdot n(\mathrm{Al}) \\ &=1\times 0.18528\; \rm mol \approx 0.185\; \rm mol\end{aligned}[/tex].

In other words, if [tex]\rm Al[/tex] is the limiting reactant (meaning that [tex]\rm Br_2[/tex] is in excess,) then approximately [tex]0.556\; \rm mol[/tex] of [tex]\rm AlBr_3[/tex] will be produced.

On the other hand, assume that [tex]\rm Br_2\; (g)[/tex] is the limiting reactant. Similarly, from the coefficients:

[tex]\displaystyle \frac{n(\mathrm{AlBr_3})}{n(\mathrm{Br_2})} = \frac{2}{3}[/tex].

Based on the assumption that [tex]\rm Br_2\, (g)[/tex] is the limiting reactant:

[tex]\begin{aligned}n(\mathrm{AlBr_3}) &= \frac{n(\mathrm{AlBr_3})}{n(\mathrm{Br_2})} \cdot n(\mathrm{Br_2}) \\ &= \frac{2}{3}\times 0.13767\; \rm mol \approx 0.0918\; \rm mol\end{aligned}[/tex].

Compare the [tex]n(\mathrm{AlBr_3})[/tex] value based on the two assumptions. Only the smallest value, [tex]n(\mathrm{AlBr_3}) \approx 0.0918\; \rm mol[/tex] (under the assumption that [tex]\rm Br_2\, (g)[/tex] is the limiting reactant,) would resemble the theoretical yield. The reason is that [tex]\rm Br_2\, (g)[/tex] would run out before all that [tex]\rm 5.0\; g[/tex] of [tex]\rm Al\, (s)[/tex] was converted to [tex]\rm AlBr_3\, (g)[/tex].

Apply the formula mass of [tex]\rm AlBr_3[/tex] to find the mass of that (approximately) [tex]0.0918\; \rm mol[/tex] of [tex]\rm AlBr_3[/tex] formula units:

[tex]\begin{aligned}m(\mathrm{AlBr_3}) &= n(\mathrm{AlBr_3}) \cdot M(\mathrm{AlBr_3}) \\ &= 0.0918\; \rm mol \times 266.698\; g \cdot mol^{-1} \approx 24\; \rm g\end{aligned}[/tex].

Answer:

The molar concentration of HCl in the aqueous solution is 0.0131 mol/dm3

Explanation:

To get the molar concentration of a solution we will use the formula:

Molar concentration = mass of HCl/ molar mass of HCl

Mass of HCl in the aqueous solution will be 40% of the total mass of the solution.

We can extract the mass of the solution from its density which is 1.2g/mL

We will further perform our analysis by considering only 1 ml of this aqueous solution.

The mass of the substance present in this solution is 1.2g.

The mass of HCl Present is 40% of 1.2 = 0.48 g.

The molar mass of HCl can be obtained from standard tables or by adding the masses of Hydrogen (1 g) and Chlorine (35.46 g) = 36.46g/mol

Therefore, the molar concentration of HCl in the aqueous solution is 0.48/36.46 = 0.0131 mol/dm3

Answer:

b, H2SO4 with HCl, as they are both liquid acids

Answer:

0.167M

Explanation:

Molarity, M, is an unit of concentration in chemistry defined as the ratio between moles of solute (NaOH in this case) and volume of the solution in liters.

To find molarity of 5.00 g Sodium Hydroxide in 750.0 mL of solution we need to convert mass of NaOH to moles (Using its molar mass: 40g/mol) and the mililiters of solution to liters (1L = 1000mL), thus:

Moles NaOH = 5.00g × (1mol/ 40g) = 0.125 moles NaOH = Moles solute

Liters solution = 750.0mL × (1L / 1000mL) = 0.7500L solution

And molariy is:

0.125 moles NaOH  / 0.7500L solution =

0.167M

Answer:

2MnO4^- (aq) + 3C2O4^2- (aq) + 2H2O (l) --> 2MnO2(s) +6CO3^2 -(aq) + 4H^+ (aq)

Explanation:

First, write the half equations for the reduction of MnO4^- and the oxidation of C2O4^2- respectively. Balance it.

Reduction requires H+ ions and e- and gives out water, vice versa for oxidation.

Reduction:

MnO4^- (aq) + 4H^+ (aq) + 3e- ---> MnO2(s) + 2H2O (l)

Oxidation:

C2O4^2- (aq) + 2H2O (l) ---> 2CO3^2 -(aq) + 4H^+ (aq) + 2e-

Balance the no. of electrons on both equations so that electrons can be eliminated. we can do so by multiplying the reduction eq by 2, and oxidation eq by 3.

2MnO4^- (aq) + 8H^+ (aq) + 6e- ---> 2MnO2(s) + 4H2O (l)

3C2O4^2- (aq) + 6H2O (l) ---> 6CO3^2 -(aq) + 12H^+ (aq) + 6e-

Now combine both equations and eliminate repeating H+ and H2O.

2MnO4^- (aq) + 8H^+ (aq) + 3C2O4^2- (aq) + 6H2O (l) --> 2MnO2(s) + 4H2O (l) +6CO3^2 -(aq) + 12H^+ (aq)

turns into:

2MnO4^- (aq) + 3C2O4^2- (aq) + 2H2O (l) --> 2MnO2(s) +6CO3^2 -(aq) + 4H^+ (aq)

Answer:   3.87M  of HI remains after 27.3 s

Explanation:

Using the Second order decomposition equation of

1/[H]t =K x t +1/[A]o

Given initial concentration ,[A]o = 4.78M

time, t = 27.3 s

rate of constant , k= 1.80 x 10^-3 M-1s-1

1/[H] t= 1/[A] t= concentration after time, t=?

SOLUTION

1/[A] t =kt +1/[A]o

1/[A] t =(1.80 x 10^-3 (27.3)+1/4.78

0.04914+0.2092=0.2583

1/[A] t =0.2583

[A] t =1/0.2583= 3.87M

Answer:

Ethane would have a higher boiling point.

Explanation:

In this case, for the lewis structures, we have to keep in mind that all atoms must have 8 electrons (except hydrogen). Additionally, each carbon would have 4 valence electrons, with this in mind, for methane we have to put the hydrogens around the carbon, and with this structure, we will have 8 electrons for the carbon. In ethane, we will have a bond between the carbons, therefore we have to put three hydrogens around each carbon to obtain 8 electrons for each carbon.

Now, the main difference between methane and ethane is an additional carbon. In ethane, we have an additional carbon, therefore due to this additional carbon, we will have more area of interaction for ethane. If we have more area of interaction we have to give more energy to the molecule to convert from liquid to gas, so, the ethane will have a higher boiling point.

I hope it helps!

The Lewis structure shows the valence electrons in a molecule. Ethane will have a higher boiling point than methane.

We can deduce the number of valence electrons in a molecule by drawing the Lewis structure of the molecule. The Lewis structure consists of the symbols of elements in the compound and the valence electrons in the compound.

We know that the higher the molar mass of a compound the greater its boiling point. Looking at the Lewis structures of methane and ethane, we cam see that ethane has a higher molecular mass (more atoms) and consequently a higher boiling point than methane.

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Answer: The equilibrium value of [tex]N_2O_4[/tex] is 0.379 M

Explanation:

Equilibrium constant is the ratio of the concentration of products to the concentration of reactants each term raised to its stochiometric coefficients.

Using ideal gas equation : [tex]PV=nRT[/tex]

P = pressure of gas

V = volume of gas

n = no of moles

R = gas constant

T = Temperature

pressure of [tex]N_2O_4[/tex] = [tex]\frac{1\times 0.0821Latm/Kmol\times 298}{5L}=5atm[/tex]

pressure of [tex]NO_2[/tex] = [tex]\frac{2\times 0.0821Latm/Kmol\times 298}{5L}=10atm[/tex]

[tex]N_2O_4(g)\rightleftharpoons 2NO_2(g)[/tex]

at t= 0    5 atm                                10 atm

at eqm    (5-x) atm                          (10+2x) atm

[tex]K_p=\frac{[p_NO_2]^2}{[p_N_2O_4]}[/tex]

[tex]0.1134=\frac{(10+2x)^2}{(5-x)}[/tex]

[tex]x=-4.48[/tex]

pressure of [tex]N_2O_4[/tex] at equilibrium = (5-(-4.48))= 9.48 atm

pressure of [tex]N_2O_4[/tex] = [tex]\frac{n\times 0.0821Latm/Kmol\times 298}{V}[/tex]

9.48 = [tex]{M\times 0.0821Latm/Kmol\times 298}[/tex]

[tex]M=0.379[/tex]

Thus the equilibrium value of [tex]N_2O_4[/tex] is 0.379 M

Answer:

[tex]w/v\%=3.02\frac{g}{mL} \%[/tex]

Explanation:

Hello,

In this case, we first define the formula for the calculation of weight/volume percentage considering cobalt (II) fluoride as the solute, water the solvent and the both of them as the solution:

[tex]w/v\%=\frac{mass_{solute}}{V_{solution}}*100\%[/tex]

In such a way, since the mass of the solute is given as 6.04 g and the final volume of the solution 200 mL, the weight/volume percentage turns out:

[tex]w/v\%=\frac{6.04g}{200mL}*100\%\\\\w/v\%=3.02\frac{g}{mL} \%[/tex]

Regards.

Answer:

Anode is the positively charged electrode which has the following characteristics:

1) Electrons leave anode to enter to the cathode by the battery.

2) Negatively charged ions are attracted towards cathode.

3) It is connected to the positive terminal of the battery.

Answer:

a) 1.5 x 10^-3 mol/L

b) 1.35×10^-8

c) decrease

Explanation:

The solubility of lead II iodide is given by the equation;

PbI2(s) -----> Pb^2+(aq) + 2I^-

By looking at the ICE table, I^-=2x= 3.0 x 10^-3 mol/L/2 = 1.5×10^-3 mol/L

Hence molar solubility of PbI2 = 1.5 x 10^-3 mol/L

Ksp= [Pb^2+] [2I^-]^2 =

Let the molar solubility of each ion be x, therefore;

Ksp= 4x^3

Ksp= 4(1.5 x 10^-3 mol/L)^3= 1.35×10^-8

Addition of kI to the saturated solution will shift the equilibrium position to the left thereby decreasing the solubility of the PbI2 in the system due to common ion effect. The concentration of the iodide ion is now excess in the system leading to the reverse reaction being favoured according to Le Chateliers principle.

a) The molar solubility of PbI₂ is  [tex]1.5 * 10^{-3} mol/L[/tex]

b) The solubility constant is [tex]1.35*10^{-8}[/tex]

c) The molar solubility of lead (II) will decrease.

The solubility of lead II iodide is given by the equation;

[tex]PbI_2(s) ----- > Pb^{2+}(aq) + 2I^-[/tex]

By looking at the ICE table,

[tex]I^-=2x= 3.0 * 10^{-3} mol/L/2 =[/tex]  [tex]1.5 * 10^{-3} mol/L[/tex]

Hence, molar solubility of PbI2 = [tex]1.5 * 10^{-3} mol/L[/tex]

[tex]Ksp= [Pb^{2+}] [2I^-]^2[/tex]

Let the molar solubility of each ion be x, therefore;

[tex]Ksp= 4x^3\\\\Ksp= 4(1.5 * 10^{-3} mol/L)^3\\\\Ksp= 1.35*10^{-8}[/tex]

The addition of KI to the saturated solution will shift the equilibrium position to the left thereby decreasing the solubility of the PbI₂ in the system due to common ion effect. The concentration of the iodide ion is now excess in the system leading to the reverse reaction being favoured according to Le- Ch-ateliers principle.

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Answer:

1. Increased levels of fructose-2,6-bisphosphatase : Activate gluconeogenesis Inhibit glycolysis

2. Activation of fructose-2,6-bisphosphate (FBPase-2) : Activate glycolysis Inhibit gluconeogenesis

3. Increased glucagon levels : Activate gluconeogenesis Inhibit glycolysis

4. Activation of PFK-2 : Activate glycolysis Inhibit gluconeogenesis

5. Increased levels of CAMP : Activate gluconeogenesis Inhibit glycolysis

Explanation:

Glycolysis is the breakdown of glucose molecules in order to release energy in the form of ATP in response to the energy needs of the cells of an organism.

Gluconeogenesis is the process by which cells make glucose from other molecules for other metabolic needs of the cell other than energy production.

Glycolysis and gluconeogenesis are metabolically regulated in the cell by various enzymes and molecules.

The following shows the various regulatory methods and their effects on both processes:

1. The enzyme fructose-2,6-bisphosphatase functions in the regulation of both processes. It catalyzes the breakdown of the molecule fructose-2,6-bisphosphate which is an allosteric effector of two enzymes phosphofructokinasse-1, PFK-1 and fructose-1,6-bisphosphatase, FBPase-1 which fuction in glycolysis and gluconeogenesis respectively.

Increased levels of fructose-2,6-bisphosphatase  activates gluconeogenesis and inhibits glycolysis by its breakdown of fructose-2,6-bisphosphate.

2. Fructose-2,6-bisphosphate increases the activity of PFK-1 and inhibits the the activity of FBPase-1. The effect is that glycolysis is activated while gluconeogenesis is inhibited.

3. Glucagon is a hormone that stimulates the synthesis of cAMP. It fuctions to activate gluconeogenesis and inhibit glycolysis.

4. Phosphosfructikinase-2, PFK-2 is an enzyme that catalyzes the formation of fructose-2,6-bisphosphate. Activation of PFK-2 results the activation of glycolysis and inhibition of gluconeogenesis.

5. Cyclic-AMP (cAMP) synthesis in response to glucagon release serves to activate a cAMP-dependent protein kinase which phosphorylates the bifunctional protein PFK-2/FBPase-2. This phosphorylation enhances the activity of FBPase-2 while inhibiting the activity of PFK-2, resulting in the  activation of gluconeogenesis and inhibition of glycolysis.

Answer:

Molecule C

Explanation:

In this case, on the first reaction, we will have the production of a Grignard reagent. This molecule will react with [tex]D_2O[/tex] and a deuterium atom will be transferrred to the benzene ring. Then at the top of the molecule, we will have an acetal structure. This acetal can be broken by the action of the acid [tex]DCl[/tex], In the mechanism at the end, we will obtain a carbonyl group bonded to a hydrogen atom. Therefore we will have in the final product the aldehyde group. See figure 1 to further explanations.

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Answer:

pH = 3.95

Explanation:

It is possible to calculate the pH of a buffer using H-H equation.

pH = pka + log₁₀ [HCOONa] / [HCOOH]

If concentration of [HCOONa] = [HCOOH] = 0.50M and pH = 3.77:

3.77 = pka + log₁₀ [0.50] / [0.50]

3.77 = pka

Knowing pKa, the NaOH reacts with HCOOH, thus:

HCOOH + NaOH → HCOONa + H₂O

That means the NaOH you add reacts with HCOOH producing more HCOONa.

Initial moles of 100.0mL = 0.1000L:

[HCOOH] = (0.50mol / L) ₓ 0.1000L = 0.0500moles HCOOH

[HCOONa] = (0.50mol / L) ₓ 0.1000L = 0.0500moles HCOONa

After the reaction, moles of each species is:

0.0500moles HCOOH - 0.010 moles NaOH (Moles added of NaOH) = 0.0400 moles HCOOH

0.0500moles HCOONa + 0.010 moles NaOH (Moles added of NaOH) = 0.0600 moles HCOONa

With these moles of the buffer, you can calculate pH:

pH = 3.77 + log₁₀ [0.0600] / [0.0400]

When the pH be after 0.010 mol of NaOH has been added to 100.0 mL of the buffer pH is = 3.77 + log₁₀ [0.0600] / [0.0400] = 3.95

It is possible to Computation the pH of a buffer using H-H equation.

Then pH is = pka + log₁₀ [HCOONa] / [HCOOH]

Then If concentration of [HCOONa] is = [HCOOH] then = 0.50M and pH = 3.77:

3.77 is = pka + log₁₀ [0.50] / [0.50]

After that, 3.77 = pka

Then, Knowing pKa, the NaOH reacts with HCOOH, thus:

After that,[tex]HCOOH + NaOH \rightarrow HCOONa + H2O[/tex]

Now, That means the NaOH you add reacts with HCOOH producing more HCOONa.

Then, Initial moles of 100.0mL = 0.1000L:

After that, [HCOOH] = (0.50mol / L) ₓ 0.1000L = 0.0500moles HCOOH

Then, [HCOONa] = (0.50mol / L) ₓ 0.1000L = 0.0500moles HCOONa

After that, when the reaction, moles of each species is:

Then, 0.0500moles HCOOH - 0.010 moles NaOH (Moles added of NaOH) = 0.0400 moles HCOOH

Now, 0.0500moles HCOONa + 0.010 moles NaOH (Moles added of NaOH) = 0.0600 moles HCOONa

Then, With these moles of the buffer, you can calculate pH:

pH = 3.77 + log₁₀ [0.0600] / [0.0400]

Therefore, pH = 3.95

Find more information about Formic Acid here:

https://brainly.com/question/26708431

Answer:

D) 4f

Explanation:

To determine which electron orbital that will have higher energy than a 4d orbital, we write the electron configuration starting with s-orbital.

1s

2s         2p

3s         3p          3d       3f

4s         4p          4d       4f

5s         5p           5d       5f

6s         6p          6d       6f

7s         7p           7d       7f

In ascending order, 1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p, 5s, 3f, 4d, 5p, 6s, 4f, 5d, 6p, 7s, 5f, 6d, 7p, 6f, 7d, 7f.

From the electronic configuration formula above, the electron orbitals that have higher energy than a 4d orbital are 5p, 6s, 4f, 5d, 6p, 7s, 5f, 6d, 7p, 6f, 7d, 7f.

Therefore, 4f is the correct answer.

Answer:

D) 4f

Explanation:

Answer:

AI(NO₃)₃ → Acidic   pH < 7

CH₃NH₃CN  → Neutral  pH = 7

NaCIO  → Basic   pH > 7

CH₃NH₃CI → Acidic   pH < 7

NaNO₃ → Neutral  pH = 7

Explanation:

First of all we dissociate the salts:

Nitrate anion comes from the nitric acid which is strong, so the anion is the conjugate weak base. It does not react to water, but the Al is an special case. Aluminum as a cathion comes from the Al(OH)₃ which is a base but this compound can also react as an acid, it is called amphoterous.

Al³⁺  +  H₂O  ⇄ Al(OH)²⁺  +  H⁺

Aluminium cathion reacts to water in order to produce a complex and to give protons to the medium, so the salt is acid.

Both ions come from a weak base and a strong acid, so both ions are the conjugate strong base and acid, respectively. They can make hydrolysis to water so the salt is neutral.

CH₃NH₃⁺   +  H₂O  ⇄  CH₃NH₂  +  H₃O⁺     Ka

CN⁻  +  H₂O  ⇄  HCN +  OH⁻   Kb

Sodium cathion, comes from the strong base NaOH so it is does not react to water. It is the conjugate weak acid. Hypochlorite comes from the weak acid, so it can hydrolyse to water.

ClO⁻  +  H₂O  ⇄  HClO  +  OH⁻     Kb

Hypochlorous acid is formed giving OH⁻ to medium, so the salt is basic.

Chloride comes from the strong acid HCl. It does not react to water.

Methylammonium comes from the weak base, methylamine so it can react to water in order to make hydrolysis. The salt will be acidic.

CH₃NH₃⁺   +  H₂O  ⇄  CH₃NH₂  +  H₃O⁺     Ka

NaNO₃ → Na⁺  +  NO₃⁻

Both ions come from a strong base and acid, so they are the conjugate base and acid, respectively. As they do not make hydrolisis in water, the salt will be neutral.

Answer:

2K+(aq) + CO3²¯(aq) + Ca^2+(aq) + 2F¯(aq) —› Cu2CO3(s) + 2K+(aq) + 2F¯(aq)

Explanation:

K2CO3(aq) + 2CuF(aq) → Cu2CO3(s) + 2KF(aq)

The complete ionic equation for the above equation can be written as follow:

In solution, K2CO3 and CuF will dissociate as follow:

K2CO3(aq) —› 2K+(aq) + CO3²¯(aq)

CuF(aq) —› Ca^2+(aq) + 2F¯(aq)

Thus, we can write the complete ionic equation for the reaction as shown below:

K2CO3(aq) + 2CuF(aq) —›

2K+(aq) + CO3²¯(aq) + Ca^2+(aq) + 2F¯(aq) —› Cu2CO3(s) + 2K+(aq) + 2F¯(aq)

Answer:

3.97

Explanation:

pH of buffer solution = pKa+Log(Cb/Ca)

pH of buffer solution = -log(Ka)+log(Cb/Ca)............... Equation 1

Where Ca = concentration of acid, Cb = concentration of base.

Given: Ka = 6.5×10⁻⁵, Ca = 0.25 M, Cb = 0.15 M

Substitute into equation 1

pH of buffer solution = -log(6.5×10⁻⁵)+log(0.15/0.25)

pH of buffer solution = 4.19+(0.22)

pH of buffer solution = 3.97.

Answer:

Ethyl cinnamate

Explanation:

For this question, we have to start with the IR info. If we have a peak at 1710 this indicates the presence of a carbonyl group in the molecule (C=O). Additionally, if we calculate the I.H.D (index of hydrogen deficiency), we will have a value of "6". We already know that we have a C=O group, so, this counts for 1 of the 6 additionally, we can have a benzene ring so, this counts for 4, so far we have 5. Finally, we will have a double bond outside of benzene and we will have a total of 6, so:

Benzene: 4

Carbonyl group: 1

Double bond: 1

For a total of six (that fits with the I.H.D calculation). So, so far we know that we have a benzene ring, a double bond, and a carbonyl group. In the formula we have 2 oxygens, therefore we can have a carboxylic acid or an ester. In this case, the IR info doesn't give any additional info, so our best option is the ester group.

The 1H NMR info give is:

Signal A= 1.3 (3 H, triplet)

Signal B= 4.3 (2 H, quartet)

Singal C= 6.5 (1 H, doublet)

Signal D= 7.4-7.6 (5 H, multiplet)

Signal E= 7.7 (1 H, doublet)

The molecule that fits with this NMH spectrum and the info given by the I.H.D is "ethyl cinnamate".

See figure 1

I hope it helps!

Answer:

aldehyde

carbon-1

ketone

carbon-2

Explanation:

Monosaccharides are colorless crystalline solids that are very soluble in water. Moat have a swwet taste. D-Fructose is the sweetest monosaccharide.

In the open chain form, monosaaccharides have a carbonuyl group in one of their chains. If the carbonyl group is in the form of an aldehyde group, the monosaccharide is an aldose; if the carbonyl group is in the form of a ketone group, the monosaccharide is known as a ketose. glucose is an aldose while fructose is a ketose.

In D-glucose, there is an aldehyde functional group, and the carbonyl group is at carbon-1 when looking at the Fischer projection.

In D-fructose, there is a ketone functional group, and the carbonyl group is at carbon-2 when looking at the Fischer projection.

Answer:

ΔH = -338.8kJ

Explanation:

it is possible to sum the enthalpy changes of some reactions to obtain the enthalpy change of the whole reaction (Hess's law).

Using the reactions:

R₁ O₂(g) → 2O(g) ΔH₁°= 449.2 kJ

R₂ O₃(g) + Cl(g) → O₂(g) + ClO(g) ΔH₂° = -126 kJ

R₃ ClO (g) + O (g) → O₂ (g) + Cl (g) ΔH₃°= -268 kJ

By the sum 2R₂ + 2R₃:

(2R₂ + 2R₃) = 2O(g) + 2O₃(g) → 4O₂(g)

ΔH = 2ₓ(-126kJ) + (2ₓ-268kJ) = -788kJ

Now, this reaction + R₁

2O₃(g) → 3O₂(g)

ΔH = -768kJ + 449.2kJ