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Cellulose is a polymer that makes up plant cell walls. Cellulose is made from a chain of C6H10O5 molecules.

Which monomers are most likely used to produce cellulose?

a)nucleotides and amino acids
b)glucose molecules
c)nucleotides
d)amino acids

By applying the modified Gibbs Free Energy formula with given values for equilibrium constants, atom concentrations, and other parameters, we find ΔG for nitrous acid at 25°C to be around 27.94 KJ/mol.

In chemistry, the Gibbs free energy (ΔG) is calculated using the equation ΔG = -RTlnK, where R is the gas constant, T is the temperature in Kelvin, and K is the equilibrium constant. However, since we're given Ka, the equation must be adapted. Therefore, we use ΔG = -RTlnKa + RTlnQ, where Q is the reaction quotient given by [NO₂⁻][H⁺] / [HNO₂].

Inserting the given values, such as Ka = 4.5×10⁻⁴, R (in appropriate units) as 0.0083145 KJ/(mol.K), T as 298.15K (25°C in Kelvin), and Q = ([NO₂⁻][H⁺]) / [HNO₂] = (6.3×10⁻⁴ × 5.9×10−2) / 0.21, we can now solve for ΔG. Doing the math, we find that ΔG ≈ 27.94 KJ/mol.

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To neutralize 20.0 mL of 0.250 M NaOH, you will need 50.0 mL of 0.100 M HCl.

To determine how much 0.100 M HCl is required to neutralize 20.0 mL of 0.250 M NaOH, we can use the balanced equation for the reaction between HCl and NaOH: HCl(aq) + NaOH(aq) → H2O(l) + NaCl(aq). The mole ratio between HCl and NaOH is 1:1, meaning that 1 mole of HCl reacts with 1 mole of NaOH. We can use this ratio to calculate the amount of HCl needed.

First, find the number of moles of NaOH:

0.250 M NaOH x 0.0200 L = 0.005 moles NaOH

Since the mole ratio between HCl and NaOH is 1:1, we need 0.005 moles of HCl to neutralize the NaOH.

Now, calculate the volume of 0.100 M HCl needed to contain 0.005 moles:

0.005 mol HCl / 0.100 mol/L = 0.050 L = 50.0 mL

Therefore, 50.0 mL of 0.100 M HCl is required to completely neutralize 20.0 mL of 0.250 M NaOH.

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The empirical formula of the compound is C₁₈H₂₇NO₃

From the question given above, the following data were obtained:

Carbon (C) = 70.79%

Hydrogen (H) = 8.91%

Nitrogen (N) = 4.59%

Oxygen (O) = 15.72%

The empirical formula of the compound can be obtained as follow:

C = 70.79%

H = 8.91%

N = 4.59%

O = 15.72%

C = 70.79 / 12 = 5.899

H = 8.91 / 1 = 8.91

N = 4.59 / 14 = 0.328

O = 15.72 / 16 = 0.9825

C = 5.899 / 0.328 = 18

H = 8.91 / 0.328 = 27

N = 0.328 / 0.328 = 1

O = 0.9825 / 0.328 = 3

Therefore, the empirical formula of the compound is C₁₈H₂₇NO₃

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For producing 61 L of [tex]\rm CO_2[/tex], 227.23 grams of  [tex]\rm CaCO_3[/tex]. is required.

Any gas occupies 22.4 L/mol space at STP.

So, 61.0 L of gas will be;

Moles of [tex]\rm CO_2[/tex] = [tex]\rm \frac{61.0}{22.4}[/tex]

Moles of  [tex]\rm CO_2[/tex] = 2.723 moles

From the reaction,

1 mole of [tex]\rm CO_2[/tex] has been produced by 1 mole of [tex]\rm CaCO_3[/tex].

2.723 moles of [tex]\rm CO_2[/tex] has been produced by 2.723 mole of [tex]\rm CaCO_3[/tex].

Mass = [tex]\rm moles\;\times\;molecular\;weight[/tex]

Mass of [tex]\rm CaCO_3[/tex]. = 2.723 [tex]\times[/tex] 100

Mass of [tex]\rm CaCO_3[/tex]. = 227.23 grams.

For producing 61 L of [tex]\rm CO_2[/tex], 227.23 grams of  [tex]\rm CaCO_3[/tex]. is required.

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Final answer:

Potassium iodide-starch paper is used to detect the presence of nitrogen dioxide in the preparation of diazonium salts by revealing a blue-black coloration upon reacting with iodine produced from the reaction between NO2 and potassium iodide.

Explanation:

The potassium iodide-starch paper is used to test for the presence of nitrogen dioxide (NO2), which is a byproduct of the reaction used to prepare diazonium salts. During the diazo coupling reaction, excess nitrous acid can decompose and produce nitrogen dioxide. This gas can then react with the potassium iodide (KI) in the starch paper to produce iodine (I2), which subsequently forms a blue-black complex with starch. The immediate formation of the blue color on the potassium iodide-starch paper is a positive test indicating the presence of nitrogen dioxide. It's important to monitor this because the presence of NO2 suggests that the diazonium salt solution might be unsafe due to the potential release of toxic gases.

The iodine-starch test is a well-known reaction in which iodine (I2), produced by the oxidation of iodide ions by NO2, interacts with starch to produce a characteristic blue-black color. This test provides a quick and sensitive method for detecting the presence of iodine, which, in this context, indirectly indicates the generation of nitrogen dioxide in the reaction mixture.

Carbon has a higher boiling point than [tex]CH_4[/tex].

The boiling point of a substance is defined as the temperature at which the vapor pressure of the liquid becomes equal to the pressure surrounding the liquid and the liquid changes into vapor. This point of liquid varies depending on the surrounding environmental pressure.

The boiling point is the temperature at which the pressure exerted by the surroundings on a liquid is equal to the pressure exerted by the liquid's vapor, in which case the liquid changes to vapor without additional heat raising the temperature.

For above given example,

[tex]CH_4[/tex] is a gas at room temperature having a boiling point of -258.7 F (-161.5 C) which is why it’s boiling point is so low while Carbon is a solid at room temperature having a boiling point of 8,721 F (4,827 C)

Thus, Carbon has a higher boiling point than [tex]CH_4[/tex].

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The  skeletal structures of a 2° alkyl bromide with two  stereogenic centers is shown below.

A carbon atom in a molecule that is linked to four separate substituents is referred to as a stereogenic center, also referred to as a chiral center. Because secondary carbon atoms normally have two identical alkyl groups and two hydrogen atoms linked to them.

The two of the substituents (two alkyl groups) are the same. Tetrahedral carbon compounds with four distinct substituents are frequently where chirality occurs.

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Equilibrium in a reversible reaction is established when the rate of the forward reaction (reactants transforming into products) equals the rate of the backward reaction (products reconverting into reactants). This indicates that the amounts of products and reactants are no longer changing over time.

In a reversible reaction, equilibrium is established when option (b) is correct: the rate of the forward reaction (reactants transforming into products) becomes equal to the rate of the backward (or reverse) reaction (products converting back into reactants). This does not necessarily mean the concentrations of the products and reactants are equal. Rather, it means the amounts of products and reactants are no longer changing over time, demonstrating a state of dynamic equilibrium. The equilibrium can shift depending on external factors and conditions such as temperature, pressure, or concentration changes, which is described by Le Chatelier's Principle.

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Reaction of reduction at cathode(-): Ba²⁺(l) + 2e⁻ → Ba(l).

Reaction of oxidation at anode(+): 2Cl⁻(l) → Cl₂(g) + 2e⁻.

The anode is positive and the cathode is negative.

The study of chemicals and bonds is called chemistry. There are two types of elements and these are metals and nonmetals.

The correct answer is -12.266.

Elevation in boiling point is mathematically expressed as

[tex]Tb = Kb * m[/tex]

Where

ΔTb[tex]= 2.53 * 3.47 = 8.779 oC[/tex]

But, the boiling point of benzene = 80.1 oC

Boiling point of solution = 88.879 oC

Now, Depression in freezing point = ΔTf[tex]= Kf * m[/tex]

where,

ΔTf =[tex]5.12 X*3.47 = 17.766 oC[/tex]

But the freezing point of benzene = 5.5 oC.

Freezing point of solution = -12.266 oC.

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Explanation:

The given equation is as follows.

       [tex]PbBr_2 \rightleftharpoons Pb^{2+} + 2Br^{-}[/tex]

                          s       2s

It is given that,

       [tex]K_{sp} = [Pb^{2+}][Br^{-}]^{2} = 6.60 \times 10^{-6}[/tex]

Let the solubility of given ions be "s".

Since, KBr on dissociation will given bromine ions.

Hence,  [tex]K_{sp} = [Pb^{2+}] \times ([Br^{-}])^{2}[/tex]

        [tex]6.60 \times 10^{-6}[/tex]  = [tex]s \times (2s)^{2}[/tex]

                        = [tex]1.18 \times 10^{-2}[/tex] M

Therefore, solubility of [tex][PbBr_{2}][/tex] is [tex]1.18 \times 10^{-2}[/tex] M  in KBr.

Now, we will calculate the molar solubility of [tex]PbBr_{2}[/tex] in 0.5 M KBr solution as follows.

           [tex]K_{sp} = (s) \times (2s + 0.5)^{2}[/tex]

  [tex]6.60 \times 10^{-6}[/tex]  = [tex]4s^{3} + 0.25s + 2s^{2}[/tex]

                         s = [tex]2.63 \times 10^{-5}[/tex]

Thus, we can conclude that molar solubility of [tex][PbBr_{2}][/tex] in 0.500 m KBr solution is [tex]2.63 \times 10^{-5}[/tex].

The chemical reaction is given as:

[tex]2NaDP^{+} +2H_{2}O\rightarrow 2NaDPH+2H^{+}+O_{2}[/tex]

Here, oxygen is oxidised and [tex]NaDP^{+}[/tex] is reduced. Thus, redox reaction occurs.

For cell reaction, [tex]\Delta G^{o} = -nFE^{o}_{cell}[/tex]          (2)

where, [tex]\Delta G^{o} [/tex] = standard state free energy

n= number of electrons

F= Faraday constant ([tex]96485.33 C/mol[/tex])

[tex]E^{o}_{cell}[/tex] = cell potential

Substitute the value of number of electrons i.e. 2, Faraday constant and cell potential in the formula to determine the value of  [tex]\Delta G^{o} [/tex].

Now, calculate the value of cell potential

[tex]E^{o}_{cell} = E^{o}_{cathode}- E^{o}_{anode}[/tex]           (1)

[tex]E^{o}_{cathode}[/tex] = [tex]-0.324 V[/tex] (standard reduction potential of [tex]NaDP^{+}[/tex])

[tex]E^{o}_{anode}[/tex] = [tex]1.23 V[/tex] (standard reduction potential of [tex]O_{2}[/tex])

Put the above values in formula (1), we get:

[tex]E^{o}_{cell} = -0.324 V-1.23 V[/tex]

= [tex]-1.554 V[/tex]

Now, substitute above value in formula (2)

[tex]\Delta G^{o} = -2\times 96485.33 C/mol \times(-1.554 V) [/tex]  

= [tex]299876.40564 CV/mol[/tex]

Since, one coulomb volt is equal to one joule.

Thus, value of [tex]\Delta G^{o}[/tex] is equal to [tex]299876.40564 J/mol[/tex] or [tex]299.87640564 kJ/mol[/tex]

Option B: Nuclear fission reactions rarely occur naturally while nuclear fusion reactions occur in the stars.

Nuclear fission reaction is defined as formation of two or more atoms by splitting of large atoms. On the other hand, nuclear fusion reaction is formation of large atom from small atoms.

The splitting of large atoms into small atoms (nuclear fission) generally does not occur naturally because it requires high speed neutrons and critical mass of the substance undergoing fission.

Nuclear fusion reaction occurs in stars such as sun because it requires extremely high energy to bring two more proton closer to each other by overcoming electronic repulsion.

Therefore, nuclear fission reactions rarely occur naturally while nuclear fusion reactions occur in the stars.

Answer:

The concentration of hydronium ions in the solution is [tex]4.35\times 10^{-13} M[/tex].

Explanation:

Concentration of hydroxide ions = [tex][OH^-]=2.3\times 10^{-2}[/tex]

Concentration of hydroxide ions = [tex][H_3O^+]=?[/tex]

[tex]H_2O+H_2O\rightleftharpoons H_3O^++OH^-[/tex]

The ionic product of water is given as:

[tex]K_w=[H_3O^+][OH^-][/tex]

The value of ionic product of water, [tex]K_w=1\times 10^{-14}[/tex]

[tex]K_w=1\times 10^{-14}=[H_3O^+][OH^-][/tex]

[tex]1\times 10^{-14}=[H_3O^+]\times 2.3\times 10^{-2}M[/tex]

[tex][H_3O^+]=\frac{1\times 10^{-14}}{2.3\times 10^{-2}}=4.35\times 10^{-13} M[/tex]

The concentration of hydronium ions in the solution is [tex]4.35\times 10^{-13} M[/tex].

The concentration of hydronium, ion [H₃O⁺] in the solution containing 2.3×10⁻² M concentration of hydroxide ion, [OH⁻] is 4.35×10⁻¹³ M

The following data were obtained from the question given above:

[H₃O⁺] × [OH⁻] = 10¯¹⁴

[H₃O⁺] × 2.3×10⁻² = 10¯¹⁴

Divide both side by 2.3×10⁻²

[H₃O⁺] = 10¯¹⁴ / 2.3×10⁻²

= 4.35×10⁻¹³ M

Thus, we can conclude that the concentration of hydronium ion, [H₃O⁺] in the solution is 4.35×10⁻¹³ M

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The average kinetic energy of hydrogen molecules at 100 K with three degrees of freedom can be calculated using the formula kavg = 3/2 * k * T, where k is Boltzmann's constant and T is the temperature in Kelvin.

The average kinetic energy kavg of hydrogen molecules at a temperature of 100 K with three degrees of freedom can be calculated using the formula:

kavg = 3/2 * k * T

where k is Boltzmann's constant (1.38 x 10^-23 J/K) and T is the temperature in Kelvin. Plugging in the values:

The new freezing point of the solution is -8.63 °C.

Using the ideal gas law, PV = nRT, we need to convert 4.00 L of gas at 27°C (which is 300 K) and 748 mmHg to moles of gas.

P = 748 mmHg / 760 mmHg atm = 0.984 atm

[tex]n = \frac{PV}{RT} = \frac{0.984 \, \text{atm} \times 4.00 \, \text{L}}{0.0821 \, \text{L atm / K mol} \times 300 \, \text{K}} = 0.160 \, \text{mol}[/tex]

Calculate molality:

Molality (m) = moles of solute / kg of solvent = 0.160 mol / 0.0580 kg = 2.759 m

Calculate freezing point depression (ΔTf):

ΔTf = i * Kf * m

For non-electrolytes, i = 1.

ΔTf = 1 * 5.12 °C/m * 2.759 m = 14.13 °C

Determine the new freezing point:

The freezing point of pure benzene is 5.5 °C.

New freezing point = 5.5 °C - 14.13 °C = -8.63 °C

Thus, the freezing point of the solution is -8.63 °C.

The pH of the new solution has been 4. Thus, the solution has been more acidic than the precious solution.

The pH has been defined as the negative log of hydronium ion concentration in the solution. The pH has been expressed as:

[tex]\rm pH=-\;log\;[H_3O^+][/tex]

The pH of the given solution has been 7. The hydronium ion concentration has been given as:

[tex]\rm pH=\;-log\;[H_3O^+]\\\\ 7=\;-\;log[H_3O^+]\\\\ H_3O^+=10^-^7\;M[/tex]

The concentration of the new solution has been 1000 times greater than the previous solution.

The previous solution has hydronium ion concentration of [tex]\rm 10^-^7\;M[/tex]. The concentration of new solution will be:

[tex]\rm New\;solution=10^-^7\;\times\;1000\;M\\ New\;solution=10^-^4\;M[/tex]

The concentration of the new solution has been [tex]\rm 10^-^4\;M[/tex]. The pH of the solution has been given as:

[tex]\rm pH=-log\;[H_3O^+]\\ pH=-log\;[10^-^4]\\ pH=4[/tex]

The pH of the new solution has been 4. Thus, the solution has been more acidic than the precious solution.

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