A variation of the acetamidomalonate synthesis can be used to synthesize serine. The process involves the following steps: Ethoxide ion deprotonates diethyl acetamidomalonate, forming enolate anion 1; Enolate anion 1 makes a nucleophilic attack on formaldehyde, forming tetrahedral intermediate 2; Protonation of the oxyanion forms alcohol 3; Acid hydrolysis yields dicarboxyamino alcohol 4; Decarboxylation leads to the final amino acid. Write out the mechanism on a separate sheet of paper, and then draw the structure of dicarboxyamino alcohol 4.

Answer:

Aldol condensation is possible only when their is alpha Hydrogen atom is present. It ia present only in the acetophenone and not in benzaldehyde.

Explanation:

Answer:

1. 2NaN₃(s) → 2Na(s) + 3N₂(g)

2. 14.5 g NaN₃

Explanation:

The answer is incomplete, as it is missing the required values to solve the problem. An internet search shows me these values for this question. Keep in mind that if your values are different your result will be different as well, but the solving methodology won't change.

" The airbags that protect people in car crashes are inflated by the extremely rapid decomposition of sodium azide, which produces large volumes of nitrogen gas. 1. Write a balanced chemical equation, including physical state symbols, for the decomposition of solid sodium azide (NaN₃) into solid sodium and gaseous dinitrogen. 2. Suppose 71.0 L of dinitrogen gas are produced by this reaction, at a temperature of 16.0 °C and pressure of exactly 1 atm. Calculate the mass of sodium azide that must have reacted. Round your answer to 3 significant digits. "

1. The reaction that takes place is:

2. We use PV=nRT to calculate the moles of N₂ that were produced.

P = 1 atm

V = 71.0 L

n = ?

T = 16.0 °C ⇒ 16.0 + 273.16 = 289.16 K

Now we convert N₂ moles to NaN₃ moles:

Finally we convert NaN₃ moles to grams, using its molar mass:

Answer:

Newton's three laws of motion may be stated as follows:

Every object in a state of uniform motion will remain in that state of motion unless an external force acts on it.

Force equals mass times acceleration

For every action there is an equal and opposite reaction

Explanation:

pls mark brainliest

The scaling factor is determined by dividing the compound's actual molar mass (150 g/mol) by the molar mass of its empirical formula CH2O (30 g/mol), which results in a factor of 5. The molecular formula of the compound is C5H10O5.

The student's question is asking about the scaling factor of a compound's molar mass based on its empirical formula. The empirical formula is CH2O, which has a molar mass of 30 g/mol (C = 12, H = 1 x 2 = 2, O = 16). If the compound's actual molar mass is 150 g/mol, we divide the actual molar mass by the empirical formula's molar mass to find the scaling factor.

Scaling factor = Actual molar mass / Empirical formula molar mass

= 150 g/mol / 30 g/mol

= 5

Thus, the actual compound's formula is obtained by multiplying each subscript in the empirical formula by the scaling factor of 5. Therefore, the molecular formula of the compound is C5H10O5.

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Answer:

AsH3(g) + 4H2O(l)------> H3AsO4(aq) +8H^+(aq) + 8e-

Explanation:

Now we need to work through the problem in steps.

Step1: reaction of arsine with water

AsH3(g) + 4H2O(l) -----> H3AsO4(aq) this is the molecular reaction

Step II: oxygen is balanced using hydrogen ions

AsH3(g) + 4H2O(l)------> H3AsO4(aq) +8H^+(aq)

Step III: We specify the number of electrons transferred in the redox reaction

AsH3(g) + 4H2O(l)------> H3AsO4(aq) +8H^+(aq) + 8e-

The balanced half-reaction for the oxidation of gaseous arsine AsH₃ to

aqueous arsenic acid H₃AsO₄ in acidic aqueous solution is

AsH₃(g) + 4H₂O(l)------> H₃AsO₄(aq) +8H+(aq) + 8e-

AsH₃(g) + 4H₂O(l) -----> H₃AsO₄(aq)

AsH₃(g) + 4H₂O(l)------> H₃AsO₄(aq) +8H⁺(aq)

AsH₃(g) + 4H₂O(l)------> H₃AsO4(aq) +8H⁺(aq) + 8e⁻

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Answer : The partial pressure of nitrogen gas in the mixture is, 9.88 atm

Explanation :

According to the Dalton's Law, the total pressure of the gas is equal to the sum of the partial pressure of individual gases.

Formula used :

[tex]p_T=p_{He}+p_{Ar}+p_{N_2}[/tex]

where,

[tex]p_T[/tex] = total pressure of gas  = 13.6 atm

[tex]p_{He}[/tex] = partial pressure of helium gas  = 1831 torr =  2.41 atm

[tex]p_{Ar}[/tex] = partial pressure of argon gas  = 997 torr =  1.31 atm

Conversion used: (1 atm = 760 torr)

[tex]p_{N_2}[/tex] = partial pressure of nitrogen gas  = ?

Now put all the given values in the above formula, we get:

[tex]13.6=2.41+1.31+p_{N_2}[/tex]

[tex]p_{N_2}=9.88atm[/tex]

Thus, the partial pressure of nitrogen gas in the mixture is, 9.88 atm

Answer:

(B) ethyl-propyl-ether

Explanation:

ethyl-propyl-ether

a step by step explanation

As the compound has ether as functional group the name of the compound is ethyl propyl ether.

Functional group is defined as a substituent or group of toms or an atom which causes chemical reactions.Each functional group will react similarly regardless to the parent carbon chain to which it is attached.This helps in prediction of chemical reactions.

The reactivity of functional group can be enhanced by making modifications in the functional group .Atoms present in functional groups are linked to each other by means of covalent bonds.They are named along with organic compounds according to IUPAC nomenclature.

The compound has ether as functional group,thus  the name of the compound is ethyl propyl ether.

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Answer :  The value of standard free energy is, -152.4 kJ/mol

Explanation :

The given balanced cell reaction is:

[tex]FADH_2+\frac{1}{2}O_2\rightarrow FAD+H_2O[/tex]

The half reaction will be:

Reaction at anode (oxidation) : [tex]FADH_2\rightarrow FAD+2H^++2e^-[/tex]     [tex]E^0_{Anode}=+0.03V[/tex]

Reaction at cathode (reduction) : [tex]\frac{1}{2}O_2+2H^++2e^-\rightarrow H_2O[/tex]     [tex]E^0_{Cathode}=+0.82V[/tex]

First we have to calculate the standard electrode potential of the cell.

[tex]E^o=E^o_{cathode}-E^o_{anode}[/tex]

[tex]E^o=(+0.82V)-(+0.03V)=+0.79V[/tex]

Relationship between standard Gibbs free energy and standard electrode potential follows:

[tex]\Delta G^o=-nFE^o_{cell}[/tex]

where,

[tex]\Delta G^o[/tex] = standard free energy = ?

n = number of electrons transferred = 2

F = Faraday constant = [tex]96.48kJ.mol^{-1}V^{-1}[/tex]

[tex]E^o_{cell}[/tex]  = standard electrode potential of the cell = 0.79 V

Now put all the given values in the above formula, we get:

[tex]\Delta G^o=-(2)\times (96.48kJ.mol^{-1}V^{-1})\times (0.79V)[/tex]

[tex]\Delta G^o=-152.4kJ/mol[/tex]

Therefore, the value of standard free energy is, -152.4 kJ/mol

To calculate the standard free energy change for the reduction of O2 with FADH2, we first calculate the difference in reduction potentials to find ΔE0′. We then substitute the values into the relation ΔG∘′ = −nFΔE0′, with n=2 and Faraday constant F=96.48 kJ⋅mol‾¹⋅V‾¹. The calculated ΔG∘′ is -152.4384 kJmol‾¹.

The standard free energy change, ΔG∘′, is related to the change in reduction potential, ΔE0′, by the equation ΔG∘′ = −nFΔE0′. To find the standard free energy released by the reduction of O2 with FADH2, we first need to find ΔE0′.

ΔE0′ is given by the difference in reduction potentials of the two half reactions involved. In this case, the reduction of O2 to H2O (+0.82 V) and the reduction of FAD to FADH2 (+0.03 V). Therefore, ΔE0′ = E(O2/H2O) - E(FAD/FADH2) = +0.82 V - (+0.03 V) = +0.79 V.

Inserting the values into the equation, and knowing that the number of transferred electrons (n) is 2 and the Faraday constant (F) is 96.48 kJ⋅mol‾¹⋅V‾¹, we get ΔG∘′ = −2 * 96.48 kJ⋅mol‾¹⋅V‾¹ * (+0.79 V) = -152.4384 kJmol‾¹. Thus, the standard free energy released by reducing O2 with FADH2 is -152.4384 kJmol‾¹.

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Answer:

Weak Acid solution

Explanation:

Acidic substances usually have a pH of 0.1-6.9 or it is usually less than 7. Acidic substances too have a very unique sour taste.

However it is understood that it conducts electricity poorly which means the acid doesn’t readily dissociate in water. This makes it a weak acid and an example is Acetic acid.

Answer:

57.5%

Explanation:

The mass percent of copper in malachite (Cu2(OH)2CO3) can be determined as follow:

Molar Mass of malachite (Cu2(OH)2CO3) = (2x63.5) + 2(16 +1) + 12 + (16x3) = 127 + 2(17) + 12 + 48 = 127 + 34 + 12 + 48 = 221g/mol

Mass of Cu in Cu2(OH)2CO3 = 2 x 63.5 = 127g

The percentage by mass of Cu in Cu2(OH)2CO3 is given by:

Mass of Cu/Molar Mass of Cu2(OH)2CO3 x 100

=> 127/221 x 100

=> 57.5%

Therefore, 57.5% by mass of Cu is contained in malachite Cu2(OH)2CO3

The pH of an aqueous solution at 25.0°C with [H+] = 0.0025 M is calculated using the pH formula and results in a pH of 2.60, which is answer option B.

The pH of an aqueous solution at 25.0°C with a hydrogen ion concentration [H+] of 0.0025 M can be found using the pH formula:

pH = -log [H3O+]

By substituting the given concentration into the formula, the calculation would be:

pH = -log(0.0025) = -log(2.5 x 10-3)

Using a scientific calculator:

pH = 2.60

Therefore, the correct answer is B) 2.60.

Answer:

The most appropriate structure given the sparse spectral data is 4-acetyl benzoic acid (see attached).

Explanation:

It is difficult to accurately elucidate the structure of this compound without its chemical formula. But from the 1H NMR spectral data shows a total of 8 hydrogen atoms:

The attached files show the structure and the neighboring hydrogen atoms.

The most likely structure i 4-acetyl benzoic acid

Answer:

Ksp FeS = 5.2441 E-18

Explanation:

         S          S           S

∴ Ksp = [Fe2+]*[S2-].....solubility product constant

∴ [S2-] = 2.29 E-9 M = S

⇒ Ksp = (S)(S) = S²

⇒ Ksp = (2.29 E-9)²

⇒ Ksp = 5.2441 E-18

Final answer:

The maximum possible mass of NO2 that could be formed from the given amounts of reactants is 755.20 grams, and the amount of CO2 that would be produced from the combustion of 37.3 grams of C6H6 is 126.10 grams.

Explanation:

Calculating the Maximum Amount of NO2 Formed

For the balanced reaction 2 NO + O2 → 2 NO2, the stoichiometry shows a 1:1 ratio between NO and NO2. With 16.42 mol of NO, if O2 is in excess, 16.42 mol of NO2 could theoretically be formed. The molar mass of NO2 is 46.0055 g/mol, so the maximum mass of NO2 would be 16.42 mol × 46.0055 g/mol = 755.20 grams of NO2.

Amount of CO2 Produced from C6H6 Combustion

The balanced reaction for the combustion of benzene (C6H6) is 2 C6H6 + 15 O2 → 12 CO2 + 6 H2O. One mole of C6H6 (78.1134 g/mol) produces 6 moles of CO2. With 37.3 grams of C6H6, there would be 37.3 g / 78.1134 g/mol = 0.4775 mol C6H6 which would yield 0.4775 mol C6H6 × 6 mol CO2/mol C6H6 = 2.865 mol CO2.

The molar mass of CO2 is 44.0095 g/mol, so the mass of CO2 would be 2.865 mol × 44.0095 g/mol = 126.10 grams of CO2.

Final answer:

The maximum amount of NO2 that can be formed from 16.42 mol of NO and 14.47 mol of O2 is 755.1 grams. When reacting 37.3 grams of C6H6 with 126.1 grams of O2, the amount of CO2 produced is 125.96 grams.

Explanation:

Reaction Stoichiometry and Limiting Reactants:

To determine the maximum amount of NO2 that can be formed from 16.42 mol of NO and 14.47 mol of O2, we use the balanced chemical equation 2 NO + O2 → 2 NO2. This equation shows that 2 moles of NO react with 1 mole of O2 to produce 2 moles of NO2. We can see that NO will be the limiting reactant because 14.47 mol of O2 would theoretically react with only 28.94 mol of NO, which is more than the 16.42 mol of NO present.

Since NO is the limiting reactant, we can form a maximum of 16.42 mol of NO2. Using the molar mass of NO2 (46.01 g/mol), this is equivalent to 755.1 grams of NO2. For the reaction C6H6 + O2 → CO2 + H2O, first, we would calculate the moles of C6H6 and O2, then use stoichiometry to determine the moles of CO2 produced. The molar mass of C6H6 (78.11 g/mol) means we start with 0.477 moles of C6H6.

From the balanced equation, we can see that for every 2 moles of NO, we need 1 mole of O2 to produce 2 moles of NO2. This gives us the mole ratio of NO to O2 as 2:1.

To find the limiting reagent (the reactant that is completely consumed and determines the maximum amount of product formed), we compare the number of moles of each reactant to the mole ratio.

For NO:

16.42 mol NO * (1 mol O2 / 2 mol NO) = 8.21 mol O2

For O2:

14.47 mol O2 * (2 mol NO / 1 mol O2) = 28.94 mol NO

Since the calculated amount of O2 (28.94 mol) is greater than the actual amount of O2 (14.47 mol), O2 is in excess and NO is the limiting reagent.

Now, we need to calculate the maximum amount of NO2 produced from the limiting reagent, which is NO.

From the balanced equation, we know that 2 moles of NO react to form 2 moles of NO2. Therefore, the mole ratio of NO to NO2 is 2:2.

Using the moles of NO (8.21 mol), we can calculate the moles of NO2:

8.21 mol NO * (2 mol NO2 / 2 mol NO) = 8.21 mol NO2

To convert moles of NO2 to grams, we need to use the molar mass of NO2, which is 46.01 g/mol.

8.21 mol NO2 * 46.01 g/mol = 377.27 g NO2

Therefore, the maximum amount of NO2 that could be formed from 16.42 mol of NO and 14.47 mol of O2 is 377.27 grams of NO2.

For the second question, we have the balanced chemical equation:

C6H6 + 15O2 → 6CO2 + 3H2O

From the balanced equation, we can see that for every mole of C6H6, we need 15 moles of O2 to produce 6 moles of CO2.

To determine the amount of CO2 produced, we first need to find the limiting reagent.

For C6H6:

37.3 g C6H6 * (1 mol C6H6 / 78.11 g C6H6) = 0.477 mol C6H6

For O2:

126.1 g O2 * (1 mol O2 / 32 g O2) = 3.94 mol O2

Since the calculated amount of C6H6 (0.477 mol) is less than the actual amount of O2 (3.94 mol), C6H6 is the limiting reagent.

Using the mole ratio from the balanced equation, we find that 0.477 mol of C6H6 will produce 6 moles of CO2.

To convert moles of CO2 to grams, we need to use the molar mass of CO2, which is 44.01 g/mol.

0.477 mol CO2 * 44.01 g/mol = 21.0 g CO2

Therefore, 37.3 grams of C6H6 reacted with 126.1 grams of O2 will produce 21.0 grams of CO2.

Answer:

Explanation:

find the solution below

Answer:

The chemical equation is given as:

[tex]4NH_3(g)+5O_2(g)\rightarrow 4NO(g)+6H_2O(g)[/tex]

Explanation:

When gaseous ammonia reacts with gaseous oxygen it gives nitrogen monoxide gas and water vapors as product.

The chemical equation is given as:

[tex]4NH_3(g)+5O_2(g)\rightarrow 4NO(g)+6H_2O(g)[/tex]

According to reaction, 4 moles of ammonia reacts with 5 moles of oxygen gas to give 4 moles of nitrogen monoxide gas and 6 moles of water vapor.

To write the skeletal equation, begin by writing the chemical formula for each reactant and product.

Reactants:

1. Ammonia's chemical formula is given as  NH3.

2. Gaseous oxygen exists as a diatomic molecule with the chemical formula  O2.

Products:

1. Nitrogen monoxide has the chemical formula of  NO.

2. The chemical formula for water is  H2O.

Therefore, the skeletal equation is written as:

NH3(g)+O2(g)→NO(g)+H2O(g)

Now count the number of each atom on each side of the equation to determine if the equation is balanced.

Reactants

1N atom

3H atoms

2O atoms

Products

1N atom

2H atoms

2O atoms

Begin by balancing the number of hydrogen atoms by adding a coefficient of 2 to  NH3 and a coefficient of 3 to  H2O. Next, balance the number of nitrogen atoms by adding a coefficient of 2 to  NO. Now there are two oxygen atoms on the reactant's side and five oxygen atoms on the product's side of the reaction. Since only whole number coefficients should be used, all coefficients need to be increased by a factor of two to balance the oxygen atoms. Thus the coefficient for  NH3 is 4, the coefficient for  H2O is 6, and the coefficient for  NO is 4. Finally, balance the oxygen atoms by adding a coefficient of 5 to  O2. The balanced equation is:

4NH3(g)+5O2(g)→4NO(g)+6H2O(g)

Answer:

price ($) Au = $ 19183.94

Explanation:

april 15,2000:

∴ price Au = $ 282/t.oz

∴ 1.00 t.oz = 28.4 g

∴ V Au = 100.0 cm³  ⇒  price ($) = ?

∴ δ Au = 19.32 g/cm³

⇒ mass Au = (100.0 cm³)*(19.32 g/cm³)

⇒ m Au = 1932 g

⇒ price ($) = (1932 g Au)*(1.00 t.oz/28.4 g Au)*( $ 282/t.oz)

⇒ price ($) = $ 19183.94

To calculate the cost of 100.0cm3 of gold on April 15, 2000, convert the volume to grams, then to troy ounces, and multiply by the price per troy ounce.

To calculate the cost of 100.0cm3 of gold on April 15, 2000, we need to convert the volume of gold to grams and then to troy ounces. First, convert 100.0cm3 to grams by multiplying it by the density of gold (19.3 g/cm3). This gives us 1930 grams. Next, convert grams to troy ounces by dividing by the conversion factor of 28.4 grams per troy ounce. This gives us approximately 67.96 troy ounces. Finally, multiply the number of troy ounces by the price per troy ounce to find the cost. Therefore, 100.0cm3 of gold on April 15, 2000 would have cost $19,191.12.

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Answer:1 2 4

Explanation:

Answer: 1.3.5.6.

Explanation:

Answer:

New volume is 271 mL

Explanation:

To determine the volume for a gas, when the pressure remains constant we follow this ratio:

V₁ / T₁ = V₂ / T₂

Remember the Ideal Gases Law → P . V = n . R . T

That's why we propose V / T

We need to determine the Absolute T°

25°C + 273 = 298 K

50°C + 273 = 323 K

We convert the volume from mL to L → 250 mL . 1L / 1000 mL = 0.250L

Now we replace: 0.250L / 298K = V₂ / 323K

V₂ = (0.250L / 298K) . 323K  → 0.271 L

In conclussion volume of a gas will be increased, while the temperature is also increased and the pressure remains constant.

Answer:

The final volume is 271 mL

Explanation:

Step 1: Data given

The initial pressure = 2 atm

The pressure will be kept constant

The initial volume = 250 mL = 0.250 L

The initial temperature = 25°C = 298 K

The final temperature = 50 °C = 323 K

Step 2: Calculate the final volume

V1/T1 = V2/T2

⇒with V1 = the initial volume = 0.250 L

⇒with T1 = the initial temperature = 25 °C = 298 K

⇒with V2 = the final volume = TO BE DETERMINED

⇒with T2 = the final temperature = 50 °C = 323 K

0.250 L / 298 K = V2 / 323 K

V2 = (0.250 /298) *323

V2 = 0.271 L = 271 mL

The final volume is 271 mL

Answer:

A pH greater than 7 is basic. The pH scale is logarithmic and as a result, each whole pH value below 7 is ten times more acidic than the next higher value. For example, pH 4 is ten times more acidic than pH 5 and 100 times (10 times 10) more acidic than pH 6.

Answer: mole fractions are

For n-pentane = 0.965

For n-octane = 0.035

Explanation: pressure exerted by each gas is,

n-pentane = 1.92atm

n-octane = 0.07atm

Total pressure exerted = 1.92 + 0.07

= 1.99atm.

Recall that the partial pressure exerted by each gas is the product of its mole fraction and the total pressure, that is,

Pres. n-pentane = n x pressure(total)

1.92 = n x 1.99

n = 1.92/1.99 = 0.965 for n-pentane

For n-octane,

n = 1 - 0.965 = 0.035 for n-octane.

Answer : The  pH of buffer is, 5.17

Explanation : Given,

[tex]pK_a=4.75[/tex]

Concentration of acetic acid = 1.00 M

Concentration of sodium acetate = 1.00 M

Volume of solution = 1.00 L

As, [tex]Moles=Concentration\times Volume[/tex]

So,

Moles of acetic acid = 1.00 mol

Moles of sodium acetate = 1.00 mol

Moles of NaOH added = 0.450 mol

The balanced chemical equilibrium reaction is:

                      [tex]CH_3COO+NaOH\rightleftharpoons CH_3COONa+H_2O[/tex]

Initial mole        1                0.450              1

At eqm.         (1-0.450)            0            (1+0.450)

                       = 0.55                                =1.450

Now we have to calculate the pH of buffer.

Using Henderson Hesselbach equation :

[tex]pH=pK_a+\log \frac{[Salt]}{[Acid]}[/tex]

[tex]pH=pK_a+\log \frac{[CH_3COONa]}{[CH_3COOH]}[/tex]

Now put all the given values in this expression, we get:

[tex]pH=4.75+\log (\frac{1.450}{0.55})[/tex]

[tex]pH=5.17[/tex]

Therefore, the pH of buffer is, 5.17

Answer:

Partial pressure of CO₂ is 406.9 mmHg

Explanation:

To solve the question we should apply the concept of the mole fraction.

Mole fraction = Moles of gas / Total moles

We have the total moles of the mixture, if we have the moles for each gas inside. (3.63 moles of O₂, 1.49 moles of N₂ and 4.49 moles of CO₂)

Total moles = 3.63 mol O₂ + 1.49 mol N₂ + 4.49 mol CO₂ = 9.61 moles

To determiine the partial pressure of CO₂ we apply

Mole fraction of CO₂ → mol of CO₂ / Total moles = P. pressure CO₂ / Total P

Partial pressure of CO₂ = (mol of CO₂ / Total moles) . Total pressure

We replace values: (4.49 moles / 9.61 moles) . 871 mmHg = 406.9 mmHg

Answer:

Partial pressure O2 = 329 mmHg

Partial pressure N2 = 135 mmHg

Partial pressure CO2 = 407 mmHg

Explanation:

Step 1: Data given

Number of moles oxygen (O2) = 3.63 moles

Number of moles nitrogen (N2) = 1.49 moles

Number of moles carbon dioxide (CO2) = 4.49 moles

Total pressure = 871 mmHg

Step 2: Calculate total number of moles

Total moles = moles O2 + moles N2 + moles CO2

Total moles = 3.63 + 1.49 + 4.49

Total moles =  9.61 moles

Step 3: Calculate the mol ratio

Mol ratio number of moles compound / total moles

Mol ratio O2 = 3.63 moles / 9.61 moles

Mol ratio O2 = 0.378

Mol ratio N2 = 1.49 moles / 9.61 moles

Mol ratio N2 = 0.155

Mol ratio CO2 = 4.49 moles / 9.61 moles

Mol ratio CO2 = 0.467

Step 4: Calculate partial pressure

Partial pressure = mol ratio * total pressure

Partial pressure O2 = 0.378 * 871 mmHg

Partial pressure O2 = 329 mmHg

Partial pressure N2 = 0.155 * 871mmHg

Partial pressure N2 = 135 mmHg

Partial pressure CO2 = 0.467 * 871 mmHg

Partial pressure CO2 = 407 mmHg

Answer:

Explanation:

the solution is solved below

When ethers react with HI, treatment with one equivalent of HI produces an alcohol and an alkyl iodide as major products. Treatment with excess HI yields both alkyl iodides as major products.

Ethers react with HI to form two cleavage products. When treated with one equivalent of HI, the major products that could be recovered are an alcohol and an alkyl iodide. The alcohol is formed by the substitution of the ether oxygen with a hydrogen atom from HI, and the alkyl iodide is formed by the substitution of one of the alkyl groups of the ether with iodine.

When treated with excess HI, the major products that could be recovered are both alkyl iodides. The initial products from the first step do not further react but are still recovered.

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Answer:

PLATO Answer

Explanation:

Underneath Photosynthesis Is Carbon Sink

Underneath Animals Is Carbon Sink

Underneath Combustion Is Carbon Source

Among the given options, only the title 'a' correctly identifies the processes as carbon sources and sinks. Burning fossil fuels is a carbon source, while oceans are a carbon sink. The other titles misinterpret the roles of various processes in the carbon cycle.

The correct title to location identifications for each process as carbon source or sink are as follows:

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Answer : The correct option is, (D) [tex]9.0\times 10^9J[/tex]

Explanation :

As we are given that the energy require for decomposition is, [tex]9.0\times 10^6kJ[/tex].

Now we have to calculate the energy in joules.

Conversion used :

1 kJ = 1000 J

As, 1 kJ of energy = 1000 J

So, [tex]9.0\times 10^6kJ[/tex] of energy = [tex]\frac{9.0\times 10^6kJ}{1kJ}\times 1000J[/tex]

                                          = [tex]9.0\times 10^9J[/tex]

Therefore, the energy in joules is, [tex]9.0\times 10^9J[/tex]

Thermal decomposition of 5.0 metric tons of limestone to lime and carbon dioxide requires 9.0 × 10⁹ Joules of heat.

Enegy is the quantitative property which is used by any system to perform any work.

Chemical reactions generally involves energy in the form of heat energy and given amount of energy is 9.0 × 10⁶ kJ.

We know that:

1 kJ = 1000 J

So, 9.0 × 10⁶ kJ = 9.0 × 10⁶ kJ  × 1000

9.0 × 10⁶ kJ = 9.0 × 10⁹ J

Hence, option (D) is correct i.e. 9.0 × 10⁹ J.

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Answer:

10.8 ml

Explanation:

The BOD is an empirical test to determine the molecular oxygen used during a specified incubation period (usually five days), for the biochemical degradation of organic matter (carbonaceous demand) and the oxygen used to oxidise inorganic matter.

See attached file

The maximum volume of treated wastewater that will be in the bottle is 10.8 mL.

The given parameters;

The dilution factor (P) is calculated as follows;

[tex]200 \ mg/L= \frac{9.2 \ mg/L \ - \ 2\ mg/L}{P} \\\\P = \frac{7.2 \ mg/L}{200 \ mg/L} \\\\P = 0.036[/tex]

The maximum  volume of treated wastewater that will be in the bottle to have at least 2.0 mg/L DO;

[tex]0.036 = \frac{V_w}{300 \ mL} \\\\V_w = 0.036 \times 300 \ mL\\\\V_w = 10.8 \ mL[/tex]

Thus, the maximum volume of treated wastewater that will be in the bottle is 10.8 mL.

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Answer : The final concentration of ONGP is, 0.47 mM

Explanation :

Formula used :

[tex]M_1V_1=M_2V_2[/tex]

where,

[tex]M_1\text{ and }V_1[/tex] are the initial molarity and volume

[tex]M_2\text{ and }V_2[/tex] are the final molarity and volume

We are given:

[tex]M_1=3.3mM\\V_1=1.42mL\\M_2=?\\V_2=10mL[/tex]

Now put all the given values in above equation, we get:

[tex]3.3mM\times 1.42mL=M_2\times 10mL\\\\M_2=0.47mM[/tex]

Hence, the final concentration of ONGP is, 0.47 mM