When a 1:1 mixture of ethyl propanoate and ethyl butanoate is treated with sodium ethoxide, four Claisen condensation products are possible. Draw the structure(s) of the product(s) that have an ethyl group on the chiral center

Answer:

The energy required to break a covalent bond

Explanation:

When a chemical bond is formed, energy is released. When a chemical bond is broken, energy is absorbed.

We define the bond dissociation energy as the energy required to break a covalent bond. The process of covalent bond cleavage is endothermic hence energy is absorbed for the process to occur.

Answer:

There is no difference between the two.

Explanation:

They both show the same volume. But, adding decimal places shows the least count of the instrument used and is more acceptable when recording values in scientific experiments

Answer:

1. Pyruvate reacts with TPP and is decarboxylated forming hydroxyethyl-TPP.

2. The lipoamide arm of E2 moves to the active site of E1, enabling the transfer of the acetyl group to the lipoamide.

3. The acetyl lipoamide arm of E2 moves to the active site of E2, where the acetyl group is transferred to CoA, forming acetyl-CoA and the reduced form of lipoamide.

4. The lipoamide arm moves to the active site of E3 where the reduced lipoamide is oxidized by FAD forming the active lipoamide and FADH2.

5. FADH2 is reoxidized to FAD reducing NAD+ to NADH.

Explanation:

The oxidation of pyruvate to AcetylCoA is catalyzed by the pyruvate dehydrogenase complex. The reaction is an irreversible oxidative decarboxylation process in which the carboxyl group of pyruvate is removed as a molecule of carbon dioxide, CO₂, while the remaining two carbons are attached to a CoASH molecule to form acetylCoA.

The pyruvate dehydrogenase complex contains three enzymes - Pyruvate dehydrogenase known as E₁, dihydrolipoyl transacetylase known as E₂, and dihydrolipoyl dehydrogenase known as E₃. It also requires five coenzymes namely: thiamine pyrophosphate (TPP), flavine adenine dinucleotide (NAD), coenzyme A (CoA-SH), nicotinamide adenine dinucleotide (NAD) and lipoate.

Oxidative decorbyxylation of pyruvate takes place in the pyruvate dehydrogenase complex in five steps:

1. Pyruvate reacts with TPP and is decarboxylated forming hydroxyethyl-TPP.

2. The lipoamide arm of E2 moves to the active site of E1, enabling the transfer of the acetyl group to the lipoamide.

3. The acetyl lipoamide arm of E2 moves to the active site of E2, where the acetyl group is transferred to CoA, forming acetyl-CoA and the reduced form of lipoamide.

4. The lipoamide arm moves to the active site of E3 where the reduced lipoamide is oxidized by FAD forming the active lipoamide and FADH2.

5. FADH2 is reoxidized to FAD reducing NAD+ to NADH.

Answer:

the rate of a reaction is measured by how fast a REACTANT is used up or how fast a PRODUCT is formed

N₂ + 3H₂ ⇄ 2NH₃ + heat

In the given equilibrium, we notice that the heat is on the right. which means that if the heat requirements don't meet, the reactants on the right will no longer react due to the lack of heat

but because the reactants on the left don't have such weaknesses, they will keep reacting hence producing more and more ammonia until a new equilibrium is reached

where there will be more ammonia and less nitrogen and hydrogen as compared to the equilibrium we had initially

Answer:

Explanation:

heat is given out as 1 of the products, along w/ NH3 in the forward reaction. so its an exothermic reaction

decreasing temperature favors exothermic reaction as more heat can be absorbed by the environment

so equilibrium will shift towards the products

Answer:

-76.3 kJ

Explanation:

Here is the complete question

Given the standard enthalpy changes for the following two reactions:

(1) 2Fe(s) + O₂(g) → 2FeO(s)......ΔH° = -544.0 kJ

(2) 2Zn(s) + O₂(g) → 2ZnO(s)......ΔH° = -696.6 kJ. What is the standard enthalpy change for the reaction:

(3) FeO(s) + Zn(s) → Fe(s) + ZnO(s)......ΔH° = ?

Solution

Since (1) 2Fe(s) + O₂(g) → 2FeO(s)......ΔH° = -544.0 kJ

reversing the reaction, we have

2FeO(s) → 2Fe(s) + O₂(g) ......ΔH° = +544.0 kJ (4)

Adding reactions (2) and (3), we have

2FeO(s) → 2Fe(s) + O₂(g) ......ΔH° = +544.0 kJ (4)

2Zn(s) + O₂(g) → 2ZnO(s)......ΔH° = -696.6 kJ  (2)

This gives

2FeO(s) + 2Zn(s) → 2Fe(s) + 2ZnO(s)......ΔH° =

The enthalpy change for this reaction is the sum of enthalpy changes for reaction (2) and (3) = ΔH° = +544.0 kJ + (-696.6 kJ)

= +544.0 kJ - 696.6 kJ)

= -152.6 kJ

Since the required reaction is (3) which is FeO(s) + Zn(s) → Fe(s) + ZnO(s)

we divide the enthalpy change for reaction (4) by 2 to obtain the enthalpy change for reaction (3).

So, ΔH° = -152.6 kJ/2 = -76.3 kJ

So, the standard enthalpy change for the reaction

FeO(s) + Zn(s) → Fe(s) + ZnO(s) is -76.3 kJ

Answer:

11.76 g/cm^3

Explanation:

Mass of empty flask and stopper = 64.232g

Mass of flask filled with water = 153.617 g

Mass of water = 153.617 g - 64.232g = 89.385 g

Mass of flask, stopper and metal = 143.557 g

Mass of metal = 143.557 g - 64.232g = 79.325 g

Mass of water, flask, stopper and metal = 226.196 g

Mass of water = 226.196 g - 143.557 g = 82.639 g

Since mass of water =volume of water

Volume occupied by metal = 89.385 cm^3 - 82.639 cm^3 = 6.746 cm^3

Density of metal = mass/volume = 79.325 g/6.746 cm^3

= 11.76 g/cm^3

Answer:

Moles of water are 0.868

Explanation:

Answer:

CuSO4(s) + 5H2O(l) ----> CuSO4.5H2O(s)

Explanation:

Hydration is the process by which anhydrous CuSO4 acquires molecules of water of crystalization to form the pentahydrate.

The water of crystalization becomes attached go the crystals of the CuSO4 to form the hydrated salt.

Beginning with solid anhydrous CuSO4 we have;

CuSO4(s) + 5H2O(l) ----> CuSO4.5H2O(s)

Answer:

"[tex]6.7\times 10^{-4} \ atm[/tex]" is the right answer.

Explanation:

Given:

Partial pressure of [tex]N_2[/tex],

= 0.20 atm

Partial pressure of [tex]H_2[/tex],

= 0.15 atm

[tex]K_p = 1.5\times 10^3[/tex] at [tex]400^{\circ} C[/tex]

As we know,

⇒ [tex]K_p = \frac{pN_2\times pH_2^3}{pNH_3^2}[/tex]

By putting the values, we get

    [tex]1.5\times 10^3=\frac{0.20\times (0.15)^3}{pNH_3^2}[/tex]

        [tex]pNH_3^2 = \frac{0.000675}{1.5\times 10^3}[/tex]

                    [tex]=6.7\times 10^{-4} \ atm[/tex]

Answer:

[tex]\boxed {\boxed {\sf 11 \ grams \ NaCl}}[/tex]

Explanation:

We are asked to find how many grams of sodium chloride are in a solution.

Molarity is a measure of concentration in moles per liter.

[tex]molarity= \frac{ moles \ of \ solute}{liters \ of \ solution}[/tex]

We know the molarity is 0.75 molar. 1 molar is the same as 1 mole per liter, so the solution contains 0.75 moles of sodium chloride per liter.

There are 250 milliliters of solution but molarity uses liters for volume. We must convert milliliters to liters. Remember that 1 liter contains 1000 milliliters. Set up a ratio and use dimensional analysis to convert.

[tex]250 \ mL * \frac{1 \ L} {1000\ mL} = \frac{ 250}{1000} \ L = 0.250 \ L[/tex]

Now we know the molarity and the liters of solution, but the moles of solute are unknown.

Substitute the values into the formula.

[tex]0.75 \ mol \ NaCl/L = \frac{x}{0.250 \ L}[/tex]

We are solving for the moles of solute, so we must isolate the variable x. It is being divided by 0.250 liters. The inverse of division is multiplication, so multiply both sides of the equation by 0.250 L.

[tex]0.250 \ L *0.75 \ mol \ NaCl/L = \frac{x}{0.250 \ L} * 0.250 \ L[/tex]

[tex]0.250 \ L *0.75 \ mol \ NaCl/L = x[/tex]

The units of liters cancel.

[tex]0.250 * 0.75 \ mol \ NaCl[/tex]

[tex]\bold {0.1875 \ mol \ NaCl}[/tex]

Now that we have calculated the moles of solute, we must convert this to grams. We use the molar mass or the mass of 1 mole of a substance. Sodium chloride's molar mass is given and it is 58.44 grams per mole. This means there are 58.44 grams of sodium chloride in 1 mole of sodium chloride.

Set up a ratio so we can convert using dimensional analysis.

[tex]\frac {58.44 \ g \ NaCl}{1 \ mol \ NaCl}[/tex]

Multiply by the number of moles we calculated.

[tex]0.1875 \ mol \ NaCl *\frac {58.44 \ g \ NaCl}{1 \ mol \ NaCl}[/tex]

The units of moles of sodium chloride cancel.

[tex]0.1875 *\frac {58.44 \ g \ NaCl}{1}[/tex]

[tex]\bold {10.9575 \ g \ NaCl}[/tex]

The original measurements of molarity and volume have 2 significant figures, so our answer must have the same. For the number we calculated, that is the ones place. The 9 in the tenths place tells us to round the 0 up to a 1.

[tex]11 \ g \ NaCl[/tex]

There are approximately 11 grams of sodium chloride in 250 mL of a 0.75 molar solution.

Answer:

1. Because the rules will keep you safe it prevents you from getting hurt.

2.i) don't taste chemical

ii) Always wear protective gears

iii) be careful with tool

iv) wear protective gloves

Explanation:

help everyone get out quickly

Answer:

LiOH(aq) → Li⁺(aq) + OH⁻(aq). 

Calories, we know that fat burn is calories.

Answer:

nucleus is a collection of particles called protons,which are positively charged..and neutrons which are electrically neutral..electrons which are negatively charged..and neutrons are in turn made up of particles called quarks ..

Explanation:

hope this helps u ...

Answer:

The Nucleus is made up of protons and neutrons.

Answer:

The answer is "17200 years".

Explanation:

Given:

[tex]A = 20 \ \frac{counts}{minute}\\\\A_{o} = 160\ \frac{counts}{minute}[/tex]

Let the half-life of carbon-14, is beta emitter, is [tex]T = 5730\ years[/tex]

Constant decay [tex]\ w = \frac{0.693}{ T}[/tex]

[tex]= 1.209 \times 10^{-4} \ \frac{1}{year}\\[/tex]

The artifact age [tex]t= ?[/tex]

[tex]A = A_{o} e^{-wt} \\\\e^{-wt} = \frac{A}{A_{o}}\\\\-wt = \ln \frac{A}{A_{o}}\\\\= -2.079\\\\t = 1.7199 \times 10^{4} \ years\\\\\sim \ 17200\ years\\[/tex]

Answer:

This is an acid-base reaction (neutralization): Ba(OH) 2 is a base, H 2SO 4 is an acid. This is a precipitation reaction: BaSO 4 is the formed precipitate.

The question is incomplete. The complete question is :

An analytical chemist is titrating 148.9 mL of a 1.100 M solution of benzoic acid [tex]$HC_6H_5CO_2$[/tex]  with a 0.3600 M solution of KOH. The [tex]pK_a[/tex]  of benzoic acid is 4.20. Calculate the pH of the acid solution after the chemist has added 232.0 mL of the KOH solution to it. Note for advanced students: you may assume the final volume equals the initial volume of the solution plus the volume of KOH solution added.

Solution :

Number of moles of [tex]$C_6H_5OCOOH$[/tex] [tex]$=148.9 \ mL \times \frac{L}{1000\ mL} \times \frac{1.100 \ mol}{L}$[/tex]

                                                         = 0.16379 mol

Number of moles of NaOH added [tex]$=232.0 \ mL \times \frac{L}{1000\ mL} \times \frac{0.3600 \ mol}{L}$[/tex]

                                                         = 0.08352 mol

ICE table :

                     [tex]C_6H_5OCOOH \ \ \ + \ \ \ OH^- \ \ \ \rightarrow \ \ C_6H_5OCOO^- \ \ \ \ + \ \ H_2O[/tex]

I (mol)              0.16379                 0.08352              0

C (mol)           -0.08352              -0.08352            +0.08352

E (mol)            0.08027                    0                    0.08352

Total volume = (148.9 + 232) mL

                      = 380.9 mL

                     = 0.3809 L

Concentration of [tex]$C_6H_5OCOOH, [C_6H_5OCOOH]$[/tex] [tex]$=\frac{0.08027 \ mol}{0.3809 \ L}$[/tex]

                                                                               = 0.211 M

Concentration of [tex]$C_6H_5OCOO^- , [C_6H_5OCOO^-] =\frac{0.08352 \ mol}{0.3809 \ L}[/tex]

                                                                              = 0.219 M

[tex]pK_a[/tex] of [tex]C_6H_5OCOOH = 4.20[/tex]

According to Henderson equation,

[tex]$pH = pK_a + \log \frac{[C_6H_5OCOO^-]}{[C_6H_5OCOOH]}[/tex]

     [tex]$=4.20 + \log \frac{0.219}{0.211}$[/tex]

     = 4.22

Therefore, the pH of the acid solution is 4.22

Answer:

21.4g of HBr is the minimum mass that could be left over.

Explanation:

Based on the reaction:

HBr + NaOH → NaBr + H2O

1 mole of HBr reacts per mole of NaOH

To solve this question we need to find the moles of both reactants. If moles NaOH > moles HBr, the difference in moles represents the minimum moles of HBr that could be left over because this reaction is 1:1. Using the molar mass we can find the minimum mass of HBr that could be left over, as follows:

Moles NaOH -40.0g/mol-

17g * (1mol/40.0g) = 0.425 moles NaOH

Moles HBr -Molar mass: 80.91g/mol-

55.8g * (1mol/80.91g) = 0.690 moles HBr

The difference in moles is:

0.690 moles - 0.425 moles =

0.265 moles of HBr could be left over

The mass is:

0.265 moles * (80.91g/mol) =

Answer

The properties of fatty acids and of lipids derived from them are markedly dependent on chain length and degree of saturation. Unsaturated fatty acids have lower melting points than saturated fatty acids of the same length. For example, the melting point of stearic acid is 69.6°C, whereas that of oleic acid (which contains one cis double bond) is 13.4°C. The melting points of polyunsaturated fatty acids of the C18 series are even lower. Chain length also affects the melting point, as illustrated by the fact that the melting temperature of palmitic acid (C16) is 6.5 degrees lower than that of stearic acid (C18). Thus, short chain length and unsaturation enhance the fluidity of fatty acids and of their derivatives.

Answer:

C . Kinetic Molecular Theory

Answer:

hello? are you still here? reply if you are

Answer:

The various uses of water :

1. Water is used for daily purpose like cooking , bathing , cleaning and drinking.

2. Water is used as a universal solvent.

3. water maintains the temperature of our body.

4. Water helps in digestion in our body.

5 .water is used in factories and industries.

6. Water is used to grow plants , vegetables and crops.

Answer:

The various use of water are;

I) Cooking.

ii) Drinking

III) Bathing

iv) Generating hydro- electricity

v) Construction work etc

The half life of a radioactive isotope refers to the time taken for half of the number of original number of atoms present in the sample to decay.

The equation below gives the number of atoms present at time t

[tex]N=Noe^-kt[/tex]

N = Number of atoms present at time t

No = Number of atoms initially present

k = decay constant

t = time taken

Given that;

t1/2 = 0.693/k

where t1/2 = half life

k = 0.693/t1/2

k = 0.693/ 55.6 s

k = 0.0125 s-1

Substituting values;

N = 16,000 e^-0.0125(0)

N = 16,000 atoms

At 50 s

N = 16,000 e^-0.0125(50)

= 8564 atoms

At 100 s

N = 16,000 e^-0.0125(100)

= 4584 atoms

At 200 s

N = 16,000 e^-0.0125(200)

= 1313 atoms

https://brainly.com/question/2998270

Answer:

We slip when we step on a banana peel because the inner side of banana peel being smooth and slippery reduces the friction between the sole of our shoe and the surface of road.

Answer:

4.54 atm

Explanation:

Step 1: Calculate the total number of gaseous moles

We will calculate the moles of each gas using its molar mass.

He: 5.25 g × 1 mol/4.00 g = 1.31 mol

N₂: 3.25 g × 1 mol/28.01 g = 0.116 mol

The total number of moles is:

n = 1.31 mol + 0.116 mol = 1.43 mol

Step 2: Convert 27 °C to Kelvin

We will use the following expression.

K = °C + 273.15 = 27 + 273.15 = 300 K

Step 3: Calculate the total pressure of the mixture

We will use the ideal gas equation.

P × V = n × R × T

P = n × R × T / V

P = 1.43 mol × (0.0821 atm.L/mol.K) × 300 K / 7.75 L = 4.54 atm

Explanation:

here's the answer to the question

Answer:

a. LiF(s) ⇄ Li⁺(aq) + F⁻(aq)

b. [Li⁺] = [F⁻] = 6.2 x 10⁻² M

c. Ksp = [Li⁺] [F⁻]

d. Ksp = 3.8 × 10⁻³

Explanation:

The solubility (S) of lithium fluoride, LiF, is 1.6 g/L, or 6.2 x 10⁻² M.

a. The balanced solubility equilibrium equation for LiF is:

LiF(s) ⇄ Li⁺(aq) + F⁻(aq)

b. We will make an ICE chart.

        LiF(s) ⇄ Li⁺(aq) + F⁻(aq)

I                       0           0

C                     +S         +S

E                       S           S

Then, [Li⁺] = [F⁻] = S = 6.2 x 10⁻² M

c. The solubility product constant, Ksp​, is the equilibrium constant for a solid substance dissolving in an aqueous solution.

Ksp = [Li⁺] [F⁻]

d.

Ksp = [Li⁺] [F⁻] = (6.2 x 10⁻²)² = 3.8 × 10⁻³