generally speaking, are fatty acid molecules polar or nonpolar or both? what type of forces hold them together in the solid? (1 pt) g

Answer:

cellular respiration. C6H12O6 + 6O2 → 6CO2 + 6H2O.

Explanation:

the process where cells convert chemical energy and/or nutrients into ATP and then the cells release the waste.

At pH 9.0, the solution is basic and the ionizable groups in the peptide will be ionized. The correct answer is option b: n-terminus: amine; lysine r group: ammonium; c-terminus: carboxylate; aspartic acid r group: carboxylate.

At pH 9.0, the ionizable groups in a peptide are likely to be ionized. The pKa values of the different ionizable groups in the peptide determine which form they will be in at a given pH. The amino group at the N-terminus of the peptide has a pKa value of about 9.0, which means that it will mostly exist in the ionized form as an amine (NH2) at pH 9.0. The lysine R group has a pKa value of approximately 10.8, which means that it will exist in the ionized form as ammonium (NH3+) at pH 9.0. The carboxyl group at the C-terminus of the peptide has a pKa value of around 2.2, which means that it will exist in the ionized form as a carboxylate (COO-) at pH 9.0. The aspartic acid R group has a pKa value of about 3.9, which means that it will also exist in the ionized form as a carboxylate (COO-) at pH 9.0. Therefore, at pH 9.0, the major forms of each ionizable group in a peptide are N-terminus - amine (NH2), Lysine R group - ammonium (NH3+), C-terminus - carboxylate (COO-), and Aspartic acid R group - carboxylate (COO-). Understanding the ionization of peptides at different pH values is important in many biochemical and biophysical studies that involve peptides and proteins.

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By observing the reactions and the formation of precipitates, we can determine whether the unknown solution is strontium nitrate or magnesium nitrate.

What is an  unknown solution?

There are two possible reactions that could occur when potassium carbonate or potassium sulfate are added to the unknown solution:

To determine which of these reactions occurred, we need to observe which reaction produced a precipitate (an insoluble solid) and which did not.

If a precipitate forms when potassium carbonate is added to the unknown solution, then the unknown solution must have been magnesium nitrate. The balanced molecular equation for this reaction is:

\(Mg(NO_{3}){2}\) (aq) + \(K_{2} CO_{3}\) (aq) → \(MgCO_{3}\) (s) + 2\(KNO_{3}\) (aq)

If no precipitate forms when potassium carbonate is added, but a precipitate forms when potassium sulfate is added, then the unknown solution must have been strontium nitrate. The balanced molecular equation for this reaction is:

\(Sr(NO_{3})_{2}\) (aq) + \(K_{2}SO_{4}\) (aq) → \(SrSO_{4}\) (s) + \(2KNO_{3}\) (aq)

By observing the reactions and the formation of precipitates, we can determine whether the unknown solution is strontium nitrate or magnesium nitrate.

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A 3.4 × 10⁶ L swimming pool must have a mass of 1.0 × 10⁷ mg Cl₂ to maintain a concentration of 3.0 ppm.

"ppm" of "parts per million" is a unit of concentration equivalent to milligrams of solute per liters of solution.

A pool must maintain a chlorine concentration of 3.0 ppm (3.0 mg/L). The mass of chlorine in 3.4 × 10⁶ L is:

3.0 mg Cl₂/L × 3.4 × 10⁶ L = 1.0 × 10⁷ mg Cl₂

A 3.4 × 10⁶ L swimming pool must have a mass of 1.0 × 10⁷ mg Cl₂ to maintain a concentration of 3.0 ppm.

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Answer:

Explanation:

i may be wrong but i think its b

Answer:

B Ndaira is right for real

Answer:

An Atom

Explanation:

This is the smallest unit which still has the properties of an element.

Answer: 22.5mL

Explanation:

Answer:

Digestive system

Explanation:

..........

The magnitude of the angular acceleration of the disk is -203.9 rad/s^2 (negative sign is indicating the direction of angular acceleration is opposite to the direction of applied forces).

The magnitude of the angular acceleration of the disk can be determined by using the equation for torque:

τ = Iα

where τ is the torque, I is the moment of inertia, and α is the angular acceleration.

The moment of inertia of a solid disk is given by the equation:

I = (1/2) * m * r^2

where m is the mass and r is the radius of the disk.

So we can substitute the values of mass and radius in the equation of the moment of inertia:

I = (1/2) * 24.3 kg * (0.314 m)^2 = 0.03898 kg*m^2

Now we can calculate the torque caused by the forces of 90 n and 125 n. These forces are applied in opposite direction so the net torque caused by them is:

τ = 90 N * 0.314 m - 125 N * 0.314 m = -7.922 Nm

So we can now calculate the angular acceleration using the torque equation:

-7.922 Nm = 0.03898 kgm^2 * α

α = -7.922 Nm / 0.03898 kgm^2 = -203.9 rad/s^2

Therefore, The magnitude of the angular acceleration of the disk is -203.9 rad/s^2 (negative sign is indicating the direction of angular acceleration is opposite to the direction of applied forces).

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C = m V . The unit of concentration is typically given as g/mL because the solute mass is frequently given in grammes and the volume is frequently given in milliliters. However, there are a tone of other mass and volume unit combinations that are possible.

Given that a solution consists of both a solute and a solvent, its total volume is equal to the sum of the volumes of both the solute and the solvent it contains.

A solution's concentration is an indicator of how much solute has dissolved in a specific volume of solvent or solution. When there is a significant amount of dissolved solute in a solution, it is said to be concentrated. When a dissolved solute is present in a solution, it is said to be diluted.

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Approximately 0.15 moles of iodine were consumed in the reaction. Use the stoichiometry of the balanced equation to find the moles of iodine.

To find the moles of iodine consumed in the reaction, we can use the given heat absorbed and the heat of the reaction (AH). The heat absorbed is given as 7.93x10 J, which we need to convert to kilojoules (kJ) by dividing it by 1000.

7.93x10 J ÷ 1000 = 7.93x10 kJ

Now, we can use the stoichiometry of the balanced equation to relate the heat of the reaction (53.0 kJ) to the moles of iodine consumed. According to the balanced equation:

H₂ (g) + I₂ (s) → 2 HI (g)

The stoichiometric coefficient in front of I₂ is 1, which means that 1 mole of iodine is consumed in the reaction.

Using the ratio of heat of reaction to moles of iodine, we can set up a proportion:

53.0 kJ ÷ 1 mole of iodine = 7.93x10 kJ ÷ x moles of iodine

Cross-multiplying and solving for x, we get:

x = (7.93x10 kJ)(1 mole of iodine) ÷ 53.0 kJ

x ≈ 0.15 moles of iodine

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A. none of the above

The answer would be 180 m/s

Hope this helps! Brainliest if you can!

Answer:

2x(hn)po

Explanation:

Use the community property to reorder the terms

Remove the unnecessary parentheses

2hnpo

2hnop

use the community property to reorder the terms

2hnop

To show the acid-catalyzed mechanism for acetic anhydride reacting with 2-methylpropanol to form an ester, we can follow the Fischer Esterification process as mentioned in the Klein text (sections 21.9 and 21.10). Here's a step-by-step explanation of the mechanism:

1. Protonation of the carbonyl oxygen: The oxygen of the acetic anhydride's carbonyl group is protonated by the sulfuric acid (H2SO4) catalyst. This increases the electrophilicity of the carbonyl carbon.
O=C(OC(O)C)R + H2SO4 -> O=C(+)(OC(O)C)R-OH2 + HSO4(-)

2. Nucleophilic attack by 2-methylpropanol: The oxygen from the 2-methylpropanol molecule acts as a nucleophile, attacking the carbonyl carbon of the protonated acetic anhydride.
O=C(+)(OC(O)C)R-OH2 + CH3CH(OH)CH3 -> O=C(-)(OCH(CH3)CH3)R-OH2 + CH3CH(OH2+)CH3

3. Proton transfer: A proton is transferred from the protonated 2-methylpropanol to the anhydride oxygen.
CH3CH(OH2+)CH3 -> CH3CH(OH)CH3 + H+

O=C(-)(OCH(CH3)CH3)R-OH2 + H+ -> O=C(OCH(CH3)CH3)R-OH3+

4. Removal of the leaving group: The protonated anhydride oxygen acts as a leaving group, breaking the bond between the carbonyl carbon and the oxygen.
O=C(OCH(CH3)CH3)R-OH3+ -> O=C(OCH(CH3)CH3)R + H2O

5. Deprotonation of the carbonyl oxygen: The carbonyl oxygen gets deprotonated by the bisulfate ion (HSO4-) to regenerate the H2SO4 catalyst and form the ester product.
O=C(+)(OCH(CH3)CH3)R + HSO4(-) -> O=C(OCH(CH3)CH3)R + H2SO4

The final ester product is formed by the reaction between acetic anhydride and 2-methylpropanol through the acid-catalyzed Fischer Esterification mechanism.

As for the second part of your question, since I cannot physically print, analyze, or staple the NMR spectra to your MOW, I recommend you do so with the provided files. Once you have printed the spectra, analyze the chemical shifts, splitting patterns, and integrations for each signal to identify the starting alcohol and draw the corresponding ester products on the appropriate spectrum.

Answer:

Explanation: Every electron has a single negative(-) charge, the addition of two electrons results in an oxygen ion with a charge of negative(−)2. This is  in fact true of every element located under oxygen in the periodic table.

P.S. hope this helps :)

The overall AGO for the two successive reactions is -5.4 kJ/mol so, the reaction is exergonic. That is, the correct answer is option d: -5.4 kJ/mol and is exergonic.

During glycolysis, glucose 1-phosphate is converted to fructose 6-phosphate in two successive reactions. To determine the overall ∆G for the reaction and whether it is exergonic or endergonic, you need to add the individual ∆G values:

1. Glucose 1-phosphate → Glucose 6-phosphate; ∆G = -7.1 kJ/mol
2. Glucose 6-phosphate → Fructose 6-phosphate; ∆G = +1.7 kJ/mol

Overall reaction: Glucose 1-phosphate → Fructose 6-phosphate
∆G = (-7.1 kJ/mol) + (+1.7 kJ/mol) = -5.4 kJ/mol

Since the overall ∆G is negative, the reaction is exergonic. Therefore, the correct answer is d. -5.4 kJ/mol and is exergonic.

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Answer:

(CH3COO)2 + redP + Cl2 → ClCH2COOH + HCl

Explanation:

This is an example of halogenation of carboxylic acids at alpha carbon atom. In this reaction, red phosphorus and chlorine are treated with carboxylic acids having alpha hydrogen atom followed by hydrolysis to form alpha chloro carboxylic acid.

0.034 mol of carbon dioxide gas, are there in 0.667 L.

We can assume that the volume of one mole of any gas is 22.4 L we know that the pressure of the gas is 1 atmosphere and that the temperature of a gas is 0C

This type is called standard temperature and pressure (STP).

This can be calculated by using the formula in which we can find the moles:

PV =nRT------(1)

Where P is the pressure

V is volume

T is the temperature in Kelvin

R is the ideal gas constant consistent with the units of volume and pressure.

0.667 L x 1 mol/22.4 L = 0.034 mol

And 0.667 liters of CO2 or any other gas at STP would contain 0.034 moles.

If the gas was at some other temperature or pressure, assuming they were “moderate,” you could calculate the number of moles by solving the following formula for n.

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Answer:

0.880 (0.88039867109 to be exact)

Explanation:

To convert from molecules to moles, simply divide by Avogadro's number which is 6.02 x 10^23

So, 5.30x10^23/6.02x10^23 = 0.880

The energy levels present in argon, aluminum and chlorine is 3 , with two electrons on the first energy level , 8 in the second level and third level have 8 electrons of argon.

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When 2-pentyne reacts with excess HCl, the expected organic products are pent-2-ene-1-chloride, 2,2-dichloropentane, and 3,3-dichloropentane.

The reaction can be considered an electrophilic addition of HCl to the carbon-carbon triple bond in 2-pentyne.

Step 1: Write the reaction equation:2-pentyne + HCl →

Step 2: Add the first HCl molecule to the carbon-carbon triple bond to produce a carbocation intermediate.Hence, the reaction proceeds as follows: 2-pentyne + HCl → 3-chloro-2-pentyne (Carbocation intermediate)

Step 3: The carbocation intermediate then reacts with another HCl molecule to produce a second carbocation intermediate.

Hence, the second reaction proceeds as follows:3-chloro-2-pentyne + HCl → 2-chloro-3-chloromethylpentane (Carbocation intermediate)

Step 4: In the final step, water (H2O) acts as a nucleophile and attacks the carbocation intermediate to produce the final organic products, which are: pent-2-ene-1-chloride, 2,2-dichloropentane, and 3,3-dichloropentane.

3-chloro-2-pentyne + H2O → pent-2-ene-1-chloride + HCl2-chloro-3-chloromethylpentane + H2O → 2,

2-dichloropentane + HCl2-chloro-3-chloromethylpentane + H2O → 3,

3-dichloropentane + HCl

Hence, the organic products formed from the reaction of 2-pentyne with excess HCl are pent-2-ene-1-chloride, 2,2-dichloropentane, and 3,3-dichloropentane.

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Answer:

it keeps the heat in and the bugs and flys out

Explanation:

The conformation of globular proteins is determined by a delicate balance of different molecular interactions and entropy effects. There may be more than one answer to this question. The correct answers are:

a. The main driving force opposing the folding of globular proteins is the loss of configurational entropy. When a protein folds, it loses its freedom of movement, which leads to a decrease in its configurational entropy. This decrease in entropy is the main driving force opposing protein folding.
b. A major driving force favoring the folding of many globular proteins is the electrostatic attraction between oppositely charged amino acid groups. This is true for proteins that have charged amino acids on their surface.
c. A major driving force favoring the folding of many globular proteins is the hydrophobic effect (reduction in contact area between non-polar groups and water). This is true for proteins that have non-polar amino acids on their surface.
d. After a protein has folded into a globular structure, the polypeptide chains often form ordered regions due to intramolecular hydrogen bond formation (secondary structure). This is true for many proteins, as hydrogen bonds stabilize the secondary structure.

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The atomic radius of Argon is smaller than the atomic radius of sodium because the effective nuclear charge increases as we move left to right in a period.

The atomic radius can be described as the shortest distance between the center of the nucleus of the atom and the outermost shell of the atom. An atomic radius can also be defined as half the distance between two atoms of the same element bonded together.

In general, the atomic radius of an atom reduces as we move from left to right in a period on the periodic table because while moving from left to right in a period, there is an increase in the effective nuclear charge. In periods, the valence electrons enter in the same outermost shell.

The atomic radius of atom increases when we move down a group because of the addition of a new principle shell.

Sodium is present at beginning of the third period and the Argon is present at the end of the same period. Therefore, Argon experiences a more effective nuclear charge than sodium so it has a smaller atomic radius.

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