Problem Title: Peptide Structure

#1:

Draw a formula for Phe-Ala-Val (F-A-V) in its predominant ionic form at pH 7.3. You may assume for the purposes of this question that the pKa values of the acidic groups of amino acid residues in the peptide are the same as in the amino acid itself.

You do not have to consider stereochemistry.

You do not have to explicitly draw H atoms.

Do not include lone pairs in your answer. They will not be considered in the grading.

_______________________

#2:

Draw a formula for Ile-Leu-Arg (I-L-R) in its predominant ionic form at pH 7.3. You may assume for the purposes of this question that the pKa values of the acidic groups of amino acid residues in the peptide are the same as in the amino acid itself.

You do not have to consider stereochemistry.

You do not have to explicitly draw H atoms.

Do not include lone pairs in your answer. They will not be considered in the grading.

Answer:

1.6×10^3 KJ

Explanation:

We must first calculate the number of moles corresponding to 650.50g of sulphur trioxide.

Number of moles = mass of sulphur trioxide/ molar mass of sulphur trioxide

Molar mass of sulphur trioxide= 80.066 g/mol

Given mass of sulphur trioxide= 650.50g

Number of moles = 650.50g / 80.066 g/mol

Number of moles = 8.124 moles of sulphur trioxide

Now from the balanced reaction equation;

Formation of 2 moles of sulphur trioxide evolves 396.8KJ of energy.

Therefore formation of 8.124 moles of sulphur trioxide will evolve;

8.124moles × 396.8 KJ /2 moles = 1611.8KJ or 1.6×10^3 KJ

Answer:

I don't know sorry

Explanation:

ask tutor

Answer:

See below

Explanation:

The glycosidic bond forms between the carbon atom on the glucose molecule and the nitrogen atom present on the nitrogenous base. A diagram has been attached to show this particular bond. There is also a phosphate molecule bonded on the sugar molecule at the other end.

A balanced equation obey the law of conservation of mass, the mass can neither be converted nor  be destroyed but can converted from one form to another. Here the given reaction indicates Haber process.

The ratio of the product of concentrations of the products to that of the reactants is also known as the concentration quotient and it is denoted as Q. At equilibrium Q becomes equal to the equilibrium constant.

The Haber process is:

N₂ + 3H₂ → 2NH₃

Q = [NH₃]² / [N₂] [H₂]³

Q = [0.0596]² / [0.600] [0.420]

Q = 0.014

Here Q is less than K, so the reaction proceeds in the forward direction.

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To find the mass percent composition of an element, divide the mass contribution of the element by the total mass then multiply by 100.

The total mass = 200.0 g

Phosphorus mass = 87.28 g

Oxygen mass = 112.72 g

%composition of P:

=> (87.28 g/200.0g)*100

=> 43.64 %

%composition of O:

=> (112.72g/200.0 g)*100

=> 56.36 %

hope this is correct

Answer:

D

Explanation:

Iron is the fourth most abundant element, by mass, in the Earth's crust. The core of the Earth is thought to be largely composed of iron with nickel and sulfur. The most common iron-containing ore is haematite, but iron is found widely distributed in other minerals such as magnetite and taconite.

After 5 half-lives, approximately 3.125 grams of the Protactinium-234 sample would be left.

The half-life of Protactinium-234 is given as 6.7 hours, which means that after every 6.7 hours, the amount of Protactinium-234 in a sample is halved.

To determine how much of the sample would be left after 5 half-lives, we need to calculate the remaining fraction of the original sample. Each half-life reduces the amount of the sample to half of its previous value.

Number of half-lives = 5

Fraction remaining after each half-life = 1/2

To calculate the remaining mass, we can multiply the fraction remaining by the original sample mass:

Remaining mass = Original sample mass * (Fraction remaining)^(Number of half-lives)

Remaining mass = 150 g * (1/2)⁵

≈ 150 g * (1/32)

≈ 4.6875 g

≈ 3.125 g (rounded to three significant figures)

Therefore, approximately 3.125 grams of the Protactinium-234 sample would be left after 5 half-lives.

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The answer is (70.35, 84.57) mL.

The three volumes of NaOH required to attain a pH of 7 in each of the three titrations are: 82.16, 75.79, and 75.43 mL. To estimate the mean number of milliliters required to neutralize 1 gram of the acid, a 99 percent confidence interval will be used. Let's calculate the sample mean, sample standard deviation, and margin of error using the provided data.

Sample standard deviation:

Sample Mean The sample mean of a dataset is defined as the sum of all the data points divided by the number of data points. So the sample mean will be: (82.16+75.79+75.43) / 3 = 77.46 mL. Sample Standard Deviation The sample standard deviation (s) is defined as the square root of the sample variance. To calculate s, we need to first compute the sample variance (s²):s² = ∑(x - μ)² / (n - 1)where x is the value of the observation, μ is the sample mean, and n is the sample size.s² = [(82.16 - 77.46)² + (75.79 - 77.46)² + (75.43 - 77.46)²] / (3 - 1)s² = [20.4 + 6.74 + 5.84] / 2s² = 16.49s = sqrt(16.49) = 4.06 mL.

Marginal Error The formula for the margin of error for a confidence interval for the mean is:

margin of error = t (α/2) * (s / sqrt(n)) where t(α/2) is the critical value of the t-distribution with n-1 degrees of freedom and a level of significance of α/2 (in this case, α/2 = 0.005).s is the sample standard deviation that we computed earlier. n is the sample size (in this case, n = 3). margin of error = t(α/2) * (s / sqrt(n))margin of error = 3.182 * (4.06 / sqrt(3)) = 7.11 mL. The margin of error is 7.11 mL. Confidence Interval The confidence interval formula for a population mean is: sample mean - margin of error < μ < sample mean + margin of error where μ is the population mean and sample mean is the value obtained from the sample.μ = 77.46 - 7.11 < μ < 77.46 + 7.11Thus, the 99% confidence interval for the mean number of milliliters needed to neutralize 1 gram of the acid is (70.35, 84.57) mL (rounded to three decimal places).Therefore, the answer is  (70.35, 84.57) mL.

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the splitting of the uranium atom is called nuclear fission.

Fission occurs when a neutron slams into a larger atom, forcing it to excite and split into two smaller atoms also known as fission products. Additional neutrons are also released that can initiate a chain reaction. When each atom splits, a tremendous amount of energy is released.

Nuclear fission is more dangerous than fusion as it produces harmful weapons-grade radioactive waste in the fuel rods that need to be stored safely away for thousands of years.

Fission is the splitting of a heavy, unstable nucleus into two lighter nuclei, and fusion is the process where two light nuclei combine together releasing vast amounts of energy. While different, the two processes have an important role in the past, present and future of energy creation.

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Explanation:

Because the place on the map can be changed to another shape by natural effects like faulting,folding , volcanism

Answer:

Maps must change over time as the geographical structures change from time o time. The roads and highways also change often due to the development activities .

The entropy of system 1, consisting of box A alone, will remain constant, while the entropy of system 2, consisting of both boxes A and B, will be greater due to the increase in the number of available microstates.

When two identical boxes of matter, A and B, are in thermal contact but cannot exchange materials, they will eventually come to thermal equilibrium. At thermal equilibrium, the two boxes will have the same temperature, and their individual entropies will be equal.

If we consider system 1, which consists of box A alone, its entropy will remain constant as it is an isolated system. However, if we consider system 2, which consists of both boxes A and B, the total entropy of the system will increase because there are now more ways for energy to be distributed between the two boxes. In other words, the addition of box B provides more microstates or configurations for the system to occupy, and hence the entropy of the combined system will be greater than that of system 1.

Therefore, we can say that the entropy of system 2, consisting of both boxes A and B, will be greater than the entropy of system 1, consisting of box A alone, as a result of the greater number of available microstates in system 2. This increase in entropy is a manifestation of the second law of thermodynamics, which states that the entropy of an isolated system will always tend to increase over time.

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Answer: The heat flows into Earth's atmosphere which changes as the Sun rises and sets, which increases during the day.

Explanation:

volume of the liquid sample, composition of the liquid sample, and The data we require to calculate the liquid sample's density is the mass of the liquid sample.

The substance's density is defined as its mass per unit of volume (volumetric mass density or specific mass). Although the Latin letter D may also be used, the symbol for density that is most usually used is (the lower case Greek letter rho). Mass divided by volume gives the density value:

\({\displaystyle \rho ={\frac {m}{V}}}\)

where \(V\) is the volume, is the density, and \(m\) is the mass. Weight per unit volume is a common general definition of density, however this is wrong technically; the actual term is specific weight.

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complete question: what information do we need to determine the density of the liquid sample? (select all that apply.)

a)composition of the liquid sample

b)volume of the liquid sample

c)mass of the liquid sample

d)height of the liquid sample in the container

e)color of the liquid sample

Hybrid orbitals form on the central atom and can either be sp, sp2, sp3, sp3d, or sp3d2. The type of hybrid orbitals that are used is dictated by the electron domain geometry.

Sigma bonds are formed by the end-to-end overlap of bonding orbitals. Pi bonds are formed by the side-to-side overlap of p orbitals.

Hybrid orbitals are the result of a model which combines atomic orbitals on a single atom in ways that lead to a new set of orbitals that have geometries appropariate to form bonds in the directions predicted by the VSEPR model. The VSEPR model predicts geometries that are very close to those seen in real molecules.

Nitrogen trifluoride (NF3) lewis structure contains three sigma bonds and one lone pair around nitrogen atom. Therefore, there are total of four electrons regions around nitrogen atom. So, hybridization of center atom, nitrogen is sp3.

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Answer:

C. ZnOH - zinc(II) hydroxide is incorrect.

Explanation:

Formula: Zn(OH)2

Name: Zinc hydroxide

Hope this help

143.25 grams of Fe2O3 will be produced.

Mole is an SI unit used to measure the amount of any substance.

First we can use the stoichiometry of the reaction to find the amount of Fe2O3 that will be produced if a certain amount of SO2 is produced.

From the balanced equation, it can be seen that for every 4 moles of FeS2 that react, 2 moles of Fe2O3 are produced, and for every 8 moles of SO2 that are produced, 2 moles of Fe2O3 are also produced.

Since 3.5861 moles of SO2 are produced, we can calculate the number of moles of Fe2O3 produced by dividing by 8 and multiplying by 2:

3.5861 moles SO2 / 8 moles Fe2O3/8 moles SO2 * 2 moles Fe2O3/2 moles SO2 = 0.8965 moles Fe2O3.

Finally, to convert moles of Fe2O3 to grams, we multiply by its molar mass (159.69 g/mol).

0.8965 moles Fe2O3 * 159.69 g Fe2O3/1 mole Fe2O3 = 143.25 g Fe2O3.

So, 143.25 grams of Fe2O3 will be produced.

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Answer:

0.1169g .

Explanation:

1. Multiply the concentration (0.5 mols/Liters) by the volume of solution you want (0.5 Liters) to find the moles of NaCl you need. 2. Multiply the moles of NaCl by its molar mass (58.44 g/mol) to find the grams of solute needed.

If the age of the lowermost layer of a rock sample is more than 2 million years, then the following which could be the age of the upper layers is less than one million and is denoted as option A.

This is referred to as a solid mineral which is formed from the accumulation and cementing of sand grains and other substances. Rocks are usually broken down into simpler substances through the process known as weathering.

The law of Superposition states that in a sequence of rocks, the rocks that are at its base are oldest, while those that are found at the apex portion of the sequence are the youngest rocks.

Therefore the age of the upper layers of the rocks which means that they will be younger which is why option A was chosen as the correct choice.

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The options are:

Answer:

The first thing we have to do is change and state all the units so that we can use our ideal gas law equation (\(PV = nRT\)).

650 mmHg is a pressure unit, we have to convert this to kiloPascals. We know that 760 mmHg gives us 101 kPa.

\(650 \ mmHg \ * \ \frac{101kPa}{760 mmHg} = 86 \ kPa\)

P = 86kPa

T = 15°C + 273K = 288K

R (Gas constant) = 8.31 kj/mol  

Molar mass of Ammonia (\(NH_{3}\)) = (1 x 3) + (14) = 17g/mol

n (moles) = \(\frac{mass}{molar \ mass}\) \(= \frac{56.8}{17} =\) 3.34 mol

V = ?

Rearrange the equation to solve for Volume:

\(V = \frac{nRT}{P}\)

Substitute the values inside:

V = \(\frac{(3.34)(8.31)(288)}{(86)} = 93 L (rounded)\)

Therefore 93 L of volume is occupied by the ammonia gas.

The number of the shared pairs in the molecule is two.

The term molecular geometry has to do with the actual arrangement of the electrons that are in the molecule. We know that the arrangement of the molecules that surround the atom is dependent on the nature of the electron pairs that surround the central atom.

Now we are told that  a central atom has 5 lone pairs and the molecular geometry is linear. The fact that the molecular geometry is linear shows that the number of the shared pairs is two.

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Answer:

68 degrees fahrenheit

Explanation:

Answer:

Arcuate: A fan-shaped delta, bowed or curved.

Cuspate: V-shaped delta.

Bird-foot: Outstretched claws delta

Inverted: The narrow end of delta, or river delta.

Waves: Large or regular shaped waves delta.

Tires: Round shaped delta

River estuary: A landform deposited by sediment

Explanation:

Answer:

Maritime tropical air mass

Continental polar air mass

Answer:

A)

Explanation:

Entre todos esas materias el hidrógeno es el más fuerte

Answer:

Ester

Explanation:

Got it right on the quiz, hope this helps :)

Answer:

do we have the volume ?

Explanation:

Answer:

Explanation:chlorine

The reaction of 1,3-cyclohexadiene and maleic acid (cis-2-butenedioic acid) leads to the formation of a cycloadduct known as a Diels-Alder adduct. The structure of the product is a cyclohexene ring fused with a carboxylic acid group.

The Diels-Alder reaction is a powerful organic transformation that involves the reaction of a conjugated diene (such as 1,3-cyclohexadiene) with a dienophile (such as maleic acid). In this reaction, the diene acts as a nucleophile and attacks the electron-deficient dienophile, resulting in the formation of a cyclic product.

When 1,3-cyclohexadiene reacts with maleic acid, the diene's double bonds undergo cycloaddition with the double bonds of maleic acid, forming a new six-membered ring. The resulting product is a cyclohexene ring fused with a carboxylic acid group. The stereochemistry of the product depends on the orientation of the dienophile. In the case of cis-2-butenedioic acid (maleic acid), the product would have a cis configuration.

The exact structure of the product would require more specific information about the reaction conditions and any other substituents present. However, based on the reactants provided, the reaction of 1,3-cyclohexadiene and maleic acid would yield a Diels-Alder adduct, consisting of a fused cyclohexene ring and a carboxylic acid group in a cis configuration.

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The main site for water reabsorption along the nephron is the renal tubules

Particularly the proximal tubule and the descending limb of the loop of Henle. As filtrate flows through the renal tubules, water and solutes are selectively reabsorbed into the bloodstream, with the majority of water reabsorption occurring in the proximal tubule. In this region, water is reabsorbed via osmosis, facilitated by the presence of aquaporin channels in the apical and basolateral membranes of the tubule cells. The descending limb of the loop of Henle is also important for water reabsorption, as it is permeable to water but not solutes, allowing for the creation of a concentration gradient that facilitates water reabsorption in the later parts of the nephron.

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