Asparagine has pKa1 = 2.02 and pKa2 = 8.80. Use the Henderson-Hasselbalch equation to calculate the ratio neutral form/protonated form at pH = 1.82.
Calculate the ratio, deprotonated form/neutral form, at pH = 9.25

Answer: 2/3

Explanation:

Answer:

Electron gain enthalpy becomes more and more negative from left to right in a period. As we move across a period from left to right the atomic size decreases and the nuclear charge increases

Answer: Variation in electron gain enthalpy in the period : in the modern periodic table, on moving from left to right across a period the atomic size of elements decreases and the effective nuclear charge increases.

hope it helps you

Explanation:

Answer:

Explanation:

as we know that 1dm3 =1000cm3

therefore x dm3 =120 cm3

1000X =1*120

X dm3 =120/1000=0.12

so 120 cm3 =0.12 dm3

now the number of moles of carbon dioxide at RTP is

moles =given volume(dm3)/standard volume at RTP

moles=0.12dm3/24

moles =5*10^-3

Answer:

Subatomic particles

Explanation:

I hope it helped

Answer:

pre lab information would be the thing you to ressearch the lab to get the information

Limiting Reactant and Percent Yield Lab Report attached

To calculate the volume of the reactor and space-time for a reversible gas phase reaction, A+2B, conducted at 540 °F and 3 atm with a feed rate of 75 lb mol/h and 40% A, and an equilibrium conversion of 75%, we need to consider the specific reaction rate and the concentration equilibrium constant.

The space-time for a reactor is defined as the volume of the reactor divided by the feed rate. To calculate the volume of the reactor, we first determine the molar flow rate of component A, which is 75 lb mol/h * 0.40 = 30 lb mol/h. Then, we divide the molar flow rate of A by the specific reaction rate to obtain the volume: Volume = 30 lb mol/h / (1.6 s * 3600 s/h) = 5.2083 ft³.

To calculate the space-time, we divide the volume by the feed rate: Space-time = 5.2083 ft³ / 75 lb mol/h = 0.0694 ft³/lb mol/h.

Therefore, the volume of the reactor is 5.2083 ft³ and the space-time is 0.0694 ft³/lb mol/h.

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To determine the maximum number of NH3 molecules that could form, we need to know the quantities of N2 and H2 available and identify the limiting reactant using the balanced chemical equation.

To determine the maximum number of NH3 molecules that could form, we need to consider the balanced chemical equation for the reaction. Without specific reactants mentioned in the question, I will assume we are referring to the synthesis of ammonia (NH3) from its elements.

The balanced chemical equation for the reaction is:

N2 + 3H2 → 2NH3

From this equation, we can see that 1 molecule of nitrogen gas (N2) reacts with 3 molecules of hydrogen gas (H2) to form 2 molecules of ammonia (NH3). Therefore, the stoichiometry of the reaction indicates that the ratio of N2 to NH3 is 1:2.

Given that we want to find the maximum number of NH3 molecules that can form, we need to determine the limiting reactant. The limiting reactant is the reactant that is completely consumed first, thus limiting the amount of product formed.

To do this, we need to know the quantities of N2 and H2 available. Without this information, we cannot provide a specific numerical answer. However, it is important to note that the maximum number of NH3 molecules that could form will be determined by the quantity of the limiting reactant.

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The ingestion of one Tums Ultra 1000 tablet, containing 1000 mg of calcium carbonate, can cause a pH change in the stomach acid due to the antacid properties of calcium carbonate.

Calcium carbonate is a common ingredient in antacid tablets like Tums Ultra 1000. It works by neutralizing excess stomach acid, raising the pH level and reducing the acidity. The pH scale measures the acidity or alkalinity of a solution, with lower pH values indicating higher acidity.

The exact pH change resulting from the ingestion of one Tums Ultra 1000 tablet depends on several factors such as the concentration of the stomach acid and the buffering capacity of the tablet. However, in general, calcium carbonate reacts with stomach acid (hydrochloric acid) to form water, carbon dioxide, and calcium chloride. This reaction reduces the concentration of hydrochloric acid, thereby increasing the pH of the stomach acid.

The specific calculation of the pH change requires more information, such as the initial pH of the stomach acid and the exact concentration of the tablet's active ingredient. Nevertheless, the antacid properties of calcium carbonate in Tums Ultra 1000 can effectively raise the pH of the stomach acid and provide relief from symptoms of acidity.

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Answer:

c because this is the one hundred all the time

Explanation:

Answer:

Si tú puedes 1-2-3-4-5-6-7-8-9-10 eso debes de ser más bien dicho sacaran Club Ojalá que te ayude Chau

Explanation:

de Ojalá que te ayude No te olvides de sacar todas las preguntas esas lo que tú dijiste y si tú promoción 123 saquen globo que quieras tú

The IUPAC name for the given compound \((CH_3)_2C=CHCH_2CH_2OH\) is 3-hydroxy-2-methylbut-2-ene.

First, we identify the longest continuous carbon chain, which consists of five carbon atoms. This forms the parent chain, which is a butene.

Next, we locate the double bond, which is between the second and third carbon atoms. Since the double bond starts at the second carbon atom, we indicate it as "but-2-ene."

Moving on, we need to indicate the presence of substituents. In this case, we have a methyl group attached to the second carbon atom. Therefore, we include the prefix "2-methylbut-2-ene."

Lastly, there is a hydroxyl group (-OH) attached to the fourth carbon atom. We denote this as "hydroxy" and specify the position of the hydroxyl group, resulting in the full name "3-hydroxy-2-methylbut-2-ene."

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The reason why 4-aminophenol is acetylated at the amine group rather than the phenol group is due to the difference in reactivity and nucleophilicity between the amine and phenol functional groups.

The amine group (–NH2) is more reactive and nucleophilic compared to the phenol group (–OH). This is because the lone pair of electrons on the nitrogen atom in the amine group is more available for nucleophilic attack.

When an acetylating agent, such as acetic anhydride (CH3CO)2O or acetyl chloride (CH3COCl), is used to acetylate 4-aminophenol, the amine group acts as a stronger nucleophile and readily reacts with the electrophilic acetylating agent. This results in the substitution of the hydrogen atom on the amine group with an acetyl group (–COCH3), leading to the formation of N-acetyl-4-aminophenol.

On the other hand, the phenol group in 4-aminophenol is less reactive and nucleophilic. The lone pair of electrons on the oxygen atom in the phenol group is less available for nucleophilic attack. Therefore, under the same conditions, the amine group is more likely to undergo acetylation instead of the phenol group.

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Which of the following explains why 4-aminophenol is acetylated at the amine group rather than the phenol group? the nitrogen is less electronegative than the oxygen the nitrogen is more electronegative than the oxygen the nitrogen is less sterically hindered than the oxygen the nitrogen is smaller than the oxygen Question 2 1 pts Starting with 3.00 g of 4-aminophenol (109.13 g/mol) and excess acetic anhydride (102.09 g/mol), what is the theoretical yield (in g) of acetaminophen ( 151.16 g/mol)?

mass divided by volume

Answer: This may not work for you

Explanation: Density, mass of a unit volume of a material substance. The formula for density is d = M/V, where d is density, M is mass, and V is volume. Density is commonly expressed in units of grams per cubic centimetre.

The concentration of perchlorate ion in a perchloric acid solution with a pH of 3.158 can be calculated using the Henderson-Hasselbalch equation.Therefore, the concentration of perchlorate ion in the solution is 4.7 x 10^{-4} M.

Perchloric acid (HClO4) is a strong acid. A strong acid is a substance that almost entirely dissociates into ions when it dissolves in water. As a result, perchloric acid can be thought of as a source of hydrogen ions (H+) in solution.The concentration of hydrogen ions in solution can be determined using the pH of the solution. pH is the negative logarithm of the hydrogen ion concentration in moles per liter (mol/L) of solution.A pH of 3.158 corresponds to a hydrogen ion concentration of:10^{-3.158} = 6.59 x 10^{-4}  mol/LIn a perchloric acid solution, the perchlorate ion (ClO4) is the conjugate base of the acid. It can be concluded that the concentration of perchlorate ions (ClO4-) is equal to the concentration of hydrogen ions (H+) in the solution because the acid completely dissociates. Therefore, the concentration of perchlorate ions in a solution with a pH of 3.158 is 6.59 x 10^{-4} mol/L.Therefore, the answer is that the concentration of perchlorate ion in perchloric acid solution with a pH of 3.158 is 6.59 x 10^{-4 }mol/L.

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Answer:

It is c I am doing the test I hope this helps ; )

Explanation:

Answer:

c

Explanation:

Answer:

Ag one dot

Explanation:

i hope it helps you

Answer:

D) Temperature,

Explanation:

Answer:

2 Na

Explanation:

2 Na atoms per 1 Sulfide atom

hope this helps

Answer:

I think its b

Explanation:

Sorry if wrong! Hope this helps tho! maybe mark me brainliest if I am?

Answer:

Whats the question? I can try and help you

Explanation:

Answer:

Its the first one

Explanation:

4 mol is the highest of the Bunch and 4.0L is the lowest making the first answer the correct one

If you have 10 kg each of a radioactive sample a with a half-life of 100 years, and another sample b with a half-life of 1000 years, sample A has a higher decay constant and higher activity.

The activity of a radioactive sample refers to the number of decays occurring per unit time. It is measured in units of becquerels (Bq) or curies (Ci). The activity of a sample is proportional to the number of radioactive nuclei present in the sample.

The decay rate of a radioactive sample is determined by its half-life. The shorter the half-life, the higher the decay rate and the higher the activity. Therefore, sample A with a half-life of 100 years will have a higher decay rate and higher activity than sample B with a half-life of 1000 years.

To calculate the activity of a sample, we use the following formula

Activity = λN

where λ is the decay constant, and N is the number of radioactive nuclei present in the sample.

Since the two samples have the same mass, the number of radioactive nuclei will be the same. Therefore, the sample with the higher decay constant (λ) will have the higher activity.

The decay constant is related to the half-life by the following formula:

λ = ln(2) / \(t^{\frac{1}{2} }\)

where ln(2) is the natural logarithm of 2, and \(t^{\frac{1}{2} }\) is the half-life.

Using this formula, we can calculate the decay constants for samples A and B

\(\lambda_{A}\)= ln(2) / 100 years = 0.00693 per year

\(\lambda_{B}\) = ln(2) / 1000 years = 0.000693 per year

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The carbon-carbon bonds where rotation can occur on 2-methylhexane are the C-1−C-2 bond, C-2−C-3 bond, and C-5−C-6 bond.

Rotation is possible around single bonds, and in 2-methylhexane, these bonds are all single bonds. The C-1−C-2 bond, C-2−C-3 bond, C-4−C-5 bond, and C-5−C-6 bond are all single bonds, allowing for free rotation. On the other hand, the C-2−C-7 bond is not present in 2-methylhexane and therefore rotation cannot occur on that specific bond.

In 2-methylhexane, rotation can occur around the following carbon-carbon bonds:

C-1−C-2 bond: Yes, rotation can occur around this bond.

C-2−C-3 bond: Yes, rotation can occur around this bond.

C-4−C-5 bond: No, rotation cannot occur around this bond because it involves the methyl group attached to the second carbon, which creates a hindered rotation.

C-5−C-6 bond: Yes, rotation can occur around this bond.

C-2−C-7 bond: No, rotation cannot occur around this bond because it involves the methyl group attached to the second carbon, which creates a hindered rotation.

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The average mass of the copper atoms in the sample is 63.4622232 amu.

The average mass of the copper atoms in the sample can be calculated by multiplying the mass of each isotope by its percentage abundance and summing the results.

Copper-63: 69.2% x 62.92960 amu = 43.480832 amu
Copper-65: 30.8% x 64.92779 amu = 19.9813912 amu

Adding these values together:

43.480832 amu + 19.9813912 amu = 63.4622232 amu

Therefore, the average mass of the copper atoms in the sample is 63.4622232 amu.

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Answer:

a

barium

b

calcium

e

aluminum

f

magnesium

g

copper

Forms cation

Explanation:

The final molarity of the acetate anion in the solution is 0.04 mM.

To calculate the final molarity of acetate anion in the solution, we need to know the mass of barium acetate that was dissolved and the number of moles of acetate anion that it contains.

Barium acetate has the formula Ba(C2H3O2)2, which means that it contains 1 mole of Ba, 2 moles of C2H3O2, and a total of 3 moles of ions. Since we are interested in the molarity of acetate anion, we need to calculate the number of moles of C2H3O2 that are present in 0.270 g of barium acetate.

The molecular weight of barium acetate is 244.26 g/mol, which means that 0.270 g of barium acetate contains 0.270 g / 244.26 g/mol = 0.0011 moles of barium acetate. Since each mole of barium acetate contains 2 moles of C2H3O2, this means that 0.0011 moles of barium acetate contain 0.0011 moles × 2 moles/mole = 0.0022 moles of C2H3O2.

The initial molarity of the solution is 37.0 mM, which means that it contains 37.0 moles/L of sodium chromate. The volume of the solution is 50.0 mL, which means that it contains 50.0 mL × 37.0 moles/L = 1.85 moles of sodium chromate.

When 0.270 g of barium acetate is dissolved in the solution, the total number of moles of ions increases by 0.0022 moles. This means that the final molarity of the acetate anion in the solution is:

Molarity = (total number of moles of ions) / (volume of solution)

= (1.85 moles + 0.0022 moles) / (50.0 mL)

= 0.037 mM

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In order to properly balance an equation, we need to make sure that the same amount of elements on the reactants side matches the number of elements on the products side, we can do that by increasing the number in front of each molecule, the so called stoichiometric coefficient. In the reaction from the question we can properly balance by adding the following stoichiometric coefficients

1. 2 H2 + O2 -> 2 HO2

2. FeCl2 + H2O -> FeO + 2 HCl

3. C4H8 + 6 O2 -> 4 CO2 + 4 H2O

4. ?

5. 2 NaOH + MgCl2 -> 2 NaCl + Mg(OH)2

These orbital configurations represent the arrangement of electrons within the different energy levels and subshells of the respective elements.

The orbital configurations of the given elements are as follows:

Helium: 1s² - Helium has two electrons that occupy the 1s orbital.

Nitrogen: 1s² 2s² 2p³ - Nitrogen has two electrons in the 1s orbital, two electrons in the 2s orbital, and three electrons in the 2p orbital (specifically, 2p³ indicates three electrons in the 2p subshell).

Silicon: 1s² 2s² 2p⁶ 3s² 3p² - Silicon has two electrons in the 1s orbital, two electrons in the 2s orbital, six electrons in the 2p orbital, two electrons in the 3s orbital, and two electrons in the 3p orbital (specifically, 3p² indicates two electrons in the 3p subshell).

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