# The Generic Metal Hydroxide M(OH)2 Has Ksp = 7.651012. (NOTE: In This Particular Problem, Because Of (2023)

Chemistry College

Solubility will be = 1.24 X 10⁻⁴ M

Explanation:

When the concentration of the salt dissolved in water is very low the concentration of hydroxide ions generated by the water can be ignored.

The dissociation of metal hydroxide can be expressed as:

M^{+2}+2OH^{-}" alt="M(OH)_{2}--->M^{+2}+2OH^{-}" align="absmiddle" class="latex-formula">

Ksp =

Let the solubility of the metal hydroxide is "s"

Ksp = (s)(2s)²=4s³

Ksp = 7.65X10⁻¹² = 4s³

Therefore

s = 1.24 X 10⁻⁴ M

## Related Questions

A 0.1946 g piece of magnesium metal is burned in a constant-volume calorimeter that has a heat capacity for the bomb only of 1349 J/°C. The calorimeter contains 5.00 x 102g of water and the temperature rise is 1.40°C. Calculate the heat of combustion of magnesium metal in kJ/g, given that thespecific heat of water is 4.184 J/g·°C.

The heat of combustion of magnesium metal is 24.76 kJ/gram

Explanation:

Step 1: Data given

Mass of magnesium sample = 0.1946 grams

Molar mass of magnesium = 24.3 g/mol

bomb calorimeter that has a heat capacity of 1349 J/°C

Mass of water = 500 grams

Temperature change = 1.40 °C

Step 2: Calculated heat released

Q = (1349 J/°C * 1.40 °C) + (500 grams * 4.184 J/g°C * 1.40 °C)

Q =4817.4 J = 4.82 kJ

Step 3: Calculate the heat given off by the burning Mg, in kJ/g

4817.4 J / 0.1946 grams = 24755.4 J/ gram = 24.76 kJ/ gram

The heat of combustion of magnesium metal is 24.76 kJ/gram

State the Pauli exclusion principle. Choose one: A. No two electrons in an atom can have the same set of four quantum numbers. B. Each electron goes into the lowest-energy orbital available. C. An attraction between two electrons occurs due to vibrations in the crystal lattice. D. The lowest-energy electron configuration of an atom has the maximum number of unpaired electrons, all of which have the same spin, in degenerate orbitals. E. Electrons orbiting and protons within the nucleus of the atom have a balanced charge, excluding attraction to charged species. F. We cannot determine both the position and the momentum of an electron in an atom at the same time.

The Pauli exclusion principle has been part of the electron quantum mechanics. It states that no two electrons in an atom can occupy the same set of four quantum numbers. Thus option A is correct.

Quantum mechanics helps in providing the address to the electrons. Since the address has been provided to the electrons, it has been unique and can not be allocated to other electrons in an atom. This address has been in the form of a quantum number.

Thus, Pauli's exclusion principle states that no two electrons in an atom can occupy the same set of four quantum numbers. Thus option A is correct.

The other options are incorrect because:

• Option B states that each electron goes into the lowest-energy orbital available. This has been the statement for Aufbau's Principle. Thus option B is incorrect.
• Option C states that an attraction between two electrons occurs due to vibrations in the crystal lattice. The statement has been true, but it did not correspond to Pauli's exclusion principle. Thus option C is incorrect.
• Option D states that the lowest-energy electron configuration of an atom has the maximum number of unpaired electrons, all of which have the same spin, in degenerate orbitals. The statement has been Hund's rule of maximum multiplicity. It does not correspond to Pauli's principle. Thus option D is incorrect.
• Option E states that electrons orbiting and protons within the nucleus of the atom have a balanced charge, excluding attraction to charged species. The statement has been true for the neutral species but is incorrect for the charged species. The statement has no relation with Puli's principle. Thus option E is incorrect.
• Option F states that we cannot determine both the position and the momentum of an electron in an atom at the same time. The statement has been the part of the Heisenberg Uncertainty principle, and not Pauli's principle. Thus option F is incorrect.

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A. No two electrons in an atom can have the same set of four quantum numbers.

Explanation:

State the Pauli exclusion principle.

A. No two electrons in an atom can have the same set of four quantum numbers. Electrons are characterized by 4 quantum numbers. According to Pauli exclusion principle, there cannot be 2 electrons having the same 4 quantum numbers.

B. Each electron goes into the lowest-energy orbital available. This is known as Aufbau principle.

C. An attraction between two electrons occurs due to vibrations in the crystal lattice. This is true but it does not correspond to Pauli principle.

D. The lowest-energy electron configuration of an atom has the maximum number of unpaired electrons, all of which have the same spin, in degenerate orbitals. This is known as Hund's rule of maximum multiplicity.

E. Electrons orbiting and protons within the nucleus of the atom have a balanced charge, excluding attraction to charged species. The difference between the number of electrons and protons leads to the charge of the species.

F. We cannot determine both the position and the momentum of an electron in an atom at the same time. This is known as Heisenberg uncertainty principle.

One mole of an ideal gas is sealed in a 22.4-L container at a pressure of 1 atm and a temperature of 273 K. The temperature is then increased to 304 K , but the container does not expand. What will the new pressure be? Part A

The most appropriate formula for solving this problem includes only which variables?

Enter the required variables, separated by commas (e.g., P,V,T).

Q2)A sample of nitrogen gas in a 1.69-L container exerts a pressure of 1.37 atm at 17 ∘C.

-What is the pressure if the volume of the container is maintained constant and the temperature is raised to 327 ∘C?

Q3)A gas mixture with a total pressure of 770 mmHgcontains each of the following gases at the indicated partial pressures: 120 mmHg CO2, 227mmHg Ar, and 190 mmHg O2. The mixture also contains helium gas

-What mass of helium gas is present in a 14.0-L sample of this mixture at 282 K ?

Q4)

A

Calculate the density of oxygen, O2, under each of the following conditions:

STP

1.00 atm and 35.0 ∘C

B

To identify a diatomic gas (X2), a researcher carried out the following experiment: She weighed an empty 4.1-L bulb, then filled it with the gas at 2.00 atm and 24.0 ∘C and weighed it again. The difference in mass was 9.5 g . Identify the gas.

1.11 atm

(P, T)

2.83 atm

0.740 g

0.179 g/L, 0.158 g/L

N₂

Explanation:

One mole of an ideal gas is sealed in a 22.4-L container at a pressure of 1 atm and a temperature of 273 K. The temperature is then increased to 304 K, but the container does not expand. What will the new pressure be?

Assuming ideal behavior, we can calculate the new pressure (P₂) using Gay-Lussac's law.

The most appropriate formula for solving this problem includes only which variables?

Gay-Lussac's law includes pressure (P) and absolute temperature (T).

Q2) A sample of nitrogen gas in a 1.69-L container exerts a pressure of 1.37 atm at 17 °C. What is the pressure if the volume of the container is maintained constant and the temperature is raised to 327 °C?

Initially the system is at 17°C (290 K) and the temperature is raised to 327°C (600 K). We can calculate the new pressure using Gay-Lussac's law.

Q3) A gas mixture with a total pressure of 770 mmHg contains each of the following gases at the indicated partial pressures: 120 mmHg CO₂, 227mmHg Ar, and 190 mmHg O₂. The mixture also contains helium gas .

What mass of helium gas is present in a 14.0-L sample of this mixture at 282 K?

First, we have to calculate the pressure of Helium. We know that the total pressure is the sum of partial pressures.

Ptotal = pCO₂ + pAr + pO₂ + pHe

pHe = Ptotal - pCO₂ - pAr - pO₂

pHe = 770mmHg - 120mmHg - 227mmHg - 190mmHg=233mmHg

We can calculate the moles of Helium using the ideal gas equation.

The molar mass of He is 4.00g/mol.

Calculate the density of oxygen, O₂, under each of the following conditions:

• STP
• 1.00 atm and 35.0 ∘C

STP stands for Standard Temperature and Pressure. The standard temperature is 273 K and the standard pressure is 1 atm.

We can calculate the density using the following expression:

At 1.00 atm and 35.0 °C (308 K)

To identify a diatomic gas (X₂), a researcher carried out the following experiment: She weighed an empty 4.1-L bulb, then filled it with the gas at 2.00 atm and 24.0 ∘C and weighed it again. The difference in mass was 9.5 g . Identify the gas. Express your answer as a chemical formula.

We will look for the molar mass of the compound using the ideal gas equation.

If the molar mass of X₂ is 28 g/mol, the molar mass of X is 14 g/mol. Then, X is nitrogen and X₂ is N₂.

Red gold is a gold‑copper alloy used to make jewelry. A piece of jewelry made of red gold weighs 7.67 g and has a volume of 0.430 cm3. Gold has a density of 19.3 g/cm3 and copper has a density of 8.96 g/cm3 . Calculate the percentage by mass of each metal in the jewelry. Assume the total volume of the jewelry is the sum of the volumes of the two metals it contains. gold: % copper: % Pure gold is defined as having 24 carats. When mixed in an alloy, the carats of gold are given as a percentage of this value. For example, a piece of jewelry made with 50% gold has 12 carats. State the purity of this piece of red gold jewelry in carats. purity: carats

% gold = 92.83 %

% copper = 7.17 %

Given:

Mass of the piece = 7.67 grams

Volume of the piece = 0.430 cm³

Density of gold = 19.3 g/cm³

Density of copper = 8.96 g/cm³

To find:

% gold and % copper=?

Since the total volume is the sum of both volumes (gold, copper) we can say:

Vgold + Vcopper = 0.430 cm³

Density:

For Vgold = Mass of gold / Density of gold

For Vcopper = Mass of copper / Density of copper

On substituting the values in the above formula we will get:

• For gold:
• For copper:

On rearranging both the equations:

• Calculation for volume:

Total volume = 0.430 cm³

Volume of gold = 0.369 cm³

Volume of copper = 0.061 cm³

Total volume = 0.369 + 0.061 = 0.430 cm³

• Calculation of mass:

Mass gold = volume gold * density

Mass gold = 0.369 cm³ * 19.3 g/cm³

Mass gold = 7.1217 grams

Mass copper = volume copper * density copper

Mass copper = 0.061 cm³ * 8.96 g/cm³

Mass copper = 0.54656 grams

Total mass = 7.12+0.55 = 7.67 grams

• Calculation of %gold and % copper

% gold = (7.12/7.67) * 100% = 92.83 %

% copper = (0.55 / 7.67) *100% = 7.17 %

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% gold = 92.83 %

% copper = 7.17 %

Explanation:

Step 1: Data given

Mass of the piece = 7.67 grams

Volume of the piece = 0.430 cm³

Density of gold = 19.3 g/cm³

Denisty of copper = 8.96 g/cm³

Step 2: define volumes

Since the total volume is the sum of both volumes (gold, copper) we can say:

Vgold + Vcopper = 0.430 cm³

with Vgold = Mass of gold / Density of gold

with Vcopper = Mass of copper / Density of copper

Step 3: rearrange formules

19.3 g/cm³ = m1/V1 → 19.3*V1 = m1 (mass of the gold in terms of it's density and volume )

8.96 g/cm³ = m2/V2 --> 8.96g * V2 = m2 (mass of the silver in terms of it's density and volume)

m1 + m2 = 7.67grams

19.3*V1 + 8.96g * V2 = 7.67 grams

Step 4: Calculate volumes

Total volume = 0.430 cm³

V1 + V2 = 0.430 cm³ --> V1 = 0.430 - V2

19.3*(0.430 - V2) + 8.96*V2 = 7.67

8.299 - 19.3V2 + 8.96V2 = 7.67

8.299 - 7.67 = 19.3 V2 - 8.96V2

0.629 =10.34V2

V2 =0.061 cm³

V1 *= 0.430 cm³ - 0.061cm³ = 0.369 cm³

Volume of gold = 0.369 cm³

Volume copper = 0.061

Total volume = 0.369 + 0.061 = 0.430 cm³

Step 5: Calculate mass

Mass gold = volume gold * density

Mass gold = 0.369 cm³ * 19.3 g/cm³

Mass gold = 7.1217 grams

Mass copper = volume copper * density copper

Mass copper = 0.061 cm³ * 8.96 g/cm³

Mass copper = 0.54656 grams

Total mass = 7.12+0.55 = 7.67 grams

Step 6: Calculate % godl and copper

% gold = (7.12/7.67) * 100% = 92.83 %

% copper = (0.55 / 7.67) *100% = 7.17 %

Suppose a 250.mL flask is filled with 0.50 mol of H2O , 1.2 mol of CO2 and 1.3 mol of H2 . The following reaction becomes possible: CO(g) +H20(g)-CO2(g)+H2(g) The equilibrium constant K for this reaction is 0.289 at the temperature of the flask. Calculate the equilibrium molarity of H2 . Round your answer to one decimal place.

2.3 M

Explanation:

First, we have to calculate the initial molarities of the substances.

In the beginning there is no CO, so the reaction must proceed to the left to attain equilibrium. In order to know the concentrations at equilibrium we will use an ICE chart. "I" stands for Initial, "C" stands for Change and "E" stands for equilibrium. We complete each row with the concentration or change in concentration in that stage.

CO(g) + H₂O(g) ⇄ CO₂(g) + H₂(g)

I 0 2.0 4.8 5.2

C +x +x -x -x

E x 2.0 + x 4.8 - x 5.2 - x

The equilibrium constant K is:

The answers are x₁ = 11.9 and x₂=2.9. Since x₁ would make concentrations negative (which is not possible) we choose x₂.

The molarity of H₂ is:

5.2 - x = 5.2 -2.9 = 2.3 M

What type of ions have names ending in -ide?

Explanation:

The that end in -ide include, hydroxide, peroxide, cyanide, ##NH_2^-## amide and hydrogen sulfide. The -ide suffix, is usually reserved for monatomic anions. Non-metal ions like oxygen, fluorine and chlorine take on the -ide and become oxide, fluoride and chloride.

Suppose that a 1.0 liter vessel contains N2O gas at 1.0 atm pressure. Suppose that a second vessel, 6.0 L in volume contains oxygen at 6.0 atm pressure. Now suppose that the two vessels are connected by a pipe of negligible volume and the two gases mix and react to produce as much nitrogen dioxide as possible. Assume that the temperature remains constant. What is the pressure in the apparatus at the end of the reaction?

P = 5.868 atm

Explanation:

• Pt = (PN2O)(XN2O) + (PO2)(XO2)

assume T = 25°C = 298.15 K

∴ PN2O = 1 atm

∴ XN2O = nN2O/nt

∴ n = PV/RT....ideal gas

⇒ nN2O = ((1atm)(1L))/((0.082atm.L/K.mol)(298.15K)) = 0.041 mol N2O

∴ PO2 = 6 atm

∴ XO2 = nO2/nt

∴ nO2 = ((6atm)(6L))/((0.082atm.L/K.mol)(298.15K)) = 1.472 mol O2

⇒ nt = nN2O + nO2 = 0.041 + 1.472 = 1.513 mol

⇒ XN2O = 0.041/1.513 = 0.03

⇒ XO2 = 1.472/1.513 = 0.973

⇒ Pt = ((1atm)(0.03)) + ((6atm)(0.973))

⇒ Pt = 5.868 atm

State Hund's rule. Choose one:

A. An attraction between two electrons occurs due to vibrations in the crystal lattice.

B. Each electron goes into the lowest-energy orbital available.

C. The lowest-energy electron configuration of an atom has the maximum number of unpaired electrons, all of which have the same spin, in degenerate orbitals.

D. We cannot determine both the position and the momentum of an electron in an atom at the same time.

E. No two electrons in an atom can have the same set of four quantum numbers.F. Electrons orbiting and protons within the nucleus of the atom have a balanced charge, excluding attraction to charged species.

C. The lowest-energy electron configuration of an atom has the maximum number of unpaired electrons, all of which have the same spin, in degenerate orbitals.

Explanation:

The Hund's rule is used to place the electrons in the orbitals is it states that:

1. Every orbital in a sublevel is singly occupied before any orbital is doubly occupied;

2. All of the electrons in singly occupied orbitals have the same spin.

So, the electrons first seek to fill the orbitals with the same energy (degenerate orbitals) before paring with electrons in a half-filled orbital. Orbitals doubly occupied have greater energy, so the lowest-energy electron configuration of an atom has the maximum number of unpaired electrons, and for the second statement, they have the same spin.

The other alternatives are correct, but they're not observed by the Hund's rule.

Glucose (C6H12O6) can be fermented to yield ethanol (CH3CH2OH) and carbon dioxide (CO2). C6H12O6⟶2CH3CH2OH+2CO2 The molar mass of glucose is 180.15 g/mol, the molar mass of ethanol is 46.08 g/mol, and the molar mass of carbon dioxide is 44.01 g/mol. What is the theoretical yield of ethanol from the fermentation of 61.5 g of glucose? theoretical yield: g If the reaction produced 23.4 g of ethanol, what is the percent yield? percent yield:

The % yield is 74.45 %

Explanation:

Step 1: The balanced equation

C6H12O6⟶2CH3CH2OH+2CO2

Step 2: Data given

Molar mass glucose = 180.15 g/mol

Molar mass of ethanol = 46.08 g/mol

Molar mass of carbon dioxide = 44.01 g/mol

Mass of glucose = 61.5 grams

Mass of ethanol = 23.4 grams

Step 3: Calculate moles of glucose

Moles glucose = Mass glucose / Molar mass of glucose

Moles glucose = 61.5 grams / 180.15 g/mol

Moles glucose = 0.341 moles

Step 4: Calculate moles of ethanol

1 mole of glucose consumed, produces 2 moles of ethanol and 2 moles of CO2

0.341 moles of glucose, will produce 2*0.341 = 0.682 moles of ethanol

Step 5: Calculate mass of ethanol

Mass ethanol = moles ethanol * Molar mass ethanol

Mass ethanol = 0.682 moles * 46.08 g/mol

Mass ethanol = 31.43 grams = theoretical mass

Step 6: Calculate % yield

% yield = actual mass / theoretical mass

% yield = (23.4 grams / 31.43 grams) * 100%

% yield = 74.45 %

The % yield is 74.45 %

The combustion of ethane in the presence of excess oxygen yields carbon dioxide and water: 2C2H6 (g) + 7O2 (g) → 4CO2 (g) + 6H2O (l)

The value of ΔG° for this reaction is __________kJ/mol. Compare results for both methods of finding ΔG as done in class. Is this difference significant?

ΔG° = - 2936.604 KJ/mol

Explanation:

• 2C2H6(g) + 7O2(g) → 4CO2(g) + 6H2O(l)

⇒ ΔG°t = 4ΔG°fCO2(g) + 6ΔG°fH2O(l) - 7ΔG°fO2 - 2ΔG°fC2H6(g)

∴ ΔG°fCO2(g) = - 394.359 KJ/mol....from literature

∴ ΔG°fH2O(l) = - 237.178 KJ/mol

∴ ΔG°fO2(g) = 0.00 KJ/mol

∴ ΔG°fC2H6(g) = - 31.95 KJ/mol

⇒ ΔG°t = 4(- 394.359 KJ/mol) + 6(- 237.178 KJ/mol) - 7(0.00 KJ/mol) - 2( -31.95 KJ/mol)

⇒ ΔG°t = - 1577.436 KJ/mol - 1423.068 KJ/mol + 63.9 KJ/mol

⇒ ΔG°t - 2936.604 KJ/mol

The enzyme urease catalyzes the breakdown of urea in the body. Urease breaks urea down to 2NH3+CO2. This is an example of a hydrolysis reaction (urea plus water). For each equivalent of carbon dioxide (CO2), two equivalents of ammonia (NH3) are produced. Label the enzyme, substrate, enzyme–substrate complex, enzyme–product complex, and product in the enzymatic reaction for the breakdown of urea by urease using the induced-fit model.

Enzyme = Urease

Substrate= Urea

Urease-urea complex = Enzyme-substrate complex

Urease-Ammonia, Carbon dioxide complex = Enzyme-product complex

Ammonia and Carbon Dioxide = Products

Explanation:

In order to answer the query, we need to review some of the definitions.

Enzymes are proteins which speed up chemical reactions. The substance on which enzyme exerts its effect is called Substrate.

The combination of Enzyme and substrate is called Enzyme-Substrate complex.

After the chemical reaction, the enzyme-substrate complex breaks into product and enzyme.

Induced-fit model was proposed by Koshland in which he stated that during the reaction of enzymes and substrate, the active site of enzyme modifies (changes shape) which helps it to perform its catalytic activity more efficiently.

Coming back to the scenario in question,

Enzyme = Urease (It is stated that it catalyzes the breakdown of urea)

Substrate= Urea (It is the substance on which urease enzyme acts)

Urease-urea complex = Enzyme-substrate complex

Urease-Ammonia and Carbon dioxide = Enzyme-product complex

Ammonia and Carbon Dioxide = Products

A gas‑filled weather balloon with a volume of 58.0 L is held at ground level, where the atmospheric pressure is 771 mmHg and the temperature is 24.0 ∘C . The balloon is released and rises to an altitude where the pressure is 0.0571 bar and the temperature is −6.94 ∘C . Calculate the weather balloon's volume at the higher altitude.

The balloon's volume at higher altitude is 935.4 L

Explanation:

Step 1: Data given

Volume = 58.0 L

Atmospheric pressure = 771 mmHg = 1.01447 atm

Temperature = 24.0 °C = 297.15 Kelvin

When the balloon is released , it has a pressure of 0.0571 bar (= 0.0563533185 atm)

Temperature = -6.94 °C ( = 266.21 Kelvin)

Step 2: Calculate the volume

P*V=n*R*T with n and R are constant

(P1*V1)/ T1 = (P2*V2)/ T2

with P1 = 771 mmHg = 1.01447 atm

with V1 = 58.0 L

with T1 = 24.0 °C = 297.15 Kelvin

with P2 = 0.0571 bar (= 0.0563533185 atm)

with V2 = TO BE DETERMINED

with T2 = -6.94 °C ( = 266.21 Kelvin)

V2 = (P1 * V1 * T2 )/( T1 * P2)

V2 = (1.01447 atm * 58.0 L * 266.21 K) / (297.15 K * 0.0563533185 atm)

V2 =935.4 L

The balloon's volume at higher altitude is 935.4 L

The rate of reaction in terms of the "rate law expression" includes the rate constant (k), the concentration of the reactants, and the orders of the reaction with respect to the different reactants. Consider the following reaction: A+B→C+D The initial concentrations of the reactants A and B are 0.400 M and 0.290 M, respectively. The rate of reaction is 0.060 M⋅s−1, and the orders of the reaction, with respect to reactants A and B, are 1 and 2, respectively. Determine the rate constant (k) for the reaction using the rate law. Express your answer in M−2⋅s−1 to three significant figures. View Available Hint(s) The rate constant(k) k (k) = nothing M−2⋅s−1

1.78 M⁻²s⁻¹

Explanation:

Let's consider the following reaction:

A + B → C + D

The reaction order for A is 1 and the reaction order for B is 2.

The rate law is:

r = k.[A].[B]²

where,

r is the rate of the reaction

k is the rate constant

[A] and [B] are the molar concentrations of the reactants

Then, we can find the value of k.

The value of Δ????°′ΔG°′ for the conversion of glucose-6-phosphate to fructose-6-phosphate (F6P) is +1.67 kJ/mol+1.67 kJ/mol . If the concentration of glucose-6-phosphate at equilibrium is 1.85 mM1.85 mM , what is the concentration of fructose-6-phosphate? Assume a temperature of 25.0°C25.0°C .

Answer : The concentration of fructose-6-phosphate is 0.275 mM

Explanation :

First we have to calculate the value of equilibrium constant.

The relation between the equilibrium constant and standard Gibbs free energy is:

where,

= standard Gibbs free energy = +1.67 kJ/mol = +1670 J/mol

R = gas constant = 8.314 J/K.mol

T = temperature =

K = equilibrium constant = ?

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

Now we have to calculate the concentration of fructose-6-phosphate.

The expression of equilibrium constant is:

Therefore, the concentration of fructose-6-phosphate is 0.275 mM

What is the catalytic triad of chymotrypsin, a type of serine protease? the enzyme-cofactor-intermediate complex the amino acids serine, histidine, and aspartate the amino acids cysteine, histidine, and aspartate the enzyme-cofactor-substrate complex the amino acids serine, histidine, and glutamate

The correct answer is: Serine, Histidine, Aspartate

Explanation:

The catalytic triad of an enzyme is composed of three aminoacid residues which are the most important for its catalytic activity. They are located in the catalytic site of the enzyme. In the case of chymotrypsin- a serine protease, the catalytic triad is composed by serine, histidine and aspartate (Ser-His-Asp). Serine proteases hydrolyse peptidic bonds in proteins and peptides. To do that, the histidine-which interacts with the aspartate by a hydrogen bond so its pKa increases- take a proton from the serine. Thus, deprotonated serine is able to attack the peptide bond and to perform hydrolysis.

Using the Henderson-Hasselbalch equation, calculate the amount of HEPES (sodium salt) and HEPES (free acid) required to prepare 50 ml of a 100 mM buffer that is pH = 7.20. The pKa of HEPES is 7.55 at 20° C. The formula weight of the sodium salt is 260.31. The formula weight of the free acid is 238.31. . Weigh out the appropriate amounts of the HEPES (sodium salt) and HEPES (free acid), transfer to a 100 mL beaker, dissolve in deionized water to an approximate volume of 40 mL

Explanation:

According to the Henderson-Hasselbalch equation,

pH =

where, [Salt] = molar concentration of sodium salt

[Acid] = molar concentration of free acid

As per the given problem, pH = 7.20 and = 7.55.

Hence, putting the given values into the above formula as follows.

pH =

7.20 = 7.55 +

= -0.35

or, = 0.4467

Now, the concentration of buffer is 100 mM.

Hence, [Salt] + [Acid] = 100 mM

= 100 mM

[Acid](1 + 0.4467) = 100 mM

[Acid] =

= 69.123 mM

Therefore, concentration of salt will be calculated as follows.

[Salt] = 0.4467 × [Acid]

=

= 30.877 mM

For calculating the amount of acid:

Desired concentration = 69.123 mM

So, in 1000 ml amount of acid should be .

In 50 ml, the amount of acid should be as follows.

= 0.824 g

or, = 824 mg (as 1 g = 1000 mg)

Calculation for the amount of sodium salt is as follows.

Desired concentration = 30.877 mM

In 1000 ml, amount of salt should be .

In 50 ml, amount of salt should be as follows.

= 0.402 g

or, = 402 mg

Thus, we can conclude that we need to weigh out 824 mg of HEPES (free acid) and 402 mg of HEPES (sodium salt) that is required to be dissolved in 40 ml of water. After than, we have to adjust the volume of 50 ml by adding water drop-wise.

When chemical, transport, or mechanical work is done by an organism, what happens to the heat generated? a. It is captured to store energy as more ATP.
b. It is used to generate ADP from nucleotide precursors.
c. It is lost to the environment.
d. It is used to power yet more cellular work.

It is lost to the environment.

Explanation:

The energy currency of our body is ATP

The ATP undergoes hydrolysis to give ADP and inorganic phosphate with generation of energy.

The equation is

The energy being generated is used for different work done by an organism.

The energy thus generated is finally lost to the environment.

Which of the following is a chemical reaction? a. Making a hydrogen bond between a water molecule and a sugar molecule.
b. The melting of ice.
c. Changing a carbon atom to a nitrogen atom by radioactive decay.
d. The formation of a covalent bond between two amino acids

d. The formation of a covalent bond between two amino acids

Explanation:

In a chemical reaction, a substance transforms into another with different chemical properties and a different formula. Bonds are created and/or destroyed. It must obey the law of conservation of mass and the law of conservation of elements.

Which of the following is a chemical reaction?

a. Making a hydrogen bond between a water molecule and a sugar molecule. NO. This is a physical change in which sugar dissolves in water but both their chemical identities remain the same. The hydrogen bond is a force of attraction and not a real bond in which electrons are gained, lost or shared.

b. The melting of ice. NO. This is a physical change. Solid water turns into liquid water but the identity of the molecules is the same.

c. Changing a carbon atom to a nitrogen atom by radioactive decay. NO. This is a nuclear reaction in which a new element appears but it is not a chemical reaction because it does not fulfill the law of conservation of elements.

d. The formation of a covalent bond between two amino acids. YES. A new bond is created and a protein dimer is formed.

What is ATP?

Explanation:

Atmospheric Temperature and Pressure

Adenosine triphosphate is a complex organic chemical that provides energy to drive many processes in living cells, e.g. muscle contraction, nerve impulse propagation, and chemical synthesis. Found in all forms of life, ATP is often referred to as the "molecular unit of currency" of intracellular energy transfer. Wikipedia

Formula: C10H16N5O13P3

Molar mass: 507.18 g/mol

IUPAC ID: [[(2R,3S,4R,5R)-5-(6-aminopurin-9-yl)-3,4-dihydroxyoxolan-2-yl]methoxy-hydroxyphosphoryl] phosphono hydrogen phosphate

Acidity (pKa): 6.5

Soluble in: Water

UV-vis (λmax): 259 nm

Explanation:

Which of the following statements describes a common characteristic of catabolic pathways? a. They require energy from ATP hydrolysis to break down polymers into monomers.
b. They are endergonic and release energy that can be used for cellular work.
c. They are exergonic and provide energy that can be used to produce ATP from ADP and Pi.
d. They combine small molecules into larger, more energy-rich molecules.

C. They are exergonic and provide energy that can be used to produce ATP form ADP and Pi.

Explanation:

There are 2 types of metabolic reactions.

1. Anabolic reactions (Anabolism)
2. Catabolic reactions (Catabolism)

CATABOLIC REACTIONS:

They consist of reactions in which larger molecules are broken down into simpler ones. They are exergonic i.e. energy producing reactions. The released energy can be stored in the form of ATP (energy currency of cell).

ANABOLIC REACTIONS:

They consist of reactions in which larger molecules are synthesized from simpler ones. They are endergonic reactions i.e. energy is absorbed.

MEMORY AID:

An easy way to remember catabolic reactions is to think of them as "catastrophic reactions" i.e. larger molecules are destructed to produce simpler ones.

Coming back to the question,

a. They require energy from ATP hydrolysis to break down polymers into monomers.

Energy requiring reactions are anabolic and so this is the incorrect answer.

b. They are endergonic and release energy that can be used for cellular work.

Endergonic reactions are anabolic and hence this is the incorrect answer.

c. They are exergonic and provide energy that can be used to produce ATP from ADP and Pi

This statement correctly fits the definition of Catabolic reactions and is therefore the correct answer.

d. They combine small molecules into larger, more energy-rich molecules.

This statement describes anabolic reactions and hence the incorrect choice.

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