Which of the following statements about the ionization of the side chain of glutamine in water is true (select all that apply)? Glutamine has a positively charged form with a pK, around 10 The nitrogen has a slight positive charge from resonance so will not gain a third proton Glutamine is usually deprotonated but will gain a proton at low pH The amide nitrogen will not lose a proton because the conjugate base is very strong Glutamine is usually protonated but will lose a proton at high pH

However, it is commonly approximated as 3.14159.

An irrational number is a number that cannot be expressed as a simple fraction or ratio of two integers. It is a non-repeating, non-terminating decimal. Examples of irrational numbers include pi (π), the square root of 2 (√2), and the golden ratio (∅).

In mathematics, a terminating decimal is a decimal number that has a finite number of digits after the decimal point, i.e., the decimal representation ends in a finite number of zeroes. For example, 0.75, 2.0, and 0.0625 are terminating decimals.

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Answer:

1. Equilibrium expressions:

a. K = [HSO4-][H3O+]/[H2SO4][H2O]

b. K = [NO]^4[H2O]^6/[NH3]^4[O2]^5

c. K = [NH3][HCl]/[NH4Cl]

d. K = [NO2]^2/[N2O4]

2. Since K = 0.050, the concentrations of the reactants (N2 and H2) are larger than the concentrations of the products (NH3).

3. Since K = 6.8 x 10^3, the concentrations of the products (SO2 and O2) are larger than the concentrations of the reactant (SO3).

4. The Ksp expression for each of the reactions is:

a. Ksp = [Na+][Cl-]

b. Ksp = [Ba2+][SO42-]

5. Brønsted-Lowry acids and bases:

a. Acid: HNO3; Conjugate base: NO3-; Base: H2O; Conjugate acid: H3O+

b. Acid: HCN; Conjugate base: CN-; Base: H2O; Conjugate acid: HCN

c. Acid: H2SO4; Conjugate base: HSO4-; Base: Cl-; Conjugate acid: HCl

d. Acid: NH3; Conjugate base: NH2-; Base: H2O; Conjugate acid: NH4+

e. Acid: H2O; Conjugate base: OH-; Base: O2-; Conjugate acid: OH-

6. Conjugate acids and bases:

a. Acid: H2O; Conjugate base: OH-

b. Acid: H3O+; Conjugate base: H2O

c. Acid: H2CO3; Conjugate base: HCO3-

d. Acid: NH4+; Conjugate base: NH3

e. Acid: HSO4-; Conjugate base: SO42-

7. The strongest acid is HIO4 (highest K value), followed by HCN, HC2H3O2, and H2CO3 (lowest K value). The K values represent the degree to which the acids dissociate in solution. HIO4 is a strong acid, meaning it dissociates almost completely in solution, while H2CO3 is a weak acid, meaning it only dissociates partially.

8. pH and pOH calculations:

a. pH = -log[H3O+] = -log(0.200) = 0.699; pOH = -log[OH-] = -log(1.0 x 10^-14/0.200) = 12.301

b. pOH = -log[OH-] = -log(0.0143) = 1.844; pH = 14.000 - pOH = 12.156

c. pH = -log[H3O+] = -log(3.0) = 0.522; pOH = 13.478

d. pOH = -log[OH-] = -log(0.0062) = 2.206; pH = 14.000 - pOH = 11.794

9. Hydronium and hydroxide ion concentrations:

pH = 3.6; hydronium ion concentration = 10^-pH = 3.98 x 10^-4 M; hydro

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The empirical formula of a molecule containing 65.5% carbon, 5.5% hydrogen, and 29% oxygen is CH2O.

To calculate this, you need to first convert the percentage composition into mass composition. This is done by multiplying the percentages by the molecular weight of each element.

Carbon: 65.5% x 12 g/mol = 0.786 g/mol
Hydrogen: 5.5% x 1 g/mol = 0.055 g/mol
Oxygen: 29% x 16 g/mol = 0.464 g/mol
Now that you have the mass composition, you can calculate the moles of each element by dividing the mass of each element by its molar mass.
Carbon: 0.786 g/mol / 12 g/mol = 0.065 mol
Hydrogen: 0.055 g/mol / 1 g/mol = 0.055 mol
Oxygen: 0.464 g/mol / 16 g/mol = 0.029 mol
Finally, divide each element's moles by the smallest moles to get the empirical formula.

Carbon: 0.065 mol / 0.029 mol = 2.24 = 2 mol
Hydrogen: 0.055 mol / 0.029 mol = 1.90 = 1 mol
Oxygen: 0.029 mol / 0.029 mol = 1.00 = 1 mol
Therefore, the empirical formula of the molecule is CH2O.

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Answer:

First, we need to calculate the change in temperature:

ΔT = final temperature - initial temperature

ΔT = 85 °C - 35 °C

ΔT = 50 °C

Next, we can use the following formula to calculate the heat energy required:

Q = m·C·ΔT

where Q is the heat energy in calories, m is the mass of the sample in grams, C is the specific heat in cal/g °C, and ΔT is the change in temperature in °C.

Plugging in the given values, we get:

Q = 35.0 g · 0.108 cal/g °C · 50 °C

Q = 189.0 calories

Therefore, 189.0 calories are required to raise the temperature of the sample from 35 °C to 85 °C

Answer:

lusterous metal

Explanation:

ex gold, iron etc hope it helps

Answer: It is not necessarily advantageous for a predator to prey exclusively on a single prey species, as this can limit their options and make them vulnerable if the population of that prey species declines or becomes extinct. Predators that are more flexible and able to switch between different prey species may be better equipped to survive and thrive in changing environments.

However, there are some advantages to specializing in a single prey species. For example, a predator that is well adapted to hunting a particular prey species may be more efficient and successful at capturing and consuming that prey, which could provide a reliable source of energy. Additionally, if the predator and prey have co-evolved, the predator may have adaptations that specifically allow it to exploit the weaknesses or vulnerabilities of its prey, giving it an advantage over predators that are less specialized.

At has a higher initial ionisation energy than Br, while Br has a bigger atomic radius. Se has a bigger atomic radius than Br, and Br has a higher initial ionisation energy than Se.

The most loosely bound electron is further from the nucleus and thus easier to remove in bigger atoms. Hence, the ionisation energy should decrease as size (atomic radius) increases.

This is because the outer electrons aren't bound as strongly because they are farther from the nucleus.

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The standard change in Gibbs energy at 25 degree Celsius is  490.6 °C. for given equilibrium partial pressure  .

The Gibbs energy is the thermodynamic potential that is minimized when a system reaches chemical equilibrium at constant pressure and temperature when not driven by an applied electrolytic voltage. Its derivative with respect to the reaction coordinate of the system then vanishes at the equilibrium point.

Using the formula

ΔG° =  - R × T ln K

WHERE R=  8.3144598 J⋅mol⁻¹⋅K⁻¹.

T = 298 K

K = 0.82

SOLVING ,

The standard change in Gibbs energy at 25 degree Celsius is  490.6 °C.

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According to the Bronsted-Lowry concept of acids and bases, the statement "If a base is strong, then its conjugate acid will be relatively weaker" is NOT true.

The Bronsted-Lowry concept defines an acid as a proton (H+) donor and a base as a proton acceptor. The strength of an acid or base is determined by the extent to which it is willing to donate or accept a proton. Therefore, when a strong acid donates a proton, the resulting conjugate base is also strong; likewise, when a strong base accepts a proton, the resulting conjugate acid is also strong.
Other characteristics of bases include the presence of a hydroxide group (OH-), and the ability to accept a proton (H+) to form a salt. However, the strength of a base does not necessarily depend on its presence or absence of a hydroxide group.
In conclusion, the Bronsted-Lowry concept of acids and bases states that the strength of a base does not determine the strength of its conjugate acid; therefore, the statement "If a base is strong, then its conjugate acid will be relatively weaker" is NOT true.

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(S)-2-butanol will undergo an SN2 reaction with HBr to produce a racemic mixture of alkyl bromides. Here option B is the correct answer.

When optically active alcohol is treated with HBr, the reaction follows an SN1 or SN2 mechanism. In the case of SN1, a carbocation intermediate is formed, and in SN2, a backside attack by the nucleophile occurs. The stereochemistry of the product depends on the configuration of the intermediate and the direction of attack.

In the case of (S)-2-butanol, the hydroxyl group is attached to the second carbon atom, which makes it a primary alcohol. When treated with HBr, it undergoes an SN2 reaction, where the hydroxyl group is replaced by the bromine atom. The nucleophile attacks from the backside of the molecule, leading to an inversion of configuration.

This results in the formation of a racemic mixture of alkyl bromides, as both enantiomers have an equal chance of being attacked from either side. On the other hand, (R)-2-butanol, being the enantiomer of (S)-2-butanol, will also undergo the same reaction and produce the same racemic mixture of alkyl bromides.

In the case of (R)-1-phenyl ethanol and (S)-1-phenyl ethanol, they are secondary alcohols and can undergo either SN1 or SN2 reactions depending on the reaction conditions. However, the reaction mechanism will lead to the formation of a mixture of diastereomers, rather than a racemic mixture of enantiomers.

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Complete question:

Which of the following optically active alcohols, when treated with HBr, results in a racemic mixture of alkyl bromides?

a) (R)-2-butanol

b) (S)-2-butanol

c) (R)-1-phenyl ethanol

d) (S)-1-phenyl ethanol

Answer:

7.68 x 1025 atoms of phosphorous correspond to 1.06 mole of diphosphorous pentoxide. This can also be written as 1.06 mol of P2O5.

Answer:

0.8 L

Explanation:

The volume of the stock solution needed to make 2.0 L of 0.20M MgSO4 is 0.8 L.

Answer:

Explanation:

2-bromopropane is the major product because the reaction mechanism involves the formation of the most stable carbocation intermediate. When hydrogen bromide reacts with propene, the hydrogen atom from HBr adds to the carbon atom of the double bond that has fewer hydrogen atoms attached, resulting in the formation of a carbocation intermediate. The intermediate can either form 1-bromopropane or 2-bromopropane depending on the position of the carbocation. The 2-bromopropane is the major product because the secondary carbocation formed in this case is more stable than the primary carbocation formed in the case of 1-bromopropane.

To test for an alkene, bromine water can be used. When an alkene reacts with bromine water, the bromine molecule adds across the double bond, forming a colorless dibromoalkane product. The mechanism for the reaction between hex-1-ene and bromine involves the formation of a cyclic bromonium ion intermediate, followed by the attack of water on the intermediate, resulting in the formation of the dibromoalkane product.

a) Addition polymerization is a process in which unsaturated monomers are joined together to form a polymer. The process involves breaking the double bond of the monomer and joining the monomers together to form a long-chain polymer. The process requires a catalyst to initiate the reaction.

b) The repeating unit of the polymer formed from the addition polymerization of chloroethene monomers is -CH2-CHCl-. This polymer is called polyvinyl chloride (PVC), and it has a wide range of uses, including pipes, electrical cables, and vinyl flooring.

Answer: Manganese

Explanation:

With titanium, it only has two d electrons, so it can't form different high and low spin complexes. It doesn't matter because it will never fill the higher-energy orbitals. The total spin state turns out to be +1 (two unpaired d electrons, no matter what). Therefore, manganese will form both a high and low spin complex.

Answer: There are 3019.2 atoms of iron.

Answer:

Explanation:

Iron is a ductile, malleable, silver-white metallic element, scarcely known in a pure condition, but much used in its crude or impure carbon-containing forms for making tools, implements, machinery, etc. Symbol: Fe; atomic weight: 55.847; atomic number: 26; specific gravity 7.86 at 20°C.

1 mol contains [tex]=6.02\times10^{23}[/tex] particles, whether it be atoms, ions, molecules or whatever (Avogadro's number).

So you just divide:

[tex]\frac{2.96\times10^{20}}{6.02\times10^{23}}[/tex] = = 4.9169435215947 × 10^-4

a balanced equation for the redox reaction: 2 Ca(s) + O2(g) → 2 CaO(s)

A redox reaction is a type of chemical reaction that involves the transfer of electrons between species. One species undergoes oxidation (loses electrons) while another species undergoes reduction (gains electrons).

In the reaction between calcium metal and oxygen gas, the calcium metal is being oxidized (loses electrons) and the oxygen gas is being reduced (gains electrons). This can be seen in the balanced equation where the calcium atoms go from having an oxidation state of 0 to +2, while the oxygen atoms go from having an oxidation state of 0 to -2.

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If we use the Gizmo to find the freezing, melting, and boiling points of water at 5,000 meters (16,404 feet) then,

Freezing point: 32 ºF (0ºC)

Melting point: 32 ºF (0ºC)

Boiling point: 203°F (95°C)

The freezing point is defined as the temperature at which a liquid becomes a solid. Increased pressure usually raises the freezing point with the melting point of the solid. The boiling point of a pure substance is defined as the temperature at which the substance transitions from a liquid to the gaseous phase. At the boiling point the vapor pressure of the liquid is equal to the applied pressure on the liquid. The melting point of a substance is defined as the temperature at which the substance changes from a solid to a liquid.

Melting occurs at a single temperature for the pure substances. The normal and average melting point and boiling point of water at 1 atmospheric pressure are 0°C and 100°C respectively. Decreasing the pressure under 1 atm. will lower the boiling point since the external pressure will be lower so it will become equal with the vapor pressure at a lower temperature.

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Answer:

4897 J

Explanation:

The heat transferred to the surroundings, q_surr, can be calculated using the equation:

q_surr = -q_rxn = -CmΔT

where C is the specific heat capacity of the mixture (assumed to be the same as water, 4.184 J/g°C), m is the mass of the mixture (which we can calculate using the density, assuming that the volumes are additive), and ΔT is the change in temperature (in Celsius).

First, let's calculate the mass of the mixture:

density of water = 1.00 g/cm^3

volume of mixture = volume of acid + volume of base = 106 mL + 157 mL = 263 mL = 0.263 L

mass of mixture = density of water x volume of mixture = 1.00 g/cm^3 x 0.263 L = 263 g

Next, let's calculate the change in temperature:

ΔT = final temperature - initial temperature = 27.29°C - 22.94°C = 4.35°C

Now we can calculate the heat transferred to the surroundings:

q_surr = -CmΔT

q_surr = -(4.184 J/g°C) x (263 g) x (4.35°C)

q_surr = -4897 J

Note that the negative sign indicates that heat is lost by the system to the surroundings. Therefore, the heat transferred to the surroundings, q_surr, is 4897 J.

The pH of 0.335 M trimethylammonium chloride, (CH3)3NHCI is approximately 5.676.

What is pH?

To find the pH of the solution, we need to first determine if (CH3)3NHCI acts as an acid or base. Since (CH3)3NHCI is a salt composed of a weak base, trimethylamine, and a strong acid, hydrochloric acid (HCI), it will undergo hydrolysis in water.

The hydrolysis reaction is given by:

(CH3)3NH+ (aq) + H2O (l) ⇌ (CH3)3N (aq) + H3O+ (aq)

The Kb expression for the equilibrium reaction is:

Kb = [ (CH3)3N ] [ H3O+ ] / [ (CH3)3NH+ ]

Since (CH3)3NH+ and HCl dissociate completely in water, the initial concentration of (CH3)3NH+ is equal to the concentration of (CH3)3NHCI, which is 0.335 M.

Using the Kb value given, we can solve for the concentration of H3O+:

Kb = 6.3 x [tex]10^{-5}[/tex] = [ (CH3)3N ] [ H3O+ ] / 0.335

[ H3O+ ] = Kb x (CH3)3NH+ / (CH3)3N

[ H3O+ ] = 6.3 x [tex]10^{-5}[/tex] x 0.335 / 1

[ H3O+ ] = 2.1095 x [tex]10^{-6}[/tex] M

Finally, we can calculate the pH using the expression:

pH = -log [H3O+]

pH = -log (2.1095 x [tex]10^{-6}[/tex])

pH = 5.676

Therefore, the pH of the solution is approximately 5.676.

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Complete question is: The pH of 0.335 M trimethylammonium chloride, (CH3)3NHCI is approximately 5.676.

Answer:

See below.

Explanation:

The atomic number of lead (Pb) is 82, which means it has 82 electrons. The electronic configuration of lead is

1s² 2s² 2p⁶ 3s² 3p⁶ 3d¹⁰ 4s² 4p⁶ 4d¹⁰ 5s² 5p⁶ 4f¹⁴ 5d¹⁰ 6s² 6p²

The orbital diagram for the valence electrons of lead (Pb) is

↑↓ ↑↓ ↑↓ ↑↓ ↑↓ ↑↓ ↑↓ ↑↓

s s p p p p d d

2 1 6 2 6 2 10 10

|||||||||

1 2 3 4 5 6 7 8

The notation ↑↓ represents a pair of electrons with opposite spins.

To determine if lead (Pb) is diamagnetic or paramagnetic, we need to look at whether there are any unpaired electrons. Based on the orbital diagram, we can see that all the electrons in the valence shell are paired, meaning that lead (Pb) is diamagnetic.

Answer: NaAlF6

Explanation:

1) divide by the smallest number of moles

0.953/.322 =2.96 round to 3 Na

.322/.322 = 1 Al

1.93/.322 =5.99 round to 6 F

2) write in order numbers were give

NaAlF6

The  atoms of lithium that  are in 18.7 g is 16 × 10²³ atoms . This is taken out by mole concept .

The mole is a unit of measurement similar to the pair, dozen, gross, and so on. It provides a precise count of the atoms or molecules in a bulk sample of matter. A mole is the amount of substance that contains the same number of discrete entities (atoms, molecules, ions, etc.)

if 7 grams of lithium contain 6 × 10²³ atoms

then 18.7 will contain 16 × 10²³ atoms

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Answer:

Carbonothioyl dibromide, also known as CBr2S, has a bond angle of approximately 109.5 degrees, which is the typical tetrahedral bond angle for molecules with sp3 hybridization.

The molecular shape of CBr2S is also tetrahedral, with the two bromine atoms and the sulfur atom arranged at the corners of a tetrahedron, and the carbon atom at the center.

Answer:

The correctly written thermochemical equation is:

C3H8 (g) + 5O2 (g) → 3CO2 (g) + 4H2O (l), ΔH = -2,220 kJ/mol

This equation represents the combustion of propane (C3H8) in the presence of oxygen (O2) to produce carbon dioxide (CO2) and water (H2O), with a heat release of -2,220 kJ/mol. The state symbols (g) for gases and (l) for liquids indicate the physical state of each substance at standard conditions.

Explanation:

ABOVE

122.5 grams of oxalic add dihydrate (MW = 126.07 g/mole) and disodium oxalate (MW = 133.99 g/mole) were required to prepare this buffer if the total oxalate concentration is 0.115 M.

Weak acids are defined as acids that don't completely dissociate in solution. It can be explained as any acid that is not a strong acid. The strength of a weak acid depends on how much it gets dissociates and the more it dissociates, the stronger the acid. The mass of the weak acid in a solution of a certain pH can be determined by calculating the original concentration of the acid after calculating the concentration of the hydrogen ions with the help of the pH value of the solution.

The Concentration of oxalate ion is  0.115 M.

pKa1 is 1.250.

pKa2 is 4.266.

pH is 5.193.

Molarity = (mass / molar mass) / 1 / volume in liter

The molar mass is 126.07g/mole.

Mass = Molarity × molar mass × Volume in liter

Mass=0.972 M × 126.07 g/mole × 1.00 L

        = 122.5 gram

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The complete question is,

A buffer prepared by dissolving oxalic add dihydrate (H2C2O4⋅2H2O) and disodium oxalate (Na2C2O4) in 1.00 L of water has a pH of 5.193. How many grams of oxalic add dihydrate (MW = 126.07 g/mole) and disodium oxalate (MW = 133.99 g/mole) were required to prepare this buffer if the total oxalate concentration is 0.115 M? Oxalic acid has pKa values of 1.250 (pKa1) and 4.266 (pKa2).

The empirical formula, CH2O9(menthol) is multiplied by 5 to get the molecular formula, C10H20O.The empirical and molecular formula for menthol are CH2O and C10H20O, respectively.

The molecular formula for menthol is C5H10O.

This can be determined by dividing the molar mass of the empirical formula (156.26 g/mol) by the molar mass of CO2 (44.01 g/mol). This gives a ratio of 3.55, which is equal to the ratio of C atoms in the empirical formula, C10H20O.

Therefore, the molecular formula is C5H10O.
Given:

Menthol is composed of C, H, and O0.1005g sample of menthol is combusted and produces0.2829g of CO2 0.1159g H2O

1. Find: Empirical and molecular formula for menthol.

Let's first calculate the number of moles of CO2 produced. The balanced equation for combustion of menthol is:

C10H20O(s) + 12O2(g) → 10CO2(g) + 10H2O(l)

From the above equation, we can see that for 10 moles of CO2 produced 1 mole of menthol is required.

2. By taking the number of moles of CO2 produced, we can calculate the number of moles of menthol burned.

Moles of CO2 = 0.2829g / 44.01g/mol= 0.00643 mol

C10H20O(s) + 12O2(g) → 10CO2(g) + 10H2O(l)

From the balanced equation,1 mole of C10H20O requires 10 moles of CO2

Moles of C10H20O burned = 10 * 0.00643= 0.0643 mol

Next, we can calculate the number of moles of H2O produced.

Moles of H2O = 0.1159g / 18.015g/mol= 0.00643 mol

C10H20O(s) + 12O2(g) → 10CO2(g) + 10H2O(l)

From the balanced equation,1 mole of C10H20O requires 10 moles of H2O

Moles of C10H20O burned = 10 * 0.00643= 0.0643 mol

3. Now we can calculate the empirical formula of menthol. The empirical formula can be calculated as follows:

Empirical formula = CH2O (Divide all moles by smallest moles) The molecular weight of CH2O = 30 g/mol

The empirical formula mass of the compound is:

mass = (12.011 + 2*1.008 + 15.999) = 30.026

Empirical formula mass of CH2O is 30.026g/mol, and the given sample weighs 0.1005 g.
The number of empirical formula units in the sample is 0.1005 g / 30.026 g/mol = 0.003348Units.

Empirical formula = CH2OThe empirical formula weight of menthol is CH2O, which is equal to 30.026g/mol.

4. To find the molecular formula, we need to know the molecular weight of the menthol. We can calculate it as follows:

Molecular formula mass = Empirical formula mass x n

Where n = integer Molecular formula mass of menthol is 156 g/mol, and the empirical formula mass is 30.026 g/mol.

So, n = 156 g/mol ÷ 30.026 g/mol = 5.192

Thus the empirical formula, CH2O is multiplied by 5 to get the molecular formula, C10H20O.The empirical and molecular formula for menthol are CH2O and C10H20O, respectively.

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Answer:

when the solution of potassium hydroxide and zinc chloride are mixed,the double-displacement reaction occur ,resulting in precipitation and the reaction forms potassium chloride and zinc hydroxide .