b. Imagine an unusual life form in which the N atom in an amino acid is changed to a C atom. Could a hydrogen bond in this unusual alpha helix occur? Why or why not?

The question is incomplete, the complete question is:

For a certain chemical reaction, the standard Gibbs free energy of reaction is 144. kJ. Calculate the temperature at which the equilibrium constant K = [tex]5.9\times 10^{-26}[/tex] .

Round your answer to the nearest degree.

Answer:

25°C is the temperature at which the equilibrium constant is [tex]5.9\times 10^{-26}[/tex].

Explanation:

[tex]\Delta G^o=-RT\ln K[/tex]

where,

[tex]\Delta G^o[/tex] = standard Gibbs free energy = 144.0 kJ=144,000 J  (Conversion factor: 1kJ = 1000J)

R = Gas constant = [tex]8.314 J/K mol[/tex]

T = temperature at which reaction is occurring = ?

K = Equilibrium constant of the reaction =[tex]5.9\times 10^{-26}[/tex]

Putting values in above equation, we get:

[tex]144,000 J/mol=-(8.3145J/Kmol)\times T\times \ln [5.9\times 10^{-26}][/tex]

[tex]T=\frac{144,000 J/mol}{-(8.314 J/Kmol)\times \ln [5.9\times 10^{-26}]}[/tex]

T = 298.15 K

T = 298.15 - 273 °C = 25°C

25°C is the temperature at which the equilibrium constant is [tex]5.9\times 10^{-26}[/tex].

Using the equation ΔG° = -RTlnK, which relates the Gibbs free energy change to the equilibrium constant, you can solve for temperature by rearranging the formula. Ensure that the units for Gibbs energy and the gas constant match. Plugging the given values into the rearranged formula will provide the temperature.

The student's question asked how to calculate the temperature at which the equilibrium constant K equals 5.9 x 10 with a standard Gibbs free energy of reaction of 144 kJ.

The relationship between the Gibbs free energy change and the equilibrium constant is defined by the equation ΔG° = -RTlnK, where R is the gas constant (8.314 J/K mol), T is the absolute temperature in Kelvin, and K is the equilibrium constant.

To solve for temperature, rearrange the equation as T = -ΔG / (R * lnK). But first, you must ensure that the units for Gibbs energy and the gas constant match. If ΔG is given in kJ, it should be converted to J (1 kJ = 1000 J).

So, T = -(144,000 J) / (8.314 J/K mol * ln(5.9 x 10)), which should give you the answer.

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Answer: option B and option C

Explanation:

Answer: A. contains 18 electrons

Explanation:

Atomic number is defined as the number of protons or number of electrons that are present in an atom. It is characteristic of an element.

Atomic number of phosphorous is 15.

The electronic configuration of phosphorous (P) will be,

[tex]P:15:1s^22s^22p^63s^23p^3[/tex]

Atomic number = Number of electrons = Number of protons = 15

As the phosphorous atom has gained 3 electrons, it will have 15+3= 18 electrons , the phosphorous anion will be having a charge of -3.

The electronic configuration of [tex]P^{3-}[/tex] will be,

[tex]P^{3-}:18:1s^22s^22p^63s^23p^6[/tex]

Thus the correct statement is this ion contains 18 electrons

The phosphide ion contains 18 electrons. The correct answer is Option A.

The ion formed by phosphorus, called phosphide, has the formula P3−. To determine its properties, we can look at its electron configuration. Phosphorus has an atomic number of 15, meaning it has 15 electrons in its neutral state. When phosphorus forms the phosphide ion (P3−), it gains three extra electrons to achieve a stable electron configuration. So, the phosphide ion contains a total of 18 electrons.

Therefore, the correct answer is A. The ion contains 18 electrons.

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The question is incomplete, here is the complete question:

A chemist adds 345.0 mL of a 0.0013 mM (MIllimolar) copper(II) fluoride [tex]CuF_2[/tex] solution to a reaction flask.

Calculate the mass in micrograms of copper(II) fluoride the chemist has added to the flask. Be sure your answer has the correct number of significant digits.

Answer: The mass of copper (II) fluoride is 0.13 mg

Explanation:

We are given:

Millimolarity of copper (II) fluoride = 0.0013 mM

This means that 0.0013 millimoles of copper (II) fluoride is present in 1 L of solution

Converting millimoles into moles, we use the conversion factor:

1 moles = 1000 millimoles

So, [tex]0.0013mmol\times \frac{1mol}{1000mmol}=1.3\times 10^{-6}mol[/tex]

To calculate the number of moles, we use the equation:

[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex]

We are given:

Moles of copper (II) fluoride solution = [tex]1.3\times 10^{-6}mol[/tex]

Molar mass of copper (II) fluoride = 101.5 g/mol

Putting values in above equation, we get:

[tex]1.3\times 10^{-6}mol=\frac{\text{Mass of copper (II) fluoride}}{101.5g/mol}\\\\\text{Mass of copper (II) fluoride}=(1.3\times 10^{-6}mol\times 101.5g/mol)=1.32\times 10^{-4}g[/tex]

Converting this into milligrams, we use the conversion factor:

1 g = 1000 mg

So,

[tex]\Rightarrow 1.32\times 10^{-4}g\times (\frac{1000mg}{1g})=0.13mg[/tex]

Hence, the mass of copper (II) fluoride is 0.13 mg

Answer:

The mass of copper(II) fluoride is 45.54 micrograms

Explanation:

A chemist adds 345.0 mL  of a 0.0013 mM copper(II) fluoride solution to a reaction flask. Calculate the mass in micrograms of copper(II) fluoride the chemist has added to the flask. Round your answer to significant digits.

Step 1: Data given

Molarity of the copper(II) fluoride solution = 0.0013 mM =

Volume of the solution 345.0 mL = 0.345 L

Molar mass copper(II) fluoride = 101.54 g/mol

Step 2: Calculate moles of copper(II) fluoride

Moles CuF2 = molarity * volume

Moles CuF2 = 0.0000013 M * 0.345 L

Moles CuF2 = 0.0000004485 moles

Step 3: Calculate mass of  CuF2

Mass CuF2 = moles * Molar mass

Mass CuF2 = 0.0000004485 moles * 101.54 g/mol

Mass CuF2 = 0.00004554 grams = 0.04554 miligrams = 45.54 micrograms

The mass of copper(II) fluoride is 45.54 micrograms

Answer: the new volume will be 0.245L

Explanation:Please see attachment for explanation

Answer:

The volume will be 238 mL or 0.238 L

Explanation:

Step 1: Data given

Pressure = constant = 5.80 atm

The initial moles of helium = 0.04 moles

Temperature = 240.00 K

Volume = 0.14L

The number of moles increases to 0.07 moles

Step 2: Calculate the new volume

p*V = n*R*T

V = (n*R*T)/p

⇒ with n = the number of moles = 0.07 moles

⇒ with R = the gas constant = 0.08206 L*atm/mol*K

⇒ with T = the temperature = 240.00 K

⇒ with p = the pressure = 5.80 atm

V = (0.07 * 0.08206 * 240.00) / 5.80

V = 0.238 L = 238 mL

The volume will be 238 mL or 0.238 L

Molality  is the measure of concentration of solute in 1 kg of solution. The molality of the solution is 3.05 mol/kg.

10% of HCl (by mass) means 10 g of HCl and in 90 g of water.

Molar mass of HCl = 36.5 g/mol

Molality:

It is the measure of concentration of solute in 1 kg of solution. It can be calculated by the formula.

[tex]\bold {m = \dfrac {n }{w}\times 1000}[/tex]

Where,

m- molality

n - number of moles

w - weight of solvent in grams

Number of moles of HCl

[tex]\bold {n = \dfrac w{m} = \dfrac {10}{36.5} = 0.274 g}[/tex]

put the value in molality formula,

[tex]\bold {m = \dfrac {0.274 }{90}\times 1000}\\\\\bold {m = 3.05\ g/mol}[/tex]

Therefore, the molality of the solution is 3.05 mol/kg.

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Final answer:

To calculate the molarity of a 10.0% HCl solution, convert 10 grams of HCl to moles, and then divide by the volume of the solution in liters. This results in a molarity of 2.74 M.

Explanation:

To calculate the molarity of a 10.0% (by mass) aqueous solution of hydrochloric acid (HCl), start by understanding that a 10.0% solution means there are 10 grams of HCl in 100 grams of the solution. First, since the solution is aqueous, we can assume the density is close to that of water, which is approximately 1 g/mL, so 100 grams of the solution is roughly equivalent to 100 mL (0.1 L) of solution.

Next, convert the mass of HCl to moles. The molar mass of HCl is about 36.46 g/mol:

10 grams HCl × (1 mol HCl / 36.46 grams) = 0.274 moles HCl

Then, divide the moles of HCl by the volume of the solution in liters to find the molarity:

Molarity = Moles of solute / Volume of solution in liters

Molarity = 0.274 moles HCl / 0.1 L = 2.74 M

The molarity of the 10.0% HCl solution is therefore 2.74 M.

Explanation:

A)

We know, each mole contains [tex]N_A=[/tex] [tex]6.023 \times 10^{23}[/tex] atoms.

It is given that mass of one oxygen atom is m= [tex]2.66\times 10^{-23}\ g[/tex].

Therefore, mass of one mole of oxygen, [tex]M=m\times N_A[/tex].

Putting value of n and [tex]N_A[/tex],

[tex]M=2.66\times 10^{-23}\times 6.023\times 10^{23} \ gm\\M=16.0\ gm[/tex]

B)

Given,

Mass of water in glass=0.050 kg = 50 gm.

From above part mass of one mole of oxygen atoms = 16.0 gm.

Therefore, number of mole of oxygen equivalent to 50 gm oxygen[tex]=\dfrac{50}{16}=3.1 \ moles.[/tex]

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The mass of 1 mole of oxygen atoms is 16.00 grams. There are 3125 moles of oxygen atoms with a mass equal to that of a glass of water.

To find the mass of 1 mole of oxygen atoms, we need to use the molar mass of oxygen. The molar mass is the mass of one mole of a substance, and it is equal to the atomic mass in grams. The atomic mass of oxygen is 16.00 g/mol, so the mass of 1 mole of oxygen atoms is 16.00 grams.

To determine how many moles of oxygen atoms have a mass equal to the mass of a glass of water, we need to use the given mass of the water and the molar mass of oxygen. The molar mass of oxygen is 16.00 g/mol, and the mass of a glass of water is 0.050 kg (or 50,000 grams). Using the molar mass, we can set up a proportion to find the number of moles of oxygen atoms.

Moles of oxygen atoms = (Mass of water / Molar mass of oxygen) = (50,000 g / 16.00 g/mol) = 3125 moles of oxygen atoms.

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Answer:

10.6%

Explanation:

The determined percent mass of water can be calculated from the formula of the hydrate by  

dividing the mass of water in one mole of the hydrate by the molar mass of the hydrate and  

multiplying this fraction by 100.

Manganese(ii) sulphate monohydrate is MnSO4 . H2O

1. Calculate the formula mass. When determining the formula mass for a hydrate, the waters of  

hydration must be included.

1 Manganes  52.94 g = 63.55 g  

1 Sulphur  32.07 g =  

32.07 g 2 Hydrogen is  = 2.02 g

4 Oygen       =  

64.00 g 1 Oxygen 16.00 = 16.00 g

151.01 g/mol  18.02 g/mol

Formula Mass = 151.01 + (18.02) = 169.03 g/mol

2. Divide the mass of water in one mole of the hydrate by the molar mass of the hydrate and  

multiply this fraction by 100.

Percent hydration = (18.02 g /169.03 g) x (100) = 10.6%

The final result is 10.6% after the two steps calculations

The mass percentage of the water in hydrated magnesium sulfate (MnSO4 . H2O) is 10.6%.

The percentage mass is the ratio of the mass of the element or molecule in the given compound.

The percentage can be given as:

[tex]\text{Percent Mass} = \frac{\text{Mass of molecule}}{\text{total mass of compound}} \times 100 \%[/tex]

The mass of the water is 18.02 g/mol and the molar mass of hydrated magnesium sulfate (MnSO4 . H2O) is 169.03 g/mol.

Thus,

[tex]\text{Percent Mass} = \frac{\text{18.02}}{\text{169.03 }} \times 100 \%\\\\\text{Percent Mass} = 10.6 \%}[/tex]

Therefore, the mass percentage of the water in hydrated magnesium sulfate (MnSO4 . H2O) is 10.6%.

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