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vladimir1956 [14]
3 years ago
12

Which of the following would have the largest pKa?

Chemistry
1 answer:
garri49 [273]3 years ago
5 0

Answer:

CH3CH2NH3+/CH3CH2NH2 would have the largest pKa

Explanation:

To answer this question we must know Kb of CH3CH2NH2 is 5.6x10⁻⁴, and for C6H5NH2 is 4.0x10⁻¹⁰. And the CH3CH2NH3+ and C6H5NH3+ are related with these substances because are their conjugate base. That means:

pKa of  CH3CH2NH3+ =  CH3CH2NH2;  C6H5NH3+ =  C6H5NH2

Also, Kw / Kb = Ka

Thus:

pKa of CH3CH2NH3+/CH3CH2NH2 is:

Kw / kb = Ka = 1.79x10⁻¹¹

-log Ka = pKa

pKa = 10.75

pKa of C6H5NH3+/ C6H5NH2 is:

Kw / kb = Ka = 2.5x10⁻⁵

-log Ka = pKa

pKa = 4.6

That means CH3CH2NH3+/CH3CH2NH2 would have the largest pKa

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Define the ground state of an atom
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Explanation:

The nucleus of an atom is surround by electrons that occupy shells, or orbitals of varying energy levels. The ground state of an electron, the energy level it normally occupies, is the state of lowest energy for that electron. There is also a maximum energy that each electron can have and still be part of its atom.

8 0
3 years ago
Name:_____________________________________________________ Date:___________ Period:_________ 3/23 - 3/27 Assignment 1: Gas Law P
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Answer:

The answers are;

1. 8.2 liters

2. 1214.84 ml

3. 318.027 K

4. 4.00 l.

Explanation:

1. Boyle's law states that the volume of  given mass of gas is inversely proportional to its pressure at constant temperature

that is

P₁·V₁ = P₂·V₂

Where:

P₁ = Initial pressure = 40.0 mm Hg

V₁ = Initial volume = 12.3 liters

P₂ = Final pressure = 60.0 mm Hg

V₂ = Final volume = Required

From P₁·V₁ = P₂·V₂, V₂ is given by

V_2=\frac{P_1\cdot V_1}{P_2} = \frac{40.0 mm Hg\cdot 12.3 l}{60.0 mm Hg} =  8.2 l

The volume reduces to V₂ = 8.2 liters

2. Here Charles law states that

\frac{T_1}{V_1} =\frac{T_2}{V_2}

T₁ = Initial temperature = 27.0 °C = 300.15 K

V₁ = Initial volume = 900.0 mL

T₂ = Final temperature = 132.0 °C = 405.15 K

V₂ = Final volume = Required

Therefore  V_2 =\frac{T_2\cdot V_1}{T_1} = \frac{405.15 K\times 900.0 mL}{300.15 K} = 1214.84 ml

V₂ = 1214.84 ml

3.  Gay-Lussac's Law states that

\frac{T_1}{P_1} =\frac{T_2}{P_2}

Where:

P₁ = Initial pressure = 15.0 atmospheres

T₁ = Initial temperature = 25.0 °C = 298.15 K

P₂ = Final pressure = 16.0 atmospheres

T₂ = Final temperature = Required

∴ T_2 = \frac{T_1\times P_2}{P_1}

=  \frac{298.15 K\times 16.0atm}{15.0atm} = 318.027 K

T₂ = 318.027 K

4. Avogadro's law states that,

Equal volume of all gases at the same temperature and pressure contain equal number of molecules.

Therefore if 5.00 moles of gas occupies 2.00 l volume, then

1 moles will occupy 2.00/5 l volume and

10 moles will occupy 2.00/5 × 10 or 4.00 l volume.

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Organic compounds and their uses​
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Organic compounds essential to human functioning include carbohydrates, lipids, proteins, and nucleotides. These compounds are said to be organic because they contain both carbon and hydrogen
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Hydrogen iodide decomposes according to the equation: 2HI(g) H 2(g) + I 2(g), K c = 0.0156 at 400ºC A 0.660 mol sample of HI was
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Answer:

[HI] = 0.264M

Explanation:

Based on the equilibrium:

2HI(g) ⇄ H₂(g) + I₂(g)

It is possible to define Kc of the reaction as the ratio between concentration of products and reactants using coefficients of each compound, thus:

<em>Kc = 0.0156 = [H₂] [I₂] / [HI]²</em>

<em />

As initial concentration of HI is 0.660mol / 2.00L = <em>0.330M, </em>the equlibrium concentrations will be:

[HI] = 0.330M - 2X

[H₂] = X

[I₂] = X

<em>Where X is reaction coefficient.</em>

<em />

Replacing in Kc:

0.0156 = [X] [X] / [0.330M - 2X]²

0.0156 = X² / [0.1089 - 1.32X + 4X² ]

0.00169884 - 0.020592 X + 0.0624 X² = X²

0.00169884 - 0.020592 X - 0.9376 X² = 0

Solving for X:

X = - 0.055 → False solution, there is no negative concentrations

X = 0.0330 → Right solution.

Replacing in HI formula:

[HI] = 0.330M - 2×0.033M

<h3>[HI] = 0.264M</h3>

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An aqueous solution has a normal boiling point of 103.0°
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About 5 degrees  that is

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