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EleoNora [17]
3 years ago
11

3) (2 points) The phenolic indicator (In-OH) has approximately the same pKa as a carboxylic acid. Which H is the most acidic pro

ton in In-OH

Chemistry
1 answer:
PilotLPTM [1.2K]3 years ago
7 0

Explanation:

In a phenolic indicator, the hydrogen atom is attached to more electronegative atom which is oxygen atom (O atom). So, when a phenolic indicator loses a proton then a conjugate base is formed.

This conjugate base is much more stable due to the resonance stabilization.

Therefore, we can conclude that proton attached to oxygen atom is much more stabilized than any other protons in a phenolic indicator.

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A dialysis unit requires 70000 mL of distilled water. How many gallons of water are needed? (1 gal = 4 qt)
Lady bird [3.3K]

Answer: 19.25 gallons

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5 0
3 years ago
Read 2 more answers
Antique paintings will deteriorate more rapidly in a humid environment. Art museums will combat this by placing a desiccant into
stira [4]

233.1g

Explanation:

Given:

Formula units of FeCl₃ = 8.65 x 10²³ formula units

Unknown:

Mass of the FeCl₃ = ?

Solution:

The formula unit is the same as the amount of subatomic particles in an atom.

  1 mole of a substance = 6.02 x 10²³ formula units

 For  8.65 x 10²³ formula units; the number of moles;

  Number of moles = \frac{8.65 x 10 ^{23} }{6.02 x 10^{23} } = 1.44moles

Now to find the mass of  FeCl₃;

   Mass of  FeCl₃ = 1.44 x 162.2 = 233.1g

learn more:

Number of moles brainly.com/question/1841136

#learnwithBrainly

3 0
2 years ago
Sulfuric acid is produced in larger amounts by weight than any other chemical. It is used in manufacturing fertilizers, oil refi
Fed [463]

Answer:

A. -166.6 kJ/mol

B. -127.7 kJ/mol

C. -133.9 kJ/mol

Explanation:

Let's consider the oxidation of sulfur dioxide.

2 SO₂(g) + O₂(g) → 2 SO₃(g)     ΔG° = -141.8 kJ

The Gibbs free energy (ΔG) can be calculated using the following expression:

ΔG = ΔG° + R.T.lnQ

where,

ΔG° is the standard Gibbs free energy

R is the ideal gas constant

T is the absolute temperature (25 + 273.15 = 298.15 K)

Q is the reaction quotient

The molar concentration of each gas ([]) can be calculated from its pressure (P) using the following expression:

[]=\frac{P}{R.T}

<em>Calculate ΔG at 25°C given the following sets of partial pressures.</em>

<em>Part A  130atm SO₂, 130atm O₂, 2.0atm SO₃. Express your answer using four significant figures.</em>

[SO_{2}]=[O_{2}]=\frac{130atm}{(0.08206atm.L/mol.K).298K} =5.32M

[SO_{3}]=\frac{2.0atm}{(0.08206atm.L/mol.K).298K} =0.0818M

Q=\frac{[SO_3]^{2} }{[SO_{2}]^{2}.[O_{2}] } =\frac{0.0818^{2} }{5.32^{3} } =4.44 \times 10^{-5}

ΔG = ΔG° + R.T.lnQ = -141.8 kJ/mol + (8.314 × 10⁻³ kJ/mol.K) × 298 K × ln (4.44 × 10⁻⁵) = -166.6 kJ/mol

<em>Part B  5.0atm SO₂, 3.0atm O₂, 30atm SO₃  Express your answer using four significant figures.</em>

<em />

[SO_{2}]=\frac{5.0atm}{(0.08206atm.L/mol.K).298K}=0.204M

[O_{2}]=\frac{3.0atm}{(0.08206atm.L/mol.K).298K}=0.123M

[SO_{3}]=\frac{30atm}{(0.08206atm.L/mol.K).298K}=1.23M

Q=\frac{[SO_3]^{2} }{[SO_{2}]^{2}.[O_{2}] } =\frac{1.23^{2} }{0.204^{2}.0.123 } =296

ΔG = ΔG° + R.T.lnQ = -141.8 kJ/mol + (8.314 × 10⁻³ kJ/mol.K) × 298 K × ln 296 = -127.7 kJ/mol

<em>Part C Each reactant and product at a partial pressure of 1.0 atm.  Express your answer using four significant figures.</em>

<em />

[SO_{2}]=[O_{2}]=[SO_{3}]=\frac{1.0atm}{(0.08206atm.L/mol.K).298K}=0.0409M

Q=\frac{[SO_3]^{2} }{[SO_{2}]^{2}.[O_{2}] } =\frac{0.0409^{2} }{0.0409^{3}} =24.4

ΔG = ΔG° + R.T.lnQ = -141.8 kJ/mol + (8.314 × 10⁻³ kJ/mol.K) × 298 K × ln 24.4 = -133.9 kJ/mol

7 0
3 years ago
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