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k0ka [10]
3 years ago
15

Consider the equation below. Upper C a upper C upper O subscript 3 (s) double-headed arrow upper C a upper O (s) plus upper C up

per O subscript 2 (g). What is the equilibrium constant expression for the given reaction? K subscript e q equals StartFraction bracket upper C a upper O EndBracket over upper C a upper C upper O subscript 3 EndBracket EndFraction. K subscript e q equals StartFraction StartBracket upper C upper O subscript 2 EndBracket StartBracket upper C a upper O EndBracket over StartBracket upper C a upper C upper O subscript 3 EndBracket EndFraction. K subscript e q equals StartBracket upper C upper O subscript 2 EndBracket. K subscript e q equals StartFraction 1 over StartBracket upper C upper O subscript 2 EndBracket EndFraction.
Chemistry
2 answers:
stiks02 [169]3 years ago
8 0

Answer:K subscript e q equals StartFraction StartBracket upper C upper O subscript 2 EndBracket StartBracket upper C a upper O EndBracket over StartBracket upper C a upper C upper O subscript 3 EndBracket EndFraction

Explanation: the answer has it's root in Law of mass action which states that; the rate of a chemical reaction is directly proportional to the product of the concentrations of the reactants raised to their respective stoichiometric coefficients.

Len [333]3 years ago
8 0

Answer: is A

Explanation:pij

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How many milligrams of product can be produced from the complete Diels-Alder reaction of 180. mg of anthracene and 100. mg of ma
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Answer: It will be produced 276,3 mg of product

Explanation: The reaction of anthracene (C14H10) and maleic anhydride (C4H2O3) produce a compound named 9,10-dihydroanthracene-9,10-α,β-succinic anhydride (C18H12O3), as described below:

C14H10 + C4H2O3 → C18H12O3

The reaction is already balanced, which means to produce 1 mol of C18H12O3 is necessary 1 mol of anthracene and 1 mol of maleic anhydride.

1 mol of C14H10 equals 178,23 g. As it is used 180 mg of that reagent, we have 0,001 mol of anthracene. With it, the reaction produces 0,001 mol of C18H12O3.

As 1 mol of C18H12O3 equals 276,3 g, the mass produced is 276,3 mg.

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What is the pH of a solution that has a hydroxide ion concentration of 1.0×10−9 M?
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Answer:

\boxed{\text{pH = 5.00}}

Explanation:

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7 0
3 years ago
During the following chemical reaction, 46.3 grams of C3H6O react with 73.2 grams of O2
ra1l [238]

Answer:

a) O2 is the limiting reactant

b) 75.70 grams CO2 (theoretical yield)

c) There remains 12.81 grams of C3H6O

d) The actual yield CO2 is 34.29 grams

Explanation:

Step 1: Data given

Mass of C3H6O = 46.3 grams

Mass of O2 = 73.2 grams

Molar mass of C3H6O = 58.08 g/mol

Molar mass  of O2 = 32 g/mol

Step 2: The balanced equation

C3H6O + 4O2 → 3 CO2 + 3H2O

Step 3: Calculate moles C3H6O

Moles C3H6O = mass C3H6O / molar mass C3H6O

Moles C3H6O = 46.3 grams / 58.08 g/mol

Moles C3H6O = 0.793 moles

Step 4: Calculate moles O2

Moles O2 = 73.2 grams / 32 g/mol

Moles O2 = 2.29 moles

Step 5: Calculate limiting reactant

For 1 mol C3H6O we need 4 moles of O2 to produce 3 moles CO2 and 3 moles H2O

O2 is the limiting reactant. It will completely be consumed. (2.29 moles).

C3H6O is in excess. There will react 2.29/4 = 0.5725 moles C3H6O

There will remain 0.793 - 0.5725 = 0.2205 moles C3H6O

This is 0.2205 moles * 58.08 g/mol =<u> 12.81 grams</u>

Step 6:  Calculate moles of CO2

For 1 mol C3H6O we need 4 moles of O2 to produce 3 moles CO2 and 3 moles H2O

For 2.29 moles O2 we need 3/4 * 2.29 = 1.72 moles CO2

This is 1,72 moles * 44.01 g/mol = <u>75.70 grams CO2</u>

Step 7: Calculate actual yield

% yield = 45.3 % = 0.453 = (actual yield / theoretical yield)

actual yield = 0.453 * 75.70 = <u>34.29 grams</u>

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