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Novosadov [1.4K]
3 years ago
10

I need help! In an equilibrium reaction with a Keq of 1 × 108 A. products are favored. B. the reaction is nonspontaneous. C. the

reaction is endothermic. D. reactants are favored.
Chemistry
2 answers:
scZoUnD [109]3 years ago
7 0
<span>When K is > than 1 products are favored. When q is < less than 1 reactants are favored. 1 x 108 = 108 which is > 1 so products are favored.</span>
Amiraneli [1.4K]3 years ago
3 0

<u>Answer:</u> The correct answer is Option A.

<u>Explanation:</u>

Equilibrium constant is defined as the ratio of concentration of products to the concentration of reactants. It is represented as K_{eq}

For the general reaction:

A+B\rightarrow C+D

Expression for equilibrium constant is given as:

K_{eq}=\frac{[C][D]}{[A][B]}

Conditions for K_{eq} are:

K_{eq}>1, then products are favored

K_{eq}=1, then forward reaction is equal to backward reaction

K_{eq}, then reactants are favored

As, K_{eq}=1\times 10^8 and is very much greater than 1, so products will be favored.

Hence, the correct answer is Option A.

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the cell enentually dies because it has no way to survive

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2 years ago
A titration of 25.0 mL of a solution of the weak base aniline, C6H5NH2, requires 25.67 mL of 0.175 M HCl to reach the equivalenc
Lorico [155]

Answer:

a. 0.180M of C₆H₅NH₂

b. 0.0887M C₆H₅NH₃⁺

c. pH = 2.83

Explanation:

a. Based in the chemical equation:

C₆H₅NH₂(aq) + HCl(aq) → C₆H₅NH₃⁺(aq) + Cl⁻(aq)

<em>1 mole of aniline reacts per mole of HCl</em>

Moles required to reach equivalence point are:

Moles HCl = 0.02567L ₓ (0.175mol / L) = 4.492x10⁻³ moles HCl = moles C₆H₅NH₂

As the original solution had a volume of 25.0mL = 0.0250L:

4.492x10⁻³ moles C₆H₅NH₂ / 0.0250L = 0.180M of C₆H₅NH₂

b. At equivalence point, moles of C₆H₅NH₃⁺ are equal to initial moles of C₆H₅NH₂, that is 4.492x10⁻³ moles

But now, volume is 25.0mL + 25.67mL = 50.67mL = 0.05067L. Thus, molar concentration of C₆H₅NH₃⁺ is:

[C₆H₅NH₃⁺] = 4.492x10⁻³ moles / 0.05067L = 0.0887M C₆H₅NH₃⁺

c. At equivalence point you have just 0.0887M C₆H₅NH₃⁺ in solution. C₆H₅NH₃⁺ has as equilibrium in water:

C₆H₅NH₃⁺(aq) + H₂O(l) → C₆H₅NH₂ + H₃O⁺

Where Ka = Kw / Kb = 1x10⁻¹⁴ / 4.0x10⁻¹⁰ =

<em>2.5x10⁻⁵ = [C₆H₅NH₂] [H₃O⁺] / [C₆H₅NH₃⁺]</em>

When the system reaches equilibrium, molar concentrations are:

[C₆H₅NH₃⁺] = 0.0887M - X

[C₆H₅NH₂] = X

[H₃O⁺] = X

Replacing in Ka formula:

2.5x10⁻⁵ = [X] [X] / [0.0887M - X]

2.2175x10⁻⁶ - 2.5x10⁻⁵X = X²

0 = X² + 2.5x10⁻⁵X - 2.2175x10⁻⁶

Solving for X:

X = -0.0015 → False solution. There is no negative concentrations.

X = 0.001477 → Right solution.

As [H₃O⁺] = X, [H₃O⁺] = 0.001477

Knowing pH = -log [H₃O⁺]

pH = -log 0.001477

<h3>pH = 2.83</h3>

7 0
2 years ago
The boiling points of different substances are given in the table below.
Oksi-84 [34.3K]

Answer:

Water

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What molarity of nitric acid (HNO3) was used if 2.00 L must be used to prepare 4.5 L of a 0.25 M HNO3 solution?
Ymorist [56]

The   molarity  of (HNO₃) that was used  if 2.00 L must  be used   to prepare 4.5 L  of a 0.25M HNO₃ solution is   0.563 M


 <u><em>calculation</em></u>

  This is calculated  usind  M₁V₁=M₂V₂  formula

where,

         M₁(  molarity ₁) = ?

         V₁( volume ₁) = 2.00 L

        M₁ (molarity ₂) = 0.25M

         V₂( volume₂) = 4.5 L

make M₁ the subject  of the formula by  diving both side of the formula  by V₁

   M₁  is therefore = M₂V₂/V₁

M₁ =[ (0.25 M  x 4.5 L) / 2.00 L ]  =0.563 M

5 0
2 years ago
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