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pantera1 [17]
3 years ago
11

Which of these would most likely be one of the final steps when performing a strong acid-base titration? Prepare the burette. Re

cord the initial burette volume. Prepare to measure pH change. Observe any changes as base is added.
Chemistry
2 answers:
DanielleElmas [232]3 years ago
8 0

Answer is: Prepare to measure pH change.

For example for strong acid-base titration, sodium hydoxide and hydrochloric can be used.

Balanced chemical reaction: HCl + NaOH → NaCl + H₂O.

In this reaction pH of equivalence point will be always 7.

Equivalence point is the point which there is stoichiometrically equivalent amounts of acid and base.  

Chemist can draw pH curve (graph showing the change in pH of a solution, which is being titrated) for titration and determine equivalence point.  

Near equivalence point indicator should change color, so we must pick indicator who change color near pH of equivalence point.

anzhelika [568]3 years ago
5 0

Answer: C on edg

Explanation:

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The activation energy of a certain uncatalyzed biochemical reaction is 50.0 kJ/mol. In the presence of a catalyst at 37°C, the r
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Answer:

E₁ ≅ 28.96 kJ/mol

Explanation:

Given that:

The activation energy of a certain uncatalyzed biochemical reaction is 50.0 kJ/mol,

Let the activation energy for a catalyzed biochemical reaction = E₁

E₁ = ??? (unknown)

Let the activation energy for an uncatalyzed biochemical reaction = E₂

E₂ = 50.0 kJ/mol

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Temperature (T) = 37°C

= (37+273.15)K

= 310.15K

Rate constant (R) = 8.314 J/mol/k

Also, let the constant rate for the catalyzed biochemical reaction = K₁

let the constant rate for the uncatalyzed biochemical reaction = K₂

If the  rate constant for the reaction increases by a factor of 3.50 × 10³ as compared with the uncatalyzed reaction, That implies that:

K₁ = 3.50 × 10³

K₂ = 1

Now, to calculate the activation energy for the catalyzed reaction going by the following above parameter;

we can use the formula for Arrhenius equation;

K=Ae^{\frac{-E}{RT}}

If K_1=Ae^{\frac{-E_1}{RT}} -------equation 1     &

K_2=Ae^{\frac{-E_2}{RT}} -------equation 2

\frac{K_1}{K_2} = e^{\frac{-E_1-E_2}{RT}

E_1= E_2-RT*In(\frac{K_1}{K_2})

E_1= 50,000-8.314*310.15*In(\frac{3.50*10^3}{1})

E_1 = 28957.39292  J/mol

E₁ ≅ 28.96 kJ/mol

∴ the activation energy for a catalyzed biochemical reaction (E₁) = 28.96 kJ/mol

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