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aleksandrvk [35]
3 years ago
11

At the midpoint of a titration curve ________. At the midpoint of a titration curve ________. The concentration of a conjugate b

ase is twice that of the concentration of a conjugate acid The pH equals the pKa The ability of the solution to buffer is at its least effective The concentration of a conjugate base is 1/2 that of the concentration of a conjugate acid
Chemistry
1 answer:
VARVARA [1.3K]3 years ago
8 0

Answer:

The concentration of a conjugate base is twice that of the concentration of a conjugate acid

Explanation:

This is because at the midpoint, the number of moles of acid equals half the number of moles of base at the midpoint. This means that at the midpoint, half the analyte has been titrated.

Since the concentration of the conjugate acid is half that of the conjugate base at the midpoint, this implies that the concentration of the conjugate base is twice that of the conjugate acid.

<u>Thus, at the midpoint of a titration curve the concentration of a conjugate base is twice that of the concentration of a conjugate acid </u>

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Answer: The given statement is TRUE.

Explanation:

An equilibrium reaction is one in which rate of forward reaction is equal to the rate of backward reaction.

Equilibrium constant is defined as the ratio of the product of the concentration of products to the product of the concentration of reactants each raised to their stochiometric coefficient.

For example for the given equilibrium reaction;

2H_2O(g)\leftrightharpoons 2H_2(g)+O_2(g)

K_{eq}=\frac{[H_2]^2[O_2]}{[H_2O]^2}

Thus the given statement that in calculating the equilibrium constant for a reaction, the coefficients of the chemical equation are used as exponents for the factors in the equilibrium expression is True.

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3 years ago
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How many electrons in an atom can share the quantum numbers n = 4 and l = 3?
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Answer:

\boxed{\text{14}}

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4 years ago
When there is an imbalance in a body system, and the body cannot maintain homeostasis, how might other systems respond?
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8 0
3 years ago
The boiling point elevation of an aqueous sucrose solution is found to be 0.39°C.
SVEN [57.7K]

Answer:

130.4 grams of sucrose, would be needed to dissolve in 500 g of water.

Explanation:

Colligative property of boiling point elevation:

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In this case, i = 1 (sucrose is non electrolytic)

ΔT = Kb . m

0.39°C = 0.512°C/m . m

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0.762 m (molality means that this moles, are in 1kg of solvent)

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342.30 g/m . 0.381 m = 130.4 g

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3 years ago
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