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3 years ago

The ph of a 0.30 m solution of an acid, ha, is 5.20. calculate ka of ha.

1 answer:

5 0

When the acid is hydrolyzed, the reaction is

HA --> H⁺ + A⁻

So,

Ka = [H⁺][A⁻]/[HA]

Apply the ICE approach:

HA --> H⁺ + A⁻
I 0.3 0 0
C -X +X +X
E 0.3-X X X

Ka = [X][X]/[0.3-X]

X = H⁺ = A⁻
The formula for pH is:
pH = -log[X]
5.2 = -log[X]
Solving for X,
X = 6.31*10⁻⁶

Therefore,
Ka = [6.31*10⁻⁶][6.31*10⁻⁶]/[0.3-6.31*10⁻⁶]
<span>Ka = 1.33×10⁻¹⁰</span>

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3 years ago

Complete and balance the following redox equation using the set of smallest whole– number coefficients. Now sum the coefficients

Elena L [17]

Answer : The balanced chemical equation in a acidic solution is,

$BrO_3^-(aq)+6H^+(aq)+3Sb^{3+}(aq)\rightarrow Br^-(aq)+3H_2O(l)+3Sb^{5+}(aq)$

The sum of the coefficients is, 17

Explanation :

Redox reaction or Oxidation-reduction reaction : It is defined as the reaction in which the oxidation and reduction reaction takes place simultaneously.

Oxidation reaction : It is defined as the reaction in which a substance looses its electrons. In this, oxidation state of an element increases. Or we can say that in oxidation, the loss of electrons takes place.

Reduction reaction : It is defined as the reaction in which a substance gains electrons. In this, oxidation state of an element decreases. Or we can say that in reduction, the gain of electrons takes place.

Rules for the balanced chemical equation in acidic solution are :

First we have to write into the two half-reactions.

Now balance the main atoms in the reaction.

Now balance the hydrogen and oxygen atoms on both the sides of the reaction.

If the oxygen atoms are not balanced on both the sides then adding water molecules at that side where the less number of oxygen are present.

If the hydrogen atoms are not balanced on both the sides then adding hydrogen ion $(H^+)$ at that side where the less number of hydrogen are present.

Now balance the charge.

The given chemical reaction is,

$BrO_3^-(aq)+Sb^{3+}(aq)\rightarrow Br^-(aq)+Sb^{5+}(aq)$

The oxidation-reduction half reaction will be :

Oxidation : $Sb^{3+}\rightarrow Sb^{5+}$

Reduction : $BrO_3^-\rightarrow Br^-$

First balance the main element in the reaction.

Oxidation : $Sb^{3+}\rightarrow Sb^{5+}$

Reduction : $BrO_3^-\rightarrow Br^-$

Now balance oxygen atom on both side.

Oxidation : $Sb^{3+}\rightarrow Sb^{5+}$

Reduction : $BrO_3^-\rightarrow Br^-+3H_2O$

Now balance hydrogen atom on both side.

Oxidation : $Sb^{3+}\rightarrow Sb^{5+}$

Reduction : $BrO_3^-+6H^+\rightarrow Br^-+3H_2O$

Now balance the charge.

Oxidation : $Sb^{3+}\rightarrow Sb^{5+}+2e^-$

Reduction : $BrO_3^-+6H^++6e^-\rightarrow Br^-+3H_2O$

The charges are not balanced. Now multiplying oxidation reaction by 3 and then adding both equation, we get the balanced redox reaction.

Oxidation : $3Sb^{3+}\rightarrow 3Sb^{5+}+6e^-$

Reduction : $BrO_3^-+6H^++6e^-\rightarrow Br^-+3H_2O$

The balanced chemical equation in acidic medium will be,

$BrO_3^-(aq)+6H^+(aq)+3Sb^{3+}(aq)\rightarrow Br^-(aq)+3H_2O(l)+3Sb^{5+}(aq)$

The sum of the coefficients = 1 + 6 + 3 + 1 + 3 + 3

The sum of the coefficients = 17

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3 years ago

In the sixteenth century, the geocentric theory was a daim with substantial evidence. Which of the following best describes why

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Answer:

The heliocentric theory was better supported by data explaining the rotation of the planets and other bodies in the solar system.

Explanation:

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3 years ago

Rank the following compounds from highest boiling to lowest boiling. To rank items as equivalent, overlap them.

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1-pentanol > 1-butanol > 1-chlorobutane > pentane

Explanation:

You my find in the attached picture the structures of the compounds, names and also the variation of their boiling points.

Alcohols have higher boiling points because they form hydrogen bonds between the molecules.

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3 years ago

onsider the reaction, NO2(g) + CO(g) → NO(g) + CO2(g), for which the rate law has been determined to be: Rate = k[NO2]2[CO]. Whi

nlexa [21]

Answer:

The correct answer is option D.

Explanation:

Rate of the reaction is a change in the concentration of any one of the reactant or product per unit time.

$NO_2(g) + CO(g)\rightarrow NO(g) + CO_2(g)$

Rate of the reaction:

$R=-\frac{1}{1}\times \frac{d[NO_2]}{dt}=-\frac{1}{1}\times \frac{d[CO]}{dt}$

Rate of decrease in nitrogen dioxide concentration is equal to the rate of decrease in carbon monoxide.

Given rate expression of the reaction:

$R = k[NO2]^2[CO]$

Rate of the reaction on doubling concentration of nitrogen dioxide and carbon monoxide : R'

$R'=k(2\times [NO_2])^2(2\times [CO])=8\times k[NO2]^2[CO]=8R$

Doubling the concentrations of nitrogen dioxide and carbon monoxide simultaneously will increase the rate of the reaction by a factor of eight.

Hence, none of the given statements are true.

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3 years ago

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