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Lena [83]
3 years ago
13

Calculation of Molar Ratios of Conjugate Base to Weak Acid from pH For a weak acid with a pKa of 6.0, calculate the ratio of con

jugate base to acid at a pH of 5.0.
Choice of Weak Acid for a Buffer Which of these com-pounds would be the best buffer at pH 5.0: formic acid (pKa 5 3.8), acetic acid (pKa 5 4.76), or ethylamine (pKa 5 9.0)? Briefly justify your answer.
Chemistry
1 answer:
Free_Kalibri [48]3 years ago
7 0

Explanation:

According to the Handerson equation,  

          pH = pK_{a} + log \frac{\text{salt}}{\text{acid}}

or,      pH = pK_{a} + log \frac{\text{conjugate base}}{\text{acid}}

Putting the given values into the above equation as follows.

     pH = pK_{a} + log \frac{\text{conjugate base}}{\text{acid}}

       5.0 = 6.0 + log \frac{\text{conjugate base}}{\text{acid}}[/tex]

      log \frac{\text{conjugate base}}{\text{acid}} = -1.0

or,      \frac{\text{conjugate base}}{\text{acid}} = 10^{-1.0}

                            = 0.1

Therefore, we can conclude that molar ratios of conjugate base to weak acid for given solution is 0.1.

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A gas is collected in a 34.3L container at a temperature of 31.5°C. Later, the container has a volume of 29.2L, a temperature of
goldenfox [79]

Answer:

108 kPa  

Step-by-step explanation:

To solve this problem, we can use the <em>Combined Gas Laws</em>:

p₁V₁/T₁ = p₂V₂/T₂             Multiply each side by T₁

   p₁V₁ = p₂V₂ × T₁/T₂      Divide each side by V₁

      p₁ = p₂ × V₂/V₁ × T₁/T₂

Data:

p₁ = ?;                 V₁ = 34.3 L; T₁ = 31.5 °C

p₂ = 122.2 kPa; V₂ = 29.2 L; T₂ = 21.0 °C

Calculations:

(a) Convert temperatures to <em>kelvins </em>

T₁ = (31.5 + 273.15) K = 304.65 K

T₂ = (21.0 + 273.15) K = 294.15 K

(b) Calculate the <em>pressure </em>

p₁ = 122.2 kPa × (29.2/34.3) × (304.65/294.15)  

   = 122.2 kPa × 0.8542 × 1.0357

   = 108 kPa

4 0
3 years ago
A sample of 2.45g aluminum oxide decomposes into 1.3g of aluminum and 1.15g of oxygen. What is the percentage composition of the
Vitek1552 [10]

Answer:

%Al = 53.1%%

%O = 46.9%

Explanation:

If we know the grams of a chemical compound in a specific reaction, it is possible to know the percentage of each atom that composes it.

For the Aluminum Oxide in this problem, we know its total weight and the grams of each component.

therefore we can determine the percentage ratio of its components through:

For Al

%Al = \frac{mass of Al}{mass of Aluminium oxide} . 100%

% Al = \frac{1,3 g}{2,45 g} . 100%

%Al = 53.1%%

In the same way for oxygen

%O = \frac{mass of O}{mass of Aluminium oxide} . 100%

%O = \frac{1,5 g}{2,45 g} . 100%

%O = 46.9%

5 0
3 years ago
If a metal rod has a density of 7.85 g/cm and a volume of 4 cm, then what is its mass?
Bond [772]

Answer:

<h3>The answer is 31.4 g</h3>

Explanation:

The mass of a substance when given the density and volume can be found by using the formula

<h3>mass = Density × volume</h3>

From the question

volume of metal = 4 cm³

density = 7.85 g/cm³

The mass is

mass = 7.85 × 4

We have the final answer as

<h3>31.4 g</h3>

Hope this helps you

8 0
3 years ago
During a chemical reaction, how do the substances that form differ from the substances that react?
cricket20 [7]

Answer:

A physical change, such as a state change or dissolving, does not create a new substance, but a chemical change does. ... In a chemical reaction, reactants contact each other, bonds between atoms in the reactants are broken, and atoms rearrange and form new bonds to make the products.

Explanation:

8 0
3 years ago
Read 2 more answers
True or false atoms with low ionization energy give up their valence electrons with ease
My name is Ann [436]
This statement is TRUE.

The ionization energy presents the amount of energy required for the release of the valence electron, the electron farthest from the core. The smaller the amount of energy needed for the release is, the lower the ionization energy is.
Therefore, the atoms with low ionization require little energy to release their valence electrons and they do it easier that the atoms with high ionization energy.
6 0
4 years ago
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