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Advocard [28]
3 years ago
14

What is the ph of a solution that has 0.075 m ch3cooh and 0.050m nach3coo present? ka for acetic acid = 1.80 x 10-5?

Chemistry
1 answer:
Scilla [17]3 years ago
7 0
Hello!

For solving this problem we first have to determine the pKa of CH₃COOH.

pKa=-log(Ka)=-log( 1,80*10^{-5} )=4,74

Now, we use the Henderson-Hasselbach's equation to determine the pH of the buffer solution:

pH=pKa+log (\frac{NaCH_3COO}{CH_3COOH} )=4,74+log( \frac{0,050M}{0,075M} )=4,56

So, the pH of this solution is 4,56

Have a nice day!
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The use of fertilizers in agriculture has significantly altered several nutrient cycles including:
Lana71 [14]

Answer:

potassium, nitrogen and phosphorous cycle

Explanation:

A fertilizer is a substance which is applied on the plants by farmers to increase the supply of nutrients for the plants. Fertilizers have known to be toxic in many ways such as they alter the potassium, nitrogen and phosphorus cycles. Nitrogen, potassium and phosphorus are present in abundant amounts in the fertilizers. Draining of these fertilizers into rivers and ponds is toxic for the aquatic life. Hence, the use of fertilizers disrupts the natural cycles and is toxic for many aquatic plants and animals.

4 0
3 years ago
A student analyzed an unknown sample that contained a single anion. The sample gave a yellow precipitated upon addition of a sol
kenny6666 [7]
<h3>Answer:</h3>

Anion present- Iodide ion (I⁻)

Net ionic equation- Ag⁺(aq) + I⁻(aq) → AgI(s)

<h3>Explanation:</h3>

In order to answer the question, we need to have an understanding of insoluble salts or precipitates formed by silver metal.

Additionally we need to know the color of the precipitates.

Some of insoluble salts of silver and their color include;

  • Silver chloride (AgCl) - white color
  • Silver bromide (AgBr)- Pale cream color
  • Silver Iodide (AgI) - Yellow color
  • Silver hydroxide (Ag(OH)- Brown color

With that information we can identify the precipitate of silver formed and identify the anion present in the sample.

  • The color of the precipitate formed upon addition of AgNO₃ is yellow, this means the precipitate formed was AgI.
  • Therefore, the anion that was present in the sample was iodide ion (I⁻).
  • Thus, the corresponding net ionic equation will be;

Ag⁺(aq) + I⁻(aq) → AgI(s)

4 0
3 years ago
The life cycle represents the process of
slega [8]

Answer:

The correct answer would be metamorphosis.

Metamorphosis is a biological process by which an immature form transforms into an adult form passing through numerous distinct stages.

It is mainly observed in amphibians and insects. For example, frog and butterfly.

The life cycle of a butterfly can be summarized as shown below.

3 0
3 years ago
Which of the following phase changes only occurs in solids possessing vapor pressures above atmospheric pressure and at ambient
GuDViN [60]

<u>Answer:</u> The phase change process in which solids gets converted to gases is sublimation.

<u>Explanation:</u>

For the given options:

<u>Option a:</u>  Condensation

It is a type of process in which phase change occurs from gaseous state to liquid state at constant temperature.

Gas\rightleftharpoons Liquid

<u>Option b:</u>  Melting

It is a type of process in which phase change occurs from solid state to liquid state at constant temperature.

Solid\rightleftharpoons Liquid

<u>Option c:</u>  Sublimation

It is a type of process in which phase change occurs from solid state to gaseous state without passing through the liquid state at constant temperature.

Solid\rightleftharpoons Gas

<u>Option d:</u>  Deposition

It is a type of process in which phase change occurs from gaseous state to solid state without passing through the liquid state at constant temperature.

Gas\rightleftharpoons Solid

Hence, the phase change process in which solids gets converted to gases is sublimation.

4 0
3 years ago
The barium isotope 133ba has a half-life of 10.5 years. a sample begins with 1.1×1010 133ba atoms. how many are left after (a) 5
Sladkaya [172]
<span>a) 7.9x10^9 b) 1.5x10^9 c) 3.9x10^4 To determine what percentage of an isotope remains after a given length of time, you can use the formula p = 2^(-x) where p = percentage remaining x = number of half lives expired. The number of half lives expired is simply x = t/h where x = number of half lives expired t = time spent h = length of half life. So the overall formula becomes p = 2^(-t/h) And since we're starting with 1.1x10^10 atoms, we can simply multiply that by the percentage. So, the answers rounding to 2 significant figures are: a) 1.1x10^10 * 2^(-5/10.5) = 1.1x10^10 * 0.718873349 = 7.9x10^9 b) 1.1x10^10 * 2^(-30/10.5) = 1.1x10^10 * 0.138011189 = 1.5x10^9 c) 1.1x10^10 * 2^(-190/10.5) = 1.1x10^10 * 3.57101x10^-6 = 3.9x10^4</span>
4 0
3 years ago
Read 2 more answers
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