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ELEN [110]
1 year ago
7

The presence of a large population of _____________ indicates water depleted of oxygen and a high level of organic pollution.

Chemistry
1 answer:
blondinia [14]1 year ago
6 0

The presence of a large population of sludgeworms indicates water depleted of oxygen and a high level of organic pollution.

The sludgeworms Tubifex is harvested and sold as tropical fish food since it is common near sewer outlets and serves as a sign of water or organic pollution. Polychaetes are crucial in rearranging the silt on the ocean floor.

Organic contamination happens when too much organic material, such sewage or manure, gets into the water.

There will be more decomposers and sludgeworms in a pond as the organic content accumulates. These decomposers and sludgeworms develop quickly and consume a lot of oxygen as they do so.

Hence, Sludgeworms are a sign of an abundance of oxygen-depleted water and significant levels of organic pollutants.

Learn more about Water pollution here brainly.com/question/1235358

#SPJ4

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A chemist dissolves 240mg of pure barium hydroxide in enough water to make up of solution. Calculate the pH of the solution. (Th
goblinko [34]

Answer:

pH = 12.22

Explanation:

<em>... To make up 170mL of solution... The temperature is 25°C...</em>

<em />

The dissolution of Barium Hydroxide, Ba(OH)₂ occurs as follows:

Ba(OH)₂ ⇄ Ba²⁺(aq) + 2OH⁻(aq)

<em>Where 1 mole of barium hydroxide produce 2 moles of hydroxide ion.</em>

<em />

To solve this question we need to convert mass of the hydroxide to moles with its molar mass. Twice these moles are moles of hydroxide ion (Based on the chemical equation). With moles of OH⁻ and the volume we can find [OH⁻] and [H⁺] using Kw. As pH = -log[H⁺], we can solve this problem:

<em>Moles Ba(OH)₂ molar mass: 171.34g/mol</em>

0.240g * (1mol / 171.34g) = 1.4x10⁻³ moles * 2 =

2.80x10⁻³ moles of OH⁻

<em>Molarity [OH⁻] and [H⁺]</em>

2.80x10⁻³ moles of OH⁻ / 0.170L = 0.01648M

As Kw at 25°C is 1x10⁻¹⁴:

Kw = 1x10⁻¹⁴ = [OH⁻] [H⁺]

[H⁺] = Kw / [OH⁻] = 1x10⁻¹⁴/0.01648M = 6.068x10⁻¹³M

<em>pH:</em>

pH = -log [H⁺]

pH = -log [6.068x10⁻¹³M]

<h3>pH = 12.22</h3>
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