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1 year ago

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

1 answer:

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

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A 45-g aluminium spoon(specific heat 0.80 / J/gdegree Celsius) at 24 degree celsius is placed in 180 ml(180 grams) of coffee at

Vlad [161]

Explanation:

a) The amount of heat released by coffee will be absorbed by aluminium spoon.

Thus, $heat_{absorbed}=heat_{released}$

To calculate the amount of heat released or absorbed, we use the equation:

$Q=m\times c\times \Delta T=m\times c\times (T_{final}-T_{initial})$

Also,

$m_1\times c_1\times (T_{final}-T_1)=-[m_2\times c_2\times (T_{final}-T_2)]$ ..........(1)

where,

q = heat absorbed or released

$m_1$ = mass of aluminium = 45 g

$m_2$ = mass of coffee = 180 g

$T_{final}$ = final temperature = ?

$T_1$ = temperature of aluminium = $24^oC$

$T_2$ = temperature of coffee = $85^oC$

$c_1$ = specific heat of aluminium = $0.80J/g^oC$

$c_2$ = specific heat of coffee= $4.186 J/g^oC$

Putting all the values in equation 1, we get:

$45 g\times 0.80J/g^oC\times (T_{final}-24^oC)=-[180 g\times 4.186J/g^oC\times (T_{final}-83^oC)]$

$T_{final}=80.30^oC$

80.30 °C is the final temperature.

b) Energy flows from higher temperature to lower temperature.Whenever two bodies with different energies and temperature come in contact. And the resulting temperature of both bodies will less then the body with high temperature and will be more then the body with lower temperature.

So, is our final temperature of both aluminium and coffee that is 80°C less than initial temperature of coffee and more than the initial temperature of the aluminum.

8 0

3 years ago

a) One kmol of N2O4 dissociates at 25℃, 1 atm to form an equilibrium ideal gas mixture of N2O4 and NO2 in which the amount of N2

horsena [70]

Answer:

neq N2O4 = 0.9795 mol.....P = 0.5 atm; T = 25°C

Explanation:

ni change eq.

N2O4 1 1 - x 0.8154.....P = 1 atm; T = 25°C

NO2 0 0 + x x

∴ x = neq = Peq.V / R.T.....ideal gas mix

if P = 0.5 atm, T = 25°C; assuming: V = 1 L

⇒ x = neq = ((0.5 atm)(1 L))/((0.082 atm.L/K.mol)(298 K))

⇒ x = neq = 0.0205 mol

⇒ neq N2O4 = 1 - x = 1 - 0.0205 = 0.9795 mol

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