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3 years ago

8

Which would increase the amount of dissolved oxygen in a pond

1 answer:

AlexFokin [52]3 years ago

6 0

Answer:

Dissolved oxygen levels are increased by supplementing wind and wave action, adding plants to water and exposing water to purified oxygen. Also removing fish could increase the amount of dissolved oxygen

Explanation:

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igomit [66]

Answer: The scientific method is an empirical method of acquiring knowledge that has characterized the development of science since at least the 17th century. It involves careful observation, applying rigorous skepticism about what is observed, given that cognitive assumptions can distort how one interprets the observation.

Explanation:

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3 years ago

Tema: Métodos de Separação de Misturas – Homogêneas e Heterogêneas;

seraphim [82]

Answer:

fjskeowkcnekvo Dee five votes come vote for dog even r

4 0

3 years ago

How many miles are in 215 grams of water

Irina18 [472]

You're off to a good start, now find the mass of H2O and put it under I mol,

then multiply 1 mol over the mass of H2O by 215 grams

3 0

3 years ago

Which of the following (with specific heat capacity provided) would show the smallest temperature change upon gaining 200.0 J of

Brut [27]

<u>Answer:</u> The smallest temperature change is shown by water.

<u>Explanation:</u>

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

$q=mC\times \Delta T$ ......(1)

where,

q = heat absorbed = 200.0 J

m = mass of the substance

C = specific heat of substance

$\Delta T$ = change in temperature

As, the same amount of heat is getting absorbed in all the cases. So, the change in temperature will depend on the product of mass and specific heat.

For the given options:

<u>Option a:</u> 50.0 g Fe, $C_{Fe}=0.449J/g^oC$

We are given:

$m=50.0g\\C_{Fe}=0.449J/g^oC$

Putting values in equation 1, we get:

$200.0J=50.0g\times 0.449J/g^oC\times \Delta T\\\\\Delta T=\frac{200}{50\times 0.449}=8.99^oC$

Change in temperature = 8.99°C

<u>Option b:</u> 50.0 g water, $C_{water}=4.18J/g^oC$

We are given:

$m=50.0g\\C_{water}=4.18J/g^oC$

Putting values in equation 1, we get:

$200.0J=50.0g\times 4.18J/g^oC\times \Delta T\\\\\Delta T=\frac{200}{50\times 4.18}=0.96^oC$

Change in temperature = 0.96°C

<u>Option b:</u> 25.0 g Pb, $C_{Pb}=0.128J/g^oC$

We are given:

$m=50.0g\\C_{Pb}=0.128J/g^oC$

Putting values in equation 1, we get:

$200.0J=25.0g\times 0.128J/g^oC\times \Delta T\\\\\Delta T=\frac{200}{25\times 0.128}=62.5^oC$

Change in temperature = 62.5°C

<u>Option d:</u> 25.0 g Ag, $C_{Ag}=0.235J/g^oC$

We are given:

$m=25.0g\\C_{Ag}=0.235J/g^oC$

Putting values in equation 1, we get:

$200.0J=25.0g\times 0.235J/g^oC\times \Delta T\\\\\Delta T=\frac{200}{25\times 0.235}=34.04^oC$

Change in temperature = 34.04°C

<u>Option e:</u> 25.0 g granite, $C_{granite}=0.79J/g^oC$

We are given:

$m=25.0g\\C_{Fe}=0.79J/g^oC$

Putting values in equation 1, we get:

$200.0J=25.0g\times 0.79J/g^oC\times \Delta T\\\\\Delta T=\frac{200}{25\times 0.79}=10.13^oC$

Change in temperature = 10.13°C

Hence, the smallest temperature change is shown by water.

5 0

3 years ago

Can somebody please help me ?

larisa [96]

Answer:

I belive it is either a element or co mpound Hope this helped :D :3

Explanation:

4 0

3 years ago