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emmainna [20.7K]

3 years ago

9

State relationship between the level of dissolved oxygen and water temperature

1 answer:

Artemon [7]3 years ago

8 0

More dissolved oxygen is present in water with a lower temperature compared to water with a higher temperature. The reason for this inverse relationship between dissolved oxygen and temperature is that the solubility of a gas in a liquid is an equilibrium phenomenon.

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50 Points!! 25 Each Points Please Help ASAP!! View Attached Image!! Will Mark Brainliest If All Are Answered!!

adell [148]

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7 0

2 years ago

What is the average of the two weights of a nickel and dime. Dime is 2.26g and nickel is 4.97g?

ANTONII [103]

Dime=2.26g
Nickel=4.97g

$\\ \rm\Rrightarrow Avg\:Weight=\dfrac{Sum\:of\:Weights}{No\:of\:weights}$

$\\ \rm\Rrightarrow Avg\:weight=\dfrac{2.26+4.96}{2}$

$\\ \rm\Rrightarrow Avg\;weight=\dfrac{7.23}{2}$

$\\ \rm\Rrightarrow Avg\:Weight=3.6g$

5 0

2 years ago

3. What is the primary consumer in the following food chain, sun - tomato plant - tomato worm - wasp?

Julli [10]

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Tomato worm

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5 0

2 years ago

Read 2 more answers

If gas in a sealed container has a pressure of 50 kpa at 300 k, what will the pressure be if the temperature rises to 360 k? 161

Illusion [34]

$60 \; \text{kPa}$

The pressure an ideal gas exerts on a sealed container of fixed volume is directly related to its temperature in degrees Kelvin. That is:

$T_1 / T_2 = P_1 / P_2$

In this scenario:

$T_1 = 300 \; \text{K}$
$P_1 = 50 \; \text{kPa}$
$T_2 = 360 \; \text{K}$

Rearranging gives

$\begin{array}{lll}P_2 &=& P_1 \cdot T_2 / T_1 \\ &=& 50 \; \text{kPa} \times (360 \; \text{K})/ (300 \; \text{K})\\ &=& 60 \; \text{kPa}\end{array}$

6 0

2 years ago

What is the energy released in this β − β − nuclear reaction 40 19 K → 40 20 C a + 0 − 1 e 19 40 K → 20 40 C a + − 1 0 e ? (The

Effectus [21]

<u>Answer:</u> The energy released in the given nuclear reaction is 1.3106 MeV.

<u>Explanation:</u>

For the given nuclear reaction:

$_{19}^{40}\textrm{K}\rightarrow _{20}^{40}\textrm{Ca}+_{-1}^{0}\textrm{e}$

We are given:

Mass of $_{19}^{40}\textrm{K}$ = 39.963998 u

Mass of $_{20}^{40}\textrm{Ca}$ = 39.962591 u

To calculate the mass defect, we use the equation:

$\Delta m=\text{Mass of reactants}-\text{Mass of products}$

Putting values in above equation, we get:

$\Delta m=(39.963998-39.962591)=0.001407u$

To calculate the energy released, we use the equation:

$E=\Delta mc^2\\E=(0.001407u)\times c^2$

$E=(0.001407u)\times (931.5MeV)$ (Conversion factor: $1u=931.5MeV/c^2$ )

$E=1.3106MeV$

Hence, the energy released in the given nuclear reaction is 1.3106 MeV.

6 0

2 years ago