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3 years ago

What is the temperature of a water body containing a 90% saturation and 11.5 ppm of dissolved oxygen?

1 answer:

ozzi3 years ago

6 0

Using the chart that has been provided, we may determine water temperature. We do this by drawing a straight line form the bottom scale which has the ppm of oxygen dissolved to the middle scale which has the percentage saturation.

The line starts from 11.5 ppm on the bottom scale and goes to 90% on the middle scale. Next, we continue this line, without changing its slope, to the third scale showing temperature. We see that it crosses the temperature scale at 4°C.

The temperature of the water is 4 °C.

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3 years ago

What is one beneift of using scientific notation instead of regular notation

kotykmax [81]

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2 years ago

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alexdok [17]

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3 years ago

Why do living things need carbon

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2 years ago

A solution containing a mixture of 0.0351 M potassium chromate ( K 2 CrO 4 ) and 0.0537 M sodium oxalate ( Na 2 C 2 O 4 ) was ti

ki77a [65]

Answer:

Explanation:

As it was stated in the comments by other user, the question is incomplete but luckily it was posted the rest of the question, so, I'm gonna answer it with the data of that. If you have another questions there, please submit it again or put it in the comments.

a) Which barium salt will precipite first?

In order to know this, we need to take a look at the Ksp of both salts. The given Ksp are:

Ksp 1: 2.1x10^-10 (BaCrO4)

Ksp 2: 1.3x10^-6 (BaC2O4)

Now, we can see that Ksp1 < Ksp 2, but what's this Ksp value means? a Ksp value means that an aqueous solution will form a precipite (So the solid formed, it's not soluble in water), As Ksp 1 is a smaller value than Ksp2, means that the concentrations of Ba and CrO4 are too small, and therefore, it takes more time to precipite. Therefore, the BaC2O4 will precipite first.

b) Concentration of Ba2+ present to BaCrO4 precipite

In this, we know that the reaction taking place is the following:

BaCrO4(s) <--------> Ba^2+ + CrO4^2- Ksp = 2.1x10^-10

The expression for Ksp is:

Ksp = [Ba][CrO4]

We know the concentration of CrO4 cause this comes from the K2CrO4 so, replacing here, we solve for Ba:

[Ba^2+] = Ksp / [CrO4^2-]

[Ba^2+] = 2.1x10^-10 / 0.0351

[Ba^2+] = 5.98x10^-9 M

c) Concentration of Ba^2+ to reduce 10% oxalate concentration

The 10% concentration of oxalate is:

[C2O4] = 0.0537 * 0.1 = 0.00537 M

So, we do the same thing we did in part b) but solving for C2O4:

BaC2O4(s) <--------> Ba^2+ + C2O4^2- Ksp = 1.3x10^-6

[Ba^2+] = 1.3x10^-6 / 0.00537

[Ba^2+] = 2.42x10^-4 M

d) ratio of oxalate and chromate when Ba concentration is 0.005 M

In this case, we calculate concentration of CrO4 and C2O4 with the value of Ba and it's respective Ksp, and then, calculate the ratio:

[CrO4^2-] = Ksp1 / [Ba^2+] = 2.1x10^-10 / 0.0050 = 4.2x10^-8 M

[C2O4^2-] = Ksp2 / [Ba^2+] = 1.3x10^-6 / 0.0050 = 2.6x10^-4 M

So the ratio is:

[C2O4^2-] / [CrO4^2-] = 2.6x10^-4 / 4.2x10^-8

[C2O4^2-] / [CrO4^2-] = 6.19x10^3

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3 years ago