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Arturiano [62]
3 years ago
15

Most lakes have rivers flowing out of them, carrying water to the ocean. However, some lakes, including Great Salt Lake and the

Dead Sea, have no outlet. Water flowing into these lakes leaves only through evaporation. Considering that the water flowing into the lakes contains many dissolved substances, how does the lack of an outlet affect the composition of these lakes? What would you expect to happen after many more years of inflow and evaporation?
100 PTS
Chemistry
1 answer:
Kaylis [27]3 years ago
4 0

Since the only way of water flow to these lakes or bodies of water is through evaporation, I would expect an increase in unknown substances in the composition of the lakes due to the amount of contamination that globalization produces and affects terribly the surroundings when these unknown substances travel through evaporation as the outlet of these bodies of water. Therefore I think continuous contamination is what to expect after many more years of inflow and evaporation.

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B density

Explanation:

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Pu is a nuclear waste byproduct with a half-life of 24,000 y. What fraction of the 239Pu present today will be present in 1000 y
olya-2409 [2.1K]

Answer:

0.9715 Fraction of Pu-239 will be remain after 1000 years.

Explanation:

\lambda =\frac{0.693}{t_{\frac{1}{2}}}

A=A_o\times e^{-\lambda t}

Where:

\lambda= decay constant

A_o =concentration left after time t

t_{\frac{1}{2}} = Half life of the sample

Half life of Pu-239 = t_{\frac{1}{2}}=24,000 y[

\lambda =\frac{0.693}{24,000 y}=2.8875\times 10^{-5} y^{-1]

Let us say amount present of  Pu-239 today = A_o=x

A = ?

A=x\times e^{-2.8875\times 10^{-5} y^{-1]\times 1000 y}

A=0.9715\times x

\frac{A}{A_0}=\frac{A}{x}=0.9715

0.9715 Fraction of Pu-239 will be remain after 1000 years.

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3 years ago
How many grams are equal to 54.7 Mg?
neonofarm [45]

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Explanation:

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Answer the question below pls
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Determine the mass of each sample 1.55 kmol sulfur trioxide
xxMikexx [17]

Explanation:

Given parameters:

Number of moles of the sulfur trioxide = 1.55kmol = 1.55 x 10³mole

Unknown:

Mass of the sulfur trioxide = ?

Solution:

To solve for the mass of the sample of sulfur trioxide:

  •  Find the molar mass of the compound i.e SO₃

          atomic mass of S = 32g

                                    O = 16g

              molar mass = 32 + 3(16) = 80g/mol

  • Use the formula below:

           mass of SO₃ = number of moles x molar mass

           mass of SO₃ =  1.55 x 10³ x 80 = 124000g or 124kg

Learn more:

mole calculation  brainly.com/question/13064292

#learnwithBrainly

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3 years ago
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