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Sauron [17]
3 years ago
7

Suppose a species of snake is living in a certain area. Some of the snakes in the population are black and some are orange. Over

time, the area changes and the orange snakes have a harder time blending into the environment. This makes it difficult for the orange snakes to catch food.
According to Darwin’s theory of evolution, what will most likely happen to this snake population over time?

Most orange snakes will survive and reproduce, passing on their traits to their offspring. Few black snakes will remain in the population.
Most black snakes will survive and reproduce, passing on their traits to their offspring. The number of orange snakes in the population will not change.
The black snakes will survive and reproduce, passing on their traits to their offspring. Few orange snakes will remain in the population.
Most orange snakes will survive and reproduce, passing on their traits to their offspring. The number of black snakes in the population will not change.
Chemistry
2 answers:
dybincka [34]3 years ago
6 0

Answer:

The black snakes will survive and reproduce while the orange snakes die out.

Explanation:

The orange snakes will be easier to see by their prey and predators, and therefore will die out while the black snakes thrive because of their camouflage.

juin [17]3 years ago
3 0

Answer:

the black snakes will survive and reproduce, therefore, passing their traits to their offspring, while few orange snakes will remain in the population.  

Explanation:

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Balance this chemical reaction H2+N2–> NH3
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You are given a solid that is a mixture of Na2SO4 and K2SO4. A 0.205-g sample of the mixture is dissolved in water. An excess of
Svetach [21]

Answer:

Mass of SO₄⁻² = 0.123 g.

Mass percentage of SO₄⁻² = 41.2%

Mass of Na₂SO₄ = 0.0773 g

Mass of K₂SO₄ = 0.1277 g

Explanation:

Here we have

We place Na₂SO₄ = X and

K₂SO₄ = Y

Therefore

X +Y = 0.205 .........(1)

Therefore since the BaSO₄ is formed from BaCl₂, Na₂SO₄ and K₂SO₄ we have

Amount of BaSO₄ from Na₂SO₄ is therefore;

X\times\frac{Molar \, Mass \, of \, BaSO_4}{Molar \, Mass \, of \, Na_2SO_4}

Amount of BaSO₄ from K₂SO₄ is;

Y\times\frac{Molar \, Mass \, of \, BaSO_4}{Molar \, Mass \, of \, K_2SO_4}

Molar mass of

BaSO₄ = 233.38 g/mol

Na₂SO₄ = 142.04 g/mol

K₂SO₄ = 174.259 g/mol

X\times\frac{Molar \, Mass \, of \, BaSO_4}{Molar \, Mass \, of \, Na_2SO_4} = X\times\frac{233.38 }{142.04} = 1.643·X

Y\times\frac{Molar \, Mass \, of \, BaSO_4}{Molar \, Mass \, of \, K_2SO_4} = Y\times\frac{233.38 }{174.259 } = 1.339·Y

Therefore, we have

1.643·X + 1.339·Y = 0.298 g.....(2)

Solving equations (1) and (2) gives

The mass of SO₄⁻² in the sample is given by

Mass of sample = 0.298

Molar mass of BaSO₄ = 233.38 g/mol

Mass of Ba = 137.327 g/mol

∴ Mass of SO₄ = 233.38 g - 137.327 g = 96.05 g

Mass fraction of SO₄⁻² in BaSO₄ = 96.05 g/233.38 g = 0.412

Mass of SO₄⁻² in the sample is 0.412×0.298 = 0.123 g.

Percentage mass of SO₄⁻² = 41.2%

Solving equations (1) and (2) gives

X = 0.0773 g and Y = 0.1277 g.

8 0
3 years ago
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