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3 years ago

Soil comes in different colors.

Chemistry

1 answer:

ollegr [7]3 years ago

7 0

Answer: yes they come in different colors.

Explanation:

Most shades of soil is mostly black,brown,red,gray,and white the color of soil and other properties including texture, structure, and consistency are used to distinguish and identify soil.

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10 points Which of these is the best example of heat transfer by radiation? *

insens350 [35]

Answer:

a) A satellite is warmed by sunlight.

Explanation:

Heat transfer by radiation mostly involves heat gain or heat loss from the Sun. In this case, Option A is the only option where sunlight is involved so it is the best example of heat transfer by radiation.

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3 years ago

The statement 2 2 = 4 would be considered a _____ in the science world

mars1129 [50]

The statement would be considered a law.

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3 years ago

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The rainforest contains what percentage of all of the species in the world?

Anuta_ua [19.1K]

Answer:

I think 50%

Explanation:

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3 years ago

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Curiosity is important in the progress of science because it

Simora [160]

Answer:

Explanation:

because it makes you want to know what happens in an experiment

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4 years ago

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A solution is prepared by dissolving 0.26 mol of hydrazoic acid and 0.26 mol of sodium azide in water sufficient to yield 1.00 L

Pavel [41]

Answer:

The pH does not decrease drastically because the HCl reacts with the <u>sodium azide (NaN₃)</u> present in the buffer solution.

Explanation:

The buffer solution is formed by 0.26 moles of the weak acid, hydrazoic acid (HN₃), and by 0.26 moles of sodium azide (NaN₃). The equilibrium reaction of this buffer solution is the following:

HN₃(aq) + H₂O(l) ⇄ N₃⁻(aq) + H₃O⁺(aq)

The pH of this solution is:

$pH = pka + log(\frac{[N_{3}^{-}]}{[HN_{3}]}) = -log(2.5 \cdot 10^{-5}) + log(\frac{0.26 mol/1 L}{0.26 mol/1 L}) = 4.60$

When 0.05 moles of HCl is added to the buffer solution, the following reaction takes place:

H₃O⁺(aq) + N₃⁻(aq) ⇄ HN₃(aq) + H₂O(l)

The number of moles of NaN₃ after the reaction with HCl is:

$\eta_{NaN_{3}} = \eta_{i} - \eta_{HCl} = 0.26 moles - 0.05 moles = 0.21 moles$

Now, the number of moles of HN₃ is:

$\eta_{HN_{3}} = \eta_{i} + \eta_{HCl} = 0.26 moles + 0.05 moles = 0.31 moles$

Then, the pH of the buffer solution after the addition of HCl is:

$pH = pka + log(\frac{[N_{3}^{-}]}{[HN_{3}]}) = -log(2.5 \cdot 10^{-5}) + log(\frac{0.21 mol/V_{T}}{0.31 mol/V_{T}}) = 4.43$

The pH of the buffer solution does not decrease drastically, it is 4.60 before the addition of HCl and 4.43 after the addition of HCl.

Therefore, the pH does not decrease drastically because the HCl reacts with the sodium azide (NaN₃) present in the buffer solution.

I hope it helps you!

6 0

3 years ago