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tresset_1 [31]
3 years ago
14

Animals adapted for lack of water and extremes in temperature are found in which biome?

Chemistry
2 answers:
nika2105 [10]3 years ago
8 0
The desert because there is a huge lack of water and very extreme temperatures of heat.
Arturiano [62]3 years ago
7 0

Answer:

Tundra.

Explanation:

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The green areas on the diagram show the world's rainforests. What can you conclude about the location of rainforests?
Ber [7]

Answer:

Option A

They are located near the equator

Explanation:

From the diagram, we can observe that all the rain forest zones occur near the equator.

This is because on an annual basis, the equator receives the highest concentration of solar radiation in the whole Earth. This implies that the rate of evaporation of water from the water bodies present  is high, and consequently, the amount of rainfall is high also. This abundant sunshine and high rainfall  leads to the growth of the tropical rain forests at those regions

6 0
3 years ago
A teacher makes the following statement.
melamori03 [73]
The correct answer among the choices given is the first option.The teacher most likely is talking about distillation of a mixture. Distillation is a unit operation that separates component substances from a liquid mixture which is shown by the teacher. Also, the most common purifying technique in the production of gasoline is by this process.
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3 years ago
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What must be the molarity of an aqueous solution of trimethylamine, (ch3)3n, if it has a ph = 11.20? (ch3)3n+h2o⇌(ch3)3nh++oh−kb
Stolb23 [73]

0.040 mol / dm³. (2 sig. fig.)

<h3>Explanation</h3>

(\text{CH}_3)_3\text{N} in this question acts as a weak base. As seen in the equation in the question, (\text{CH}_3)_3\text{N} produces \text{OH}^{-} rather than \text{H}^{+} when it dissolves in water. The concentration of \text{OH}^{-} will likely be more useful than that of \text{H}^{+} for the calculations here.

Finding the value of [\text{OH}^{-}] from pH:

Assume that \text{pK}_w = 14,

\begin{array}{ll}\text{pOH} = \text{pK}_w - \text{pH} \\ \phantom{\text{pOH}} = 14 - 11.20 &\text{True only under room temperature where }\text{pK}_w = 14 \\\phantom{\text{pOH}}= 2.80\end{array}.

[\text{OH}^{-}] =10^{-\text{pOH}} =10^{-2.80} = 1.59\;\text{mol}\cdot\text{dm}^{-3}.

Solve for [(\text{CH}_3)_3\text{N}]_\text{initial}:

\dfrac{[\text{OH}^{-}]_\text{equilibrium}\cdot[(\text{CH}_3)_3\text{NH}^{+}]_\text{equilibrium}}{[(\text{CH}_3)_3\text{N}]_\text{equilibrium}} = \text{K}_b = 1.58\times 10^{-3}

Note that water isn't part of this expression.

The value of Kb is quite small. The change in (\text{CH}_3)_3\text{N} is nearly negligible once it dissolves. In other words,

[(\text{CH}_3)_3\text{N}]_\text{initial} = [(\text{CH}_3)_3\text{N}]_\text{final}.

Also, for each mole of \text{OH}^{-} produced, one mole of (\text{CH}_3)_3\text{NH}^{+} was also produced. The solution started with a small amount of either species. As a result,

[(\text{CH}_3)_3\text{NH}^{+}] = [\text{OH}^{-}] = 10^{-2.80} = 1.58\times 10^{-3}\;\text{mol}\cdot\text{dm}^{-3}.

\dfrac{[\text{OH}^{-}]_\text{equilibrium}\cdot[(\text{CH}_3)_3\text{NH}^{+}]_\text{equilibrium}}{[(\text{CH}_3)_3\text{N}]_\textbf{initial}} = \text{K}_b = 1.58\times 10^{-3},

[(\text{CH}_3)_3\text{N}]_\textbf{initial} =\dfrac{[\text{OH}^{-}]_\text{equilibrium}\cdot[(\text{CH}_3)_3\text{NH}^{+}]_\text{equilibrium}}{\text{K}_b},

[(\text{CH}_3)_3\text{N}]_\text{initial} =\dfrac{(1.58\times10^{-3})^{2}}{6.3\times10^{-5}} = 0.040\;\text{mol}\cdot\text{dm}^{-3}.

8 0
3 years ago
What are four ways to measure matter?
Cerrena [4.2K]

Answer:

C

The answer is C pleaaaaaaase

6 0
3 years ago
A pressure of 125,400 Pa is equal to<br> kPa.
dalvyx [7]

Answer:

it is equal to 125 kPa

Explanation:

move the decimal 3 to the left

3 0
3 years ago
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