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AfilCa [17]
4 years ago
6

In which biome would you most likely find plants that grow far apart from one another, with narrow leaves and shallow roots? Why

do these plants have these adaptions?
Please help!?
Chemistry
2 answers:
mr Goodwill [35]4 years ago
8 0

Answer: Answer. A plant with narrow leaves and shallow roots would probably be found in the desert, where there is not a lot of water.

Explanation:

alexdok [17]4 years ago
8 0

Answer:

A plant with narrow leaves and shallow roots would probably be found in the desert, where there is not a lot of water. Narrow leaves help prevent water loss.

Explanation:

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The___ of an element is the number of
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Answer:atomic mass, neutrons in the nucleus

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What occurs when a solid substance dissolves into a liquid? 1 It stays solid and sinks to the bottom of the liquid substance. 2
katrin [286]

Explanation:

When a solid substance is dissolved in a liquid such as water, the intermolecular forces of attraction between the molecules in the solid substance are overcome and they exist as ions in the solution. Hence the answer is 2.

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Match each term to its description. (3 points)
madreJ [45]

Answer:   Limiting reactant = 3  

                Theoretical Yield= 1

                  Excess reactant=2

Explanation:  The theoretical yield is the maximum possible mass of a product that can be made in a chemical reaction. It can be calculated from: the balanced chemical equation. the mass and relative formula mass of the limiting reactant , and. the relative formula mass of the product.

An excess reactant is a reactant present in an amount in excess of that required to combine with all of the limiting reactant. It follows that an excess reactant is one remaining in the reaction mixture once all the limiting reactant is consumed.

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3 0
3 years ago
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The buffer solution is used to control the pH to insure that it does not become too high because excessively basic solutions cou
oksian1 [2.3K]

Answer:

pH = 12.7

Explanation:

First, we have to calculate the [Ca²⁺] in a solution of about 250 ppm CaCO₃.

\frac{250mgCaCO_{3}}{L} .\frac{1gCaCO_{3}}{1000mgCaCO_{3}} .\frac{1molCa^{2+} }{100gCaCO_{3}} =2.5 \times 10^{-3} M

Now, let's consider the dissolution of Ca(OH)₂ in water.

Ca(OH)₂(s) ⇄ Ca²⁺(aq) + 2 OH⁻(aq)

The solubility product Ksp is:

Ksp = [Ca²⁺] × [OH⁻]²

[OH⁻] = √(Ksp/[Ca²⁺]) = √(6.5 × 10⁻⁶/2.5 × 10⁻³) = 5.1 × 10⁻² M

Finally, we can calculate pOH and pH.

pOH = -log [OH⁻] = -log (5.1 × 10⁻²) = 1.3

pH + pOH = 14 ⇒ pH = 14 - pOH = 14 - 1.3 = 12.7

5 0
3 years ago
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