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2 years ago

What effect do tube dwelling worms have on mudflat ecosystems?

1 answer:

8 0

Earthworms influence (and benefit) the soil ecosystem in a number of ways: Recycling organic material: Earthworms, along with bacteria and fungi, decompose organic material. ... Improving soil structure: Earthworm burrows alter the physical structure of the soil. They open up small spaces, known as pores, within the soil.

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During an endothermic phase change, what happens to potential energy and the kinetic energy

polet [3.4K]

During this phase Change heat energy is being absorbed by the molecules, and as a result the molecules possess a greater ability to move around and possess higher kinetic energy because of this. The molecules also possess a higher potential energy.

7 0

2 years ago

Read 2 more answers

Two substances in a mixture differ in density and particle size. These properties can be used to

lianna [129]

Answer:

(1) separate the substances

(2) chemically combine the substances

(3) determine the freezing point of the mixture

(4) predict the electrical conductivity of the mixture

Explanation:

3 0

2 years ago

A compound that is composed of carbon, hydrogen, and oxygen contains 70.6% C, 5.9% H, and 23.5% O by mass. The molecular weight

zhannawk [14.2K]

Answer: The molecular formula will be $C_8H_8O_2$

Explanation:

If percentage are given then we are taking total mass is 100 grams.

So, the mass of each element is equal to the percentage given.

Mass of C= 70.6 g

Mass of H = 5.9 g

Mass of O = 23.5 g

Step 1 : convert given masses into moles.

Moles of C = $\frac{\text{ given mass of C}}{\text{ molar mass of C}}= \frac{70.6g}{12g/mole}=5.9moles$

Moles of H = $\frac{\text{ given mass of H}}{\text{ molar mass of H}}= \frac{5.9g}{1g/mole}=5.9moles$

Moles of O = $\frac{\text{ given mass of O}}{\text{ molar mass of O}}= \frac{23.5g}{16g/mole}=1.5moles$

Step 2 : For the mole ratio, divide each value of moles by the smallest number of moles calculated.

For C = $\frac{5.9}{1.5}=4$

For H = $\frac{5.9}{1.5}=4$

For O = $\frac{1.5}{1.5}=1$

The ratio of C : H: O= 4: 4:1

Hence the empirical formula is $C_4H_4O$

The empirical weight of $C_4H_4O$ = 4(12)+4(1)+1(16)= 68g.

The molecular weight = 136 g/mole

Now we have to calculate the molecular formula.

$n=\frac{\text{Molecular weight }}{\text{Equivalent weight}}=\frac{136}{68}=2$

The molecular formula will be= $2\times C_4H_4O=C_8H_8O_2$

4 0

2 years ago

If the 3.90 m solution from Part A boils at 103.45 ∘C, what is the actual value of the van't Hoff factor, i? The boiling point o

erma4kov [3.2K]

<span>Some of the solutions exhibit colligative properties. These properties depend on the amount of solute dissolved in a solvent. For boiling point elevation, we calculate the increase in temperature by the equation:

</span><span>ΔT(boiling point) = (Kb)mi

where Kb is a constant, m is the molality of the solution, i is the van't Hoff factor.

From the given data, we can easily calculate for i as follows:

</span>ΔT(boiling point) = (Kb)mi
103.45 - 100 = (0.512)3.90i
i = 1.73 <-------van't Hoff factor

7 0

2 years ago

An excess of NiCO3, 12 g, was added to 40cm cube of sulpheric acid, 2.0 mol/dm cube. The unreacted nickel carbonate was filtered

Zepler [3.9K]

Calculate the number of moles reacted - 0.08 moles
Calculate the mass of nickel carbonate reacted - 0.08 * 119g
Calculate the mass of unreacted nickel carbonate - 12.0 - (0.08 * 119)g

1 mole of NiCO3 reacts with 1 mole of H2SO4

3 0

2 years ago