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2 years ago

Which is an appropriate layering process for a compost pile?

Chemistry

2 answers:

jarptica [38.1K]2 years ago

6 0

Answer: O a layer of grass clippings, a layer of fruit scraps, some soil, and some water

Explanation:

The compost pile is made up of decomposing organic materials that are obtained from dead and decaying vegetable or plant matter, animal materials like dung, dead skins and others. The decomposition is done by the soil microbes that recycle the matter into nutrients that are absorbed by the plants. All the components of the compost pile are degradable and completely converted into organic matter. The grass clipping and fruit scraps mixed with soil and water can be easily decomposed by the soil microbes easily.

seropon [69]2 years ago

3 0

Answer: a layer of paper napkins, a layer of grass clippings, some soil, and some water

Explanation: Grass clippings and fruit scraps are both greens so this is the correct answer.

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In order to determine if 2 atoms are cooper what must be the same for each

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-number of electrons
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2 years ago

Determine the basic oxidation number for elements in s and p orbitals

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Answer:

Explanation:

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The term oxidation was first used by Antoine Lavoisier to signify reaction of a substance with oxygen. Much later, it was realized that the substance, upon being oxidized, loses electrons, and the meaning was extended to include other reactions in which electrons are lost, regardless of whether oxygen was involved.

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2 years ago

For two variables in a direct proportion, what is the result of doubling one variable?

Airida [17]

Answer:

the other variable is also doubled

Explanation:

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3 years ago

Chemistry question! Please help!

Illusion [34]

When a solid (solute) comes in contact with the liquid (solvent), the solute goes about C) dissolution, in which the solid dissolves into the liquid.

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3 years ago

Read 2 more answers

Three 1.0-l flasks, maintained at 308 k, are connected to each other with stopcocks. initially the stopcocks are closed. one of

Licemer1 [7]

Answer:

0.6103 atm.

Explanation:

We need to calculate the vapor pressure of each component after the stopcocks are opened.

Volume after the stopcocks are opened = 3.0 L.

1) For N₂:

P₁V₁ = P₂V₂

P₁ = 1.5 atm & V₁ = 1.0 L & V₂ = 3.0 L.

P₂ of N₂ = P₁V₁ / V₂ = (1.5 atm) (1.0 L) / (3.0 L) = 0.5 atm.

2) For H₂O:

Pressure of water at 308 K is 42.0 mmHg.

we need to convert from mmHg to atm: (1.0 atm = 760.0 mmHg).

P of H₂O = (1.0 atm x 42.0 mmHg) / (760.0 mmHg) = 0.0553 atm.

We must check if more 2.2 g of water is evaporated,

n = PV/RT = (0.0553 atm) (3.0 L) / (0.082 L.atm/mol.K) (308 K) = 0.00656 mole.

m = n x cmolar mass = (0.00656 mole) (18.0 g/mole) = 0.118 g.

It is lower than the mass of water in the flask (2.2 g).

3) For C₂H₅OH:

Pressure of C₂H₅OH at 308 K is 102.0 mmHg.

we need to convert from mmHg to atm: (1.0 atm = 760.0 mmHg).

P of C₂H₅OH = (1.0 atm x 102.0 mmHg) / (760.0 mmHg) = 0.13421 atm.

We must check if more 0.3 g of C₂H₅OH is evaporated,

n = PV/RT = (0.13421 atm) (3.0 L) / (0.082 L.atm/mol.K) (308 K) = 0.01594 mole.

m = n x molar mass = (0.01594 mole) (46.07 g/mole) = 0.7344 g.

It is more than the amount in the flask (0.3 g), so the pressure should be less than 0.13421 atm.

We have n = mass / molar mass = (0.30 g) / (46.07 g/mole) = 0.00651 mole.

So, P of C₂H₅OH = nRT / V = (0.00651 mole) (0.082 L.atm/mole.K) (308.0 K) / (3.0 L) = 0.055 atm.

So, total pressure = P of N₂ + P of H₂O + P of C₂H₅OH = 0.5 atm + 0.0553 atm + 0.055 atm = 0.6103 atm.

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2 years ago