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Margaret [11]
3 years ago
5

Mineral matter and organic matter together make up about 50 percent of soil. What two substances make up the other 50 percent?

Biology
1 answer:
ale4655 [162]3 years ago
4 0

The particles that make up soil are categorized into three groups by size: sand, silt, and clay. Sand particles are the largest and clay particles the smallest. Although a soil could be all sand, all clay, or all silt, that's rare. Instead most soils are a combination of the three.

The relative percentages of sand, silt, and clay are what give soil its texture. A loamy texture soil, for example, has nearly equal parts of sand, silt, and clay.

SOIL HAS STRUCTURE

Soil structure is the arrangement of soil particles into small clumps, called "peds". Much like the ingredients in cake batter bind together to form a cake, soil particles (sand, silt, clay, and organic matter) bind together to form peds. Peds have various shapes depending on their “ingredients” and the conditions under which the peds formed: getting wet and drying out, freezing and thawing--even people walking on or farming the soil affects the shapes of peds.

Ped shapes roughly resemble balls, blocks, columns, and plates. Between the peds are spaces, or pores, in which air, water, and organisms move. The sizes of the pores and their shapes vary from soil structure to soil structure.

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A) Nucleus and Ribosome

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What type of properties change in a physical change?Give an example to support the answer?​
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3 years ago
Gold has a specific heat of 0.129 J/(g×°C). How many joules of heat energy are required to raise the temperature of 15 grams of
Elodia [21]

Answer:

\boxed {\boxed {\sf 121.905 \ J  }}

Explanation:

We are asked to find the energy given mass, specific heat, and change in temperature. Therefore, we must use this formula;

q= mc \Delta T

The mass is 15 grams and the specific heat is 0.129 J/(g×°C). Let's calculate the change in temperature.

  • ΔT= final temperature - initial temperature
  • ΔT= 85 °C- 22°C = 63°C

Now we know all the values:

m= 15 \ g \ \\c= 0.129 \ J / (g* \textdegree C) \\\Delta T= 63 \ \textdegree C

Substitue the values into the formula.

q= (15 \ g)( 0.129 \ J / (g* \textdegree C)) ( 63 \ \textdegree C)

Multiply the first numbers together. The grams will cancel.

q= (1.935 \ J/ \textdegree C) ( 63 \ \textdegree C)

Multiply again, this time the degrees Celsius cancels.

q= 121.905 \ J

<u>121.905 Joules</u> are required.

5 0
2 years ago
Read 2 more answers
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