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Airida [17]
3 years ago
5

3 A soil has 70% sand, 20% silt

Chemistry
1 answer:
Kisachek [45]3 years ago
7 0

Today, as part of the series of posts on soils, we are going to look at ‘soil texture’. Soil forms the basis for all life but it’s important to know about its mineral constitution as well as its biological profile.

Texture refers to the ‘feel’ of the soil. This is affected by the constituent materials found within it, specifically sand, silt and clay particles. A coarse sand will feel gritty but a wet clay will feel heavy and sticky. The texture of a soil has a direct impact on the way the soil reacts to certain environmental conditions – for example, towards drought or heavy rain (with sandy soils more freely draining).

There is a big difference in the size of the different particles.

Coarse sand = diameter 2-0.2mm

Fine sand = diameter 0.2-0.02mm

Silt = diameter 0.02-0.002mm

Clay = diameter less than 0.002mm

Note how the clay particles are much smaller than the sand particles – this is important as it means the total surface area of a clay soil is much greater and so the capacity to hold water is also much greater.

Between the sand, silt and clay particles there are lots of pores. In fact a soil as a whole is generally 45% mineral, 5% organic matter (depending on the soil) and 50% pore space through which air and water can pass.

Sand –

Made up of weathered primary rock minerals.

The particles are irregular in outline.

They are large and so do not pack together easily.

Large pore spaces in between.

Air gets in very easily and water flows rapidly through it.

Silt –

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Scrat [10]

Answer:

                      252.68 K  or   -20.46 °C

Explanation:

                    According to Gay-Lussac's Law, "Pressure and Temperature at given volume are directly proportional to each other".

Mathematically,

                                              P₁ / T₁  =  P₂ / T₂   ---- (1)

Data Given:

                  P₁  =  30.7 kPa

                  T₁  =  0.00 °C  =  273.15 K

                  P₂  =  28.4 kPa

                  T₂  =  <u>???</u>

Solving equation for T₂,

                  T₂  =  P₂ T₁ / P₁

Putting values,

                  T₂  =  28.4 kPa × 273.15 K / 30.7 kPa

                  T₂  =  252.68 K  or   -20.46 °C

8 0
3 years ago
Matter can be a compound and a _______.
ratelena [41]

Answer:

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Matter can be classified as a compound and a mixture.

4 0
2 years ago
C12H22O11 + 11 H2SO4 → 12 C + 11 H2SO4 + 11 H2O
BlackZzzverrR [31]
Answer:
            T<span>he gaseous product of this reaction is water (Option-A).

Explanation:
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5 0
3 years ago
(a) Calculate the total volume (in liters) of air an adult breathes in a day. (b) In a city with heavy traffic, the air contains
Furkat [3]

Answer:

a) V air/day = 8640 L air  an adult breaths / day

b) 0.0181 L CO intake a person / day

Explanation:

a) one average person has 12 breaths for min:

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in each breath it take an average  of 500 mL on air.

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⇒ 12 breath / min * 500mL air / breath = 6000 mL air / min

the average air volume per day of a person is:

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4 0
2 years ago
Consider the reaction: P(s) + 5/2 Cl2(g)PCl5(g) Write the equilibrium constant for this reaction in terms of the equilibrium con
Pani-rosa [81]

Answer: The equilibrium constant for the overall reaction is K_a\times K_b

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Equilibrium constant is defined as the ratio of concentration of products to the concentration of reactants each raised to the power their stoichiometric ratios.

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K_a=\frac{[PCl_3]}{[Cl_2]^{\frac{3}{2}}}

b) PCl_3(g)+Cl_2(g)\rightarrow PCl_5(g)

K_b=\frac{[PCl_5]}{[Cl_2]\times [PCl_3]}

For overall reaction on adding a and b we get c

c) P(s)+\frac{5}{2}Cl_2(g)\rightarrow PCl_5(g)

K_c=\frac{[PCl_5]}{[Cl_2]^\frac{5}{2}}

K_c=K_a\times K_b=\frac{[PCl_3]}{[Cl_2]^{\frac{3}{2}}}\times \frac{[PCl_5]}{[Cl_2]\times [PCl_3]}

The equilibrium constant for the overall reaction is K_a\times K_b

4 0
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