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Lemur [1.5K]
3 years ago
12

Which term names the underground region between where water saturates pores and where it does not? A. water zone B. water table

C. aqua zone D. porous zone
Chemistry
2 answers:
Amiraneli [1.4K]3 years ago
5 0

Water table is the underground region between where water saturates pores and where it does not.

Answer: B

Explanation

It is known that water is present below the ground. The rocky terrain inside the ground contains water and the water will be filled in the pores between the rocks.

The water filled in the saturated region below the ground is separated from the unsaturated region which is the region above the ground by a surface termed as water table.

The unsaturated region is the soil above the ground and the plants get water from this region.

The water is present in unsaturated region occupy between the pores of rocks and soils but they are loosely filled so that plants can extract water from them.

Then below the ground water table region will be present and below the water table the saturated water region is present.

So water table is the region dividing the saturated and unsaturated water pore regions.

Pani-rosa [81]3 years ago
3 0
A)Water Table :)


Enjoy
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Given the following data:N2(g) + O2(g)→ 2NO(g), ΔH=+180.7kJ2NO(g) + O2(g)→ 2NO2(g), ΔH=−113.1kJ2N2O(g) → 2N2(g) + O2(g), ΔH=−163
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The Hess' Law states that the enthalpy of the overall reaction is the sum of the enthalpy of the step reactions. To do the addition of the reaction, we first must reorganize them, to disappear with the intermediaries (substances that are not presented in the overall reaction).

If the reaction is inverted, the signal of the enthalpy changes, and if its multiplied by a constant, the enthalpy must be multiplied by the same constant. Thus:

N₂(g) + O₂(g) → 2NO(g) ΔH = +180.7 kJ

2NO(g) + O₂(g) → 2NO₂(g) ΔH = -113.1 kJ

2N₂O(g) → 2N₂(g) + O₂(g) ΔH = -163.2 kJ

The intermediares are N₂ and O₂, thus, reorganizing the reactions:

N₂(g) + O₂(g) → 2NO(g) ΔH = +180.7 kJ

NO₂(g) → NO(g) + (1/2)O₂(g) ΔH = +56.55 kJ (inverted and multiplied by 1/2)

N₂O(g) → N₂(g) + (1/2)O₂(g) ΔH = -81.6 kJ (multiplied by 1/2)

------------------------------------------------------------------------------------

N₂O(g) + NO₂(g) → 3NO(g)

ΔH = +180.7 + 56.55 - 81.6

ΔH = +155.6 kJ

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