Hope this helps :) I didn’t know how to write subscripts so I wrote it down on some paper.
A-leads to the abrasion of rocks and minerals
A-dense vegetation cover
True
Explanation:
Weathering is the physical disintegration and chemical decomposition of rocks to form sediments and soils.
Agent of weathering are wind, water and glacier.
Chemical weathering contributes to physical weathering in that it leads to the abrasion of rocks and minerals.
During chemical weathering, a rock chemically combines with materials in the environment and weakens it.
When physical weathering processes are induced, grains produced independently weakening of bonds in rocks grind against one another and wears each other off.
An area with a dense vegetation cover undergoes rapid chemical weathering:
- Plant roots penetrates deep into the rock and increases the surface area of chemical action.
- Plants produce chemicals that combines with rocks and causes them to decay.
- Since the area is always moist, chemical action becomes more severe.
Buildings and statues made of stone are subjected to the same degree of weathering as rocks exposed naturally.
This is true.
Statues and buildings weather just like rocks we find in nature.
It is the same sunshine and rain that impacts rocks that also impacts buildings and statues.
So they degrade at the same rate except they are protected.
learn more:
Erosion brainly.com/question/2473244
#learnwithBrainly
Board up your windows
Go to higher ground
Get food and water
Answer:
the concentration of bicarbonate is <em>[HCO₃⁻] = 0,03996 M </em>and carbonate is <em>[CO₃²⁻] = 3,56x10⁻⁵ M.</em>
Explanation:
Carbonate-bicarbonate is:
HCO₃⁻ ⇄ CO₃²⁻ + H⁺ With pka = 10,25
Using Henderson-Hasselbalach formula:
pH = pka + log₁₀![\frac{[CO_{3}^{2-}]}{[HCO_{3}^-]}](https://tex.z-dn.net/?f=%5Cfrac%7B%5BCO_%7B3%7D%5E%7B2-%7D%5D%7D%7B%5BHCO_%7B3%7D%5E-%5D%7D)
7,2 = 10,25 + log₁₀
8,91x10⁻⁴ = <em>(1)</em>
Also:
0,040 M = [CO₃²⁻] + [HCO₃⁻] <em>(2)</em>
Replacing (2) in 1:
<em>[HCO₃⁻] = 0,03996 M</em>
Thus:
<em>[CO₃²⁻] = 3,56x10⁻⁵ M</em>
I hope it helps.
