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
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Answer:
9g/cm^3 is the density
Explanation:
P = m/V
P = 18/2 = 9g/cm^3
(This is more of a physics question than chem btw)
Answer:
Explanation:
Hello there!
In this case, since the equation for the calculation of dilutions is:
Whereas M is the molarity and V the volume, because the final concentration is lower than the initial. Thus, since we are asked to calculate the final volume, we solve for V2 as follows:
Best regards!
Answer:
Approximately 75%.
Explanation:
Look up the relative atomic mass of Ca on a modern periodic table:
There are one mole of Ca atoms in each mole of CaCO₃ formula unit.
- The mass of one mole of CaCO₃ is the same as the molar mass of this compound:
. - The mass of one mole of Ca atoms is (numerically) the same as the relative atomic mass of this element:
.
Calculate the mass ratio of Ca in a pure sample of CaCO₃:
.
Let the mass of the sample be 100 g. This sample of CaCO₃ contains 30% Ca by mass. In that 100 grams of this sample, there would be 
of Ca atoms. Assuming that the impurity does not contain any Ca. In other words, all these Ca atoms belong to CaCO₃. Apply the ratio 
:
.
In other words, by these assumptions, 100 grams of this sample would contain 75 grams of CaCO₃. The percentage mass of CaCO₃ in this sample would thus be equal to:
.
