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Lina20 [59]
3 years ago
6

How can chemical weathering contribute to physical weathering?

Chemistry
1 answer:
lora16 [44]3 years ago
5 0

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

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Trava [24]

Answer:

<h3>1.A                                                                                                                  2.P waves are the fastest kind of seismic wave. a longitudinal P wave has the ability to move through solid rock and fluid rock, like water or the semi-liquid layers of the earth. It pushes and pulls the rock it moves through in the same way sound waves push and pull the air.                                                                     3.The second type of body wave is the S wave or secondary wave, which is the second wave you feel in an earthquake. An S wave is slower than a P wave and can only move through solid rock, not through any liquid medium. It is this property of S waves that led seismologists to conclude that the Earth’s outer core is a liquid.                                                                        4.P Waves The first kind of body wave is the P wave or primary wave. This is the fastest kind of seismic wave, and, consequently, the first to 'arrive' at a seismic station. The P wave can move through solid rock and fluids, like water or the liquid layers of the earth. </h3><h3 />

Explanation:

4 0
2 years ago
What does an algae cell, tree,mushroom, and animal have in common
wel
Each one is a living organism.
5 0
3 years ago
3. Hydrogen reacts with nitrogen to produce ammonia according to the equation:
lys-0071 [83]

Mass of ammonia produced : 121.38 g

<h3>Further explanation</h3>

Given

Reaction

3H₂(g) + N₂(g) ⇒ 2NH₃(g)

100g of N₂

Required

Ammonia produced

Solution

mol of N₂ :

\tt mol=\dfrac{mass}{MW}\\\\mol=\dfrac{100}{28}\\\\mol=3.57

From the equation, mol ratio of N₂ and NH₃ = 1 : 2, so mol NH₃ :

\tt \dfrac{2}{1}\times 3.57=7.14~moles

mass of NH₃(MW=17 g/mol) :

\tt mass=mol\times MW\\\\mass=7.14\times 17\\\\mass=121.38~g

8 0
3 years ago
Which rule for assigning oxidation numbers IS correct?
Vilka [71]

Answer:

B.) Oxygen is usually -2

Explanation:

Hydrogen is usually +1.

A pure group 1 element is not always +1.

A monoatomic ion can be a range of numbers. However, it must be a charge other than 0.

3 0
1 year ago
What is the wavelength of radiation emitted when an electron goes from the n = 7 to the n = 4 level of the Bohr hydrogen atom? G
Phantasy [73]

Answer:

the wavelength of radiation emitted  is \mathbf{\lambda= 2169.62 \ nm}

Explanation:

The energy of the Bohr's hydrogen atom can be expressed with the formula:

\mathtt{E_n =- \dfrac{13.6\ ev}{n^2}}

For n = 7:

\mathtt{E_7 =- \dfrac{13.6\ ev}{7^2}}

\mathtt{E_7 =-0.27755 \ eV}

For n = 4

\mathtt{E_4=- \dfrac{13.6\ ev}{4^2}}

\mathtt{E_4 =- 0.85\ eV}

The  electron goes from the n = 7 to the n = 4, then :

\mathtt{E_7-E_4 = (-0.27755 - (-0.85) ) \ eV}

\mathtt{= 0.57245\ eV}

Wavelength of the radiation emitted:

\mathtt{\lambda= \dfrac{hc}{0.57245 \ eV}}

where;

hc  = 1242 eV.nm

\mathtt{\lambda= \dfrac{1242 \ eV.nm }{0.57245 \ eV}}

\mathbf{\lambda= 2169.62 \ nm}

4 0
3 years ago
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