All categories

2 years ago

Using the drop-down menus, label the parts of the rock cycle. A~ B~ C~ D~

Chemistry

2 answers:

enyata [817]2 years ago

7 0

Answer:

A- Sediment

B-Metamorphic Rocks

C-Sedimentary

D-Erosion

Explanation:

Keith_Richards [23]2 years ago

5 0

Answer:

A is Sediments

B is metamorphic rock

C is sedimentary rocks

D is weathering

Explanation:

A represent smaller particles of rocks known as sediments. From the picture above sedimentary rocks, igneous rocks and metamorphic rocks were broken into smaller constituent due to events like weathering(D). Weathering disintegrated these rocks into smaller particles (sediments).

B is rock known as metamorphic rocks. Metamorphic rocks are formed from igneous, sedimentary or even metamorphic rocks. The major factor behind metamorphic rocks formation is temperature and pressure. Generally, the parents rocks never melts, the transformation usually occur in a solid state. Metamorphism is a process that is usually accompanied with change in mineral assemblage , physical attribute of the rocks or both. From the picture above the sedimentary rocks(C) was converted to metamorphic rocks due to temperature and pressure . Same occurrence was present in the conversion of igneous rocks to metamorphic rocks(B)

C is a sedimentary rocks. Sedimentary rocks are formed from smaller particles of the weathered rocks. The smaller particles are known as sediments. This sediments are then cemented and compacted by cementing agents to form sedimentary rocks.

D is a process, which is known as weathering. Weathering is the disintegration of rocks into smaller particles to form sediments . Weathering may involve change in the chemical composition of rocks or just a physical change . From the picture above igneous rocks, metamorphic rock and sedimentary rocks were weathered to form sediments(A).

You might be interested in

Explain whether or not you expect the chaparral biome to be sensitive to the loss of a single species.

aleksley [76]

Answer:

See the answer below

Explanation:

The chaparral biome is a temperate biome with a characteristic high temperature and dryness during summer and mild rainy winters and springs. The biome can be found in relatively small amounts in the major continents of the world with its rich plant and animal diversity who have successfully adapted to the conditions of the biome.

Due to the high biodiversity of the chaparral biome, <u>one would expect it to be resilient to the loss of a single species.</u> <em>The more the biodiversity of a biome or community, the more resilient such biome or community would be to the loss of species and lower the biodiversity, the more sensitive the community would be to the loss of species. </em>

3 0

2 years ago

The standard cell potential Ec for the reduction of silver ions with elemental copper is 0.46V at 25 degrees celsius. calculate

Cloud [144]

Answer : The $\Delta G$ for this reaction is, -88780 J/mole.

Solution :

The balanced cell reaction will be,

$Cu(s)+2Ag^+(aq)\rightarrow Cu^{2+}(aq)+2Ag(s)$

Here, magnesium (Cu) undergoes oxidation by loss of electrons, thus act as anode. silver (Ag) undergoes reduction by gain of electrons and thus act as cathode.

The half oxidation-reduction reaction will be :

Oxidation : $Cu\rightarrow Cu^{2+}+2e^-$

Reduction : $2Ag^++2e^-\rightarrow 2Ag$

Now we have to calculate the Gibbs free energy.

Formula used :

$\Delta G^o=-nFE^o$

where,

$\Delta G^o$ = Gibbs free energy = ?

n = number of electrons to balance the reaction = 2

F = Faraday constant = 96500 C/mole

$E^o$ = standard e.m.f of cell = 0.46 V

Now put all the given values in this formula, we get the Gibbs free energy.

$\Delta G^o=-(2\times 96500\times 0.46)=-88780J/mole$

Therefore, the $\Delta G$ for this reaction is, -88780 J/mole.

7 0

3 years ago

What is the reason you can smell food cooking even when you're not in

Dmitry_Shevchenko [17]

Answer:

the answer should be A

5 0

2 years ago

What is the volume of 0.640 grams of Oz gas at Standard Temperature and Pressure (STP)?

Ilia_Sergeevich [38]

Answer: The volume of 0.640 grams of $O_{2}$ gas at Standard Temperature and Pressure (STP) is 0.449 L.

Explanation:

Given: Mass of $O_{2}$ gas = 0.640 g

Pressure = 1.0 atm

Temperature = 273 K

As number of moles is the mass of substance divided by its molar mass.

So, moles of $O_{2}$ (molar mass = 32.0 g/mol) is as follows.

$No. of moles = \frac{mass}{molar mass}\\= \frac{0.640 g}{32.0 g/mol}\\= 0.02 mol$

Now, ideal gas equation is used to calculate the volume as follows.

PV = nRT

where,

P = pressure

V = volume

n = no. of moles

R = gas constant = 0.0821 L atm/mol K

T = temperature

Substitute the values into above formula as follows.

$PV = nRT\\1.0 atm \times V = 0.02 mol \times 0.0821 L atm/mol K \times 273 K\\V = 0.449 L$

Thus, we can conclude that the volume of 0.640 grams of $O_{2}$ gas at Standard Temperature and Pressure (STP) is 0.449 L.

6 0

3 years ago

The standard free-energy changes for the reactions below are given. Phosphocreatine → creatine + Pi ∆ G'° = –43.0 kJ/mol ATP → A

Jet001 [13]

Answer:

-12.5 kJ/mol

Explanation:

The free-energy predicts if a reaction is spontaneous or not. If it is, ΔG < 0. When a reaction happens by steps, the free-energy of the global reaction can be calculated by the sum of the free-energy of the steps (Hess law). If it's needed to operations at the reaction the same operation must be done in the value of ΔG (if the reaction is inverted, the signal of ΔG must be inverted).

Phosphocreatine → creatine + Pi ∆G'° = –43.0 kJ/mol

ATP → ADP + Pi ∆G'° = –30.5 kJ/mol (x-1)

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

Phosphocreatine → creatine + Pi ∆G'° = –43.0 kJ/mol

Pi + ADP → ATP ∆G'° = 30.5 kJ/mol

The bold compounds are in opposite sides, so they'll be canceled in the sum of the reactions:

Phosphocreatine + ADP → creatine + ATP

∆G'° = -43.0 + 30.5

∆G'° = -12.5 kJ/mol

5 0

2 years ago