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Vinvika [58]
3 years ago
9

Astronomers studying the planet of Rhombus have detected sedimentary rock on its surface. One astronomer wonders if material in

this sedimentary rock used to be in igneous rock deep in Rhombus's interior. Can igneous rock become sedimentary rock? Explain your answer
Chemistry
1 answer:
Fed [463]3 years ago
5 0

Answer:

igneous rock CAN become sedimentary rock through a process called ROCK CYCLE.

Explanation:

Rocks can be defined as solid structures of minerals that are formed naturally over a period of time. They are grouped into three main types which includes the following:

- igneous rock

- sedimentary rocks and

- metamorphic rocks.

Rocks are capable of transforming from one type to another through a process known as rock cycle. There are two forces that brings about this process which includes:

- The internal force : this is the Earth’s internal heat engine, which moves material around in the core and the mantle and leads to slow but significant changes within the crust.

- The external force: this is the the hydrological cycle, which is the movement of water, ice, and air at the surface, and is powered by the sun.

Molten magma cools to form either extrusive igneous rock or intrusive igneous rock. With time they undergo weathering, eroded, transported, and then deposited as sediments which are being compressed and cemented into SEDIMENTARY ROCKS. Again through the above mentioned forces, different kinds of rocks are either uplifted, to be re-eroded, or buried deeper within the crust where they are heated up, squeezed, and changed into METAMORPHIC ROCK.

Therefore the material in this sedimentary rock found in Rhombus planet used to be in igneous rock deep in Rhombus's interior due to continuous rock cycling on the planet. I hope this helps, thanks.

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A(g) + 2B(g) → C(g) + D(g)
WITCHER [35]

Answer:

0.169

Explanation:

Let's consider the following reaction.

A(g) + 2B(g) ⇄ C(g) + D(g)

We can find the pressures at equilibrium using an ICE chart.

       A(g) + 2 B(g) ⇄ C(g) + D(g)

I       1.00     1.00        0        0

C       -x        -2x        +x       +x

E    1.00-x  1.00-2x     x         x

The pressure at equilibrium of C is 0.211 atm, so x = 0.211.

The pressures at equilibrium are:

pA = 1.00-x = 1.00-0.211 = 0.789 atm

pB = 1.00-2x = 1.00-2(0.211) = 0.578 atm

pC = x = 0.211 atm

pD = x = 0.211 atm

The pressure equilibrium constant (Kp) is:

Kp = pC × pD / pA × pB²

Kp = 0.211 × 0.211 / 0.789 × 0.578²

Kp = 0.169

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