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.
The receive very little rainfall
Eta Carinae could be as large as 180 times the radius of the Sun, and its surface temperature is 36,000-40,000 Kelvin. Just for comparison, 40,000 Kelvin is about 72,000 degrees F. So it's the blue hypergiants, like Eta Carinae, which are probably the hottest stars in the Universe.
First, we need to get the value of Ka:
when Ka = Kw / Kb
we have Kb = 1.8 x 10^-5
and Kw = 3.99 x 10^-16 so, by substitution:
Ka = (3.99 x 10^-16) / (1.8 x 10^-5) = 2.2 x 10^-11
by using the ICE table :
NH4+ + H2O →NH3 + H+
intial 0.013 0 0
change -X +X +X
Equ (0.013-X) X X
when Ka = [NH3][H+] / [NH4+]
by substitution:
2.2 x 10^-11 = X^2 / (0.013 - X) by solving this equation for X
∴X = 5.35 x 10^-7
∴[H+] = X = 5.35 x 10^-7
∴PH = - ㏒[H+]
= -㏒(5.35 x 10^-7)
= 6.27
