In addition to sample identification, type your answers to the following questio

Question

In addition to sample identification, type your answers to the following questions in complete sentences. 1. Olivine: Olivine is the most abundant mineral in the mantle, but its atomic structure changes with increasing pressure (depth). Use the diagram below to answer the following questions a. Mg Sio, TTTITII 2000 Temperature C i. At 1000°C, at what pressure (in GPa) does olivine transform to wadsleyite? I aPa spinel? Pressure increases within the Earth at a rate of approximately 1 GPa per 35 km. At what depth does each transformation occur? What layer of the Earth do these transitions represent? ii. ili. Hawaii is one of only a few locations in the world where green sand beaches exist. Research Papakolea Beach on the Big Island of Hawaii and answer the following questions: b. i. Why is the sand green? beans m. saute dupos4 .oli rans ii. How do green sand beaches form? has iron ha», daniaytarrr~ha ill. Why are these beaches so rare? pecause as vo eath clmunt

Explanation / Answer

1.Olivine

a. i. Olivine transforms to Wadseliyite at 14.1 to 20 GPa

ii. The transformation occurs at depth of 410-520 Km from the Earth's surface.

iii. The zone where olivine transforms to wadseliyte represents the upper part of the transition zone of the upper mantle of the Earth i.e. the region between 410-520 Km. depth from the surface of the Earth. Thus, it represents the MTZ- Mantle Transition Zone.

b. i. The green colour of the sand is due to olivine. Olivine gets its green colour from the ferrous iron that is a constituent of olivine.

ii. The green sands form as a result of olivine which remains in the sand above all other minerals because it is heavier than all other minerals. The olivine sand is derived from the adjacent volcanic cone (tuff ring) which is eroded and subsequenlty, this olivine enriches the sand of the beach. The volcanic cone encloses the the beach. Olivine remains in the sand as it is denser than the other volcanic constituents- ash, dust, black olivine.

iii. There are only 4 islands in the world where green beaches are found. Thus, green beaches are rare. This is because olivine is decomposed faster than other minerals (as seen in the Goldsmidt mineral stability series). Olivine undergoes tranfromations faster by the weatherimg agents of carbon dioxide and water. Thus, it is very rare to find olivine remainining. The above 4 cases are exceptions as olivine is constantly supplied from the volcanic cones and tuff rings and so, it does not matter when olivine is constantly decomposed.

2. a. Anorthite is calculated from Plagioclase by the ratio Calcium / (Calcium + Sodium + Potassium)

c. Plagioclase is a tectosilicate.

3. a. Augite is a single-chained Ionosilicate or Chain silicate.

b. Augite is a Clino-pyroxene. Orthopyroxene has interference colour of first order, orthorhombic and parallel extinction whereas clinopyroxene has interference colour of second order, monoclinic and inclined extinction.

c. The number of cleavages in augite are 2. They meet at 90 degress.

4. a. Hornblende is a double-chained Ionosilicate/ Chain-silicate.

b. Hornblend has two perfect cleavage in directions 124 degrees and 56 degrees.

5. a. Biotite is a Phyllosilicate/ Sheet Silicate

b. Biotite exhibits basal cleavage i.e. it is a pinacoidal cleavage.

c. Biotite exhibits this type of cleavage because there is only one cleavage plane.

6. a. Muscovite is a Sheet-silicate/.Phyllosicate.

b. 3 out of 4 oxygens in silicate tetrahedra are shared with neighbouring tetrahedra. The coordination number is 4 i.e. there exists tetrahedral coordination.

c. There is one perfect cleavage in muscovite.

7. a. Potassium is the akali element present in alkali feldspar.

b. High temperature varieties of alkali feldspar are sanidine and microcline.

c. The 3 K rich alkali feldspars are:

Sanidine

Orthoclase - Monoclinic symmetry

Microcline - Triclinic symmetry

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