Shale, and Sandstone are the rocks formed by precipitation of calcium carbonate from sea water.
Option B and D
<u>Explanation:</u>
Calcium carbonate can get precipitated from sea water and it forms rocks termed as limestones. These limestones are a variety of sedimentary rocks. Along with it, shale and sandstones are also varieties of sedimentary rocks and they also contain some of the precipitation of calcium carbonate.
The composition of sandstones consists of quartz and feldspar along with some of the composition of calcium carbonate. The same concept goes for Shale also. Basically, Shale is a mixuture of Clay but they can combine with limestone.
Thus, Shale, Sandstone and limestones are the rocks formed by precipitation of calcium carbonate from sea water.
Well since covalent bonds are strong and diamonds contains a lot of covalent bonds, it makes the diamond's melting point and boiling point very high.
<h3>
Answer:</h3>
150 g Si
<h3>
General Formulas and Concepts:</h3>
<u>Math</u>
<u>Pre-Algebra</u>
Order of Operations: BPEMDAS
- Brackets
- Parenthesis
- Exponents
- Multiplication
- Division
- Addition
- Subtraction
<u>Chemistry</u>
<u>Atomic Structure</u>
- Reading a Periodic Table
- Avogadro's Number - 6.022 × 10²³ atoms, molecules, formula units, etc.
<u>Stoichiometry</u>
<h3>
Explanation:</h3>
<u>Step 1: Define</u>
[Given] 3.2 × 10²⁴ atoms Si
[Solve] grams Si
<u>Step 2: Identify Conversions</u>
Avogadro's Number
[PT] Molar Mass of Si - 28.09 g/mol
<u>Step 3: Convert</u>
- [DA] Set up:

- [DA] Multiply/Divide [Cancel out units]:

<u>Step 4: Check</u>
<em>Follow sig fig rules and round. Instructed to round to 2 sig figs.</em>
149.266 g Si ≈ 150 g Si
Continental plates are much thicker that Oceanic plates. At the convergent boundaries the continental plates are pushed upward and gain thickness. The rocks and geological layers are much older on continental plates than in the oceanic plates. The Continental plates are much less dense than the Oceanic plates.

As long as the equation in question can be expressed as the sum of the three equations with known enthalpy change, its
can be determined with the Hess's Law. The key is to find the appropriate coefficient for each of the given equations.
Let the three equations with
given be denoted as (1), (2), (3), and the last equation (4). Let
,
, and
be letters such that
. This relationship shall hold for all chemicals involved.
There are three unknowns; it would thus take at least three equations to find their values. Species present on both sides of the equation would cancel out. Thus, let coefficients on the reactant side be positive and those on the product side be negative, such that duplicates would cancel out arithmetically. For instance,
shall resemble the number of
left on the product side when the second equation is directly added to the third. Similarly
Thus
and

Verify this conclusion against a fourth species involved-
for instance. Nitrogen isn't present in the net equation. The sum of its coefficient shall, therefore, be zero.

Apply the Hess's Law based on the coefficients to find the enthalpy change of the last equation.
