During which eon was the earth made of land masses all clumped together in one supercontinent called Rodinia?
A) Proterozoic
B) Phanerozoic
C) Hadean
D) Archean
(GIVING 20 POINTS)

Answers

Answer 1

During the Proterozoic eon was the earth made of land masses all clumped together in one supercontinent called Rodinia. Therefore, option A is correct.

Why is it called the Proterozoic eon?

The Archean Eon is followed by the Proterozoic Eon, which spans 2.5 billion years to 541 million years old and is known as the "early life" eon. The majority of the continents' center regions were developed by this time, and the plate tectonic process had begun.

The Proterozoic is currently understood based on geological, geochemical, and fossil evidence. The Cambrian epoch and an explosion of various species at the beginning of the Phanerozoic eon both originated at the conclusion of this eon.

During the Proterozoic Era, Rodinia was a supercontinent that included nearly all of Earth's landmasses for around 450 million years (2.5 billion to 541 million years ago).

Thus, option A is correct.

To learn more about the Proterozoic eon, follow the link:

https://brainly.com/question/20700965

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I believe the Correct answer would be A

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A; sand and water
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Explanation:

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Answer:

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Plz help thank you!!!

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Answer:

1. Covalent Bond

2. Ionic Bond

Explanation:

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Ionic bonds are formed when valence electrons are transferred to other atoms and form oppositely charged ions. In ionic bond formation, the atoms that gain electrons become negatively charged and the atoms that loses electrons become positively charged. In the given example of Ca Cl2, Ca is also giving its 2 valence electrons to each Cl and there is no stable balance of attraction and repulsion between atoms.

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Answer:

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Explanation:

help me plz i’ll give extra points

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False is the answer :) hope I helped

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Answer:

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Answer:

The answer is:

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Hope this helps!!

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Answer:

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Answer:

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Calculate the density (g/ml) of mineral that has volume (total space occupied by something) of 3.8 ml and a mass of 22.4 g.

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Answer:

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Answer:

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Answers

Answer:

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General Formulas and Concepts:

Math

Pre-Algebra

Order of Operations: BPEMDAS

Brackets Parenthesis Exponents Multiplication Division Addition Subtraction Left to Right

Chemistry

Atomic Structure

Reading a Periodic TableAvogadro's Number - 6.022 × 10²³ atoms, molecules, formula units, etc.

Stoichiometry

Using Dimensional AnalysisExplanation:

Step 1: Define

1.9 × 10²² molecules CaCO₃

Step 2: Identify Conversions

Avogadro's Number

[PT] Molar Mass of Ca - 40.08 g/mol

[PT] Molar Mass of C - 12.01 g/mol

[PT] Molar Mass of O - 16.00 g/mol

Molar Mass of CaCO₃ - 40.08 + 12.01 + 3(16.00) = 100.09 g/mol

Step 3: Convert

Set up:                              [tex]\displaystyle 1.9 \cdot 10^{22} \ molecules \ CaCO_3(\frac{1 \ mol \ CaCO_3}{6.022 \cdot 10^{23} \ molecules \ CaCO_3})(\frac{100.09 \ g \ CaCO_3}{1 \ mol \ CaCO_3})[/tex]Multiply/Divide:                                                                                                    [tex]\displaystyle 3.15794 \ g \ CaCO_3[/tex]

Step 4: Check

Follow sig fig rules and round. We are given 2 sig figs.

3.15794 g CaCO₃ ≈ 3.2 g CaCO₃

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Answer:

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