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Tresset [83]
3 years ago
15

How are fossils used in the scientific process to support or disprove hypothesis on the origin of life on earth

Chemistry
1 answer:
ella [17]3 years ago
5 0

Answer:

Using radiometric dating they SUPPORT the hypothesis on the origin of life on Earth

Explanation:

A fossil is a preserved remnant of a living organism from the past, this fossilization can be achieved by many processes including carbonization, recrystallization, permineralization, etc. These fossils can be used to obtain different information about the past and the origin of life on earth, for example,<em> chemical fossils are a kind of fossil found in rocks and fuels like coal or petroleum, they represent one of the oldest kind of fossils found and through the use of radiometric dating, where carbon isotopes are measured, their existence implies the presence of life as early as 4.1 billion years ago.</em>

I hope you find this information useful and interesting! Good luck!

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Which type of fuel can only be used in certain geographic areas?
Lisa [10]
Answer:

A. Nuclear

Explanation:

This is because it could cause severe environmental damage in areas inhabited by the public

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2 years ago
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Identify the correct coefificients to balance it<br> -C3H8+O2 to -CO2 +-H2O
Iteru [2.4K]

Answer:

{\rm 1\; C_3H_8} + {\rm 5\; O_2} \to {3\; \rm CO_2} + {4\; \rm H_2O}.

Explanation:

{\rm ?\; C_3H_8} + {\rm ?\; O_2} \to {?\; \rm CO_2} + {?\; \rm H_2O}.

Among the four species in this reaction, \rm C_3H_8 is species with the largest number of atoms per molecule. Assume that the coefficient of this compound is 1.

{\rm 1\; C_3H_8} + {\rm ?\; O_2} \to {?\; \rm CO_2} + {?\; \rm H_2O}.

Number of atoms on the left-hand side of the reaction:

  • \rm C: 1 \times 3 = 3.
  • \rm H: 1 \times 8 = 8.
  • \rm O: not found yet.

By the conservation of atoms, the number of atoms on the right-hand side of the reaction should match those on the left-hand side. In this reaction, \rm CO_2 is the only product with carbon atoms, whereas \rm H_2O is the only product with hydrogen atoms. These 3 carbon atoms and 8 hydrogen atoms would correspond to:

  • 3 / 1 = 3 \rm CO_2 molecules, and
  • 8 / 2 = 4 \rm H_2O molecules.

{\rm 1\; C_3H_8} + {\rm ?\; O_2} \to {3\; \rm CO_2} + {4\; \rm H_2O}.

Number of atoms on the right-hand side of the reaction:

  • \rm C: 3 \times 1 = 3.
  • \rm H: 4 \times 2 = 8.
  • \rm O: 3 \times 2 + 4 \times 1 = 10.

The number of \rm O atoms on the left-hand side should match those on the right-hand side. In this reaction, \rm O_2 is the only reactant with \rm O\! atoms. These 10 \rm \! O atoms would correspond to:

  • 5 \rm O_2 molecules.

{\rm 1\; C_3H_8} + {\rm 5\; O_2} \to {3\; \rm CO_2} + {4\; \rm H_2O}.

5 0
2 years ago
What does the second law of thermodynamics say about entropy
alukav5142 [94]

Answer:

The entropy and the systems surrounding it tend to increase.

5 0
3 years ago
Read 2 more answers
URGENT CHEMISTRY EXPERT!
vovangra [49]

Answer:

Part 1: 7.42 mL; Part 2: 3Cu²⁺(aq) + 2PO₄³⁻(aq) ⟶ 2Cu₃(PO₄)₂(s)

Explanation:

Part 1. Volume of reactant

(a) Balanced chemical equation.

\rm 2Na_{3}PO_{4} + 3CuCl_{2} \longrightarrow Cu_{3}(PO_{4})_{2} + 6NaCl

(b) Moles of CuCl₂

\text{Moles of CuCl}_{2} =\text{ 16.7 mL CuCl}_{2} \times \dfrac{\text{0.200 mmol CCl}_{2}}{\text{1 mL CuCl}_{2}} =  \text{3.340 mmol CuCl}_{2}

(c) Moles of Na₃PO₄

The molar ratio is 2 mmol Na₃PO₄:3 mmol CuCl₂

\text{Moles of Na$_{3}$PO}_{4} =  \text{3.340 mmol CuCl}_{2} \times \dfrac{\text{2 mmol Na$_{3}$PO}_{4}}{\text{3 mmol CuCl}_{2}} =\text{2.227 mmol Na$_{3}$PO}_{4}

(d) Volume of Na₃PO₄

V = \text{2.227 mmol Na$_{3}$PO}_{4}\times \dfrac{\text{1 mL Na$_{3}$PO}_{4}}{\text{0.300 mmol Na$_{3}$PO}_{4}} = \text{7.42 mL Na$_{3}$PO}_{4} \\\\\text{The reaction requires $\large \boxed{\textbf{7.42 mL Na$_{3}$PO}_{4}}$}

Part 2. Net ionic equation

(a) Molecular equation

\rm 2Na_{3}PO_{4}(\text{aq}) + 3CuCl_{2}(\text{aq}) \longrightarrow Cu_{3}(PO_{4})_{2}(\text{s}) + 6NaCl(\text{aq})

(b) Ionic equation

You write molecular formulas for the solids, and you write the soluble ionic substances as ions.

According to the solubility rules, metal phosphates are insoluble.

6Na⁺(aq) + 2PO₄³⁻(aq) + 3Cu²⁺(aq) + 6Cl⁻(aq) ⟶ Cu₃(PO₄)₂(s) + 6Na⁺(aq) + 6Cl⁻(aq)  

(c) Net ionic equation

To get the net ionic equation, you cancel the ions that appear on each side of the ionic equation.

<u>6Na⁺(aq)</u> + 2PO₄³⁻(aq) + 3Cu²⁺(aq) + <u>6Cl⁻(aq)</u> ⟶ Cu₃(PO₄)₂(s) + <u>6Na⁺(aq)</u> + <u>6Cl⁻(aq)</u>  

The net ionic equation is

3Cu²⁺(aq) + 2PO₄³⁻(aq) ⟶ Cu₃(PO₄)₂(s)

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3 years ago
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sertanlavr [38]
Mass of molecule (g) = Mr of substance over avarogado constant
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