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Lostsunrise [7]
3 years ago
6

Study the graph about oxygen content of Earth’s atmosphere. Oxygen Content of Earth apostrophe s Atmosphere line graph. X axis i

s Millions of Years Before Present. Y axis is Oxygen in volume percent. The line starts at 1000 millions of years before present at 4 percent oxygen, remains level until 625 Millions of Years Before Present, increases quickly to 13 percent oxygen, then increases slowly to 18 percent oxygen. The line jumps to 35 percent oxygen near 300 millions of years before present, drops quickly to 15 percent oxygen, then increases to 31 percent oxygen near 100 millions of years before present, before dropping down to 20 percent oxygen. Which describes oxygen content as Earth evolved over time? Oxygen levels sharply declined about 400 million years ago. Oxygen levels stay above 15 percent starting 200 million years ago Oxygen levels remain below 5 percent starting 1000 million years ago. Oxygen levels raise significantly between 1000 and 700 million years ago.
Chemistry
2 answers:
expeople1 [14]3 years ago
6 0

Answer: Oxygen levels stay above 15 percent starting 200 million years ago

Explanation:

Anika [276]3 years ago
3 0

Explanation:

from the graph study about oxygen content of Earth's atmosphere, we can understand that  

1)

4 billions year ago = None, 3 billions year ago = Cyanobacteria and Archaea , 2 and 1 billions year ago = Bacteria and Green algae , 500 Ma = invertebrate fossils started to existence. Early land plants came in to existence around 398 MA that is Devonian. Dinosaurs are came in to existence during Triassic and Jurassic that is around 251 Ma. Man and animals are recent organism came under Holocene that is 11000 years ago.

2)

The first cells on the earth are anaerobic microorganisms, as the CO2 level is too high they survive by using CO2.

3)

Starting around 2.7 billion years ago, photosynthesis by Cyanobacteria and later plants , pumped “OXYGEN” in to the atmosphere. This caused the decline of anaerobic bacteria and allows the diversification of animals as seen in “CAMBRIAN” around 500 millions year ago.  

Early vascular plants “CAPTURED” CO2 starting before the Carboniferous period that began around 350 millions year.Leading to lower temperatures and allowing and allowing the seed plants to outcompetes seedless plants.

Modern human activities has raised both “CO2 and METHANE” level in the atmosphere to over leading to higher temperature and extinction of other species.

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How many chiral centers does clavulanic acid have?
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Answer:

Clavulanic acid has two (2) chiral centers.

Explanation:

A chiral center is a center (usually carbon) with four different substituents.

The structure of clavulanic acid is shown in the attachment below.

Consider the labeled diagram in the attachment,

Carbon A is not a chiral carbon because it has two hydrogen atoms attached to it

Carbon B is not a chiral carbon because it has only three substituents

Carbon C is a chiral carbon because it has four different substituents

Carbon D is a chiral carbon because it has four different substituents

Carbon E is not a chiral carbon because it has only three atoms directly attached to it

Carbon F is not a chiral carbon because it has only three atoms directly attached to it

Carbon G is not a chiral carbon because it has two hydrogen atoms attached to it

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Then, only carbons C and D are chiral carbons.

Hence, clavulanic acid have two (2) chiral centers.

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3 years ago
What is the average binding energy per nucleon for a U-238 nucleus with a mass defect of 0.184 amu? (1 amu= 1.66 x 10-27 kg; 1 J
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What does the nucleus of the cell control?
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The nucleus controls and regulates the activities of the cell (e.g., growth and metabolism) and carries the genes, structures that contain the hereditary information. Nucleoli are small bodies often seen within the nucleus.

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3 years ago
The osmotic pressure of a solution containing 2.04 g of an unknown compound dissolved in 175.0 mLof solution at 25 ∘C is 2.13 at
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<u>Answer:</u> The molecular formula of the compound is C_4H_{10}O_4

<u>Explanation:</u>

To calculate the concentration of solute, we use the equation for osmotic pressure, which is:

\pi=iMRT

Or,

\pi=i\times \frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (in mL)}}\times RT

where,

\pi = osmotic pressure of the solution = 2.13 atm

i = Van't hoff factor = 1 (for non-electrolytes)

Given mass of compound = 2.04 g

Volume of solution = 175.0 mL

R = Gas constant = 0.0821\text{ L atm }mol^{-1}K^{-1}

T = temperature of the solution = 25^oC=[273+25]=298K

Putting values in above equation, we get:

2.13atm=1\times \frac{2.04\times 1000}{\text{Molar mass of compound}\times 175.0}\times 0.0821\text{ L.atm }mol^{-1}K^{-1}\times 298K\\\\\text{Molar mass of compound}=\frac{1\times 2.04\times 1000\times 0.0821\times 298}{2.13\times 175.0}=133.9g/mol

  • <u>Calculating the molecular formula:</u>

The chemical equation for the combustion of compound having carbon, hydrogen and oxygen follows:

C_xH_yO_z+O_2\rightarrow CO_2+H_2O

where, 'x', 'y' and 'z' are the subscripts of carbon, hydrogen and oxygen respectively.

We are given:

Mass of CO_2=36.26g

Mass of H_2O=14.85g

We know that:

Molar mass of carbon dioxide = 44 g/mol

Molar mass of water = 18 g/mol

<u>For calculating the mass of carbon:</u>

In 44 g of carbon dioxide, 12 g of carbon is contained.

So, in 36.26 g of carbon dioxide, \frac{12}{44}\times 36.26=9.89g of carbon will be contained.

<u>For calculating the mass of hydrogen:</u>

In 18 g of water, 2 g of hydrogen is contained.

So, in 14.85 g of water, \frac{2}{18}\times 14.85=1.65g of hydrogen will be contained.

Mass of oxygen in the compound = (22.08) - (9.89 + 1.65) = 10.54 g

To formulate the empirical formula, we need to follow some steps:

  • <u>Step 1:</u> Converting the given masses into moles.

Moles of Carbon = \frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{9.89g}{12g/mole}=0.824moles

Moles of Hydrogen = \frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{1.65g}{1g/mole}=1.65moles

Moles of Oxygen = \frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{10.54g}{16g/mole}=0.659moles

  • <u>Step 2:</u> Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 0.659 moles.

For Carbon = \frac{0.824}{0.659}=1.25\approx 1

For Hydrogen = \frac{1.65}{0.659}=2.5

For Oxygen = \frac{0.659}{0.659}=1

Converting the mole fraction into whole number by multiplying the mole fraction by '2'

Mole fraction of carbon = (1 × 2) = 2

Mole fraction of oxygen = (2.5 × 2) = 5

Mole fraction of hydrogen = (1 × 2) = 2

  • <u>Step 3:</u> Taking the mole ratio as their subscripts.

The ratio of C : H : O = 2 : 5 : 2

The empirical formula for the given compound is C_2H_5O_2

For determining the molecular formula, we need to determine the valency which is multiplied by each element to get the molecular formula.

The equation used to calculate the valency is:

n=\frac{\text{Molecular mass}}{\text{Empirical mass}}

We are given:

Mass of molecular formula = 133.9 g/mol

Mass of empirical formula = 61 g/mol

Putting values in above equation, we get:

n=\frac{133.9g/mol}{61g/mol}=2

Multiplying this valency by the subscript of every element of empirical formula, we get:

C_{(2\times 2)}H_{(5\times 2)}O_{(2\times 2)}=C_4H_{10}O_4

Hence, the molecular formula of the compound is C_4H_{10}O_4

4 0
3 years ago
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