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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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Organic compounds are compounds that contain carbon atoms.

Organic matter is any waste product that occurs naturally (i.e. snake skin, feces, and other traces of life).

Inorganic matter would lack carbon compounds. Radioactive waste at nuclear power plants is inorganic.
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3 years ago
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CC14
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Answer:

Na and Cl

Explanation:

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2 years ago
A 100 gram glass container contains 200 grams of water and 50.0 grams of ice all at 0°c. a 200 gram piece of lead at 100°c is ad
ASHA 777 [7]

0 \; \textdegree{\text{C}}

Explanation:

Assuming that the final (equilibrium) temperature of the system is above the melting point of ice, such that all ice in the container melts in this process thus

  • E(\text{fusion}) = m(\text{ice}) \cdot L_{f}(\text{water}) = 66.74 \; \text{kJ} and
  • m(\text{water, final}) = m(\text{water, initial}) + m(\text{ice, initial}) = 0.250 \; \text{kg}

Let the final temperature of the system be t \; \textdegree{\text{C}}. Thus \Delta T (\text{water}) = \Delta T (\text{beaker}) = t(\text{initial})  - t_{0} = t \; \textdegree{\text{C}}

  • Q(\text{water}) &= &c(\text{water}) \cdot m(\text{water, final}) \cdot \Delta T (\text{water})= 1.047 \cdot t\; \text{kJ} (converted to kilojoules)
  • Q(\text{container}) &= &c(\text{glass}) \cdot m(\text{container}) \cdot \Delta T (\text{container})= 0.0837 \cdot t \; \text{kJ}
  • Q(\text{lead}) &= &c(\text{lead}) \cdot m(\text{lead}) \cdot \Delta T (\text{lead})= 0.0255 \cdot (100 - t)\; \text{kJ}

The fact that energy within this system (assuming proper insulation) conserves allows for the construction of an equation about variable t.

E(\text{absorbed} ) = E(\text{released})

  • E(\text{absorbed} ) = E(\text{fushion}) + Q(\text{water}) + Q(\text{container})
  • E(\text{released}) =  Q(\text{lead})

Confirm the uniformity of units, equate the two expressions and solve for t:

66.74 + 1.047 \cdot t + 0.0837 \cdot t = 0.0255 \cdot (80 - t)

t \approx -55.95\; \textdegree{\text{C}} < 0\; \textdegree{\text{C}} which goes against the initial assumption. Implying that the final temperature does <em>not</em> go above the melting point of water- i.e., t \le 0 \; \textdegree{\text{C}}. However, there's no way for the temperature of the system to go below 0 \; \textdegree{\text{C}}; doing so would require the removal of heat from the system which isn't possible under the given circumstance; the ice-water mixture experiences an addition of heat as the hot block of lead was added to the system.

The temperature of the system therefore remains at 0 \; \textdegree{\text{C}}; the only macroscopic change in this process is expected to be observed as a slight variation in the ratio between the mass of liquid water and that of the ice in this system.

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3 years ago
A 36.0−g sample of an unknown metal at 99°C was placed in a constant-pressure calorimeter containing 70.0 g of water at 24.0°C.
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Answer: 0.52849 j /g °C

Explanation:

Given the following :

Mass of metal = 36g

Δ Temperature of metal = (28.4 - 99)°C = - 70.6°C

Mass of water = 70g

Δ in temperature of water = (28.4 - 24.0) = 4.4°C

Heat lost by metal = (heat gained by water + heat gained by calorimeter)

Quantity of heat(q) = mcΔT

Where; m = mass of object ; c = specific heat capacity of object

Heat lost by metal:

- (36 × c × - 70.6) = 2541.6c - - - - (1)

Heta gained by water and calorimeter :

(70 × 4.184 × 4.4) + (12.4 × 4.4) = 1288.672 + 54.56 = 1343.232 - - - - (2)

Equating (1) and (2)

2541.6c = 1343.232

c = 1343.232 / 2541.6

c = 0.52849 j /g °C

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Answer:

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