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Lostsunrise [7]
2 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]2 years ago
6 0

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

Explanation:

Anika [276]2 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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En un balneario necesitan calentarse 1 millón de litros de agua anuales, subiendo la temperatura desde 15 ºC a 50 ºC y para ello
MAXImum [283]

Answer:

a) m_{CH_4}=2630kg

b) 1657 €

Explanation:

Hola,

a) En este problema, vamos a considerar el millón de litros de agua anuales, ya que con ellos podemos calcular el calor requerido para dicho calentamiento, sabiendo que la densidad del agua es de 1 kg/L:

Q_{H_2O}=m_{H_2O}Cp(T_2-T_1)=1x10^6LH_2O*\frac{1kgH_2O}{1LH_2O}*4.18\frac{kJ}{kg\°C}(50-15) \°C\\Q_{H_2O}=146.3x10^6kJ

Luego, usamos la entalpía de combustión del metano para calcular su requerimiento en kilogramos, sabiendo que la energía ganada por el agua, es perdida por el metano:

Q_{H_2O}=-Q_{CH_4}=-146.3x10^5kJ=m_{CH_4}\Delta _cH_{CH_4}

m_{CH_4}= \frac{Q_{CH_4}}{\Delta _cH_{CH_4}} =\frac{-146.3x10^5kJ}{-890kJ/molCH_4} *\frac{16gCH_4}{1molCH_4} \\\\m_{CH_4}=2630112.36g=2630kg

b) En este caso, consideramos que a condiciones normales de 1 bar y 273 K, 1 metro cúbico de metano cuesta 0,45 €, con esto, calculamos las moles de metano a dichas condiciones:

n_{CH_4}=\frac{PV}{RT}=\frac{1atm*1000L}{0.082\frac{atm*L}{mol*K}*273K} =44.67mol

Con ello, los kilogramos de metano que cuestan 0,45 €:

44.67molCH_4*\frac{16gCH_4}{1molCH_4}*\frac{1kg}{1000g} =0.715kgCH_4

Luego, aplicamos la regla de tres:

0.715 kg ⇒ 0.45 €

2630 kg ⇒ X

X = (2630 kg x 0.45 €) / 0.715 kg

X = 1657 €

Regards.

3 0
2 years ago
Some fruits and vegetables are preserved by pickling them. Nandini got confused
stepan [7]

Answer:

ye

Explanation:

ye

3 0
1 year ago
aleksA sailor on a trans-Pacific solo voyage notices one day that if he puts of fresh water into a plastic cup weighing , the cu
finlep [7]

Answer:

40 g

See explaination

Explanation:

Archimedes' principle states that the upward buoyant force that is exerted on a body immersed in a fluid, whether fully or partially submerged, is equal to the weight of the fluid that the body displaces.

Check attachment for the detailed step by step solution of the given problem.

3 0
1 year ago
Octane is a liquid component of gasoline. Given the following vapor pressures of octane at various temperatures, estimate the bo
Hitman42 [59]

Answer:

110.8 ºC

Explanation:

To solve this problem we will make use of the Clausius-Clayperon equation:

lnP = - ΔHºvap/RT + C

where P is the pressure, ΔHºvap is the enthalpy of vaporization, R is the gas constant, T is the temperature, and C is a constant of integration.

Now this equation has a form y = mx + b where

y = lnP

x = 1/T

m = -ΔHºvap/R

Now we have to assume that ΔHºvap remains constant which is a good asumption given the narrow range of temperatures in the data ( 104-125) ºC

Thus what we have to do is find the equation of the best fit for this data using a  software as excel or your calculator.

T ( K)               1/T                  ln P

377               0.002653       5.9915

384              0.002604       6.2115

390              0.002564       6.3969

395              0.002532       6.5511

398              0.002513        6.6333

The best line has a fit:

y = -4609.5 x  + 18.218

with R² = 0.9998

Now that we have the equation of the line, we simply will substitute for a pressure of 496 mm in Leadville.

ln(496) = -4609.5(1/Tb) + 18.218

6.2066 = -4609.5(1/Tb) +18.218

⇒ 1/Tb = (18.218 - 6.2066)/4609.5 = 0.00261

Tb = 383.76 K  = (383.76 -273)K = 110.8 ºC

Notice we have touse up to 4 decimal places since rounding could lead to an erroneous answer ( i.e boiling temperature greater than 111, an impossibility given the data in the question). This is as a result of the value 496 mmHg so close to 500 mm Hg.

Perhaps that is the reason the question was flagged.

7 0
2 years ago
The lab question is “Based on a substance’s properties, how can you determine whether its bonds are ionic or covalent?” Fill in
AleksandrR [38]
If the substance has high melting/boiling point, if it requires high temperature to dissociate into simpler particles, if it's structure is hard and if it conducts heat and electricity quite frequently, then it would be "Ionic compound" otherwise, it will be covalent compound. (compound with covalent bonds).

Substance with ionic bonds, would include... (mentioned in first sentence)

Hope this helps!
8 0
2 years ago
Read 2 more answers
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