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dangina [55]
3 years ago
15

The percentages of carbon dioxide and oxygen have changed from Earth's early

Chemistry
1 answer:
Anna [14]3 years ago
5 0

Answer:

Explanation:

<u>The early atmosphere</u>

Scientists believe that the Earth was formed about 4.5 billion years ago. Its early atmosphere was probably formed from the gases given out by volcanoes. It is believed that there was intense volcanic activity for the first billion years of the Earth's existence.

The early atmosphere was probably mostly carbon dioxide, with little or no oxygen. There were smaller proportions of water vapour, ammonia and methane. As the Earth cooled down, most of the water vapour condensed and formed the oceans.

It is thought that the atmospheres of Mars and Venus today, which contain mostly carbon dioxide, are similar to the early atmosphere of the Earth.

Scientists can’t be sure about the early atmosphere and can only draw evidence from other sources. For example, volcanoes release high quantities of carbon dioxide. Iron-based compounds are present in very old rocks that could only have formed if there was little or no oxygen at the time.

Changes in the atmosphere

So how did the proportion of carbon dioxide in the atmosphere go down, and the proportion of oxygen go up?

The proportion of oxygen went up because of photosynthesis by plants.

The proportion of carbon dioxide went down because:

it was locked up in sedimentary rocks (such as limestone) and in fossil fuels

it was absorbed by plants for photosynthesis

it dissolved in the oceans

The burning of fossil fuels is adding carbon dioxide to the atmosphere faster than it can be removed. This means that the level of carbon dioxide in the atmosphere is increasing.

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A sample of He gas (3.0 L) at 5.6 atm and 25°C was combined with 4.5 L of Ne gas at 3.6 atm and 25°C at constant temperature in
LenKa [72]

Answer:

P=3.7atm

Explanation:

Hello,

In this case, it is possible to determine the pressures of both helium and neon as shown below:

n_{He}=\frac{P_{He}V_{He}}{RT}=\frac{5.6atm*3.0L}{0.082\frac{atm*L}{mol*K}*298.15K} =0.688molHe\\\\n_{Ne}=\frac{P_{Ne}V_{Ne}}{RT}=\frac{3.6atm*4.5L}{0.082\frac{atm*L}{mol*K}*298.15K}=0.663molNe

Now, one considers the total moles (addition between both neon's and helium's moles) and the total volume to compute the final pressure as shown below:

P=\frac{n_TRT}{V_T} =\frac{(0.688+0.663)mol*0.082\frac{atm*L}{mol*K}*298.15K}{9.0L}=3.7atm

Best regards.

8 0
3 years ago
Allena and Charissa were discussing whether Cl2 is an element or a compound. Allena said that it is an element because it has on
kobusy [5.1K]
Answer is: Allena is correct. It is an element because it is only made of chlorine atoms.
A chemical element bonded to an identical chemical element is not a chemical compound since it is made from only one element and not from two different elements. Chlorine is molecule, but not compound.
8 0
3 years ago
Newton's Law of Gravity says that any the effect of gravity between two objects depends on what two factors?
maw [93]
The masses of the objects and the distance between them
-hope it helps
8 0
3 years ago
A sample of an unknown metal has a mass of 58.932g. it has been heated to 101.00 degrees C, then dropped quickly into 45.20 mL o
yaroslaw [1]
<h3>Answer:</h3>

0.111 J/g°C

<h3>Explanation:</h3>

We are given;

  • Mass of the unknown metal sample as 58.932 g
  • Initial temperature of the metal sample as 101°C
  • Final temperature of metal is 23.68 °C
  • Volume of pure water = 45.2 mL

But, density of pure water = 1 g/mL

  • Therefore; mass of pure water is 45.2 g
  • Initial temperature of water = 21°C
  • Final temperature of water is 23.68 °C
  • Specific heat capacity of water = 4.184 J/g°C

We are required to determine the specific heat of the metal;

<h3>Step 1: Calculate the amount of heat gained by pure water</h3>

Q = m × c × ΔT

For water, ΔT = 23.68 °C - 21° C

                       = 2.68 °C

Thus;

Q = 45.2 g × 4.184 J/g°C × 2.68°C

    = 506.833 Joules

<h3>Step 2: Heat released by the unknown metal sample</h3>

We know that, Q =  m × c × ΔT

For the unknown metal, ΔT = 101° C - 23.68 °C

                                              = 77.32°C

Assuming the specific heat capacity of the unknown metal is c

Then;

Q = 58.932 g × c × 77.32°C

   = 4556.62c Joules

<h3>Step 3: Calculate the specific heat capacity of the unknown metal sample</h3>
  • We know that, the heat released by the unknown metal sample is equal to the heat gained by the water.
  • Therefore;

4556.62c Joules = 506.833 Joules

c = 506.833 ÷4556.62

  = 0.111 J/g°C

Thus, the specific heat capacity of the unknown metal is 0.111 J/g°C

8 0
3 years ago
To what extent do covalent compounds conduct electricity?
sertanlavr [38]

The answer is never because there's no electricity

3 0
3 years ago
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