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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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How many molecules of water will be produced if 52.6 g of methane are burned?
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Which atom has the smallest radius? Sr or I
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2 years ago
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A student performed an analysis of a sample for its calcium content and obtained the following results:
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Answer:

14.9075 g, 28.67%, 0.11%

Explanation:

The mean concentration of calcium = summation x / frequency

= ( 14.92 + 1491 + 14.88 + 14.92 ) /4 = 14.9075 g

Standard deviation = √(summation (x - μ)² /n) = √ ( ((14.92 - 14.9075)² +(14.91 - 14.9075)² + (14.88 - 14.9075)² + ( 14.92 - 14.9075)²) / 4) = 0.0164

b)  percent error = abs(14.9075 - 20.90) / 20.90 × 100 = 28.67%

c) relative standard deviation = standard deviation / mean × 100 = 0.0164 /  14.9075 × 100 = 0.11%

d) The accuracy of the measure is the measurement compared to the actual which according to the standard set by the instructor (5%error) is not very accurate because the percent error is high (28.67%) while the relative standard deviation is quite low ( 0.11%) which means the measurement precision is very high.

The student will have to redo the experiment because the experiment was not too accurate since the percent error is way higher than the set value (5%) although the precision was high.

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