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3 years ago

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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Suppose that you take 200 mg of an antibiotic every 8 hr. The half-life of the drug is 8 hr (the time it takes for half of the

DochEvi [55]

Answer:

600 mg

Explanation:

The initial amount of the drug = 200 mg

The half-life of the drug = 8 hrs

It means that:-

After 6 hours, the concentration becomes :- $\frac{200}{2}$ mg

After 12 hours, the concentration becomes :- $\frac{200}{4}$ mg

After 18 hours, the concentration becomes :- $\frac{200}{8}$ mg

And so on...

Thus,

After infinite time = $200+\frac{200}{2}+\frac{200}{4}+\frac{200}{8}+..$

Thus,

After infinite time = $200\times (1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+..)$

The sum of the infinite series is:- $1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+..$ = $\frac{1}{1+\frac{1}{2}}=2$

So,

<u>After infinite time = $200\times 2$ mg = 600 mg</u>

5 0

3 years ago

What is the IUPAC name for the compound shown below? ch3ch2ch2cnhch2ch3

Nataly_w [17]

We need the IUPAC name of the given compound.

The IUPAC name is: Hexan-3-imine.

The molecule has six carbon atoms in its skeleton. C=NH bond is attached to the skeleton at 3-position.

The functional group present in this molecule is imine (C=NH).

3 0

3 years ago

Read 2 more answers

Since the half-life of 235U (7. 13 x 108 years) is less than that of 238U (4.51 x 109 years), the isotopic abundance of 235U has

Ymorist [56]

Answer:

$\mathtt{ t_1-t_2= In(\dfrac{3}{y}) \times \dfrac{7.13 \times 10^8}{In2} \ years}$

Explanation:

Given that:

The Half-life of $^{235}U$ = $7.13 \times 10^8 \ years$ is less than that of $^{238} U = 4.51 \times 10^9 \ years$

Although we are not given any value about the present weight of $^{235}U$ .

So, consider the present weight in the percentage of $^{235}U$ to be y%

Then, the time elapsed to get the present weight of $^{235}U$ = $t_1$

Therefore;

$N_1 = N_o e^{-\lambda \ t_1}$

here;

$N_1$ = Number of radioactive atoms relating to the weight of y of $^{235}U$

Thus:

$In( \dfrac{N_1}{N_o}) = - \lambda t_1$

$In( \dfrac{N_o}{N_1}) = \lambda t_1$ --- (1)

However, Suppose the time elapsed from the initial stage to arrive at the weight of the percentage of $^{235}U$ to be = $t_2$

Then:

$In( \dfrac{N_o}{N_2}) = \lambda t_2$ ---- (2)

here;

$N_2$ = Number of radioactive atoms of $^{235}U$ relating to 3.0 a/o weight

Now, equating equation (1) and (2) together, we have:

$In( \dfrac{N_o}{N_1}) -In( \dfrac{N_o}{N_2}) = \lambda( t_1-t_2)$

replacing the half-life of $^{235}U$ = $7.13 \times 10^8 \ years$

$In( \dfrac{N_2}{N_1}) = \dfrac{In 2}{7.13 \times 10^9}( t_1-t_2)$ ( since $\lambda = \dfrac{In 2}{t_{1/2}}$ )

∴

$\mathtt{In(\dfrac{3}{y}) \times \dfrac{7.13 \times 10^8}{In2}= t_1-t_2}$

The time elapsed signifies how long the isotopic abundance of 235U equal to 3.0 a/o

Thus, The time elapsed is $\mathtt{ t_1-t_2= In(\dfrac{3}{y}) \times \dfrac{7.13 \times 10^8}{In2} \ years}$

8 0

3 years ago

Which of the following chemical equations is unbalanced? A. 2C + H2 CH4 B. 2Al2O3 4Al + 3O2 C. 2H2O2 2H2O + O2 D. 2C2H6 + 7O2 4C

Tanya [424]

Answer:

A. 2C + H₂ ⟶ CH₄

Explanation:

A. 2C + H₂ ⟶ CH₄

UNBALANCED. 2C on the left and 1C on the right

B. 2Al₂O₃ ⟶ 4Al + 3O₂

Balanced. Same number of each type of atom on each side.

C. 2H₂O₂ ⟶ 2H₂O + O₂

Balanced. Same number of each type of atom on each side.

D. 2C₂H₆ + 7O₂ ⟶ 4CO₂ + 6H₂O

Balanced. Same number of each type of atom on each side.

8 0

3 years ago

the movement of molecules from an area of low concentration to an area of high concentration occurs in what transport?

solmaris [256]

Movement of molecules from an area of higher concentration to one of lower concentration is called Diffusion

4 0

3 years ago