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Akimi4 [234]
2 years ago
13

Science Question

Geography
1 answer:
kkurt [141]2 years ago
4 0

Answer:

Iron oxides appeared after the emergence of cyanobacteria.

Explanation:

The composition of Earth's atmosphere in the first couple of billions of years of its existence was nothing like the atmosphere of today. This had a big influence on the processes that were taking place, or rather what processes and to what extent could have happened in those conditions. One big difference between the atmosphere then and after is the levels of oxygen.

Until the appearance of cyanobacteria, the oxygen levels in Earth's atmosphere were very low, and the same goes for the oceans. With a lack of oxygen, the process of oxidation was absent as well. The cyanobacteria though managed to produce oxygen, and this was on such a high scale that they changed the composition of Earth's atmosphere and the oceans. Not just that this enabled complex lifeforms to develop, but it also enabled the process of oxidation. Iron oxides for example occurred only after cyanobacteria appeared, and this can easily be seen when dating the oldest iron oxides and compare that age with the appearance of cyanobacteria.

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A balloon filled with an ideal gas initially has a temperature of 24°C and a volume of 7.0 L. If the gas in the balloon is heate
torisob [31]

Answer:

8.9L

Explanation:

\\ Using\quad Charles'\quad Law\\ \\ \frac { { V }_{ 1 } }{ { T }_{ 1 } } =\quad \frac { { V }_{ 2 } }{ { T }_{ 2 } } \\ provided\quad the\quad pressure\quad of\quad the\quad gas\quad is\quad kept\quad constant.\\ Temperature\quad should\quad be\quad in\quad Kelvins:\quad \\ Kelvin\quad Temperature\quad =\quad Celsius\quad Temperature\quad +\quad 273.15\quad \\ Given\quad information:\\ { V }_{ 1 }=\quad 7L\quad \quad \quad \quad \quad \quad { V }_{ 2}\quad is\quad unknown\quad \quad {T}_{ 1 }=\quad 24+273.15=297.5K\quad \quad \quad { T }_{ 2 }=104+273.15=377,15K\\ \therefore \quad \frac { 7L }{ 297.5K } =\frac { { V }_{ 2 } }{ 377.5K } \\ \quad \quad \quad 2640.05\quad =\quad 297.5{ V }_{ 2 }\\ \quad \quad \quad \quad \quad \quad \quad \quad { V }_{2}=\frac { 2640.05 }{ 297.5 } \quad \\ \quad \quad \quad \quad \quad \quad \quad \quad { V }_{ 2 }=8.9\quad L\quad to\quad the\quad nearest\quad tenth\quad of\quad a\quad liter.

7 0
3 years ago
TRUE / FALSE<br> The largest part of an iceberg is found above water
maria [59]

Answer:

False. The larger portions of icebergs are below the waters surface.

Explanation:

Hope this helps!

6 0
2 years ago
Read 2 more answers
Water cools from 2C to -2C. during this time what happens to the motion of the molecules
aalyn [17]
It slooowwwwssss down... the colder the slower.
5 0
3 years ago
Calculate the amount, in grams, of an original 300-gram sample of radioactive isotope Potassium-40 remaining after
Aleksandr [31]

Answer:

The current amount of the Potassium-40 sample is approximately 37.521 grams.

Explanation:

The amount of the sample of the radioactive isotope decays exponentially in time, the amount of mass of the sample as a function of time (m (t)), in grams, is described below:

m(t) = m_{o}\cdot e^{-\frac{t}{\tau} } (1)

Where:

m_{o} - Initial mass, in grams.

t - Time, in years.

\tau - Time constant, in years.

The time constant can be found from half life (t_{1/2}), in years, described in statement:

\tau = \frac{t_{1/2}}{\ln 2} (2)

If we know that m_{o} = 300\,g, t = 3.9\times 10^{9}\,yr and t_{1/2} = 1.3\times 10^{9}\,yr, then the current amount of the sample is:

\tau = \frac{1.3\times 10^{9}\,yr}{\ln 2}

\tau \approx 1.876\times 10^{9}\,yr

m = (300\,g)\cdot e^{-\frac{3.9\times 10^{9}\,yr}{1.876\times 10^{9}\,yr} }

m\approx  37.521\,g

The current amount of the Potassium-40 sample is approximately 37.521 grams.

4 0
3 years ago
Priem
sladkih [1.3K]

Answer:

A. Locations closest to the equator

Explanation:

It is this because the equator gets the most sun exposure.

3 0
3 years ago
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