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lawyer [7]
2 years ago
15

AUDIO

Chemistry
1 answer:
Eddi Din [679]2 years ago
6 0

Answer:

The ice record shows a period in Earth's history when there was an ice age and the global average temperature was

much colder. During this time, the ice record shows the amount of carbon dioxide in the atmosphere decreased, but the

amount of energy from the sun did not change. How did the the total amount of energy in the Earth system change, how

did this change happen, and how did it contribute to the ice age?

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How does convection occur in the troposphere?
Tamiku [17]

Answer:

the warm air rises and cool air sinks

Explanation:

The surface heated air expands as it warms, becomes less dense than surrounding cooler air and rises as buoyant and turbulent bubbles. This is convection and is the main process by which the troposphere mixes and heats. Although convection stirs and mixes the troposphere, the higher it is the colder it becomes.

4 0
3 years ago
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When glaciers are growing larger, the concentration of the heavier oxygen isotope in sea water increases?
Hatshy [7]
Answer : Yes, The heavier oxygen isotope in sea water is increasing as the result of growing of larger glaciers.

The global warming of the climate which is also called as global warming is helping to increase the isotopic oxygen in the ocean and sea waters. When glaciers are growing they have comparatively less isotopic oxygen but this proportion changes with the temperature.
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3 years ago
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The mineral magnesite contains magnesium carbonate, MgCO3 (molar mass= 84 g/mol), and other impurities. When a 1.26-g sample of
Leviafan [203]

The answer is 33.33 %


The explanation:


According to the reaction equation:


MgCO3(s) + 2HCl (aq) --> MgCl2(aq) + H20(l) + CO2(g)


we can see that 1 mole of MCO3 will produce → 1 mole of CO2


-Now we need o get number of mole of CO2:


and when we have 0.22 g of CO2, so number of mole = mass / molar mass


moles = 0.22 g / 44 g/mol = 0.005 mole


∴ moles of Mg = moles of CO2 = 0.005 mole


∴ mass of Mg = moles * molar mass


= 0.005 * 84 /mol = 0.42 g


∴ Percent of MgCO3 by mass of Mg = 0.42 g / 1.26 * 100


= 33.33 %



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