How does the hydrosphere help regulate the amount of radiation on earth?

Chemistry

1 answer:

MAXImum [283]2 years ago

5 0

Explanation:

The oceans and other water bodies absorb huge amounts of solar energy and radiation. Thermohaline circulation transports the absorbed heat from the equator to the poles to regulate and moderate Earth's climate.

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What is the enclosure act

VashaNatasha [74]

The Inclosure Acts, which use an old or formal spelling of the word now usually spelt "enclosure", cover enclosure of open fields and common land in England and Wales, creating legal property rights to land previously held in common.

4 0

2 years ago

The acetylene tank contains 35.0 mol C2H2, and the oxygen tank contains 84.0 mol O2.

harkovskaia [24]

Answer:- As per the question is asked, 35.0 moles of acetylene gives 70 moles of carbon dioxide but if we solve the problem using the limiting reactant which is oxygen then 67.2 moles of carbon dioxide will form.

Solution:- The balanced equation for the combustion of acetylene is:

$2C_2H_2(g)+5O_2(g)\rightarrow 4CO_2(g)+2H_2O(g)$

From the balanced equation, two moles of acetylene gives four moles of carbon dioxide. Using dimensional analysis we could show the calculations for the formation of carbon dioxide by the combustion of 35.0 moles of acetylene.

$35.0molC_2H_2(\frac{4molCO_2}{2molC_2H_2})$

= $70molCO_2$

The next part is, how we choose 35.0 moles of acetylene and not 84.0 moles of oxygen.

From balanced equation, there is 2:5 mol ratio between acetylene and oxygen. Let's calculate the moles of oxygen required to react completely with 35.0 moles of acetylene.

$35.0molC_2H_2(\frac{5molO_2}{2molC_2H_2})$

= $87.5molO_2$

Calculations shows that 87.5 moles of oxygen are required to react completely with 35.0 moles of acetylene. Since only 84.0 moles of oxygen are available, the limiting reactant is oxygen, so 35.0 moles of acetylene will not react completely as it is excess reactant.

So, the theoretical yield should be calculated using 84.0 moles of oxygen as:

$84.0molO_2(\frac{4molO_2}{5molO_2})$