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3 years ago

How do the strength and the cohesion of clay-rich regolith or soil change with the addition of water?

Chemistry

1 answer:

lisov135 [29]3 years ago

6 0

Answer:

Water lowers the strength and cohesion of clay-rich regolith or soil.

Explanation:

Water can seep into the soil or clay-rich regolith and replace the air in the pore space of the soil or regolith. Water will completely surrounds all the grains of the clay-rich regolith and breaks the bonds in between the grains, that is eliminating all grain to grain contact of the regolith. When the regolith becomes saturated with water, the angle of repose is reduced to very small values and the regolith tends to loose its form.

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Answer: The percent yield of the water is 31.98 %

Explanation:

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$ .....(1)

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Molar mass of methane = 16 g/mol

Putting values in equation 1, we get:

$\text{Moles of methane}=\frac{6.58g}{16g/mol}=0.411mol$

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Given mass of oxygen gas = 14.4 g

Molar mass of oxygen gas = 32 g/mol

Putting values in equation 1, we get:

$\text{Moles of oxygen gas}=\frac{14.4g}{32g/mol}=0.45mol$

The chemical equation for the combustion of methane is:

$CH_4+2O_2\rightarrow CO_2+2H_2O$

By Stoichiometry of the reaction:

2 moles of oxygen gas reacts with 1 mole of methane

So, 0.45 moles of oxygen gas will react with = $\frac{1}{2}\times 0.45=0.225mol$ of methane

As, given amount of methane is more than the required amount. So, it is considered as an excess reagent.

Thus, oxygen gas is considered as a limiting reagent because it limits the formation of product.

By Stoichiometry of the reaction

2 moles of oxygen gas produces 2 moles of water

So, 0.45 moles of oxygen gas will produce = $\frac{2}{2}\times 0.45=0.45$ moles of water

Now, calculating the mass of water from equation 1, we get:

Molar mass of water = 18 g/mol

Moles of water = 0.45 moles

Putting values in equation 1, we get:

$0.45mol=\frac{\text{Mass of water}}{18g/mol}\\\\\text{Mass of water}=(0.45mol\times 18g/mol)=8.1g$

To calculate the percentage yield of water, we use the equation:

$\%\text{ yield}=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100$

Experimental yield of water = 2.59 g

Theoretical yield of water = 8.1 g

Putting values in above equation, we get:

$\%\text{ yield of water}=\frac{2.59g}{8.1g}\times 100\\\\\% \text{yield of water}=31.98\%$

Hence, the percent yield of the water is 31.98 %

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