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:
108.6 g
Explanation:
- 2NaN₃(s) → 2Na(s) + 3N₂(g)
First we use the <em>PV=nRT formula</em> to <u>calculate the number of nitrogen moles</u>:
- R = 0.082 atm·L·mol⁻¹·K⁻¹
- T = 0 °C ⇒ 0 + 273.2 = 273.2 K
<u>Inputting the data</u>:
- 1.00 atm * 56.0 L = n * 0.082 atm·L·mol⁻¹·K⁻¹ * 273.2 K
Then we <u>convert 2.5 moles of N₂ into moles of NaN₃</u>, using the <em>stoichiometric coefficients of the balanced reaction</em>:
- 2.5 mol N₂ *
= 1.67 mol NaN₃
Finally we <u>convert 1.67 moles of NaN₃ into grams</u>, using its <em>molar mass</em>:
- 1.67 mol * 65 g/mol = 108.6 g
Let us say that R is the major enantiomer, while
S is the minor enantiomer, therefore the formula for enantiomeric excess (ee)
is:
ee = (R – S) * 100%
Let us further say that the fraction of R is x (R
= x), and therefore fraction of S is 1 – x (S = 1 – x), therefore:
75 = (x – (1 – x)) * 100
75 = 100 x – 100 + 100 x
200 x = 175
x = 0.875
Summary of answers:
R = major enantiomer = 0.875 or 87.5%
<span>S = minor enantiomer = (1 – 0.875) = 0.125 or
12.5%</span>
