Answer:
They become ductile and deform plastically
Explanation:
When rocks are buried by the materials up to a greater depth, then the confining pressure increases significantly. This results in the ductile behavior of the rocks at such depth. These rocks are present in the ductile region where the depth is about more than 20 to 30 km. Here the rocks are subjected to extremely high pressure and temperature conditions, which favors the transformation of rocks into more higher-grade metamorphic rocks. It is also enhanced due to the geothermal gradient.
Under such high pressure and temperature, the rocks show the behavior of plasticity, where the rocks undergo bending, buckling as well as they tend to flow, and there occurs low strain rate, resulting in the permanent deformation of rocks.
Thus, the rocks become ductile and deform plastically at such conditions.
Answer:- 171 g
Solution:- It asks to calculate the grams of sucrose required to make 1 L of 0.5 Molar solution of it.
We know that molarity is moles of solute per liter of solution.
If molarity and volume is given then, moles of solute is molarity times volume in liters.
moles of solute = molarity* liters of solution
moles of solute = 0.5*1 = 0.5 moles
To convert the moles to grams we multiply the moles by molar mass.
Molar mass of sucrose = 12(12) + 22(1) + 11(16)
= 144 + 22 + 176
= 342 grams per mol
grams of sucrose required = moles * molar mass
grams of sucrose required = 0.5*342 = 171 g
So, 171 g of sucrose are required to make 1 L of 0.5 molar solution.
Answer:
A mixture of 100. mL of 0.1 M HC3H5O3 and 50. mL of NaOH
Explanation:
The pH of a buffer solution is calculated using following relation
Thus the pH of buffer solution will be near to the pKa of the acid used in making the buffer solution.
The pKa value of HC₃H₅O₃ acid is more closer to required pH = 4 than CH₃NH₃⁺ acid.
pKa = -log [Ka]
For HC₃H₅O₃
pKa = 3.1
For CH₃NH₃⁺
pKa = 10.64
pKb = 14-10.64 = 3.36 [Thus the pKb of this acid is also near to required pH value)
A mixture of 100. mL of 0.1 M HC3H5O3 and 50. mL of NaOH
Half of the acid will get neutralized by the given base and thus will result in equal concentration of both the weak acid and the salt making the pH just equal to the pKa value.
The s sublevel has just one orbital, so can contain 2 electrons max. The p sublevel has 3 orbitals, so can contain 6 electrons max. The d sublevel has 5 orbitals, so can contain 10 electrons max. And the 4 sublevel has 7 orbitals, so can contain 14 electrons max.