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.
n = 1.5atm (15L) / .0821 (280k) = .98 mol NaCl
NaCl = 22.99g Na + 35.45g Cl = 58.44g NaCl
58.44g NaCl x .98 mol NaCl = 57.27g NaCl
Explanation:
hope you get it right :)
Answer:
1) 17.5 mL
Explanation:
Hello,
In this case, the reaction between sulfuric acid and potassium hydroxide is:

In such a way, we notice a 1:2 molar ratio between the acid and the base, therefore, at the equivalence point we have:

And in terms of concentrations and volumes:

Thus, we solve for the volume of acid:

Best regards.
<span>iron gets hotter before copper.</span>
The complete table is inserted.
A table is given,
Formulas used:
pH= -log(H⁺)
pOH= -log(OH⁻)
pH+ pOH=14
Calculations:
For A: (H⁺)=2×10⁻⁸M
Using the pH formula:
pH= -log(H⁺)=-log(2×10⁻⁸)=7.69
pOH=14 - 7.69=6.3
Calculating OH concentration,
pOH= -log(OH⁻)
6.3= -log(OH⁻)
(OH⁻)=5.011×10⁻⁷M
Hence, the nature of A is basic.
Similarily,
For B,
(OH⁻)=1×10⁻⁷
Using the pH formula:
pOH= -log(OH⁻)= -log(1×10⁻⁷)=7
pH=14-7=7
Calculating H concentration,
pH= -log(H⁺)
7= -log(H⁺)
(H⁺)=1×10⁻⁷M
Hence, the nature of B is neutral.
Similarily,
For C,
pH=12.3
Using the pH formula:
pOH=14-12.3=1.7
Calculating H concentration,
pH= -log(H⁺)
12.3= -log(H⁺)
(H⁺)=5.011×10⁻¹³M
Calculating OH concentration,
pOH= -log(OH⁻)
1.7= -log(OH⁻)
(OH⁻)=1.99×10⁻²M
Hence, the nature of C is Basic.
Similarily,
For D,
pOH=6.8
Using the pH formula:
pH=14-6.8=7.2
Calculating H concentration,
pH= -log(H⁺)
7.2= -log(H⁺)
(H⁺)=6.309×10⁻⁸M
Calculating OH concentration,
pOH= -log(OH⁻)
6.8= -log(OH⁻)
(OH⁻)=1.58×10⁻⁷M
Hence, the nature of D is basic.
Learn more about the acid and bases here:
brainly.com/question/16189013
#SPJ10