Answer:
The most likely to occur when jagged edges of rock plates grind past each other is the presence of a high degree of frictional force.
This may cause the rocks to be broken down into smaller particles.
It also implies that the energy necessary for further disintegration and movement of rocks is stored up.
The diffusion coefficient for aluminum in silicon is DAl in Si= 4 × 10-13 cm2/s at 1300 K. What is a reasonable value for DAl in Si at 1600 K ? Note: Rather than performing a specific calculation, you should be able to justify your answer from the options below based on the mathematical temperature dependence of the diffusion coefficient assuming a positive activation energy for diffusion.
Answer:
D = 4x10^-11
Explanation:
an increase in temperature would cause a resultant increase in diffusivity. as temperature rises, thermal energy of atoms would also rise and this would cause them to go faster.
4x10^-11cm²/s satisfies this condition because there is a temperature increase from 1300 to 1600.
DAI in Si = 4x10^-13cm²/sec at 1300
DAI in Si at 1600
D increases
temperature also increases
HURRY PLEASE! True or false: Making an observation is the first step of the scientific method.
Answer:
True
Explanation:
You must first observe your data then form a hypothesis.
Argon (Ar) and helium (He) are initially in separate compartments of a container at 25°C. The
Ar in compartment A, which has a volume VA of 6.00 L, has a pressure of 2.00 bar. The He in
compartment B of unknown volume V3 has a pressure of 5.00 bar. When the two compartments
are connected and the gases allowed to mix, the total pressure of gas is 3.60 bar. Assume both
gases behave ideally
(a) [4 marks) Determine the volume of compartment B.
(b) [2 marks] Determine the mole fraction of He in the mixture of gases.
Answer:
(a) [tex]V_B=11.68L[/tex]
(b) [tex]x_{He}=0.533[/tex]
Explanation:
Hello,
In this case, since the both gases behave ideally, with the given information we can compute the moles of He in A:
[tex]n_A=\frac{0.082\frac{atm*L}{mol*K}*298K}{1.974 atm*6.00L}=2.063mol[/tex]
Thus, since the final pressure is 3.60 bar, we can write:
[tex]P=x_{Ar}P_A+x_{He}P_B\\\\P=\frac{n_{Ar}}{n_{Ar}+n_{He}} P_A+\frac{n_{He}}{n_{Ar}+n_{He}} P_B\\\\3.60bar=\frac{2.063mol}{2.063mol+n_{He}} *2.00bar+\frac{n_{He}}{2.063mol+n_{He}} *5.00bar[/tex]
The moles of helium could be computed via solver as:
[tex]n_{He}=2.358mol[/tex]
Or algebraically:
[tex]3.60bar=\frac{1}{2.063mol+n_{He}} *(4.0126+5.00*n_{He})\\\\7.314+3.60n_{He}=4.013+5.00*n_{He}\\\\7.314-4.013=5.00*n_{He}-3.60n_{He}\\\\n_{He}=\frac{3.3}{1.4}=2.358mol[/tex]
In such a way, the volume of the compartment B is:
[tex]V_B=\frac{n_{He}RT}{P_B}=\frac{2.358mol*0.082\frac{atm*L}{mol*K}*298.15K}{4.935atm}\\ \\V_B=11.68L[/tex]
Finally, he mole fraction of He is:
[tex]x_{He}=\frac{2.358}{2.358+2.063}\\ \\x_{He}=0.533[/tex]
Regards.
The Ka values for several weak acids are given below. Which acid (and its conjugate base) would be the best buffer at pH 3.7?a. MES: Ka 7.9 x 10b. HEPES; Ka 3.2 x 103c. Tris; Ka 6.3 x 109d. Formic acid: K 1.8 x 10
e. Acetic acid: K 1.8 x 10
Answer:
Formic acid and Acetic acid is the best buffer at pH 3.7.
Explanation:
Given that,
The Ka values for several weak acids are given,
[tex]K_{a}\ of\ MES=7.9\times10^{-7}[/tex]
[tex]K_{a}\ of\ HEPES = 3.2\times10^{-3}[/tex]
[tex]K_{a}\ of\ Tris=6.3\times10^{-9}[/tex]
[tex]K_{a}\ of\ formic\ acid = 1.8\times10^{-4}[/tex]
[tex]K_{a}\ of\ Acetic\ acid = 1.8\times10^{-5}[/tex]
We need to calculate the pH of the weak acids with their [tex]pK_{a}[/tex] values
Using formula of [tex]pK_{a}[/tex]
For MES,
[tex]pK_{a}=-log K_{a}[/tex]
Put the value into the formula
[tex]pK_{a}=-log(7.9\times10^{-7})[/tex]
[tex]pK_{a}=7.0-log7.9[/tex]
[tex]pK_{a}=6.1[/tex]
pH range for best buffer,
[tex]pH=pK_{a}\pm 1[/tex]
Put the value into the formula
[tex]pH=6.1\pm 1[/tex]
[tex]pH=7.1, 5.1[/tex]
The pH value of the solution between 7.1 to 5.1.
This is not best buffer.
For HEPES,
[tex]pK_{a}=-log K_{a}[/tex]
Put the value into the formula
[tex]pK_{a}=-log(3.2\times10^{-3})[/tex]
[tex]pK_{a}=3.0-log3.2[/tex]
[tex]pK_{a}=2.5[/tex]
pH range for best buffer,
[tex]pH=pK_{a}\pm 1[/tex]
Put the value into the formula
[tex]pH=2.5\pm 1[/tex]
[tex]pH=3.5, 1.5[/tex]
The pH value of the solution between 3.5 to 1.5.
This is not best buffer.
For Tris,
[tex]pK_{a}=-log K_{a}[/tex]
Put the value into the formula
[tex]pK_{a}=-log(6.3\times10^{-9})[/tex]
[tex]pK_{a}=9.0-log6.3[/tex]
[tex]pK_{a}=8.2[/tex]
pH range for best buffer,
[tex]pH=pK_{a}\pm 1[/tex]
Put the value into the formula
[tex]pH=8.2\pm 1[/tex]
[tex]pH=9.2, 7.2[/tex]
The pH value of the solution between 9.2 to 7.2.
This is not best buffer.
For formic acid,
[tex]pK_{a}=-log K_{a}[/tex]
Put the value into the formula
[tex]pK_{a}=-log(1.8\times10^{-4})[/tex]
[tex]pK_{a}=4.0-log1.8[/tex]
[tex]pK_{a}=3.7[/tex]
pH range for best buffer,
[tex]pH=pK_{a}\pm 1[/tex]
Put the value into the formula
[tex]pH=3.7\pm 1[/tex]
[tex]pH=4.7, 2.7[/tex]
The pH value of the solution between 4.7 to 2.7.
This is best buffer.
For acetic acid,
[tex]pK_{a}=-log K_{a}[/tex]
Put the value into the formula
[tex]pK_{a}=-log(1.8\times10^{-5})[/tex]
[tex]pK_{a}=5.0-log1.8[/tex]
[tex]pK_{a}=4.7[/tex]
pH range for best buffer,
[tex]pH=pK_{a}\pm 1[/tex]
Put the value into the formula
[tex]pH=4.7\pm 1[/tex]
[tex]pH=5.7, 3.7[/tex]
The pH value of the solution between 5.7 to 3.7.
This is best buffer
Hence, Formic acid and Acetic acid is the best buffer at pH 3.7.
Find the density of an empty tissue box with a mass of 345 grams and a volume of 1125 cm3.
Answer:
The answer is
0.307 g/cm³Explanation:
The density of a substance can be found by using the formula
[tex]density = \frac{mass}{volume} [/tex]
From the question
mass of box is 345 g
volume = 1125 cm³
The density of the object is
[tex]density = \frac{345}{1125} \\ = 0.3066666...[/tex]
We have the final answer as
0.307 g/cm³Hope this helps you
Consider the fructose-1,6-bisphosphatase reaction. Calculate the free energy change if the ratio of the concentrations of the products to the concentrations of the reactants is 21.321.3 and the temperature is 37.0°C37.0°C ? ΔG°′ΔG°′ for the reaction is −16.7 kJ/mol−16.7 kJ/mol .
Answer:
ΔG = -8.812 kJ/mol
Explanation:
To obtain the free energy of a reaction you can use the expression:
ΔG = ΔG° + RT ln Q
Where:
ΔG° is Standard Gibbs Free energy: -16.7kJ/mol = -16700J/mol
R is gas constant: 8.314472 J/molK
T is absolute temperature (37°C + 273.15 = 310.15K)
And Q is reaction quotient: 21.3
Replacing in the formula:
ΔG = ΔG° + RT ln Q
ΔG = -16700J/mol + 8.314472J/molK*310.15K ln 21.3
ΔG = -8812.4J/mol
ΔG = -8.812 kJ/mol
general characteristics of coinage metals
Answer:
Characteristics. They are all relatively inert, corrosion-resistant metals. Copper and gold are colored. These elements have low electrical resistivity so they are used for wiring.
Explanation:
Hope this help u
️️
A beaker contains 9.80 L
of water. What is the
volume in quarts
Answer:
10.4 qt
Explanation:
Step 1: Given data
Volume of water in the beaker (V): 9.80 L
Step 2: Convert the volume of water in the beaker to US quarts (qt)
In order to convert one unit into another, we need a conversion factor. In this case, the appropriate conversion factor is 1 L = 1.06 qt. The volume of water in the beaker, in US quarts, is:
9.80 L × (1.06 qt/1 L) = 10.4 qt
[H+] for a solution is 4.59 x 10-6 M.
This solution is
A. acidic
B. basic
C. neutral
Answer:
A. acidic
Explanation:
Answer:
acidic
Explanation:
The next two questions provide some more practice on calculations using half-lives. The isotope 64Cu has t1/2 of 12.7 hours. If the initial concentration of this isotope in an aqueous solution is 845 ppm, what will the abundance be after 4.00 hours? To solve this problem, first use equation (7) to determine k for 64Cu; then use this k value in equation (6) to obtain the amount of 64Cu, A, remaining after 4.00 hours if the amount present at the start, A0, is 845 ppm.
Answer:
The value is [tex] A = 679.5 \ ppm[/tex]
Explanation:
From the question we are told that
The half life of [tex]^{64}Cu[/tex] is [tex] t_h = 12.7 \ hr [/tex]
The initial concentration is [tex] A_o = 845 \ ppm [/tex]
The time duration is [tex] t = 4 \ hr [/tex]
Generally the rate constant is mathematically represented as
[tex] k = \frac{0.693}{t_h} [/tex]
[tex] k = \frac{0.693}{12.7} [/tex]
[tex] k = 0.0545 \ hr^{-1} [/tex]
This rate constant is also mathematically represented as
[tex] k = \frac{1}{t} * ln (\frac{A_o}{A}) [/tex]
Here A is the remaining concentration after t
So
[tex] 0.0545 = \frac{1}{4} * ln (\frac{845}{A}) [/tex]
[tex] 0.218 = ln (\frac{845}{A}) [/tex]
[tex] e^{0.218} = \frac{845}{A} [/tex]
[tex] 1.2436 = \frac{845}{A} [/tex]
[tex] A = \frac{845}{1.2436} [/tex]
[tex] A = 679.5 \ ppm[/tex]
