Silica, sio2, is formed on silicon as an electrically insulating layer for microelectronic devices. silica is formed when silico
n is exposed to o2 gas at an elevated temperature. at 900˚c, it takes 90 minutes for the oxygen to diffuse from the surface to form a 0.06 micron (0.06 x 10-6 m) thick layer of sio2 on
1 answer:
Answer:
The question continues; At this same temperature, how long (total time, in minutes) would it take for oxygen to diffuse to form a 0.13 micron thick layer of SiO2? Assume that the silicon can be treated as a semi-infinite solid with a constant concentration of oxygen at the surface. (Hint: What is true of the oxygen concentration at the Si/SiO2 interface?
Explanation:
The application of Grahams law of diffusion comes in; It states that at constant temperature and pressure, the rate of diffusion of a gas is inversely proportional to the square root of its relative molecular mass or the square root of its density.
The detailed step by step explanayion and application of grahams law is as shown in the attachment.
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Answer : The molar mass of unknown substance is, 39.7 g/mol
Explanation : Given,
Mass of unknown substance = 9.56 g
Volume of solution = 100.0 mL
Molarity = 2.41 M
Molarity : It is defined as the number of moles of solute present in one liter of volume of solution.
Formula used :
Now put all the given values in this formula, we get:
Therefore, the molar mass of unknown substance is, 39.7 g/mol
Answer: (a) pH = 4.774, (b) pH = 4.811 and (c) pH = 4.681
Explanation: (a) pH of the buffer solution is calculated using Handerson equation:
pKa for acetic acid is 4.76. concentration of base and acid are given as 0.95M and 0.92M. Let's plug in the values in the equation and calculate the pH of starting buffer.
pH = 4.76 + 0.014
pH = 4.774
(b) When 0.040 mol of NaOH (strong base) are added to the buffer then it reacts with 0.040 mol of acetic acid and form 0.040 mol of sodium acetate.
Original buffer volume is 1.00 L. So, the original moles of sodium acetate will be 0.95 and acetic acid will be 0.92.
moles of acetic acid after addition of NaOH = 0.92 - 0.040 = 0.88
moles of sodium acetate after addition of NaOH = 0.95 + 0.040 = 0.99
Let's again plug in the values in the Handerson equation:
pH = 4.76 + 0.051
pH = 4.811
(c) When 0.100 mol of HCl are added then it reacts with exactly 0.100 moles of sodium acetate(base) and form 0.100 moles of acetic acid(acid).
so, new moles of acetic acid = 0.92 + 0.100 = 1.02
new moles of sodium acetate = 0.95 - 0.100 = 0.85
Let's plug in the values in the equation:
pH = 4.76 - 0.079
pH = 4.681