Answer:
The average atomic mass of sulfur is 32.065u
Answer:
P2 = 2.25 atm
Explanation:
Given:
V1 = 1.5 L. V2 = 2.0 L
P1 = 3.0 atm. P2 = ?
Use Boyle's law and solve for P2:
P1V1 = P2V2
or
P2 = (V1/V2)P1
= (1.5 L/2.0L)(3.0 atm)
= 2.25 atm
Out of the given options, the charge of an electron was not contributed by Neils Bohr.
Answer: Option 2
<u>Explanation:
</u>
The electron's charge was determined by using oil drop experiments performed by Millikan. While Neil Bohr suggested that electrons are rotating in discrete energy levels termed as orbits.
The hydrogen model of Neil Bohr dealt with the quantum energy emission when electron excite from higher energy level to lower energy level. He also stated that there is a relationship between the outer shell and the chemical properties of elements.
So, the second option that is the charge of an electron is not contributed by Bohr as it was contributed by Millikan.
?? Is that the whole question?
Answer:
Explanation:
From the information given:
oxidation of oxidized solution takes place at 950° C
For wet oxidation:
The linear and parabolic coefficient can be computed as:
![\dfrac{B}{B/A} = D_o \ exp \Big [\dfrac{-\varepsilon a}{k_BT} \Big]](https://tex.z-dn.net/?f=%5Cdfrac%7BB%7D%7BB%2FA%7D%20%3D%20D_o%20%5C%20exp%20%5CBig%20%5B%5Cdfrac%7B-%5Cvarepsilon%20a%7D%7Bk_BT%7D%20%5CBig%5D)
Using 
and 
values obtained from the graph:
Thus;
![\dfrac{B}{A} = 1.63 \times 10^8 exp \Big [ \dfrac{-2.05}{8.617 \times 10^{_-5}\times 1173}\Big] \\ \\ = 0.2535 \ \ \mu m/hr](https://tex.z-dn.net/?f=%5Cdfrac%7BB%7D%7BA%7D%20%3D%201.63%20%5Ctimes%2010%5E8%20exp%20%5CBig%20%5B%20%5Cdfrac%7B-2.05%7D%7B8.617%20%5Ctimes%2010%5E%7B_-5%7D%5Ctimes%201173%7D%5CBig%5D%20%5C%5C%20%5C%5C%20%3D%200.2535%20%5C%20%5C%20%5Cmu%20m%2Fhr)
![B= 386 \ exp \Big [-\dfrac{0.78}{8.617 \times 10^{-3} \times 1173} \Big] \\ \\ = 0.1719 \ \mu m^2/hr](https://tex.z-dn.net/?f=B%3D%20386%20%5C%20%20exp%20%5CBig%20%5B-%5Cdfrac%7B0.78%7D%7B8.617%20%5Ctimes%2010%5E%7B-3%7D%20%5Ctimes%201173%7D%20%5CBig%5D%20%5C%5C%20%5C%5C%20%20%3D%200.1719%20%5C%20%5Cmu%20m%5E2%2Fhr)
So, the initial time required to grow oxidation is expressed as:
∴
NOW;
Thus; since we will consider the positive sign, the initial thickness 
is ;
≅ 0.261 μm
