Answer:
3.52×10⁶ atoms of Si and 7.05×10⁶ atoms of O
Explanation:
It is all about unit conversions.
The area of our MOS is 1 mm². So, we know that the thickness is 160 nm. This data can give us the volume. We convert nm to mm.
160 nm . 1×10⁻⁶ mm /1nm = 1.6×10⁻⁴ mm
By the way, now we can determine the volume of MOS, in order to work with density.
1.6×10⁻⁴ mm . 1 mm² = 1.6×10⁻⁴ mm³
But density is mg/m³, so we convert mm³ to m³
1.6×10⁻⁴ mm³ . 1×10⁻⁹ m³/mm³ = 1.6×10⁻¹³ m³
Now, we apply density to determine the mass of MOS
Density = mass /volume → Density . volume = mass
1.6×10⁻¹³ m³ . 2.20mg/m³ = 3.52×10⁻¹³ mg
To make more easier the calculate, we convert mg to g.
3.52×10⁻¹³ mg . 1g /1000mg = 3.52×10⁻¹⁶ g
To count the atoms, we determine molar mass of SiO₂ → 60.08 g/mol
We need to know moles of Si and O₂ in the MOS
Firstly, we determine amount of MOS: 3.52×10⁻¹⁶ g / 60.08 g/mol = 5.86×10⁻¹⁸ moles
1 mol of SiO₂ has 1 mol of Si and 2 mol of O so:
5.86×10⁻¹⁸ mol of SiO₂ may have:
(5.86×10⁻¹⁸ . 1) /1 = 5.86×10⁻¹⁸ moles of Si
(5.86×10⁻¹⁸ .2) /1 = 1.17×10⁻¹⁷ moles of O₂
Let's count the atoms (1 mol of anything contain NA particles)
5.86×10⁻¹⁸ mol of Si . 6.02×10²³ atoms/ mol = 3.52×10⁶ atoms of Si
1.17×10⁻¹⁷ mol of O₂ . 6.02×10²³ atoms/ mol = 7.05×10⁶ atoms of O
The number of Si and O atoms present per mm² of the oxide layer are respectively; 3.52 × 10⁶ atoms of Si and 7.05×10⁶ atoms of O
What is the number of atoms present?We are given;
Area of MOS device; A = 1 mm²
Thickness of layer; t = 160 nm = 1.6 × 10⁻⁴ mm
Formula for volume is;
V = Area * thickness
V = 1.6 × 10⁻⁴ mm × 1 mm² = 1.6 × 10⁻⁴ mm³
Converting volume to m³ gives;
V = 1.6 × 10⁻¹³ m³
Now, to get the mass of MOS, we will use the formula;
Mass = Density * volume
we are given density = 2.20mg/m³
Thus;
Mass = 1.6 × 10⁻¹³ m³ × 2.20mg/m³
Mass = 3.52 × 10⁻¹³ mg = 3.52×10⁻¹⁶ g
From periodic table, the molar mass of SiO₂ = 60.08 g/mol
Then;
Number of moles of MOS = (3.52 × 10⁻¹⁶ g)/60.08 g/mol
Number of moles of MOS = 5.86 × 10⁻¹⁸ moles
Now, 1 mol of SiO₂ is formed from 1 mol of Si and 2 mol of O. Thus;
5.86×10⁻¹⁸ mol of SiO₂ will have;
5.86×10⁻¹⁸ moles of Si
and (5.86×10⁻¹⁸ .2) = 11.72 × 10⁻¹⁸ moles of O₂
From Avogadro's number that 1 mol equals 6.02×10²³ atoms , we can say that;
5.86 × 10⁻¹⁸ mol of Si × 6.02 × 10²³ atoms/mol = 3.52 × 10⁶ atoms of Si
11.72 × 10⁻¹⁸ mol of O₂ × 6.02 × 10²³ atoms/mol = 7.05×10⁶ atoms of O
Read more about number of atoms at; https://brainly.com/question/3157958
After earning a bachelor's degree, one must do which of the following before taking the PE examination to receive a Professional Engineering license?
Air at 25 m/s and 15°C is used to cool a square hot molded plastic plate 0.5 m to a side having a surface temperature of 140°C. To increase the throughput of the production process, it is proposed to cool the plate using an array of slotted nozzles with width and pitch of 4 mm and 56 mm, respectively, and a nozzle-to-plate separation of 40 mm. The air exits the nozzle at a temperature of 15°C and a velocity of 10 m/s.
Required:
a. Determine the improvement in cooling rate that can be achieved using the slotted nozzle arrangement in lieu of turbulated air at 10 m/s and 15°C in parallel flow over the plate.
b. Would the heat rates for both arrangements change significantly if the air velocities were increased by a factor of 2?
c. What is the air mass rate requirement for the slotted nozzle arrangement?
Answer:
a. 2.30
b. decreases with increasing velocity.
c. 0.179 kg/s.
Explanation:
Without mincing let's dive straight into the solution to the question above.
[a].
The improvement in cooling rate that can be achieved using the slotted nozzle arrangement in lieu of turbulated air at 10 m/s and 15°C in parallel flow over the plate can be determined by calculating turbulent flow:
The turbulent flow over the plate= 10 × 0.5/ 20.92 × 10⁻6 = 2.39 × 10⁵.
While the turbulent flow correlation = 0.037( 2.39 × 10⁵)^[tex]\frac{4}{5}[/tex] (0.7)^[tex]\frac{1}{3}[/tex] = 659.6.
Array of slot noozle = [10 × (2 × 0.004)]/ 20.92 × 10^-6] = 3824.
where A = 4/56 =0.714.
And Ar = [ 60 + 4 (40/2 × 4) - 2 ]^2 ]-1/2 = 0.1021.
N = 2/3 (0.1021)^3/4 [ 2 × 3824/ ( 0.0714 / 0.1021) + (.1021/0.0714)] (0.700)^0.42 =24.3.
h = 24.3 × 0.030/0.004 = 91.1 W/m^2k.
Therefore; 659.6 × 0.030/0.5 = 39.0 W/m²k.
The turbulent flow = 0.5 × 39.6 × 0.5( 140 -15) = 1237.5 W.
The slot noozle = 91.1 × 0.5 × 0.5 [ 140 -15] = 2846.87W.
The improvement in cooling rate = 2846.87/ 1237.5 = 2.30.
[b].
2.3 [ (2^2/3)/ 2^4/5] = 2.1
Thus, it decreases with increasing velocity
[c].
The air mass rate requirement for the slotted nozzle arrangement = 9 × 0.995 (0.5 × 0.004)10 = 0.179 kg/s.
Leland wants to work in a Production career operating heavy machinery. Which type of education or training should Leland seek?
a bachelor’s degree then a master’s degree
vocational school certificate or master’s degree
on-the-job training or vocational school certificate
associate’s degree then a bachelor’s degree
Answer:
it is indeed C
Explanation:
Answer:
c
Explanation: