Compared to carbon nanotube, carbon nanofiber (CNF) is a unique quasi-one-dimensional nanostructure with a lot of edges and flaws (CNT). Additionally, their low cost and wide availability make them a valuable nanomaterial for upcoming technology.
<h3>what are the development and characterization of Carbon Nanofiber for Additively Manufactured Piezo resistive Sensors?</h3>
In accordance with the semiconductor material's piezo resistive effect, diffusion resistance is used to manufacture piezo resistive sensors on substrates of semiconductor materials. The diffusion resistor is connected in the substrate in the form of a bridge, allowing the substrate to be employed directly as a measuring sensor element.
- Carbon nanofiber/polylactic acid filament for fused filament fabrication (FFF) and additive manufacturing (AM) strain sensors was studied for the effects of production factors.
- To investigate the effects of CNF weight fraction, extrusion temperature, and number of extrusions on sensor performance, a design of experiments (DOE) approach was used. In the initial extrusion, dry melt mixing was used to combine CNFs and powdered PLA material.
- Through the DOE procedure, it was discovered that extruding CNF/PLA material for two complete extrusions at 185 °C resulted in material with material with material with dramatically improved electrical characteristics in comparison to unmodified material.
- Piezoresistive dog-bone shaped sensors were made using the best manufacturing technique using three different sizes of 5.0, 7.5, and 10.0 wt% CNF/PLA filament.
To know more about Carbon nanofiber/polylactic acid check here:brainly.com/question/15913091
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The mass of NaCl formed is 8.307 grams
<u><em> calculation</em></u>
step 1: write the equation for reaction
Na₂CO₃ + 2HCl → 2 NaCl +CO₂ +H₂O
Step 2: find the moles of Na₂CO₃
moles = mass/molar mass
The molar mass of Na₂CO₃ is = (23 x2) + 12 + ( 16 x3) = 106 g/mol
moles = 7.5 g/106 g/mol =0.071 moles
Step 3: use the mole ratio to determine the mole of NaCl
Na₂CO₃:NaCl is 1:2 therefore the moles of NaCl =0.07 x2 =0.142 moles
Step 4: calculate mass of NaCl
mass= moles x molar mass
the molar mass of NaCl= 23 +35.5 =58.5 g/mol
mass = 0.142 moles x 58.5 g/mol =8.307 grams
Hey there!
Balance the equation:
SiCl₄ + H₂O → H₄SiO₄ + HCl
Balance H.
2 on the left, 5 on the right. Add a coefficient of 3 in front of H₂O and a coefficient of 2 in front of HCl.
SiCl₄ + 3H₂O → H₄SiO₄ + 2HCl
Balance O.
3 on the left, 4 on the right. Change the coefficient of 3 in front of H₂O to a 4.
SiCl₄ + 4H₂O → H₄SiO₄ + 2HCl
This unbalanced our H, so change the coefficient of 2 in front of HCl to a 4.
SiCl₄ + 4H₂O → H₄SiO₄ + 4HCl
Balance Cl.
4 on the left, 4 on the right. Already balanced.
Balance Si.
1 on the left, 1 on the right. Already balanced.
Our final balanced equation:
SiCl₄ + 4H₂O → H₄SiO₄ + 4HCl
Hope this helps!
Answer:
avogadro's constant
Explanation:
this is the fixed number of the atoms in the molecule of an element
avogadro's law states that equal volumes of gases<em> </em><em>at</em><em> </em><em>the</em><em> </em><em>same</em><em> </em><em>temperature</em><em> </em><em>and</em><em> </em><em>pressure</em><em> </em><em> </em><em>contain</em><em> </em><em>equal</em><em> </em><em>numbers</em><em> </em><em>of</em><em> </em><em>molecules</em><em> </em>
<em>that</em><em> </em><em>is</em><em> </em><em>all</em><em> </em><em>gases</em><em> </em><em>with</em><em> </em><em>same</em><em> </em><em>temperature</em><em> </em><em>and</em><em> </em><em>pressure</em><em> </em><em>will</em><em> </em><em>always</em><em> </em><em>have</em><em> </em><em>equal</em><em> </em><em>numbers</em><em> </em><em>of</em><em> </em><em>molecules</em><em> </em>
Answer: 
Explanation:
According to the Arrhenius equation,
or,
![\log (\frac{K_2}{K_1})=\frac{Ea}{2.303\times R}[\frac{1}{T_1}-\frac{1}{T_2}]](https://tex.z-dn.net/?f=%5Clog%20%28%5Cfrac%7BK_2%7D%7BK_1%7D%29%3D%5Cfrac%7BEa%7D%7B2.303%5Ctimes%20R%7D%5B%5Cfrac%7B1%7D%7BT_1%7D-%5Cfrac%7B1%7D%7BT_2%7D%5D)
where,
= rate constant at 
= 
= rate constant at 
= 
= activation energy for the reaction = 262 kJ/mol = 262000J/mol
R = gas constant = 8.314 J/mole.K
= initial temperature =
= final temperature = 
Now put all the given values in this formula, we get
![\log (\frac{6.1\times 10^{-8}}{K_2})=\frac{262000}{2.303\times 8.314J/mole.K}[\frac{1}{600.0K}-\frac{1}{775.0K}]](https://tex.z-dn.net/?f=%5Clog%20%28%5Cfrac%7B6.1%5Ctimes%2010%5E%7B-8%7D%7D%7BK_2%7D%29%3D%5Cfrac%7B262000%7D%7B2.303%5Ctimes%208.314J%2Fmole.K%7D%5B%5Cfrac%7B1%7D%7B600.0K%7D-%5Cfrac%7B1%7D%7B775.0K%7D%5D)
Therefore, the value of the rate constant at 775.0 K is
