Material engineering studies the physical behavior of metallic elements.
Answer: Option C
<u>Explanation:
</u>
Material Engineering is the creation and learning about the materials at an atomic level. An engineers from this branch focus on material and model its characteristics using the computer.
Also, they combine the knowledge of solid-states, metallurgy, chemistry and ceramics to the application level. It also has a great role in building the future with the advancing study in nanotechnology, biotechnology, etc. Simply, these are meant to have vivid applications in future life.
9514 1404 393
Answer:
13/80
Explanation:
The product is ...
(1 3/10)×(1/8) = (13/10)×(1/8) = (13×1)/(10×8) = 13/80
Answer:
Coat new O-rings (D) with silicone oil or polyalkyleneglycol (PAG) oil, and pull them on the injectors.
Answer:
k = 4.21 * 10⁻³(L/(mol.s))
Explanation:
We know that
k = Ae
------------------- euqation (1)
K= rate constant;
A = frequency factor = 4.36 10^11 M⁻¹s⁻¹;
E = activation energy = 93.1kJ/mol;
R= ideal gas constant = 8.314 J/mol.K;
T= temperature = 332 K;
Put values in equation 1.
k = 4.36*10¹¹(M⁻¹s⁻¹)e![^{[(-93.1*10^3)(J/mol)]/[(8.314)(J/mol.K)(332K)}](https://tex.z-dn.net/?f=%5E%7B%5B%28-93.1%2A10%5E3%29%28J%2Fmol%29%5D%2F%5B%288.314%29%28J%2Fmol.K%29%28332K%29%7D)
k = 4.2154 * 10⁻³(M⁻¹s⁻¹)
here M =mol/L
k = 4.21 * 10⁻³((mol/L)⁻¹s⁻¹)
or
k = 4.21 * 10⁻³((L/mol)s⁻¹)
or
k = 4.21 * 10⁻³(L/(mol.s))
