No you may not ask the question
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))
Answer: Even low airborne concentrations (100 ppm) of ammonia may produce rapid eye and nose irritation.
Answer:
endurance length is 236.64 MPa
Explanation:
data given:
d = 37.5 mm
Sut = 760MPa
endurance limit is
Se = 0.5 Sut
= 0.5*760 = 380 MPa
surface factor is
Ka = a*Sut^b
where
Sut is ultimate strength
for AISI 1040 STEEL
a = 4.51, b = -0.265
Ka = 4.51*380^{-0.265}
Ka = 0.93
size factor is given as
Kb =1.29 d^{-0.17}
Kb = 0.669
Se = Sut *Ka*Kb
= 380*0.669*0.93
Se = 236.64 MPa
therefore endurance length is 236.64 MPa
Answer:
Explanation:
B. you would grab the plug closest to the outlet