The control room is similar to the nucleus, because the nucleus is in charge of cell functions.
The answer is it tends to be more negative down a group. This is because as you go down the periodic table, the elements have more electron shells in their atoms. This makes the outermost shells less attracted to the nucleus due to their greater distances from the nucleus. Therefore, these shells are less likely to attract electrons (hence lower electron affinity) and are even more likely to lose electrons from their outer electron orbits.
When we have this equation:
CO(g) + Cl2(g) ↔ COCl2(g)
intial 0.147 0.175 0
change -X -X +X
final (0.147-X) (0.175-X) X
so from the ICE table, we substitute in Kc formula :(when we have Kc = 255)
Kc = [COCl2]/[CO][Cl2]
255= X / (0.147-X)(0.175-X)
255 = X / (X^2 - 0.322 X + 0.025725)
X = 0.13
∴[CO] = 0.147 - X = 0.147 - 0.13
= 0.017 m
Answer:
NA = 6.8 E-12 Kg H2(g) / hour
Explanation:
steady-state diffusion of A through non-diffuser B:
- NA = (DAB/RTz)(p*A1 - p*A2)
∴ (A): H2(g)
∴ (B): Pd
∴ DAB = 1.7 E-8 m²/s
∴ p*A1 = 2.0 Kg H2 / m³ Pd
∴ p*A2 = 0.4 Kg H2 / m³ Pd
∴ z = 6 mm = 6 E-3 m
∴ T = 600°C ≅ 873 K
∴ R = 8.314 J/mol.K = 8.314 N.m/mol.K
⇒ NA = ((1.7 E-8)/(8.314)(873)(6 E-3))(2.0 - 0.4)
⇒ NA = 6.246 E-10 mol/s.m³
for A = 0.25 m²
⇒ volume (v) = A×z = (0.25)(6 E-3) = 1.5 E-3 m³
∴ Mw H2(g) = 2.016 g/mol
⇒ NA = (6.246 E-10 mol/s.m³)(1.5 E-3 m³)(2.016 g/mol)(Kg/1000 g)(3600 s/h)
⇒ NA = 6.8 E-12 Kg H2(g)/h
For a Forty miles above the earth's surface, the temperature at 250k and the pressure at only 0.20 torr, the density of air at this altitude is mathematically given as
d=3.732*10{-4}g/l
<h3>What is the density of air at this altitude?</h3>
Generally, the equation for the ideal gas is mathematically given as
PV=(m/mw)RT
Therefore
d=m/v
d=26,664*20/83.14*250
d=0.37152
d=3.732*10{-4}g/l
In conclusion, the density
d=3.732*10{-4}g/l
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