Partial Pressure Help---Please Check
The Partial Pressure of water vapor at T = 273 K is 6.11 mbar. Find the corresponding mass concentration of water vapor. Assume P = 1 atm.
So....
ρH20 = (PH20*MH20/(R*T)
ρH20= (6.11*10^2 Pa)*(18.01 g/mol)/(8.314*273 K)
ρH20 = 11004.11/2269.722 =...
A typical global concentration of hydroxii radicals (OH) is about 10^6 moleculues/cm^3. What is the mixing ratio corresponding to this concentration at sea level P = 10^5 Pa and T=298 K?
Using the previous equations derived earlier
ndOH= (P*Av/R*T)*COH
(ndOH*R*T/(P*Av) ) = COH
(10^6...
Much better now?
Let's try this.
Please check if right.
Determine the mass concentration for N20 at a mixing ratio of 311 ppb at P = 1 atm, T= 298 K.
My work
nd N20= CN20 * ndair
P*V = nmoles air* R * T
(P*V)/(RT) = nmoles air
(P*V)/(R*T) = N molecules air/Av = nmoles...
Please check if right.
Determine the mass concentration for N20 at a mixing ratio of 311 ppb at P = 1 atm, T= 298 K.
My work
#density of N20 = Concentration of N20 * #density air
PV = #moles of air R T
PV/RT = # moles of air
PV/RT = Numberofmolecules/Avagardo # = #moles of air...
Its the part c question....which states
C) Suppose we made V = 0. Find the new current running through the 5.00 ohm resistor.
So I just took the biggest loop and did it this way
Sum of V = 0
So starting at the upper left hand corner of the circuit
-3A(15 Ohms) + 50 V + I (5) = 0
-45 + 50...
OHHHH...I think I understand now
You use OHM's Law
V = IR = IoR = Vo/R e (-t/taw) so that value x 5.00 E 3 ohms = ______ V
V(t) = _____ V (from last line) (1-e(-t/RCeq) = ______ V
Q = 3 E-6 F x ____ V (that I solved previously) = _________ C
So I plug that into Q of the energy equation to...
Ok so you are saying this:
Ceq = 1/C1 + 1/C2 = 1/12 E -6 + 1/4 E -6 but its the inverse because we are trying to find Ceq from 1/Ceq. So I got 3 E-6 F.
I dont know what you mean to find the total charge?
Ok try to review for the finals but I need help in understanding this problem.
So I can say that there is a resistance, 2 capacitors in series, and voltage source (Vo).
Vo = 20.0 V
R = 5.00 kilo ohm
C1 = 12.0 micro F
C2 = 4.0 micro F
Switch is closed at t = 0.
How much energy will...
\frac{k}{\pi}\int \frac{-Q}{R^3}\mathrm{d}l =
\frac{-k}{\pi R^2}\int {Q}\mathrm{d}l
Where do I go from here?
Because its supposed to be
E = \frac{-2kQ}{\pi R^2}