Recent content by Nimmy

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 =...

Thanks but otherwise the correct process right?

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...

Perfect that matches with what I wrote it seems. Thanks. :)

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 +5I...

Thank you! Sorry for being hard head! lol.

Can I use this instead... q = Ceq Vo (1-e^(-t/RC) To find the charge Q?

Just making sure...the answer I posted was correct right?

Huh? I revised my post...

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 don't 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}