# Search results

1. ### PH of buffer; no Ka/Kb given

Thank you; this is the point I was getting hung up on. Knowing that I'm not expected to deduce the Ka I'm willing to resort to known values.
2. ### PH of buffer; no Ka/Kb given

Homework Statement A 100.0mL buffer solution is 0.175M in HClO and 0.150M in NaClO. Find the pH of the solution. Homework Equations Ka=10^-pH ([H30+][A-])/([HA]) -log([H30+]) = pH The Attempt at a Solution [HClO] [H3O+] [ClO-] I .175 0 .150...
3. ### Equilibrium Partial Pressure

After much reworking I found my error. Thanks for your input Borek.
4. ### Equilibrium Partial Pressure

I got +/- x by solving (x*x)/(.725-x)^2 quadratically. I'm not sure where to go past this to find partial pressures.
5. ### Equilibrium Partial Pressure

Homework Statement At 600 K the equilibrium constant is 38.6 for the reaction 2 HI(g) <--> H2(g) + I2(g) Start with a pressure of 0.725 atm consisting of only HI(g) at 600 K. Calculate the equilibrium partial pressures of HI, H2 and I2. Homework Equations K=[C]^n[D]^n/[A]^n Qp =...
6. ### Circuit Power

After punching it through again I got the correct answer; I went wrong in the way I set up Hz; punching in 47.5^-1, since Hz = S^-1; but thats redundant...
7. ### Circuit Power

Homework Statement A generator with an rms voltage of 130V is connected in series to a resistor 3.65k\Omega and a capacitor 3.00\muF. Using the frequency found in part A (47.5Hz), find the average power consumed by this circuit. \phi = 17degrees Homework Equations Latex isn't...
8. ### Resistance in a light bulb

Hrmm... P = V2/R PR = V2 R = V2/P R = 160\Omega That works better huh? Thank you!
9. ### Resistance in a light bulb

Homework Statement A "90-watt" light bulb uses an average power of 90W when connected to an rms voltage of 120V . What is the resistance of the lightbulb? Homework Equations Pav=(V2rms/R) The Attempt at a Solution 90W = 120V2/R 90W/120V2 = R R = .00625\Omega
10. ### Current through resistors

Homework Statement Consider the circuit shown in the figure . Suppose the four resistors in this circuit have the values R1 = 12 \Omega, R2 = 6.4 \Omega, R3 = 7.0 \Omega, and R4 = 13 \Omega, and that the emf of the battery is \epsilon = 18V . A. Find the current through each resistor...
11. ### Electric Field & Point Charge Potential Energy

Wrong equation used: \DeltaU=q0Ed \DeltaU=8x10-6C* -8.5x105N/C * 6m = -40.8J
12. ### Electric Field & Point Charge Potential Energy

Homework Statement A uniform electric field of magnitude 8.5×105 N/C points in the positive x direction. Find the change in electric potential energy of a 8.0 \muC charge as it moves from the origin to the point (6.0 m , 0). Homework Equations 1 N/C = 1 V/m \DeltaU = q\DeltaV E =...
13. ### Electric Field Magnitude

Great, thats exactly the kick I needed. Thank you.
14. ### Electric Field Magnitude

Homework Statement The figure shows an electron entering a parallel-plate capacitor with a speed of 5.4×106m/s. The electric field of the capacitor has deflected the electron downward by a distance of 0.618cm at the point where the electron exits the capacitor. Find the magnitude of the...
15. ### Electrostatic Force between point charges

Thanks so much for the help
16. ### Electrostatic Force between point charges

So F=k((|q1||q2|)/r2) 0.67 N = k((|8.12e-6 C||Q|)/(7.26m2)) solve for Q?
17. ### Electrostatic Force between point charges

Homework Statement The attractive electrostatic force between the point charges 8.12×10−6C and Q has a magnitude of 0.670N when the separation between the charges is 7.26m. Find the sign and magnitude of the charge Q. (this is a direct cut and paste) Homework Equations...
18. ### Pulley coursework question

Thanks; that makes total sense.
19. ### Pulley coursework question

Homework Statement A string is wrapped tightly around a fixed pulley that has a moment of inertia of 0.0352kg/m2; and a radius of 12.5cm. A mass of 423g is attached to the free end of the string. The mass is allowed to fall under the influence of gravity. As the mass falls the string causes the...