suggest formula for number of electrons if given filament of 60 W electric bulb and supply is 200 Vd.c.?
924435
1 Find the current flowing in the lamp.   P  = VI\n2 Look up the charge on an electron.     q\n\nDivide 1  by 2  which will give you the number of electrons flowing per second.      n  =  I/q
333904
It depends on the Voltage.   Imagine an electron (which has known charge q = 1.6x10^-19 and mass m = 9.11x10^-31) that is inside a wire which is a perfect conductor.   Since the electron will move in a perfect conductor, all the potential energy given by the power source (Voltage * electron charge) will transform into kinetic energy (K = (m*v^2)/2), so:\n\n(m*v^2)/2 = V*q     ;     where v is the speed of the electron and V is the voltage\n\nWith this we get:   v = (2*V*q/m)^(1/2)\n\nAll the electrons would move at this same speed, and so the "speed of electricity" is given by the above formula.   For a voltage of 110V, the kind we have at home, the speed would be:\n\nv = 6'216,016.3 m/s\n\nCompared to the speed of light:\n\nc = 300'000,000 m/s\n\nThis analysis is for didactic purpouses, since for such high speeds we would need to consider relativistic effects.   Also, remember that we don't have perfect conductors, so the actual "speed of electricity" at 110V would be a lot less than the value shown above.\n\nI hope this answer was good enough.   :)
462280
E = Fd = (2E-9 N) (200E-12 m) = x J\nx J (1 kJ / 1000 J) = xE-3 kJ\n\nI think this is already per mol...
835018
about 3000 kilo J\nit depends on the atom
249869
one electron is 1.60217646 × 10-19 C\n\n1 coulomb is 6.242197253 * 10 ^ 18 electrons
1074656
[V]=J/C=W/A=Kg.m^2/(s^3.A)\n\nFor experimental reasons A is the base unit and not C, you can not count electrons one by one, not yet probably soon
1081330
I = 60 x e/s\ne = elementary charge\n60 x e/s = 60 x 1.6 x 10^-19 C/s = 9.6 aA\naA = attoampere.\natto =10^-18
25404
See "How Light Bulbs Work":\nhttp://science.howstuffworks.com/light-bulb.htm\n\nThe full answer is:\n\nLight bulbs have a very simple structure. At the base, they have two metal contacts, which connect to the ends of an electrical circuit. The metal contacts are attached to two stiff wires, which are attached to a thin metal filament. The filament sits in the middle of the bulb, held up by a glass mount. The wires and the filament are housed in a glass bulb, which is filled with an inert gas, such as argon.\n\nWhen the bulb is hooked up to a power supply, an electric current flows from one contact to the other, through the wires and the filament. Electric current in a solid conductor is the mass movement of free electrons (electrons that are not tightly bound to an atom) from a negatively charged area to a positively charged area.\n\nAs the electrons zip along through the filament, they are constantly bumping into the atoms that make up the filament. The energy of each impact vibrates an atom -- in other words, the current heats the atoms up. A thinner conductor heats up more easily than a thicker conductor because it is more resistant to the movement of electrons.\n\nBound electrons in the vibrating atoms may be boosted temporarily to a higher energy level. When they fall back to their normal levels, the electrons release the extra energy in the form of photons. Metal atoms release mostly infrared light photons, which are invisible to the human eye. But if they are heated to a high enough level -- around 4,000 degrees Fahrenheit (2,200 degrees C) in the case of a light bulb -- they will emit a good deal of visible light.\n\nThe filament in a light bulb is made of a long, incredibly thin length of tungsten metal. In a typical 60-watt bulb, the tungsten filament is about 6.5 feet (2 meters) long but only one-hundredth of an inch thick. The tungsten is arranged in a double coil in order to fit it all in a small space. That is, the filament is wound up to make one coil, and then this coil is wound to make a larger coil. In a 60-watt bulb, the coil is less than an inch long.
364536
Tungsten filament resistance changes with temperature but the change is stable and predictable. The rule of thumb for filament resistance as a function of temperature is R = R20*(1 + 0.0045*(T-20)), where R20 is resistance at 20C (room temp) and T is temperature in deg C (ref. 1). Operating temperature in lamps ranges from 2500C to 3000C (ref. 2). Thus at operating temperature, resistance will increase by a factor of 12 to 14 compared to the room-temperature value. In older electronic equipment, incandescent lamps were occasionally used as (sort of) constant-current sources.\nBecause the thermal time constant of a typical filament is a tenth of a second or more, 60 Hz AC will cause a ripple in the resistance (and the brightness), but this is stable. It might be possible to have large resistance oscillations if the resistance was proportional to a power of the temperature, and there could be a runaway condition (rapid burnout) if the resistance decreased with increasing temperature.
845374
Use the Nernst equation:\ndE = dE0 - 0.0592 V/n ln [C]^c [D]^d /([A]^a [B]^b)\n\nn is the number of electrons involved, 2 in this case.\nTherefore we have in all cases:\n dE = dE0 - 0.0296 V ln [Sn2+] /([Fe2+]\nFurthermore we use the standard electrode potentials to calculate  dE0 \nSn2+(aq) + 2e- -> Sn(s) -0.14 V\nFe2+(aq) + 2e- -> Fe(s) -0.41 V\nTherefore dE0 = 0.41 - 0.14 = 0.27 V\na) The concentration are 1 M/L and we have dE = 0.27 V\nb) dE = 0.27 - 0.0296 ln 0.01/0.2 = 0.27 + 0.09 = 0.36 V\nc) dE = 0.27 - 0.0296 ln 0.4/0.005 = 0.27 - 0.13 = 0.14 V\n\nOnly in the last case the reaction moves to the left: the concentration of Sn2+ is the highest.\n\nYou can take a look here for explanations and examples:\n\nhttp://www.science.uwaterloo.ca/~cchieh/cact/c123/nernsteq.html
713269
Faraday's Law!\n\nMass Plated = t * I * MW / (n * F)\n\nt = time\nI = current (Amps)\nMW = mol weight of metal\nn = valence charge of metal\nF = Faraday's Constant (96,500 Coulombs/mol e-)\n\nSo....\n\nt = 2 days * 24 hours/day * 3600s/hour = 172800s\nI = 7.60 A\nMW = 51.94 g/mol\nn = 3 mol e- / mol M\n\nPlug it all in and you should get 235.62g of Cr.
547995
This is completely dependant on the atomic structure of the source filament. Once at capacity, the atoms will only give off photonic energy at a specific rate that is inherant to that specific type of atom. Most household filaments are designed to handle standard electrical current,without being at full capacity (atomically).\n\nSo in a standard household bulb, it will release more photonic energy and appear brighter.\n\nIf it is already operating at max capacity (atomically)\nthen there will be no change to the appearance of the bulb, because the atoms are already filled and giving off photons as fast as possible.
1130014
V=IR\n\nWire 1:\n50 = 2 * R1 => R1 is 25 ohms\nCurrent = 2 amps\npower = V*I = 2*50 = 100 watts\n\nWire 2: \n100= 2* R2 => R2 = 50 ohms\nCurrent=2amps\npower = V*I = 100*2 = 200 watts\n\nare you looking for ratios when you say "compare" ?\nif yes: ratio of resistances is  1:2\nratio of currents is 1:1\nand ratio of power is 1:2
730469
You missed the electron changes.\n\nThe bismuth electon change is +3 to 0 for a change of 3\n\nThe Sn electron change is +2 to +4 for a change of 2\n\nSo all the Bi species need a 2  and the Sn species need a 3.\nThen fill in the O and H.
116277
the chemical formula is CaO. So total no. of electrons are \nequal to sum of electrons in Ca and O. \nElectrons in Ca = 20\nElectrons in O  = 8\nSo total electrons = 28\nsince no. of electrons is equal to that of protons,\nHence answer is = 28
892848
One of the most frequently used is E=IR    ;E =voltage    I=current   R= resistance. U can change this every way depending on the unknown.\n    Good luck
376008
The rule of thumb for tungsten filament resistance as a function of temperature is R = R20*(1 + 0.0045*(T-20)), where R20 is resistance at 20C (room temp) and T is temperature in deg C (ref. 1).  Operating temperature of incandescent bulb filaments ranges from 2500C to 3000C (ref. 2). So expect a 12- to 14-fold resistance increase from room temperature to operating temperature.