A Battery charger used to recharge the NiMH batteries used in a digital camera can deliver as much as 0.50 A of current to each battery.  If it takes 100. minutes to recharge one battery, how much Ni(OH)2 (in grams) is oxidized to NiO(OH)?
206258
First you can calculate the quantity of electricity used to charge the battery using the equation:\n\nQ = It\n\nwhere\nQ = charge in Coulombs (C)\nI = current in Amps (A)\nt = time in seconds (s)\n\nIn the above question then\n\nQ = 0.5 x 100 x60 = 3,000 C\n\nNow the oxidation state of the Ni in Ni(OH)2 is 2+, while the oxidation state of Ni in NiO(OH)2 is 4+, so the change is 2+.\n\nFaraday's constant = 96,500 Cmol-1 so \nthe moles of electrons available = 3000/96500 = 0.03109 \n\nHowever, 2 moles of electrons are required to oxidise 1 mol of Ni(OH)2 therefore moles of Ni(OH)2 that can be oxidised is 0.03109/2 = 0.015545. \n\nLastly, we can now convert the moles of Ni(OH)2 to grams by multiplying by its formula mass, which is 92.708 gmol-1, ie\n0.015545 x 92.708 = 1.44 g.
552740
Battery capacity is usually measured in mA H. e.g a 100mAH will delivery 100mA for 1 hour or 10mA for 10 hours (at the rated voltage)\n\nBigger batteries tend to have a larger capacity - i.e. bigger mAH figure
579225
Depends on the charger.  It will probably pull a constant charge on a timer.  The ones I've seen have something like a 5 hour timer.  If you presume 30 amps to run the charger for 5 hours it would be something like 1.5kwh to charge the battery.  Just a ballpark guess though.  Specs on your charger might give you more accurate info.
13412
You can't really get an accurate result with a multitester.  NiMH batteries have a fairly flat discharge curve with a sudden drop-off when they're discharged.  Like all batteries, if they’re measured open-circuit (no load), they will also display a higher voltage that will immediately drop when placed under load.  \n\nMeasuring NiMH battery voltage to determine charge levels means that you have to be able to be able to measure down to .01 or 0.001 volts under load to get an idea for how charged the battery is.  Even this method will only give you a relative result, since NiMH batteries lose capacity for each charge/discharge cycle they go through.  A fully charged battery after 200 uses may only have 90% capacity.\n\nBattery chargers for the newer battery technologies like NiMH or Li-Ion typically use several different methods at the same time when charging batteries:\n\n - Time (14 hrs) \n - Temperature (gets hot when fully charged)\n - Voltage (peaks when charged)\n - Charging current (drops when fully charged)\n\nNiMH batteries only provide about 1.2V VS 1.5V in an alkaline abttery.  A measurded no-load voltage on the battery of 1.5 volts is probably ovr 50% charge.
524653
2Mg + O2 = 2MgO  single replacement \nZn + 2HCl(aq) = ZnCl2 + H2  single replacement \n2H2O2(aq) = 2H2 + 2O2     decomposition \nFe + CuSO4 (aq) =FeSO4 + Cu    single replacement \nPb(NO3)2 (aq) +H2SO4 (aq) = PbSO4 + 2HNO3   double replacement \nCaCO3 (s) + 2HCl (aq) = H2O+CaCl2+CO2    double replacement \n\nCuSO4.5H2O (s) = CuSO4 + 5H2O   decomposition \nNa2O(s) + H2O = 2NaOH   single replacement \nFeSO4 (aq) + KMnO4 (aq) = \nNa2CO3 (aq) + 2HCl (aq) = 2NaCl + H2O + CO2   double replacement
338572
It does decrease, but the rate of decrease varies with the type of battery.  With an alkaline or carbon-zinc battery, the voltage goes down fairly quickly as the power is expended.  For a Ni-Cd or Ni-MH rechargeable, it drops just a little, until the battery is almost dead, then it drops very quickly.  Also, as the battery gets used up, its internal resistance goes up.  That limits the amount of current it can deliver, even though it may have adequate voltage when not connected to a load.  The circuit diagram below is for a real-world battery.  The battery symbol represents an ideal voltage source, ie, one that can deliver any amount of current, without a drop in voltage. The resistor is the internal resistance of the real battery.  With a fresh battery, the value of the resistance is very low, say, around 0.1 ohm for an alkaline "D" cell.  If such a new battery were shorted out, it would put out ~15 A (from Ohm's Law, I = V/R, 15 = 1.5/0.1).  As the battery is used, that resistance goes up.  In a nearly dead battery, it might be 100 ohms, so the short circuit current would be 15 ma (0.015 A).\n            \n- o---i|i|i|i|---/&#92;/&#92;/&#92;/&#92;/&#92;---o +
292320
a. 2k+ = .20moles , SO42-=.10moles\n\nb.Fe3+=.10moles,3NO3-= .30moles\n\nc.2Al3+=.20moles,3SO42-=.30moles\n\nd. Ni2+=.10moles,SO42-=.10moles
544792
C=80uF\nV=Voc1.5 volts\nI=Isc=5.0 amps (assuming short circuit current)\n\nSo thevnin's theorem implies the \nRth = Voc/Isc = 1.5/5 = 0.3ohms\n\nSo the circuit is an RL circuit,\n\nTime constant T = RC = 0.3*80uF = 24us\n\nNow you can use that formula for charging
443905
I had to read about this in order to answer your question at this web site:\n\nhttp://home.um.edu.mt/chemistry/CH230/2.Analytical/CP210a.PDF\n\nSo, two moles of DMG combines with each Ni++ ion.\n\n(0.2145 g Ni)* (1 mole Ni/58.71 g Ni) = 3.65*10^-3 mole Ni\n\nIf what you tell me is correct that DMG is 116.1194 then the Ni complex which loses 2 H+ for the Ni++ has a molecular wt of\n172.8134 (please double check)\n\n(3.65*10^-3 mole Ni)* (172.8134 g/mole) = 0.6314 g\n\nplease check for errors
468665
N=aM where a in the case of acids is the number of H+ the molecule can give (regardless if it is a strong or weak acid)\n\nso N=aM=3*0.20=0.60\n\na for a base would be the number of OH- it gives (or number of H+ that it receives), whereas for redox reactions the number of electrons gained/lost during the reaction or if you prefer the change in the oxidation number
146650
If you can get your hands on a solution of silver nitrate, it should be pretty easy.\n\nYou'll need a battery (9 volt should do it) and two lengths of wire.  Run one wire from the positive terminal of the battery and attach it to a silver electrode (available in science shops) or any piece of silver.  If you can't get the silver electrode, just use the copper wire.  This will plate some of the copper onto the ring, though.  Attach the other wire to the ring and to the negative battery terminal.  Lower the ring and the silver electrode into the silver nitrate solution (without letting them touch) and wait.\n\nThe positive silver ions in the solution will be attracted to the negative ring and should plate the ring nicely.
409531
[I assume that you want 10.35g of lead?]\n\nThe key bits of information here are that current is a measure of electrons per second and that in PbBr2 the lead (Pb) has a 2+ charge, requiring two electrons.\n\nConvert the required mass into a required number of mole, given the atomic mass of lead is 207.2 g / mol\n\n10.35 g / 207.2 g / mol = 0.05 mol.\n\nConvert the number of mole into number of atoms using Avagadro's number (6.023 x 10^23 atoms / mole)\n\n.05 x 6.023 x 10 ^23 = 3.01x10^22 atoms.\n\nYou can then work out how many electrons are needed (remember, the lead is 2+)\n\n2x 3.01 x 10^22 = 6.02 x 10^22 electrons needed\n\nFinally, you then work out how long the current needs to be passed through. Remember - Amps = coulombs per second. 1 coulomb of charge = 6.241 x 10^18 electrons...\n\n6.02 x 10^22 electrons / (10 coulombs per second x 6.241 x10^18 electrons per coulomb)\n=\n965 seconds.\n= 16 minutes 5 seconds.
538655
I sense that you have no clue what you're doing. Tell me I'm wrong.\n\nFor DC circuits, Ohm's law says: V = I * R, voltage equals current times resistance.\n\nYour circuit will consist of a battery (voltage source) and a load (resistance) plus the wires to connect them and the meter(s).\n\nIdeally you will test one battery at a time. That way the load will be identical (including the small resistance of the wires).\n\nBatteries typically have two ratings: voltage and capacity.\nVoltage is measured in volts and this decreases as the battery discharges. Capacity is measured in amp hours (current multiplied by time) which specifies how long the battery can deliver a current.\n\nAssuming the circuit resistance is constant (reasonable assumption for your purposes), when you connect the battery to the load, the voltage (independent variable) will cause a current (dependent variable) to flow. As the battery discharges, the voltage will decrease, and the current will correspondingly decrease. You will need a watch or timer as well. Time zero will be when you first connect the battery. Measure the battery voltage at that instant and at intervals. Depending on the load and how fast the battery discharges should determine how frequently you take measurements.\n\nmsg me when you have the data if you need help analyzing it.
442149
the percent yield of 91.34% means: in theory, if you calculate that you will have 100 after the reaction. However, in practical, you just have 91.34\nNow, we solve your problem\n\n4NH3 + 5O2 --> 4NO + 6H2O\n\nit is a balanced equation\n\nI call the moles of O2 as n(02), the moles of NO as n(NO)\n\nn(O2) = 64/32= 2\n\n=> n(NO)= 4/5 n(O2)= 1.6\n\n=> the weight of NO in theory is 1.6*30= 48\n\n=> the weight of NO in practical is 48* 91.34%= 43.8432
394326
you have  power = voltage*current   \n\nhere current  = 1500/120 = 12.5 amperes \n\nAn ampere = 1C/s = 6.24 10^18 electric charge of one electron\n\nso in 600 s you have  = 12.5 *600*6.24 10^18  = 4.68 10^22
139876
try out an electrolysis experiment with the battery u wil get to know abt the charges of the battery the place where the metal move is the negative charged one and where the non metal moves tht is the positive terminal
494912
Current = Charge per unit time = 14/7 Ampere = 2 A