Reduction takes place in cathode while oxidation takes place at the anode. Given that the reaction 3MnO4- +24H + +5Fe→3Mn+2+5Fe+3+12H2O Now that oxidation is termed as an increase in oxidation number or loss of electrons while reduction is a decrease in oxidation number and gain of electrons, ∴oxidation will be Fe→Fe +3+3e- reduction will MnO4- +8H+ →Mn+2+4H2O If oxidation takes place at anode then Anode: an oxidation reaction Fe→Fe+3+3e- . Then the answer is Fe(s)→Fe+3(aq)+3e-
{ 12 mol * 27 g / mol * 0.902 j/gdeg C* ( 658 -72 ) deg C + 12 * 27 * 3.95 } quantity of energy required to heat 12 mol of aluminium for 72 deg C to its melting point 658 C
This question reminds of of the solubility rules. Let us recall that all chlorides are soluble except those of lead, mercury II and silver which are insoluble in water.
The following reaction will occur leading to the formation of a precipitate;
The problem in this question can be solved using the ideal gas formula. Ideal gas formula show interaction between the pressure, volume, temperature and the number of molecules.
n= number of molecule/ avogadro number n= (2.0116 x 10^23 / 6.02*10^23)= 1/3 mole
T= celcius + 273.15 T= (41+273.15) K=314.15
P=2280 mmHg / (760mmHg/atm)= 3 atm
PV=nRT V= nrT/ P V= 1/3 moles * (<span>0.08205 L atm / mol·K) </span> * 314.15 K / 3 atm V= 2.864 L
