...equilibrium = 0.67 mol No. of moles of CH3COOCH2CH3 in new equilibrium = 0.67 + 0. 1 = 0 .77 mol Similarly, no. of moles of ...
no of mol of Na2CO3 =10/(23x2+12+16x3) =0.094mol M/V=0.1M V=0.94dm^3 the new concentration of NaCO3: =[(0. 1 )( 0 .94)]/(0.25+0.94) =0.079M
... pressure of Cl2 = 0.64PThus the % dissociation at the new pressure is 64%
... ions) So [H+]=1 So its pH =-log( 1 )= 0 for 0.5M Ca(OH)2, there will be 1M hydroxide ions...
...mol dm⁻³ [CO]o= 0.200+ 0.200 = 0.400 mol dm⁻³ [Cl2]o= 0.2 mol dm⁻³ At new eqm: [COCl2]= (0.800+ y)mol dm⁻³ [CO] = (0.400 - y) mol dm⁻³ [Cl2]= (0.200- y)mol...
...x log 2 = 2.303 x 8.314 x 298/(1 x 96500) x log 2 = 0.018 V So new potential = 1.21 + 0.018 = 1.23 V (3) Cell emf of the standard cell...
...atm of NH3 was added, let y atm of partial pressure of NH3 is decreased in the new eqm. At the new eqm: PNH3 = (0.385 + 0.1 - y) atm = (0.485 - y) atm PH2S = (0.385 - y) atm Kp...
...chloride = 3.5 mole molarity = no. of mole / volume = 3.5 / (700/1000) = 5 M 2. new volume = 42+35 = 77 cm3 molarity = 2.1M thus no. of mole of ions = 2.1...
...water into this solution, the number of moles in the solution would not change, only increase in volume such that the new concentration will be: mole = conc x vol/1000 3.85 = conc x 1000/1000 conc = 3.85M. Note that the volume ...
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