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Oh its basically like glass lid, I mean this is how I usually draw it, without knowing that tis actually existDivided flask is like this :usually used to separate SOLID reactant from the LIQUID reactant. Used in those experiments for example: CaCO3 reacting with HCl etc
YesOh its basically like glass lid, I mean this is how I usually draw it, without knowing that tis actually exist
But yah thanks
But here you don't need to pull up that lid. That BARRIER is FIXED. You just need to stir the flask to start the reaction. Pretty simple than pulling up the lid, or pulling the thread
Oww I see, thats nice!! I will label it as a divided flask right? And the examiner can later google itYes
Oww I see, thats nice!! I will label it as a divided flask right? And the examiner can later google it
yes you should write it/label it as divided flask It's actually a standard laboratory equipment.
Yeah I was just kidding
Remember what bond energy is,it is the energy released when a covalent bond of a gaseous molecule is broken to give gaseous atoms and not molecules or compounds or molecules. So in A only is this happening in the rest of the options ∆H not only contains the bond breaking energy but also bond formation so the resulting ∆H for B,C and D is lesser than the value of X----Y bond energy !
I hope you get it its pretty simple just try to understand and ask any question no matter how childish it is about this explanation !
answer for the first one is potassium?
AgF consists of Ag+ and F- ions, which are fully dissociated in aqueous solution. When solving electrolysis problems, it is important to remember that water itself may also be subject to electrolysis. Thus, determining which species is oxidized and which species is reduced depends on selecting the processes that are the most energetically favorable. In this case we have two possible reductions and two possible oxidations, listed here along with their reduction or oxidation potentials:
Thanks A lot!AgF consists of Ag+ and F- ions, which are fully dissociated in aqueous solution. When solving electrolysis problems, it is important to remember that water itself may also be subject to electrolysis. Thus, determining which species is oxidized and which species is reduced depends on selecting the processes that are the most energetically favorable. In this case we have two possible reductions and two possible oxidations, listed here along with their reduction or oxidation potentials:
Reduction:
2H+ + 2e- ==> H2, E = 0.00 V
Ag+ + e- ==> Ag, E = 0.8 V
Oxidation:
2F- ==> F2 + 2e-, E = -2.87 V
2H2O ==> O2 + 4H+ + 4e-, E = -1.23 V
(ALL THE VALUES ARE FROM THE DATA BOOKLET)
The reactions with the *most positive* potentials are most likely to proceed. Therefore, the reduction of Ag+, which occurs at the cathode, is the most favorable reduction reaction and the oxidation of H2O, which occurs at the anode, is the most favorable oxidation reaction. Consequently, the major products of this hydrolysis are Ag and O2, whereas H2 and F2 are not produced in appreciable amounts.
Same principle with all three
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