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as far as i know, high & low spin to the paramagnetic & diamagnetic(respectively) character of the complex...
you can decide on this by seeing whether the complex has all the electrons paired or not (keep in mind the strong & weak field ligands)...
plz correct me if i'm wrong somewhere...
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i guess the following should be the solution... [H+](means H+ ion concentration) =10^-2 M [OH-]=10^-2 M millimoles of H+ = 10^-2 * 200 = 2 millimoles of OH- = 10^-2 * 300 = 3 NaOH + HCl ----> NaCl + H20 3 2 0 0 ----->initially 1 0 2 2 --->after neutralisation => 1 millimole of OH- is present in 500 ml water =>[OH-] = 1/500 = 2 * 10^-3 =>pOH = 2.69 =>pH = 11.3 approx. i think option (a) is correct..
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i guess the following could be the ans. ans:: a. bcoz' Ag forms a complex with aq.NH3 i.e. [Ag(NH3)2]+
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i think the O-P directing nature is due to the +I effect of the methyl group.
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frictional impulse is observed when an impulse is given to a body executing pure rolling motion...
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i'm sorry, my mistake, didn't see the question carefully....
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see, compouds containing halides are subjected to either nucleophilic substitution or elimination reactions. now, in case od chlorobenzene, chlorine will show its +M or +R effect i.e. +ve resonance effect & send its lone pair in resonance with benzene & so, the chlorine-carbon bond will acquire a partial double-bond character, which does not happen in normal alkyl halides. due to this partial double bond, the bond strength increases & chlorobenzene becomes less reactive.
similar is the case for aniline where nitrogen sends its lone pair in resonance & therefore, its basicity decreases...
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 (a-x) + (x-b) since a > b >0 so, for domain, x>a & x>b, => domain is [a,  ) |
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domain of log x is (0,).
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impulsive forces are those forces which are applied for a very short period of time.
for example- in the game of cricket, when the batsman hit a ball for 4, then the force applied by the bat on the ball is impulsive force(applied for a very short period of time)
friction may or may not be an implusive force depending on the situation...
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see, NaCl is a salt but ROH is a very weak acid, so, they can never react to give RCl.
moreover, always the weaker base exists in the medium.
if ROH reacts with NaCl then it should possibly give RCl + NaOH but NaOH is a strong base so, the reaction never happens as a weaker base i.e. Cl- is already present...
i hope you get the concept....
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yes, carbene is the rite answer...
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for the limit of a function to exist, the LHL & RHL should exist and LHL should be equal to RHL ...
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the only reason i can think of is that the reaction might be endothermic in the backward direction so, Kc does not increase on increase in temperature...
plz correct me if i'm wrong...
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 see, if you carefully observe the mechanism then you'll see that the very 1st step is the abstraction of alpha hydrogen by a strong base. now, for the abstraction of the alpha hydrogen, the alpha hydrogen must be acidic else it will not be abstracted. in the question given by you, the alpha carbon has 2 methly groups attached to it, so, the methyl groups are giving their +I effect thereby decreasing the acidity of the alpha hydrogen. now, since the alpha hydrogen is no more acidic then the OH- ion will act as a nucleophile rather than a base & so, there will be attack of OH- as a nucleophile & the reaction will proceed for canizzaro. I hope you get the concept...
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see, this is the point where most of the people are wrong because the main condition for aldol is not alpha-hydrogen but the acidity of the alpha hydrogen. If you know the mechanism of aldol condensation then you'll know that in 1st step, the alpha hydrogen is withdrawn creating a carbanion which attacks on other molecules to proceed for aldol. As dilute NaOH is being used here & the alpha hydrogen is not acidic (in this question, due the presence of two methyl groups the alpha hydrogen will not be acidic) then OH- ion will not be able to extract the alpha hydrogen & so, there will be no aldol. therefore, it will go for canizzaro.... In this question, A VERY STRONG BASE is required for aldol which against all odds can extract the alpha hydrogen... i hope you get the concept...
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apply the canizzaro reaction in this case & you'll get the answer....
i applied canizzaro due very low acidity of the alpha-hydrogen....
i hope you'll be able to do it...
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i'm just guessin' on this. since considerable amount has the chemical has fallen so, i think the rate should increase..
am i correct?
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i'm attaching the pic of the possible ans. i could think of.. there's a high chance that it could be incorrect but do hav a look....
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