Answer:
Br- Withdraws electrons inductively
Donates electrons by resonance
CH2CH3 - Donates electrons by hyperconjugation
NHCH3- Withdraws electrons inductively
Donates electrons by resonance
OCH3 - Withdraws electrons inductively
Donates electrons by resonance
+N(CH3)3 - Withdraws electrons inductively
Explanation:
A chemical moiety may withdraw or donate electrons by resonance or inductive effect.
Halogens are electronegative elements hence they withdraw electrons by inductive effect. However, they also contain lone pairs so the can donate electrons by resonance.
Alkyl groups donate electrons by hyperconjugation involving hydrogen atoms.
-NHCH3 and contain species that have lone pair of electrons which can be donated by resonance. Also, the nitrogen and oxygen atoms are very electron withdrawing making the carbon atom to have a -I inductive effect.
+N(CH3)3 have no lone pair and is strongly electron withdrawing by inductive effects.
Answer:
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C. MEASURING CYLINDER</h3>
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Explanation:
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#BRAINLYEVERYDAY</h2>
The # 4 is the only significant # in 400.. Trailing 0's r not significant unless there is a decimal.
in 0.7000, there are 4 significant #'s.
If it is just a simple double-replacement reaction, then the balanced equation is simply:
Pb(ClO3)2(aq) + 2 NaI(aq) = PbI2 + 2 Na(ClO3)
However, if this is a precipitation reaction, we must identify which compound will precipitate (is insoluble), either NaClO3 or PbI2
NaClO3 is soluble, PbI2 is not. Only the lead cations and iodine anions will participate in the reaction.
thus, the balanced equation is: Pb^(2+) + 2I^(-) ==> PbI2
Answer: Concentration of NaOH calculated will be underestimated.
Explanation:
End point is an observational point , which tells us about the completion of reaction between the titrant (solution in burette) and titre(solution in conical flask) in titration experiment.
In this case , NaOH is titrant whose concentration is unknown.
, 
, 
....(1)
According to question a chemist overshoots the end point and adds to much of NaOH solution, which means increase in the value of 
.
Then the value of 
in equation (1), will get lowered , which means that the concentration of NaOH was lower than that of the actual value. Hence underestimated concentration of NaOH.
