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(i) Cu(s) | Cu2+ (aq) || Ag+ (aq) | Ag(s)(ii) Cu(s) | Cu2+ (aq) || Ag+ (aq) | Ag(s)
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Mg(s) | Mg2+ (0.01 M) || Sn2+ (0.1 M) | Sn(s)Mg(s) → Mg2+ (aq) + 2e–(oxidation at anode)Sn2+(aq) + 2e– → Sn(s)(reduction at cathode)
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Answer:
According to Kohlrausch’s law
Λom(H2O)  = Λom(HCl) + Λom(NaOH) – Λom(NaCl)
Hence if we
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Ag+ + e-  →  Ag1 mol                1&#
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Total no. of matches:-Total no teams -1 13-1=12 matchesTotal no of round = 2*2*2*2Digit 2 repeats four times so no of round = 4 roundsTotal no bye:- next power
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Answer;Let the mass of solution =100gxKOH = 30/56volumeof solution = mass of solutiondensity of solution= 100gd g/cm3 =&
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Gomukhasana :- This asana gets its name because while doing this asana body resembles a cow face pose. In English, it is called the cow face pose.Pre stage :- S
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The other possibility of point where total potential due to q1 and q2 are zero may be that point lies out side the segment joining q1 and q2Let required poi
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Rate = k[NO2]2 No molecules of CO are involved in slowest step because CO is not present in rate law.
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Ni(28) has outer electronic configuration 4s23d8. The oxidation state of Ni is zero . Thus electronic configuration is 4s°3d10.Since there is no any un
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