The equation Eºcell = 0.0592/n logK must be used to find n and also Eºcell
2 Al(s) + 3 Mg2+(aq) → 2 Al3+(aq) + 3 Mg(s) Al3+ +3e- --> Al Eº = -1.66 V Mg2+ +2e- -->Mg Eº = -2.37V
To balance the equation, 6 moles of electrons must be transferred (2 Al and 3 Mg). This will be the value of n in the equation.
To find Eºcell, you need the reduction potentials which should be given in a table, and given above. Eºcell = -1.66 - (-2.37) = 0.71 V log K = Eºcell x n/0.0592 = 0.71 x 6/0.0592 log K = 71.95 K = 10^71.95 K = 1.1x10^72
1) <span>NaNO3 and H2O - no reaction , it is dissolution
2) no hydrogen to make water
3) </span><span>Fe(OH)3 (base) and H2SO4(acid))
base +acid ----> salt +water
4) </span><span>Li2O and Ba(OH)2
basic oxide and base ----> no reaction
so Answer number 3)
</span> 2Fe(OH)3 +3 H2SO4 ------> Fe2(SO4)3 + 6H2O<span>
</span>
Answer:
group 16 period 2 of the periodic table
note: that is not the electronic configuration, that is the Bohr model.
Answer:
Br
Explanation:
because bromine is more reactive as reactivity increases on moving from left to right in p-block. hope this make sense :)
