Let us assume that the ring is a size 7 ring, which has a circumference of 54.3 millimeters. Converting this to centimeters, the circumference of the ring is:
54.3 mm = 5.43 cm
Now, we determine the number of gold atoms that will be present in this:
5.43 / 1 x 10⁻⁹
There will be 5.43 x 10⁹ atoms
We now determine the number of moles this is by:
one mole = 6.02 x 10²³ atoms
Moles = 5.43 x 10⁹ / 6.02 x 10²³
Moles = 9.01 x 10⁻¹⁵ moles
The molar mass of gold is 197 g/mol
The mass is 9.01 x 10⁻¹⁵ * 197
The mass of the strand is 1.76 x 10⁻¹² grams
It creates
<span>THE HALOGENATION OF ALKENES</span>
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
1st one is E
2nd one is also E
