Scoring Scheme: 3-3-2-1 When you use the half reaction method to balance the redox reaction between one mole of NBS and ascorbic
acid to give L-dehydroascorbic acid and succinimide, how many moles of electrons are transferred?
1 answer:
<span>In the redox reaction between one mole of NBS and ascorbic acid, ascorbic acid undergoes oxidation reaction to form </span><span>L-dehydroascorbic acid.
On other hand, N-bromo</span>succinimide (NBS) undergoes reduction to form <span>succinimide.
Electron transferred during this process is two. Complete reaction is attached below for quick reference.
Thus, one molecule of NBS reacts with one molecule of </span>ascorbic acid, during this redox reaction. For each molecule reacting, two electrons are transferred.
Thus, when one mole of NBS reacts with one mole of ascorbic acid, 2 x Na moles of electrons will be transferred. It may be noted here that, Na = Avagadro's number = 6.022 X 10^23.
On other hand, N-bromo</span>succinimide (NBS) undergoes reduction to form <span>succinimide.
Electron transferred during this process is two. Complete reaction is attached below for quick reference.
Thus, one molecule of NBS reacts with one molecule of </span>ascorbic acid, during this redox reaction. For each molecule reacting, two electrons are transferred.
Thus, when one mole of NBS reacts with one mole of ascorbic acid, 2 x Na moles of electrons will be transferred. It may be noted here that, Na = Avagadro's number = 6.022 X 10^23.
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Explanation:
The given data is as follows.
Mass = 27.9 g/mol
As we know that according to Avogadro's number there are atom present in 1 mole. Therefore, weight of 1 atom will be as follows.
1 atoms weight =
In a diamond cubic cell, the number of atoms are 8. So, n = 8 for diamond cubic cell.
Therefore, total weight of atoms in a unit cell will be as follows.
=
=
Now, we will calculate the volume of a lattice with lattice constant 'a' (cubic diamond) as follows.
=
=
=
Formula to calculate density of diamond cell is as follows.
Density =
=
= 2918.1
or, = 0.0029 g/cc (as 1 )
Thus, we can conclude that density of given semiconductor in grams/cc is 0.0029 g/cc.