<span>J.j thomson a british physicist was the first to identify the electron in 1987</span>
Answer: 323.61 g of 
will be produced
Explanation:
The given balanced chemical reaction is :
According to stoichiometry :
2 moles of 
require 1 mole of 
Thus 3.00 moles of will require=
of
Thus is the limiting reagent as it limits the formation of product.
As 2 moles of give = 2 moles of
Thus 3.00 moles of give =
of
Mass of 
Thus 323.61 g of will be produced from the given moles of both reactants.
Answer:
(a) ΔSº = 216.10 J/K
(b) ΔSº = - 56.4 J/K
(c) ΔSº = 273.8 J/K
Explanation:
We know the standard entropy change for a given reaction is given by the sum of the entropies of the products minus the entropies of reactants.
First we need to find in an appropiate reference table the standard molar entropies entropies, and then do the calculations.
(a) C2H5OH(l) + 3 O2(g) ⇒ 2 CO2(g) + 3 H2O(g)
Sº 159.9 205.2 213.8 188.8
(J/Kmol)
ΔSº = [ 2(213.8) + 3(188.8) ] - [ 159.9 + 3(205.) ] J/K
ΔSº = 216.10 J/K
(b) CS2(l) + 3 O2(g) ⇒ CO2(g) + 2 SO2(g)
Sº 151.0 205.2 213.8 248.2
(J/Kmol)
ΔSº = [ 213.8 + 2(248.2) ] - [ 151.0 + 3(205.2) ] J/K = - 56.4 J/K
(c) 2 C6H6(l) + 15 O2(g) 12 CO2(g) + 6 H2O(g)
Sº 173.3 205.2 213.8 188.8
(J/Kmol)
ΔSº = [ 12(213.8) + 6(188.8) ] - [ 2(173.3) + 15( 205.2) ] = 273.8 J/K
Whenever possible we should always verify if our answer makes sense. Note that the signs for the entropy change agree with the change in mol gas. For example in reaction (b) we are going from 4 total mol gas reactants to 3, so the entropy change will be negative.
Note we need to multiply the entropies of each substance by its coefficient in the balanced chemical equation.
The reduced forms of the electron carries NAD+ / NADH and FADH / FADH2 have high potential energy.
- NAD& FAD used to donate electrons as a reducing agent, they receive electrons from other molecules and then became reduced.
-FAD is called flow in adenine dinucleotide, it is a redox cofactor and it is in many important enzymatic reactions in metabolism.
Answer:
one atom of an element = 6.02 \times {10}^{23} atom
The mass of one atom of sulphur = 32g
The mass of one atom of aluminium = 27g
so one atom of aluminium = 6.02 \times {10}^{23}
27g of AL = 6.02 \times {10}^{23} atom
2.70g of AL = X atoms
Then you cross multiply ........
and get the answer
.
