This is given by Avogagro number: 1 mol = 6.02*10^23 particles
Then you can do whichever to these two relations, because they are equivalent:
- 1mol / 6.02*10^23 representative particles, and
- 6.02*10^23 representative particle /1 mol
Only the second option of the question includes one of the valid conversion factors. Then, the conversion factor of the second option is the right answer
The highest atom economy
2CO + O₂ ⇒ 2CO₂
<h3>Further explanation</h3>
Given
The reaction for the production of CO₂
Required
The highest atom economy
Solution
In reactions, there are sometimes unwanted products that can be said to be a by-product or a waste product. Meanwhile, the desired product can be said to be a useful product, which can be shown as the atom economy
of the reaction
the higher the atomic economy value of a reaction, the smaller the waste/ byproducts produced, so that less energy is wasted
The general formula:
Atom economy = (mass of useful product : mass of all reactants/products) x 100
<em>or
</em>
Atom economy = (total formula masses of useful product : total formula masses of all reactants/products) x 100
So a reaction that only produces one product will have the highest atomic value, namely the reaction in option C
Answer:
K3PO4
Explanation:
Recall that colligative properties depends on the number of particles present. The greater the number of particles present, the greater the degree of colligative properties of the solution. Let us look at each option individually;
SrCr2O7-------> Sr^2+ + Cr2O7^2- ( 2 particles)
C4H11N (not ionic in nature hence it can not dissociate into ions)
K3PO4-------> 3K^+ + PO4^3- (4 particles)
Rb2CO3-------> 2Rb^+ + CO3^2- (3 particles)
Hence K3PO4 has the greatest number of particles and will display the greatest colligative effect.
Answer:
For O: atomic number = 8 # neutrons = 8
For Al: atomic mass = 27, # electrons = 13
