•3.9g of ammonia
•molar mass of ammonia = 17.03g/mol
1st you have to covert grams to moles by dividing the mass of ammonia with the molar mass:
(3.9 g)/ (17.03g/mol) = 0.22900763mols
Then convert the moles to molecules by multiplying it with Avogadro’s number:
Avogadro’s number: 6.022 x 10^23
0.22900763mols x (6.022 x 10^23 molecs/mol)
= 1.38 x 10^23 molecules
The volume of a substance is simply the ratio of mass and
density. Therefore:
volume = mass / density
Calculating for volume of Carbon Tetrachloride that the
student has to pour out:
volume = 55.0 g / (1.59 g / cm^3)
<span>volume = 34.60 cm^3</span>
Answer:
it is possible to remove 99.99% Cu2 by converting it to Cu(s)
Explanation:
So, from the question/problem above we are given the following ionic or REDOX equations of reactions;
Cu2+ + 2e- <--------------------------------------------------------------> Cu (s) Eo= 0.339 V
Sn2+ + 2e- <---------------------------------------------------------------> Sn (s) Eo= -0.141 V
In order to convert 99.99% Cu2 into Cu(s), the equation of reaction given below is needed:
Cu²⁺ + Sn ----------------------------------------------------------------------------> Cu + Sn²⁺.
Therefore, E°[overall] = 0.339 - [-0.141] = 0.48 V.
Therefore, the change in Gibbs' free energy, ΔG° = - nFE°. Where E° = O.48V, n= 2 and F = 96500 C.
Thus, ΔG° = - 92640.
This is less than zero[0]. Therefore, it is possible to remove 99.99% Cu2 by converting it to Cu(s) because the reaction is a spontaneous reaction.
That is true b<span>ecause 1 nanometer = 1.0 × 10^-15 megameters.</span>