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.
In comparison with liquids and gases, solids are more dense. The answer is letter B. <span>The
solid has a more definite shape and volume. The particles are locked into
place. It cannot be further compressed due to the bond that exists between the
molecules. The kinetic energy of the molecules is close to none because the
molecules are so close and so compact with each other. </span>
Answer : The mass of chlorine reacted with the phosphorus is, 53.25 grams.
Explanation :
First we have to calculate the moles of phosphorus.
Now we have to calculate the moles of 
The balanced chemical reaction is:
From the balanced chemical reaction, we conclude that
As, 2 moles of phosphorous react with 3 moles of
So, 0.5 moles of phosphorous react with 
moles of
Now we have to calculate the mass of
Molar mass of = 71 g/mol
Therefore, the mass of chlorine reacted with the phosphorus is, 53.25 grams.
Answer:
A. 15859.2 L or 15900 L
B. 0.629 mol
Explanation:
At STP, one mole is equal to approximately 22.4 L
L or mL is volume, so you are attempting to solve for L or mL.
A.
708 mol x (22.4 L/1 mol) = 15859.2 L (w/ significant figures included - 15900 L)
B.
(14.1 L) x (1 mole/ 22.4 L) = 0.629 mol.
