Each mole of a substance contains 6.02 x 10²³ particles
Atoms of Fe = 4.5 x 6.02 x 10²³
= 2.709 x 10²⁹ atoms
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.
Answer:
C.)
Explanation:
I also saw this question on Brainly! Hope this helps!
1. The hypothesis for this is experiment is that the 50:50 of methanol-water mixture will not turn to solid when the temperature reaches to -40°C.
2. The procedure for this is measuring equal volumes of water and methanol using the graduated cylinder. You can measure 100 mL of water and 100 mL of methanol using the graduated cylinder. Then, mix them in the beaker. Next, measure 200 mL of water, and another 200 mL of methanol. Don't mix them. Also, make a 60:40 mixture by measuring 120 mL of water and 80 mL of methanol, then mix them together. Place them all in the refrigerator at the same time. Record the time when they would freeze to solid.
3. The controls for this experiment are the 200 mL water alone, and the 200 mL methanol alone.
4. The independent variable in here is the time, while the dependent variable is the temperature of the mixtures.
5. If the hypothesis turns out to be true, then all the mixtures prepared should freeze and become solid after a certain period of time, with the exception of the 50:50 mixture. The 50:50 mixture should still remain as a liquid even when left overnight.
Explanation:
Given parameters:
Mass of Br-79 = 78.918 amu
Abundance = 50.69%
Unknown:
Mass of Br - 81 = ?
Solution:
Relative Atomic mass of Br = 80
The proportion by which each fraction of an isotope occurs in nature is called the geonormal abundance.
RAM = Ma Ba + McBc
The formula above is used to find the average mass of the given isotopes.
Since we know the percentage abundance of Br-79, that of Br-81 = 100-50.69= 49.31%
RAM = Ma Ba + McBc
RAM = relative atomic mass
Ma Ba = mass and abundance of Br-79
McBc = mass and abundance of Br-81
80 = (78.918 x
) + (
x C)
80 = 40 + 0.49C
0.49C = 40
C =
= 81.63amu
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