Answer:
Explanation:
Froath flotaion process is based on thie wttebality properteis like hydrophillicity and hydrophobicity.
Incase of sulphide ore processing,
The metallic sulphide particles of ore are preferentially wetted by oil and the gangue particles by water. By doing thise you are sperating required sulphide ore particels from unwanted gaunge , by doing this you are incresing the concentation of Sulfide in ore by removing the not required gaunge particles.
Hope this helps you to decide your annswer
Assuming conditions are at 25c and 1 bar 101.25 Kpa 1 mole of gas is equal to 22.4 L
So if you have 500L of ar gas the you have : 500L / 22.4 × 1 Mole = 22.3 moles of ar after accounting for sig figs..Hope that helps
Answer :
The Nernst equation :
![E_{cell}=E^o_{cell}-\frac{2.303RT}{nF}\log \frac{[Anode]}{[Cathode]}](https://tex.z-dn.net/?f=E_%7Bcell%7D%3DE%5Eo_%7Bcell%7D-%5Cfrac%7B2.303RT%7D%7BnF%7D%5Clog%20%5Cfrac%7B%5BAnode%5D%7D%7B%5BCathode%5D%7D)
where,
= standard cell potential
n = number of electrons in oxidation-reduction reaction
F = Faraday constant = 96500 C
R= gas constant = 8.314 J/Kmol
T = temperature
[Anode] = anodic ion concentration
[Cathode] = cathodic ion concentration
694,563,239 rounded to the nearest thousand is 694,563.
It's because the first digit from the right is for ones, second for tens, third for hundreds and fourth for thousands and that's the one that we should take a closer look at. You can round it either to 3 or 4, depends on the digit of hundreds. In this case 3239 is clearly closer to 3000 than 4000, that's why we round it to 694,563, not 694,564.
The chemical equation given is:
<span>2x(g) ⇄ y(g)+z(s)</span>
Answer: the higher the amount of x(g) the more the forward reacton will occur and the higher the amounts of products y(g) and z(s) will be obtained at equilibrium.
Justification:
As Le Chatellier's priciple states, any change in a system in equilibrium will be compensated to restablish the equilibrium.
The higher the amount, and so the concentration, of X(g), the more the forward reaction will proceed to deal witht he high concentration of X(g), leading to an increase on the concentration of the products y(g) and z (s).
