58874889879879797fc8755555555555555555555555555555555555555555555511111111111111111111111111111111113333333333333333333222222222222220000000000000000000000000000000000000000000000..................
<span>this is a limiting reagent problem.
first, balance the equation
4Na+ O2 ---> 2Na2O
use both the mass of Na and mass of O2 to figure out how much possible Na2O you could make.
start with Na and go to grams of Na2O
55.3 gNa x (1molNa/23.0gNa) x (2 molNa2O/4 molNa) x (62.0gNa2O/1molNa2O) = 75.5 gNa2O
do the same with O2
64.3 gO2 x (1 molO2/32.0gO2) x (2 molNa2O/1 mol O2) x (62.0gNa2O/1molNa2O) = 249.2 g Na2O
now you must pick the least amount of Na2O for the one that you actually get in the reaction. This is because you have to have both reacts still present for a reaction to occur. So after the Na runs out when it makes 75.5 gNa2O with O2, the reaction stops.
So, the mass of sodium oxide is
75.5 g</span>
Answer:
1 ClO-(aq) + 1 I-(aq) ---> 1 Cl -(aq) + 1 IO-(aq).
Explanation:
1. ClO-(aq) + H2O(l) <=> HClO(aq) + OH-(aq)
2. I-(aq) + HClO(aq) <=> HIO(aq) + Cl-(aq)
3. OH-(aq) + HIO(aq) => H2O(l) + IO-(aq)
Adding all the 3 equations together gives and it gives:
ClO-(aq) + H2O(l) + I-(aq) + HClO(aq) + OH -(aq) + HIO(aq)
---> HClO(aq) + OH-(aq) + HIO(aq) + Cl -(aq) + H2O(l) + IO-(aq)
Deleting the same species on both sides of the equation gives:
1 ClO-(aq) + 1 I-(aq) ---> 1 Cl -(aq) + 1 IO-(aq)
The overall equation:
1 ClO-(aq) + 1 I-(aq) ---> 1 Cl -(aq) + 1 IO-(aq)
Answer:
It will not achieve the desired separation
Explanation:
Chromatography is a separation method that involves the use of a stationary phase and a mobile phase. The stationary phase is immobile, in the particular instance of this question, the stationary phase is paper. The mobile phase is the appropriate solvent, in this case, a salt-water solution.
If the level of solvent is above the dye spots, it will introduce error into the separation. The solvent (if volatile) may evaporate without drawing up and separating the solute. Secondly, the solvent may simply dissolve the spots without achieving any meaningful separation of the components in the system. This second reason is particularly why the salt solution must be below the dye spots in this chromatographic separation.
