A:- sn(s) => Sn +2(0.24 M) + 2e-
B:- Sn +2 (0.87 M) +2e- => Sn(s)
solution will become more concentrated and solution B become less concentrated
Sn(s)+ Sn +2(0.87 ) ----> Sn(s) + Sn +2(0.24)
E = Eo - 0.0592 / 2 * log [ (0.24 / 0.87 ) ]
E = 0.0 - 0.0592 / 2 * log ( 0.275)
( n=2 two electrons are transferred)
E = -0.0296 * ( - 0.560)
E = 0.0165 volts
Answer:
(4) concentrated and supersaturated
Explanation:
At 50.°C, 90g of KNO3 lies above the solubility curve [on the Regents Reference Table G]. This indicates that the solution is supersaturated, meaning it contains more solute than will naturally dissolve, and was formed when a saturated solution cooled. Furthermore, the percent concentration of this solution is 90% KNO3 making this solution concentrated. This can be calculated using the formula for mass percent concentration.
Percent Mass = <u>Mass of Solute (g)</u> x 100
Mass of Solution (g)
