The correct answer is 124 ✌
Answer:
double replacement MgN2O6 + KF
Explanation:
Answer:
2Cu2^+ + 2I^- ----> 2Cu^+ + I2
Explanation:
The reaction performed in the experiment is;
2 Cu(NO3)2 + 4 KI → 2 CuI (s) + 4 KNO3 + I2
The iodide ions reduces Cu^2+ to Cu^+ which is insoluble in water hence the precipitate. This is so because iodine is a good oxidizing agent seeing that it requires one electron to fill its outermost shell. Potassium on the other hand is a good reducing agent since it easily looses its one electron.
The oxidation - reduction equation is as follows;
2Cu2^+ + 2e ----> 2Cu^+ reduction half equation
2I^- ----> I2 + 2e. Oxidation half equation
Balanced redox reaction equation;
2Cu2^+ + 2I^- ----> 2Cu^+ + I2
<u>Answer:</u> The mass of water that should be added in 203.07 grams
<u>Explanation:</u>
To calculate the molality of solution, we use the equation:
Where,
m = molality of barium iodide solution = 0.175 m
= Given mass of solute (barium iodide) = 13.9 g
= Molar mass of solute (barium iodide) = 391.14 g/mol
= Mass of solvent (water) = ? g
Putting values in above equation, we get:
Hence, the mass of water that should be added in 203.07 grams
