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
Answer:
The final temperature of the solution is 44.8 °C
Explanation:
assuming no heat loss to the surroundings, all the heat of solution (due to the dissolving process) is absorbed by the same solution and therefore:
Q dis + Q sol = 0
Using tables , can be found that the heat of solution of CaCl2 at 25°C (≈24.7 °C) is q dis= -83.3 KJ/mol . And the molecular weight is
M = 1*40 g/mol + 2* 35.45 g/mol = 110.9 g/mol
Q dis = q dis * n = q dis * m/M = -83.3 KJ/mol * 13.1 g/110.9 gr/mol = -9.84 KJ
Qdis= -9.84 KJ
Also Qsol = ms * Cs * (T - Ti)
therefore
ms * Cs * (T - Ti) + Qdis = 0
T= Ti - Qdis * (ms * Cs )^-1 =24.7 °C - (-9.84 KJ/mol)/[(104 g + 13.1 g)* 4.18 J/g°C] *1000 J/KJ
T= 44.8 °C
Answer:
im trynna find it out to mamasss
Explanation:
Just do it and believe in you self come on man you got it
Answer:
3.4 mol C3H8
Explanation:
Balance: C3H8 + 5O2 → 3CO2 + 4H2O
Use a molar ratio to determine the amount needed.
13.6 mol H2O * 1 mol C3H8/4 mol H2O = 3.4 mol C3H8
