For equal moles of gas, temperature can be calculated from ideal gas equation as follows:
P×V=n×R×T ...... (1)
Initial volume, temperature and pressure of gas is 3.25 L, 297.5 K and 2.4 atm respectively.
2.4 atm ×3.25 L=n×R×297.5 K
Rearranging,
n\times R=0.0262 atm L/K
Similarly at final pressure and volume from equation (1),
1.5 atm ×4.25 L=n×R×T
Putting the value of n×R in above equation,
1.5 atm ×4.25 L=0.0262 (atm L/K)×T
Thus, T=243.32 K
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
The active site of the enzyme.
It is water soluble so is also soluble in aqueous solutions of NaOH or NaHCO3.
Answer:
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