We measure temperature in degrees of Fahrenheit
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 gas occupy 2406.4 mL at 80 K.
Explanation:
Given data:
Initial volume of gas = 752 mL
Initial temperature = 25 K
Final temperature = 80 K
Final volume = ?
Solution:
The given problem is solved by using charle's law.
V₁/T₁ = V₂/T₂
V₂ = V₁. T₂ /T₁
V₂ = 752 mL × 80 k / 25 K
V₂ = 60160 mL. k/25 K
V₂ = 2406.4 mL
Grams. liters is liquid. meters is length and atm is idk
1 cup = 2 pints
48 × 2 = 96 pints
HOPE THIS HELPED
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