The rate constant for 1st order reaction is
K = (2.303 /t) log (A0 /A)
Where, k is rate constant
t is time in sec
A0 is initial concentration
(6.82 * 10-3) * 240 = log (0.02 /A)
1.63 = log (0.02 /A)
-1.69 – log A = 1.63
Log A = - 0.069
A = 0.82
Hence, 0.82 mol of A remain after 4 minutes.
Answer:
1.35 g
Explanation:
Data Given:
mass of Potassium Permagnate (KMnO₄) = 3.34 g
Mass of Oxygen: ?
Solution:
First find the percentage composition of Oxygen in Potassium Permagnate (KMnO₄)
So,
Molar Mass of KMnO₄ = 39 + 55 + 4(16)
Molar Mass of KMnO₄ = 158 g/mol
Calculate the mole percent composition of Oxygen in Potassium Permagnate (KMnO₄).
Mass contributed by Oxygen (O) = 4 (16) = 64 g
Since the percentage of compound is 100
So,
Percent of Oxygen (O) = 64 / 158 x 100
Percent of Oxygen (O) = 40.5 %
It means that for ever gram of Potassium Permagnate (KMnO₄) there is 0.405 g of Oxygen (O) is present.
So,
for the 3.34 grams of Potassium Permagnate (KMnO₄) the mass of Oxygen will be
mass of Oxygen (O) = 0.405 x 3.34 g
mass of Oxygen (O) = 1.35 g
The bubbles indicate that there is a gas dissolved in the liquid, which may remain dissolved when the liquid is very cold, inside the refrigerator.
When the temperature of the liquid increases, it reduces its ability to dissolve the gas and the gas escapes (bubbles) from the liquid.
At the end, the behavior of the liquid permits you to conclude that the liquid is a mixture because it has a gas dissolved in it.
