A closed container contains 0.40 moles of argon gas at 25 °C and a pressure of 740 torr. The container is heated to 125 °C and t
1 answer:
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Rate law for the given 2nd order reaction is:
Rate = k[a]2
Given data:
rate constant k = 0.150 m-1s-1
initial concentration, [a] = 0.250 M
reaction time, t = 5.00 min = 5.00 min * 60 s/s = 300 s
To determine:
Concentration at time t = 300 s i.e.
Calculations:
The second order rate equation is:
substituting for k,t and [a] we get:
1/[a]t = 0.150 M-1s-1 * 300 s + 1/[0.250]M
1/[a]t = 49 M-1
[a]t = 1/49 M-1 = 0.0204 M
Hence the concentration of 'a' after t = 5min is 0.020 M
Answer:
Explanation:
Given :
Mass of a bar of lead = 115.2 g
Initial water level = 25 mL
Final water level = 35.5 mL
Difference in the water level = 35.5 - 25
= 10.5 mL
=
We know that when a body is submerged in water, it displaces its own volume of water.
Therefore, the volume of the lead bar = volume of the water displaced = 10.5 mL =
We know that mathematically, density is the ratio of mass of body to its volume.
Density of the lead bar is given by :
=