Astudent monitored the reaction rate of a 0.600m sucrose solution, at 25 oc, and determined that the reaction is first order with respect to sucrose. if at 25.0 oc it takes 3.33-hours for the concentration of the 0.600m solution to drop to 0.300m. how many hours are required for the concentration of the sucrose solution to drop to 10.0 % of its initial concentration.

11.06 hours are required

Explanation:

Because the reaction is first order with respect of sucrose, we can use the formula:

ln [A]ₓ = -k*t + ln[A]₀

Where [A]₀ is the initial concentration of sucrose, k is a constant and [A]ₓ is the concentration remaining after a time t.

Putting the data given by the exercise we can calculate k:

Now with k, we can calculate t when [A]ₓ = 0.600 * 10/100 = 0.060 M

a graduated cylinder.

explanation:

first let us lay out the formula for potential

and kinetic energy:

pe = mgh

ke = (mv^2)/2

where: m=mass, v=velocity, g=gravitational

acceleration, h=height

calculating for mass using the known values of

pe:

340j = m (9.81 m/s^2) (50ft) (1m/3.28ft)

m = 2.27 kg

the law of conservation of energy states that

energy is neither created nor destroyed. therefore, the change in kinetic

energy from 50 ft to 35 ft would be equal to the change in potential energy

from 50 ft to 35 ft. however it would be opposite in signs since one is losing

while other is gaining.

ke(35ft) – ke(50ft) = - [ pe(35ft) – pe(50ft]

ke(50ft) = 0 since the brick is initially at rest

ke(35ft) = 340j - 2.27kg (9.81 m/s^2) (35ft)

(1m/3.28 ft)

ke(35ft)

= 102.38 j