The awnser is D or the 4th one
Answer:
The half-life time, the team equired for a quantity to reduce to half of its initial value, is 79.67 seconds.
Explanation:
The half-life time = the time required for a quantity to reduce to half of its initial value. Half of it's value = 50%.
To calculate the half-life time we use the following equation:
[At]=[Ai]*e^(-kt)
with [At] = Concentration at time t
with [Ai] = initial concentration
with k = rate constant
with t = time
We want to know the half-life time = the time needed to have 50% of it's initial value
50 = 100 *e^(-8.7 *10^-3 s^- * t)
50/100 = e^(-8.7 *10^-3 s^-1 * t)
ln (0.5) = 8.7 *10^-3 s^-1 *t
t= ln (0.5) / -8.7 *10^-3 = 79.67 seconds
The half-life time, the team equired for a quantity to reduce to half of its initial value, is 79.67 seconds.
The redox reaction is
Here
Calcium undergoes reduction, and acts as cathode
Lithium undergoes oxidation and acts as anode
The reduction potential of calcium is -2.87 V
The reduction potential of lithium is - -3.05 V
We know that
Ecell = Ecathode - Eanode
Ecell = -2.87 - (-3.05) = 0.18 V
The answer would most likely be False
Hello!
Work=Force*Displacment
Work=.05N * .07m = .0035J
Any questions please feel free to ask. Thank you.
