Answer:
Atomic number of this isotope = 77
Explanation:
Given that,
Mass number = 193
No of neutrons = 116
We need to find the atomic no of this isotope.
We know that,
Atomic mass = No of protons + No. of neutrons
Also, atomic no = no of protons
So,
Atomic mass = atomic no + No. of neutrons
⇒ Atomic no = Atomic mass - no of neutrons
Atomic no = 193 - 116
Atomic no = 77
Hence, 77 is the atomic no of the isotope.
The first order rate law has the form: -d[A]/dt = k[A] where, A refers to cyclopropane. We integrate this expression in order to arrive at an equation that expresses concentration as a function of time. After integration, the first order rate equation becomes:
ln [A] = -kt + ln [A]_o, where,
k is the rate constant
t is the time of the reaction
[A] is the concentration of the species at the given time
[A]_o is the initial concentration of the species
For this problem, we simply substitute the known values to the equation as in:
ln[A] = -(6.7 x 10⁻⁴ s⁻¹)(644 s) + ln (1.33 M)
We then determine that the final concentration of cyclopropane after 644 s is 0.86 M.
Answer:
The boy is using kinetic energy. And his joules are 60
V₁ = initial Volume of the balloon after it is blown up = 365 L
V₂ = new Volume of the balloon after it is taken outside = ?
T₁ = initial temperature of the balloon = 283 K
T₂ = new temperature of the balloon = 300 K
using the equation
V₁/V₂ = T₁/T₂
365/V₂ = 283/300
V₂ = 387 L
