Answer:
(a) 0.8 s
(b) t = 7.2 s
Explanation:
(a) Half life expression for second order kinetic is:
![t_{1/2}=\frac{1}{k[A_o]}](https://tex.z-dn.net/?f=t_%7B1%2F2%7D%3D%5Cfrac%7B1%7D%7Bk%5BA_o%5D%7D)
Where,
![[A_o]](https://tex.z-dn.net/?f=%5BA_o%5D)
is the initial concentration = 1.0 M
k is the rate constant = 1.25 M⁻¹s⁻¹
So,
Half life = 0.8 s
(b) Integrated rate law for second order kinetic is:
![\frac{1}{[A_t]} = \frac{1}{[A]_0}+kt](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B%5BA_t%5D%7D%20%3D%20%5Cfrac%7B1%7D%7B%5BA%5D_0%7D%2Bkt)
Where, ![[A_t]](https://tex.z-dn.net/?f=%5BA_t%5D)
is the final initial concentration
For 90% completion, 10% is left. so,
![[A_t]=\frac {10}{100}\times 1.0=0.1\ M](https://tex.z-dn.net/?f=%5BA_t%5D%3D%5Cfrac%20%7B10%7D%7B100%7D%5Ctimes%201.0%3D0.1%5C%20M)
So,
t = 7.2 s
Nitrogen is 78.08% of the air is fills the earth's atmosphere.
Answer:
Explanation:
Since we have $N_A$ number of molecules in 22.4 L, in 15 L we have:
$\frac{15}{22.4}\times 6.023 \times 10^{23}$ molecules.
We first calculate the heat released:
Specific heat = 3.21 kJ/kgK
Heat released = heat absorbed by calorimeter
Heat absorbed = mcΔT
= 1.9 x 3.21 x 4.542
= 27.7 kJ
Now, we calculate the moles of hexane present:
Moles = mass / Mr
moles = 0.58 / (12 x 6 + 14)
= 0.0067
Heat of combustion = 27.7 / 0.0067 kJ/mol
ΔH(combustion) = 4,134.3 kJ / mol
Atoms contains small, negatively charged particles. His experiment showed that an atom contains negatively charged particles which was later known as electrons. This was confirmed by the Cathode Ray Experiment he conducted.
