Hydrogen gas, oxygen gas and water
4H2O (l) ---> 2H2O (l) + O2 (g) + 2H2 (g)
Answer:
The rate law for second order unimolecular irreversible reaction is
![\frac{1}{[A]} = k.t + \frac{1}{[A]_{0} }](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B%5BA%5D%7D%20%3D%20k.t%20%2B%20%5Cfrac%7B1%7D%7B%5BA%5D_%7B0%7D%20%7D)
Explanation:
A second order unimolecular irreversible reaction is
2A → B
Thus the rate of the reaction is
![v = -\frac{1}{2}.\frac{d[A]}{dt} = k.[A]^{2}](https://tex.z-dn.net/?f=v%20%3D%20-%5Cfrac%7B1%7D%7B2%7D.%5Cfrac%7Bd%5BA%5D%7D%7Bdt%7D%20%3D%20k.%5BA%5D%5E%7B2%7D)
rearranging the ecuation
![-\frac{1}{2}.\frac{k}{dt} = \frac{[A]^{2}}{d[A]}](https://tex.z-dn.net/?f=-%5Cfrac%7B1%7D%7B2%7D.%5Cfrac%7Bk%7D%7Bdt%7D%20%3D%20%5Cfrac%7B%5BA%5D%5E%7B2%7D%7D%7Bd%5BA%5D%7D)
Integrating between times 0 to <em>t </em>and between the concentrations of ![[A]_{0}](https://tex.z-dn.net/?f=%5BA%5D_%7B0%7D)
to <em>[A].</em>
![\int\limits^0_t -\frac{1}{2}.\frac{k}{dt} =\int\limits^A_{0} _A\frac{[A]^{2}}{d[A]}](https://tex.z-dn.net/?f=%5Cint%5Climits%5E0_t%20-%5Cfrac%7B1%7D%7B2%7D.%5Cfrac%7Bk%7D%7Bdt%7D%20%3D%5Cint%5Climits%5EA_%7B0%7D%20_A%5Cfrac%7B%5BA%5D%5E%7B2%7D%7D%7Bd%5BA%5D%7D)
Solving the integral
Answer:
The sun would appear to move more slowly across Mercury's sky.
Explanation:
This is because, the time it takes to do one spin or revolution on Mercury is 176 days (which is its period), whereas, the time it takes to do one spin or revolution on the Earth is 1 day.
Since the angular speed ω = 2π/T where T = period
So on Mercury, T' = 176days = 176 days × 24 hr/day × 60 min/hr × 60 s/min = 15,206,400 s
So, ω' = 2π/T'
= 2π/15,206,400 s
= 4.132 × 10⁻⁷ rad/s
So on Earth, T" = 1 day = 1 day × 24 hr/day × 60 min/hr × 60 s/min = 86,400 s
So, ω" = 2π/T"
= 2π/86,400 s
= 7.272 × 10⁻⁵ rad/s
Since ω' = 4.132 × 10⁻⁷ rad/s << ω" = 7.272 × 10⁻⁵ rad/s, <u>the sun would appear to move more slowly across Mercury's sky.</u>
Answer:
• The closing of the frontier
Explanation:
•• The closing of the frontier• The closing of the frontier• The closing of the frontier• The closing of the frontier• The closing of the frontier
A. Less than 1% of the alpha particles went un-deflected through the gold foil.
