Two precursor alkenes
H₃C CH₃
I I
H₂C=C-CH-CH₃ 2,3-dimethyl-1-butene
H₃C CH₃
I I
H₃C-CH=CH-CH₃ 2,3-dimethyl-2-butene
alkane
H₃C CH₃
I I
H₃C-CH-CH-CH₃ 2,3-dimethylbutane
H₃C CH₃ H₃C CH₃
I I I I
H₂C=C-CH-CH₃ + H₂ → H₃C-CH-CH-CH₃
H₃C CH₃ H₃C CH₃
I I I I
H₂C-C=CH-CH₃ + H₂ → H₃C-CH-CH-CH₃
Answer:
1.53 atm
Explanation:
From the question given above, the following data were obtained:
Volume = constant
Initial pressure (P₁) = stp = 1 atm
Initial temperature (T₁) = 273 K
Final temperature (T₂) = 144 °C = 144 °C + 273 = 417 K
Final pressure (P₂) =?
Since the volume is constant, the final pressure can be obtained as follow:
P₁ / T₁ = P₂ / T₂
1 / 273 = P₂ / 417
Cross multiply
273 × P₂ = 417
Divide both side by 273
P₂ = 417 / 273
P₂ = 1.53 atm
Therefore, the final pressure (i.e the pressure inside the hot water bottle) is 1.53 atm.
Answer:
Case 1 (energy level): In an atom, an electron jumps from energy level 1 to energy level 3. ... The energy will increase.
Explanation:
Rate law for the given 2nd order reaction is:
Rate = k[a]2
Given data:
rate constant k = 0.150 m-1s-1
initial concentration, [a] = 0.250 M
reaction time, t = 5.00 min = 5.00 min * 60 s/s = 300 s
To determine:
Concentration at time t = 300 s i.e. ![[a]_{t}](https://tex.z-dn.net/?f=%5Ba%5D_%7Bt%7D)
Calculations:
The second order rate equation is:
![1/[a]_{t} = kt +1/[a]](https://tex.z-dn.net/?f=1%2F%5Ba%5D_%7Bt%7D%20%3D%20kt%20%2B1%2F%5Ba%5D)
substituting for k,t and [a] we get:
1/[a]t = 0.150 M-1s-1 * 300 s + 1/[0.250]M
1/[a]t = 49 M-1
[a]t = 1/49 M-1 = 0.0204 M
Hence the concentration of 'a' after t = 5min is 0.020 M
Answer:
The second one
Explanation:
Gas pressure is caused by gas molecules bouncing off the container walls and each other. Every time a molecule changes direction because it hits a wall, the change in momentum results in a small push. Due to the large number of molecules involved, the pushes add up to a large amount of pressure.
