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:
1. 0.00040 calories
2. 8.57 calories
3. 0.196 calories
4. 68 calories
5. 243 calories
6. 83680 joules
7. 1,054,368 joules
8. 2.45 calories
9. 556 (it says calories to calories so it wouldn't change)
10. 28367.52 joules
11. 59.6 calories
12. 449.6 joules
13. 0.00234 calories
14. 23292.328 joules
15. 22877693.6 joules
Hope this helps!
Explanation:
Answer:
Explanation:
The relation between equilibrium constant and Ecell is given below .
E⁰cell = (RT / nF ) lnK , F is faraday constant T is 273 + 25 = 298 K
E⁰cell = 1.46 - 1.21 = .25 V
n = 2
Putting the values
.25 = (8.314 x 298 lnK) / (2 x 96485 )
lnK = 19.47
K = 2.85 x 10⁸
2 )
Change in free energy Δ G
Δ G ⁰ = nE⁰ F
n = 4
E⁰ = .4 + .83 = 1.23 V
Δ G ⁰= 4 x 1.23 x 96485
= 474706 J / mol
3 )
E⁰cell = (RT / nF ) lnK
n = 2
1.78 = 8.314 x 298 lnK / 2 x 96485
lnK = 138.638
K = 1.62 x 10⁶⁰
<h3>
Answer:</h3>
2.47 × 10^24 molecules
<h3>
Explanation:</h3>
One mole of a compound contains molecules equivalent to the Avogadro's number, 6.022 × 10^23.
That is, 1 mole of a compound = 6.022 × 10^23 molecules
Therefore,
1 mole of Na₂CO₃ = 6.022 × 10^23 molecules
Thus, we can calculate the number of molecules in 4.1 moles of Na₂CO₃
we get,
= 4.1 moles × 6.022 × 10^23 molecules
= 2.47 × 10^24 molecules
Hence, 4.1 moles of Na₂CO₃ contains 2.47 × 10^24 molecules
