Answer:
-3.28 × 10⁴ J
Explanation:
Step 1: Given data
- Pressure exerted (P): 27.0 atm
- Initial volume (Vi): 88.0 L
- Final volume (Vf): 100.0 L
Step 2: Calculate the work (w) done by the gaseous mixture
We will use the following expression.
w = -P × ΔV = -P × (Vf - Vi)
w = -27.0 atm × (100.0 L - 88.0 L)
w = -324 atm.L
Step 3: Convert w to Joule (SI unit)
We will use the conversion factor 1 atm.L = 101.325 J.
-324 atm.L × 101.325 J/1 atm.L = -3.28 × 10⁴ J
In an acidic solution, the concentration of H+ is greater than the concentration of OH-. The pH will be less than 7.
In a basic solution, the concentration of OH- is greater than the concentration of H+. The pH will be greater than 7.
In a neutral solution, the concentration of H+ ions to OH-ions will be equal, and will therefore have a pH of 7. (This is due to water autoionization, which we usually ignore because it is small in other circumstances.)
Hey <span>wademcelroy2005, thanks for submitting your question to Brainly!
The answer to your question is </span><span>B:Radiation moves from a warmer object to a cooler object</span>
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