The answer is A: When the energy transporting sediments diminishes, the sediments settle in a low-lying area; therefore, deposition always follows erosion
55.9 kPa; Variables given = volume (V), moles (n), temperature (T)
We must calculate <em>p</em> from <em>V, n</em>, and <em>T</em>, so we use <em>the Ideal Gas Law</em>:
<em>pV = nRT</em>
Solve for <em>p</em>: <em>p = nRT/V</em>
R = 8.314 kPa.L.K^(-1).mol^(-1)
<em>T</em> = (265 + 273.15) K = 538.15 K
<em>V</em> = 500.0 mL = 0.5000 L
∴ <em>p</em> = [6.25 x 10^(-3) mol x 8.314 kPa·L·K^(-1)·mol^(-1) x 538.15 K]/(0.5000 L) = 55.9 kPa
Answer:
Zero
Explanation:
Recall that;
E = q + w
Where;
q = heat, w = work done
When heat is absorbed by the system q is positive
When heat is evolved by the system q is negative
When the system does work, w is negative
When work is done on the system w is positive
Step 1
ΔE1= 60 KJ + 40 KJ = 100KJ
Step 2
ΔE2= (-30 KJ) + (-70 KJ) = (-100) KJ
ΔE1 + ΔE2= 100KJ + (-100) KJ = 0KJ
