The following first order reaction occurs in CCl4(l) at 45oC : N2O5 --> N2O4 + 1/2 O2(g). The rate constant is k = 6.2 x 10-4s-1. An 80.0 g sample of N2O5 in CCl4(l) is allowed to decompose at 45oC. How long does it take for the quantity of N2O5 to be reduced to 2.5 g?
Answers
Answered by
3
Answer:
The process by which green plants turned carbon dioxide and water into food using energy from sunlight is called photosynthesis
Answered by
0
Answer:
The following first-order reaction occurs in CCl4(l) at 45 C: N2O5 yields N2O4 + (1/2)O2. The rate constant is k=6.2 *10^{-4} s^{-1}. An 75.0-g sample
1. For a first order reaction:A = Ao*e^(-kt)(2.2 g) = (67.9 g)e^(-6.2e-4*t)t = [ln(2.2/67.9)] / (-6.2e-4)t = 5532 s = 1.54 h2.2N2O5->O2+4 NO2 is what ... More
Similar questions