Answer:
Option D, Concentration of NO2 decomposes after 4.00 s = 0.77 mol
Explanation:
Time (t) = 4.00\;s
Initial concentration of NO2 = 1.33 M
Integrated law for second order reaction:
![\frac{1}{[A]}=\frac{1}{[A]_0} =kt](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B%5BA%5D%7D%3D%5Cfrac%7B1%7D%7B%5BA%5D_0%7D%20%3Dkt)
Where, [A] = Concentration after time, t
[A]0 = Intitial concentration, k = rate constant, t = time
On substituting values in the above
![\frac{1}{[A]}=\frac{1}{1.33} =0.255 \times 4.00](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B%5BA%5D%7D%3D%5Cfrac%7B1%7D%7B1.33%7D%20%3D0.255%20%5Ctimes%204.00)
![\frac{1}{[A]} =1.772](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B%5BA%5D%7D%20%3D1.772)
[A] = 0.5644 M
Concentration of NO2 decomposes after 4.00 s = 1.33 - 0.5644 = 0.7656 M
No. of mole = Molarity * volume
= 0.7656 * 1
= 0.7656 mol 0r 0.77 mol
Answer:
200.6cm³
Explanation:
Using Charles law equation as follows:
V1/T1 = V2/T2
Where;
V1 = initial volume (cm³)
V2 = final volume (cm³)
T1 = initial temperature (K)
T2 = final temperature (K)
According to the question;
initial volume (V1) = 254cm³
final volume (V2) = ?
Initial temperature (T1) = 72.6°C = 72.6 + 273 = 345.6K
Final temperature (T2) = 273K
V1/T1 = V2/T2
254/345.6 = V2/273
Cross multiply
345.6 × V2 = 273 × 254
345.6V2 = 69342
V2 = 69342 ÷ 345.6
V2 = 200.6cm³
Answer:
endoplasmic reticulum
Explanation:
Endoplasmic reticulum will complete the chart. The Endoplasmic reticulum is one type of organelle that is present in eukaryotic cells. They form a network that is interconnected with membrane-enclosed sacs or structures that are tube shaped and are called cisternae.
Answer: 6,25 moles
Explanation: mark amount of butane x.
From equation you can calculate x in a following way:
2/x = 8/25. 8x = 50. And x = 6,25
