6.21 x 10^3 = (Move decimal point 3 spaces to the right)
6210
6210 (0.1050)
652.05
Has markings along the cylinder/beaker that indicate the volume of liquid inside them. .
Answer:
The bands are due to:
λmax = 289 nm n→π* transition (E = 12)
λmax = 182 nm π→π* transition (E=10000)
Explanation:
The two types of acetaldehyde transition are as follows:
n→π* and π→π*
From the attached diagram we have to:
ΔEn→π* < ΔEπ→π*
ΔEα(1/λ)
Thus:
λn→π* > λπ→π*
In n→π* spin forbidden, the intensity is low. Thus, the molar extinction E for n→π* is very low.
The same way, for π→π* spin allowed the intensity is high. Thus, the molar extinction coefficient E for π→π* is high too.
The bands are due to:
λmax = 289 nm n→π* transition (E = 12)
λmax = 182 nm π→π* transition (E=10000)
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Answer:
A. increased storm severity.
Please let me know if I am wrong.
~
papaguy
Answer:
The partial pressure in mm Hg for each of the species are:
PCO = 0
PH2 = 3874
PCH3OH = 347
The total pressure is 4221
Explanation:
We have to use Ideal gas equation PV = nRT and Partial pressure rule Total Pressure= ∑ Partial pressures.
We have following data:
T= 357K (85+272); nCO=0.078 (2.2/28); nH2 = 2.43 (4.86/2); R=62.36
With equation CO(g) + 2 H2(g) → CH3OH(g) we can calculate the amount of moles the reaction has finished. The limit reagent is CO because is consumed completely and moles CO in flask are 0. According to equation, every CO mol produces one CH3OH mol. That means 0.078 CO moles produces 0.078 CH3OH moles. From Ideal gas equation we have P=nRT/V.
Applying:
PH2=0.87*62.36*357/5=3874
PCH3OH =0.078*62.36*357/5=347
Total P = 3874+347=4221
I hope my answer helps you
