Chlorine and oxygen form many different compounds, including CIO, and Cl2O3. How do the names of these compounds differentiate o
1 answer:
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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)
Answer:
13.1dm³
Explanation:
Given parameters:
Pressure of gas = 880mmHg
760mmHg = 1 atm
880mmHg will give 1.16atm
Temperature = 303k
Mass of gas = 26.9g
Molar mass of CO₂ = 12 + 2(32) = 44g/mol
Number of moles = mass/molar mass = 26.9/44 = 0.61mole
Unknown:
Volume of the sample = ?
Solution:
To solve this problem, we use the ideal gas equation:
PV = nRT
P is the pressure
V is the volume
n is the number of moles
R is the gas constant = 0.082atmdm³mol⁻¹k⁻¹
T is the temperature
V = =
= 13.1dm³