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:
2P4O10 + 2H2O → 2H3PO4
Balanced Equation → P4O10 + 6H2O → 4H3PO4
Tetraphosphorus Decaoxide + Water → Phosphoric Acid
Reaction Type → Synthesis
We are given an equation 2Mg+O2-> 2MgO and a starting chemical Mg of about 212 g. In order to solve for the amount of O2 needed, we need the molecular weight of Mg and O2.
Molecular weight:
Mg=24.305 g/mol
O2=16(2)=32 g/mol
Note that for every 1 mol of O2, the amount of Mg must be 2 mol.
So,
g O2 = 212 g Mg x1mo Mgl/24.305 g Mg x1mol O2 /2 mol Mg x 32 g O2/mol O2
gO2=139.56 g
Therefore, 139.56 g of O2 is needed for every 212 g Mg.
Answer:
50 g
Explanation:
d= m/v
rearranging the above equation
m = d x v
m = 2.5 g x 20 g/cm3
m = 50 g
The answer to the selected question is Flurione
