I’m pretty sure it is A at least that’s what we did at our school to test this
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)
21.28s
Explanation:
This is a rate problem.
Number of parts produced = 225
Time taken to produce= 8hours
Rate of production is the amount of parts produced in a duration of time.
Now convert 8 hours to minute;
60 minutes = 1hr
8 hours = 60 x 8 = 480minutes
Now:
Rate of production =
= 0.47parts/minute
Now, to produce 10 parts;
It takes;
0.47 parts to be produced in 1 minute
10 parts would be produced in
= 21.28s
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Answer:
The answer to your question is: Excess oxygen = 2.3 mol
Explanation:
Data
ZnS = 5 mol
O2 = 9.8 mol
Excess reactant = ?
Balanced reaction
2ZnS(s) + 3O2(g) → 2ZnO(s) + 2SO2(g)
MW ZnS = 65 + 32 = 97 x 2 = 194 g
MW O2 = 16 x 6 = 96 g
2 mol of ZnS ------------------- 3 mol O2
Ratio from the reaction = 3 mol O2/ 2 mol ZnS
= 1.5
Ratio from the quantities in the experiment = 9.8 mol O2 / 5 mol of ZnS
= 1.96
Excess reactant = Oxygen because the ratio increases
2 mol of ZnS ------------------- 3 mol O2
5 mol of ZnS ------------------- x
x = (5 x 3) / 2
x = 7.5 mol of O2
Excess Oxygen = 9.8 mol - 7.5 mol
Excess oxygen = 2.3 mol