Um i think gold... i think?
Answer:
91.16% has decayed & 8.84% remains
Explanation:
A = A₀e⁻ᵏᵗ => ln(A/A₀) = ln(e⁻ᵏᵗ) => lnA - lnA₀ = -kt => lnA = lnA₀ - kt
Rate Constant (k) = 0.693/half-life = 0.693/10³yrs = 6.93 x 10ˉ⁴yrsˉ¹
Time (t) = 1000yrs
A = fraction of nuclide remaining after 1000yrs
A₀ = original amount of nuclide = 1.00 (= 100%)
lnA = lnA₀ - kt
lnA = ln(1) – (6.93 x 10ˉ⁴yrsˉ¹)(3500yrs) = -2.426
A = eˉ²∙⁴²⁶ = 0.0884 = fraction of nuclide remaining after 3500 years
Amount of nuclide decayed = 1 – 0.0884 = 0.9116 or 91.16% has decayed.
Answer:
Limiting reactant: O2
grams NO2 produced = 230.276 g NO2
grams of NO unused = 26.67 gNO
Explanation:
2NO + O2 --> 2NO2
Step 1: Determine the molar ratio NO:O2
molar ratio NO:O2 = 5.895: 2.503 = 2.35
stoichiometric molar ratio NO:O2 = 2:1
So, O2 is the limiting reactant.
Step2: Determine the grams of NO2:
?g NO2 = moles O2 x (2moles NO2/1 mol O2) x (MM NO2/ 1 mol NO2) = 2.503 x 2 x 46 = 230.276 g NO2
Step 3: Determine the amount of excess reagent unreacted
moles excess NO reacted = moles O2 x (2 moles NO/1 mol O2) = 2.503 x 2 = 5.006 moles NO reacted
moles NO unreacted = total moles NO - moles NO reacted = 5.895-5.006 =0.889 moles NO unreacted
mass NO unreacted = moles NO unreacted x MM NO = 0.889 x 30 =26.67 g NO unreacted
Answer:
In order to determine unequivocally which of the peaks represent Aldrin, I would run the pure sample through the chromatography equipment.
Explanation:
Gas chromatography is a technique that separates molecules based on their volatility and interaction with both the stationary phase.
The peaks on the chromatogram show how long a substance took to leave the column. Since each different substance substance will leave the column at a different time, each peak can be attributed to a substance.
Therefore, to know which of the 6 peaks represent Aldrin, it is necessary to run the pure Aldrin in the chromatography equipment and see the time of the peak. Then you just need to compare both chromatograms and indentify Aldrin.
