Answer:
a) a0 was 46.2 grams
b) It will take 259 years
c) The fossil is 1845 years old
Explanation:
<em>An unknown radioactive substance has a half-life of 3.20hours . If 46.2g of the substance is currently present, what mass A0 was present 8.00 hours ago?</em>
A = A0 * (1/2)^(t/h)
⇒ with A = the final amount = 46.2 grams
⇒ A0 = the original amount
⇒ t = time = 8 hours
⇒ h = half-life time = 3.2 hours
46.2 = Ao*(1/2)^(8/3.2)
Ao = 261.35 grams
<em>Americium-241 is used in some smoke detectors. It is an alpha emitter with a half-life of 432 years. How long will it take in years for 34.0% of an Am-241 sample to decay?</em>
t = (ln(0.66))-0.693) * 432 = 259 years
It will take 259 years
<em>A fossil was analyzed and determined to have a carbon-14 level that is 80% that of living organisms. The half-life of C-14 is 5730 years. How old is the fossil?</em>
<em />
t = (ln(0.80))-0.693) * 5730 = 1845
The fossil is 1845 years old
Answer:
<h2>
= (
1.08 /
2.2
) 100% = 49%</h2>
Explanation:
Balanced Equation: 2CH₃CH₃(g) + 5O₂(g) → 2CO₂(g) + 6H₂O(g)
Calculate moles of CH₃CH₃ and O₂
1.2 ₃₃ (
1 ₃₃/
30.0694 ₃₃
) = 0.040 ₃₃
8.6 ₂ (
1 2/
31.998 ₂
) = 0.27 ₃₃
Find limiting reagent 0.040 ₃₃ (
5 ₂/
2 ₃₃
) = 0.10 ₂
CH₃CH₃ is the limiting Reagent
CH₃CH₃ (L.R.) O₂ CO₂ H₂O
Initial (mol) 0.040 0.27 0 0
Change
(mol)
-2x=-0 -5x=
-0.10 +2x=+0.040 +6x=+0.12
Final (mol) 0 0.117 0.040 0.12
0.040 − 2 = 0 = 0.020
Determine percent yield
0.12 ₂ (
18.0148 ₂
/1 ₂
) = 2.2 ₂
= (
1.08 /
2.2
) 100% = 49%
Answer: if we lost our core it would be an entire fallout like fallout 76 we would have no way to protect ourselves from the radiation of space except for a select few that actually own astronaut suits tho we would not turn into monsters, sadly :( I wanna be a ghoul
Because there is so much suger thats how it is so sweet and people like it!
