genetics and reproduction is all about dna.
Answer:
Mn(s)/Mn^2+(aq)//Co^2+(aq)/Co(s)
Explanation:
In writing the cell notation for an electrochemical cell, the anode is written on the left hand side while the cathode is written on the right hand side. The two half cells are separated by two thick lines which represents the salt bridge.
For the cell discussed in the question; the Mn(s)/Mn^2+(aq) is the anode while the Co^2+(aq)/Co(s) half cell is the cathode.
Hence I can write; Mn(s)/Mn^2+(aq)//Co^2+(aq)/Co(s)
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:
not sure about 6 but 7 should be c
Explanation:
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Answer:
The answer to your question is: 0.028 kg of NO2
Explanation:
Data
3.7 x 10²⁰ molecules of NO2 in kg
MW of NO2 = 14 + (16 x 2) = 14 + 32 = 46 kg
1 mol of NO2 --------------------- 6.023 x 10 ²³ molecules
x --------------------- 3.7 x 10²⁰ molecules
x = 3.7 x 10²⁰ x 1 / 6.023 x 10 ²³
x = 0.00061 mol
1 mol of NO2 --------------------- 46 kg of NO2
0.00061 mol ------------------ x
x = 0.00061 x 46/1
x = 0.028 kg of NO2
