Answer:
B. decay of dead marine organisms
Explanation:
When the temperature is low, carbon dioxide is captured by the oceans, and when the temperature is high, it is released by the oceans into the atmosphere. At sea, carbon dioxide feeds phytoplankton.
Most of the carbon dioxide consumed by plant plankton (phytoplankton) returns to the atmosphere when this phytoplankton dies or is consumed, but a portion is deposited in the ocean floor sediments when these small particles sink. This process is called a "biological bomb" because carbon dioxide is transported from the atmosphere to the ocean floor.
Answer:
CH4 +2 O2 — CO2 +2 H2O
Now we see that for 1 mol i.e. 16 grams of methane results in 1 mol of CO2 or 44 grams of CO2.
That means for 3 moles of methane , we will obtain 3 moles of CO2 OR for 48 (3*16) grams of CO2 , we will obtain (44*3) 132 grams of CO2 .That's it….
Explanation:
hey .dude hope the answer was helpful ....
Answer:
The energy profile for rotation about the C-C bond in ethane is shown in the image, along with the Newman projections of the corresponding ethane conformer.
Explanation:
If you see the ethane molecule (second image) from the C-C bond axis (third image), as in the Newman projections, it's easy to draw an angle between one of the hydrogen atoms of the visible carbon, the carbon itself, and one of the hydrogens of the hidden carbon.
When you make a rotation about the C-C bond, the angle between those hydrogens will change. If you start with an eclipsed conformation, with each hydrogen of the hidden C exactly behind the hydrogens of the visible C, the angle will be 0°, or also 120° or 240°, as this rotations will be equivalent. On the other hand, if the angle is 60° (or 180°, or 300°), you will have a staggered conformation. The eclipsed conformation is less stable than the staggered one, because the interactions between hydrogens will be bigger (the repulsion between their electrons), and because of that the eclipsed conformations will be found in the maxima, while the staggered one will be found in the minima.
Answer:
0.147 billion years = 147.35 million years.
Explanation:
- It is known that the decay of a radioactive isotope isotope obeys first order kinetics.
- Half-life time is the time needed for the reactants to be in its half concentration.
- If reactant has initial concentration [A₀], after half-life time its concentration will be ([A₀]/2).
- Also, it is clear that in first order decay the half-life time is independent of the initial concentration.
- The half-life of Potassium-40 is 1.25 billion years.
- For, first order reactions:
<em>k = ln(2)/(t1/2) = 0.693/(t1/2).</em>
Where, k is the rate constant of the reaction.
t1/2 is the half-life of the reaction.
∴ k =0.693/(t1/2) = 0.693/(1.25 billion years) = 0.8 billion year⁻¹.
- Also, we have the integral law of first order reaction:
<em>kt = ln([A₀]/[A]),</em>
<em></em>
where, k is the rate constant of the reaction (k = 0.8 billion year⁻¹).
t is the time of the reaction (t = ??? year).
[A₀] is the initial concentration of (Potassium-40) ([A₀] = 100%).
[A] is the remaining concentration of (Potassium-40) ([A] = 88.88%).
- At the time needed to be determined:
<em>8 times as many potassium-40 atoms as argon-40 atoms. Assume the argon-40 only comes from radioactive decay.</em>
- If we start with 100% Potassium-40:
∴ The remaining concentration of Potassium-40 ([A] = 88.88%).
and that of argon-40 produced from potassium-40 decayed = 11.11%.
- That the ratio of (remaining Potassium-40) to (argon-40 produced from potassium-40 decayed) is (8: 1).
∴ t = (1/k) ln([A₀]/[A]) = (1/0.8 billion year⁻¹) ln(100%/88.88%) = 0.147 billion years = 147.35 million years.
The answer is somewhere, u gotta dig deeper like princess and the frog said!
