Answer:
3,964 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 the element is 5,730 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/(5,730 years) = 1.21 x 10⁻⁴ year⁻¹.
- Also, we have the integral law of first order reaction:
<em>kt = ln([A₀]/[A]),</em>
where, k is the rate constant of the reaction (k = 1.21 x 10⁻⁴ year⁻¹).
t is the time of the reaction (t = ??? year).
[A₀] is the initial concentration of the sample ([A₀] = 100%).
[A] is the remaining concentration of the sample ([A] = 61.9%).
∴ t = (1/k) ln([A₀]/[A]) = (1/1.21 x 10⁻⁴ year⁻¹) ln(100%/61.9%) = 3,964 years.
The three major types of faults are Normal, Reverse and Strike-slip faults.
Answer: FALSE
<span>ATP,O2 and NADPH are the </span>products<span>. H2O,NADP,ADP and Pi are the reactants. acts as an electron carrier between the cytochrome b6f and </span>photosystem 1 (PS1) complexes in the photosynthetic electron-transfer chain.
Photosystem II<span> (or water-plastoquinone oxidoreductase) is the first protein complex in the light-dependent reactions of oxygenic photosynthesis. It is located in the thylakoid membrane of plants, algae, and cyanobacteria.</span>
What amount of heat absorbs 50 g of steel (ce = 0.115 cal / g. ° C) that
does its temperature vary by 25 ° C?
Answer:
143.75cal
Explanation:
Given parameters:
Mass of steel = 50g
Specific heat capacity of the steel = 0.115cal/g°C
Temperature = 25°C
Unknown:
Amount of heat = ?
Solution:
The amount of heat to cause this temperature change is dependent on mass and specific heat capacity of the substance.
Amount of heat = m C (ΔT)
m is the mass
c is the specific heat capacity
ΔT is the temperature change
Now insert the parameters and solve;
Amount of heat = 50 x 0.115 x 25
Amount of heat = 143.75cal
