It's probably animal if that's what's you're asking.
<span>First:
46.7 g of N with 53.3 g of O,
=> mass ratio O to N = 53.3 / 46.7 = 1.1413
Second
17.9 g of N and 82.0 g of O.
mass ratio of O to N = 82.0 / 17.9 = 4.5810
Third
Ratio of the mass ratio of O to N in the second compound
to the mass ratio of O to N in the first compound =
= 4.5810 / 1.1413 = 4.013 ≈ 4
Answer: 4
</span>
Answer:
Specific heat is defined as the amount of heat needed to raise the temperature by one degree celsius. Therefore, in 1 kg there are 100 grams so, 10 grams equal 0.01 Kg. Thus, calculate the specific heat value as follows. Thus, we can conclude that specific heat of the given metal is 0.5 .
Explanation:
Answer:
5.0 x 10⁹ 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 K-40 = 1.251 × 10⁹ 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/(1.251 × 10⁹ years) = 5.54 x 10⁻¹⁰ 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 = 5.54 x 10⁻¹⁰ year⁻¹).
t is the time of the reaction (t = ??? year).
[A₀] is the initial concentration of (K-40) ([A₀] = 100%).
[A] is the remaining concentration of (K-40) ([A] = 6.25%).
∴ (5.54 x 10⁻¹⁰ year⁻¹)(t) = ln((100%)/( 6.25%))
∴ (5.54 x 10⁻¹⁰ year⁻¹)(t) = 2.77.
∴ t = 2.77/(5.54 x 10⁻¹⁰ year⁻¹) = 5.0 x 10⁹ years.
