Answer:
The empirical formula=CHO. To explain it is C1H1O1 but we don't put the one that's why it is CHO
Explanation:
The answer is [Ne] 3s^2 3p^5 because chlorine is the fifth element in the 3rd row of elements in in p orbital
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.
Fe+CuSO4⟶Cu+FeSO4
Given that
FeSO4 = 92.50 g
Number of moles = amount in g / molar mass
=92.50 g / 151.908 g/mol
=0.609 moles FeSO4
Now calculate the moles of CuSO4 as follows:
0.609 moles FeSO4 * 1 mole CuSO4 /1 mole FeSO4
= 0.609 moles CuSO4
Amount in g = number of moles * molar mass
= 0.609 moles CuSO4 * 159.609 g/mol
= 97.19 g CuSO4
