If you have 24 grams of an unstable atom with a half life of 5,000 years, how many grams would be left after 15,000 years

3 g.

Explanation:

The following data were obtained from the question:

Original amount (N₀) = 24 g

Half life (t½) = 5000 years

Time (t) = 15000 years

Amount remaining (N) =?

Next, we shall determine the rate constant K. This can be obtained as follow:

Half life (t½) = 5000 years

Rate constant (K) =.?

K = 0.693 / t½

K = 0.693 / 5000

K = 1.386×10¯⁴ / years

Finally, we shall determine the amount remaining (N) after 15000 years as follow:

Original amount (N₀) = 24 g

Time (t) = 15000 years

Rate constant (K) = 1.386×10¯⁴ / years

Amount remaining (N) =?

Log (N₀/N) = kt/2.3

Log (24/N) = 1.386×10¯⁴ × 15000/2.3

Log (24/N) = 2.079/2.3

Log (24/N) = 0.9039

24/N = antilog (0.9039)

24/N = 8.01

Cross multiply

24 = N × 8.01

Divide both side by 8.01

N = 24/8.01

N = 3 g

Therefore, the amount remaining after 15000 years is 3 g.

chemists can use the electron cloud model to assign electrons to different atomic orbitals. atomic orbitals also explain the patterns in the periodic table

explanation:

2fe + 3h2so4 -> fe2(so4)3 + 3h2

it is necessary to balance equations in order to get the proper amount of reactants as well as products in order not to make a miscalculation.

you cannot change the subscripts as that would change the substance itself. for instance, diatomic elements such as h2 only have the subscript two when a gas. another example would be so4, if you were to change the subscript, the element would no longer be sulfate.

we change the coefficients instead as that only changes the number of molecules rather than the substance itself.

hope this !