Answer:
91.16% has decayed & 8.84% remains
Explanation:
A = A₀e⁻ᵏᵗ => ln(A/A₀) = ln(e⁻ᵏᵗ) => lnA - lnA₀ = -kt => lnA = lnA₀ - kt
Rate Constant (k) = 0.693/half-life = 0.693/10³yrs = 6.93 x 10ˉ⁴yrsˉ¹
Time (t) = 1000yrs
A = fraction of nuclide remaining after 1000yrs
A₀ = original amount of nuclide = 1.00 (= 100%)
lnA = lnA₀ - kt
lnA = ln(1) – (6.93 x 10ˉ⁴yrsˉ¹)(3500yrs) = -2.426
A = eˉ²∙⁴²⁶ = 0.0884 = fraction of nuclide remaining after 3500 years
Amount of nuclide decayed = 1 – 0.0884 = 0.9116 or 91.16% has decayed.
It's simple, just follow my steps.
1º - in 1 L we have

of

2º - let's find the number of moles.



3º - The concentration will be

But we have this reaction

This concentration will be the concentration of

![K_{sp}=\frac{[Ba^{2+}][CO_3^{2-}]}{[BaCO_3]}](https://tex.z-dn.net/?f=K_%7Bsp%7D%3D%5Cfrac%7B%5BBa%5E%7B2%2B%7D%5D%5BCO_3%5E%7B2-%7D%5D%7D%7B%5BBaCO_3%5D%7D)
considering
![[BaCO_3]=1~mol/L](https://tex.z-dn.net/?f=%5BBaCO_3%5D%3D1~mol%2FL)
![K_{sp}=[Ba^{2+}][CO_3^{2-}]](https://tex.z-dn.net/?f=K_%7Bsp%7D%3D%5BBa%5E%7B2%2B%7D%5D%5BCO_3%5E%7B2-%7D%5D)
and
![[Ba^{2+}]=[CO_3^{2-}]=5.07\times10^{-5}~mol/L](https://tex.z-dn.net/?f=%5BBa%5E%7B2%2B%7D%5D%3D%5BCO_3%5E%7B2-%7D%5D%3D5.07%5Ctimes10%5E%7B-5%7D~mol%2FL)
We can replace it


Therefore the

is:
Answer:
No.
Materials like water get evaporated when heated, but materials like camphor get sublimed that is they directly get converted into gaseous form when heated, while materials like copper gets melted on heating