Answer:
After 190 s the concentration of X will be 0.0396 M
Explanation:
Where;
Xt is the concentration of X at a time t
X₀ is the initial concentration of X
k is rate constant = 1.7×10⁻² s⁻¹
t is time = 190s
ln(Xt/X₀) = -( 1.7×10⁻²)t
ln(Xt/1.0) = -( 1.7×10⁻²)190
ln(Xt/1.0) = -3.23
Xt = 0.0396 M
Therefore, after 190 s the concentration of X will be 0.0396 M
Answer:
The volume that this same amount of air will occupy in his lungs when he reaches a depth of 124 m is - 0.27 L.
Explanation:
Using Boyle's law
Given ,
V₁ = 3.6 L
V₂ = ?
P₁ = 1.0 atm
P₂ = 13.3 atm (From correct source)
Using above equation as:
The volume that this same amount of air will occupy in his lungs when he reaches a depth of 124 m is - 0.27 L.
Answer:
Explanation:
Hello,
In this case, we need to remember that for the required time for a radioactive nuclide as radium-226 to decrease to one half its initial amount we are talking about its half-life. Furthermore, the amount of remaining radioactive material as a function of the half-lives is computed as follows:
Therefore, for an initial amount of 100 mg with a half-life of 1590 years, after 1000 years, we have:
Best regards.