Answer:
The curve is still exponential but decreases at a lower rate for a greater half-life;
The greater the half-life, the slower radioactive decay is.
Explanation:
From the context of the actual problem, it looks like you plotted the number of non-decayed atoms against time. Since you are analyzing a radioactive decay in this problem, the number of the atoms remaining for the first-order rate law can be represented by the following equation:
![m_t = m_o e^{-kt}](https://tex.z-dn.net/?f=m_t%20%3D%20m_o%20e%5E%7B-kt%7D)
Here k is the rate constant. It is defined in terms of half-life by the following relationship:
![k = \frac{ln(2)}{T_{\frac{1}{2}}}](https://tex.z-dn.net/?f=k%20%3D%20%5Cfrac%7Bln%282%29%7D%7BT_%7B%5Cfrac%7B1%7D%7B2%7D%7D%7D)
That said, in terms of half-life, our equation becomes:
![m_t = m_o e^{-kt}=m_o e^{-\frac{ln(2)}{T_{\frac{1}{2}}}t}](https://tex.z-dn.net/?f=m_t%20%3D%20m_o%20e%5E%7B-kt%7D%3Dm_o%20e%5E%7B-%5Cfrac%7Bln%282%29%7D%7BT_%7B%5Cfrac%7B1%7D%7B2%7D%7D%7Dt%7D)
Notice that the greater the half-life is, the less negative the coefficient in front of the time variable in the exponent.
As a result, the decay for a greater half-life would occur at a lower rate. The curve would still be exponential in terms of shape but would decrease at a lower rate.
We may conclude that the greater the half-life, the slower radioactive decay is.
Answer: a) 0.0144mol/L
b) ![1.19\times 10^{-5}](https://tex.z-dn.net/?f=1.19%5Ctimes%2010%5E%7B-5%7D)
Explanation:
Solubility product is defined as the equilibrium constant in which a solid ionic compound is dissolved to produce its ions in solution. It is represented as ![K_{sp}](https://tex.z-dn.net/?f=K_%7Bsp%7D)
The equation for the ionization of the
is given as:
We are given:
Solubility of
= ![\frac{2.0g}{0.5L}=4g/L](https://tex.z-dn.net/?f=%5Cfrac%7B2.0g%7D%7B0.5L%7D%3D4g%2FL)
Molar Solubility of
= ![\frac{4g/L}{278.1g/mol}=0.0144mol/L](https://tex.z-dn.net/?f=%5Cfrac%7B4g%2FL%7D%7B278.1g%2Fmol%7D%3D0.0144mol%2FL)
1 mole of
gives 1 mole of
and 2 moles of
ions
Solubility product of
= ![[Pb^{2+}][Cl^-]^2](https://tex.z-dn.net/?f=%5BPb%5E%7B2%2B%7D%5D%5BCl%5E-%5D%5E2)
![K_{sp}=[0.0144][2\times 0.0144]^2](https://tex.z-dn.net/?f=K_%7Bsp%7D%3D%5B0.0144%5D%5B2%5Ctimes%200.0144%5D%5E2)
![K_{sp}=1.19\times 10^{-5}](https://tex.z-dn.net/?f=K_%7Bsp%7D%3D1.19%5Ctimes%2010%5E%7B-5%7D)
Thus the solubility product constant is ![1.19\times 10^{-5}](https://tex.z-dn.net/?f=1.19%5Ctimes%2010%5E%7B-5%7D)
A limiting factor could be disease
Answer: option D. CO2 and SO2
Explanation: