Answer:
[tex]\large \boxed{\text{b. 0 ng/mL}}[/tex]
Explanation:
The integrated rate law for a zero-order reaction is
A₀ - A = kt
1. Calculate the rate constant
[tex]\begin{array}{rcl}\text{300 ng/mL - 150 ng/mL} & = & k\times \text{2 h}\\\text{150 ng/mL} & = & 2k \text{ h}\\k & = & \dfrac{\text{150 ng/mL} }{ \text{2 h} }\\\\& = & \textbf{75 ng$\cdot$mL$^{-1}\cdot$ h$^{-1}$}\\\end{array}[/tex]
2. Calculate the new concentration
[tex]\begin{array}{rcl}\text{A$_{0}$ - A} & = & kt\\\text{300 ng$\cdot$mL$^{-1}$ - A} & = & \text{75 ng$\cdot$mL$^{-1}\cdot$ h$^{-1}$}\times \text{4 h}\\\text{300 ng$\cdot$mL$^{-1}$ - A} & = & \text{300 ng$\cdot$mL$^{-1}$}\\\text{A} & = &\text{300 ng/mL - 300 ng/mL}\\& = & \mathbf{0}\\\end{array}\\\text{The concentration after 4 h will be $\large \boxed{\textbf{0 ng/mL}}$}[/tex]
