Answer:
disposal of radioactive waste
Explanation:
if too much is released, it can wipe out large parts of the country
 
        
                    
             
        
        
        
A test tube of zinc oxide
        
                    
             
        
        
        
1)
 HI(aq) → H⁺(aq) + I⁻(aq)
So this is an Arrhenius acid because it releases H⁺.
2)
LiOH(s) → Li⁺ + OH⁻
So this is an Arrhenius base because it releases OH⁻
        
             
        
        
        
Ca(OH)₂ ==> Ca²⁺ + 2 OH<span>-   
Ca(OH)</span>₂ is <span>strong Bases</span><span>
</span>Therefore,  the [OH-] equals 5 x 10⁻⁴ M. For every Ca(OH)₂ you produce 2 OH⁻<span>.
</span>
pOH = - log[ OH⁻]
pOH = - log [ <span>5 x 10⁻⁴ ]
pOH = 3.30
pH + pOH = 14
pH + 3.30 = 14
pH = 14 - 3.30
pH = 10.7
hope this helps!</span>
        
             
        
        
        
Answer:
11552.45 years
Explanation:
Given that:
Half life = 5730 years
 
Where, k is rate constant
So,  
 
 
The rate constant, k = 0.00012 years⁻¹
Using integrated rate law for first order kinetics as:
![[A_t]=[A_0]e^{-kt}](https://tex.z-dn.net/?f=%5BA_t%5D%3D%5BA_0%5De%5E%7B-kt%7D) 
Where,  
![[A_t]](https://tex.z-dn.net/?f=%5BA_t%5D) is the concentration at time t
 is the concentration at time t
![[A_0]](https://tex.z-dn.net/?f=%5BA_0%5D) is the initial concentration
 is the initial concentration
Given that:
The rate constant, k = 0.00012 years⁻¹
Initial concentration ![[A_0]](https://tex.z-dn.net/?f=%5BA_0%5D) = 160.0 counts/min
 = 160.0 counts/min
Final concentration ![[A_t]](https://tex.z-dn.net/?f=%5BA_t%5D) = 40.0 counts/min
 = 40.0 counts/min
Time = ?
Applying in the above equation, we get that:-
 
 
 
