<span>The number next to the simbol of the element ions (as a superscript) means the number of charges of the ion. For example N (+),, where (+) is a superscript means that the charge of the ion is 1+. S(2-), where (2-) is a superscript, means that the charge of the ion is (2-). OH (-), where (-) is a superscript, means that the charge of OH ion is (1-) . </span>
        
             
        
        
        
Scientists expected that the law of conservation of mass would apply to nuclear fission in terms of the masses of the subatomic particles. In reality, the mass of an atom is not equal to the sum of the masses of the subatomic particles that make it up. This is because of the energy that binds the subatomic particles. This energy has mass and when the bond is broken, the mass of the energy of the bonds is lost resulting to what we now cal, a mass defect.
        
                    
             
        
        
        
Answer:
1.728 mg/kg
Explanation:
First have to convert the patient's weight to kg
1 lb = 0.454 kg
185 lb in kg = 83.9 kg
Then we have to divide the dose per the weight of the patient
145 mg for 83.9 kg =  145/83.9
= 1.728 mg/kg
 
        
             
        
        
        
Answer: The value of  for the half-cell reaction is 0.222 V.
 for the half-cell reaction is 0.222 V.
Explanation:
Equation for solubility equilibrium is as follows.
           
Its solubility product will be as follows.
        ![K_{sp} = [Ag^{+}][Cl^{-}]](https://tex.z-dn.net/?f=K_%7Bsp%7D%20%3D%20%5BAg%5E%7B%2B%7D%5D%5BCl%5E%7B-%7D%5D)
Cell reaction for this equation is as follows.
      
Reduction half-reaction:  ,
,  
Oxidation half-reaction:  ,
,    = ?
 = ?
Cell reaction: 
So, for this cell reaction the number of moles of electrons transferred are n = 1.
     Solubility product, ![K_{sp} = [Ag^{+}][Cl^{-}]](https://tex.z-dn.net/?f=K_%7Bsp%7D%20%3D%20%5BAg%5E%7B%2B%7D%5D%5BCl%5E%7B-%7D%5D)
                                                = 
Therefore, according to the Nernst equation
            ![E_{cell} = E^{o}_{cell} - \frac{0.0592 V}{n} log \frac{[AgCl]}{[Ag^{+}][Cl^{-}]}](https://tex.z-dn.net/?f=E_%7Bcell%7D%20%3D%20E%5E%7Bo%7D_%7Bcell%7D%20-%20%5Cfrac%7B0.0592%20V%7D%7Bn%7D%20log%20%5Cfrac%7B%5BAgCl%5D%7D%7B%5BAg%5E%7B%2B%7D%5D%5BCl%5E%7B-%7D%5D%7D) 
 
At equilibrium,  = 0.00 V
 = 0.00 V
Putting the given values into the above formula as follows.
          ![E_{cell} = E^{o}_{cell} - \frac{0.0592 V}{n} log \frac{[AgCl]}{[Ag^{+}][Cl^{-}]}](https://tex.z-dn.net/?f=E_%7Bcell%7D%20%3D%20E%5E%7Bo%7D_%7Bcell%7D%20-%20%5Cfrac%7B0.0592%20V%7D%7Bn%7D%20log%20%5Cfrac%7B%5BAgCl%5D%7D%7B%5BAg%5E%7B%2B%7D%5D%5BCl%5E%7B-%7D%5D%7D) 
 
         ![0.00 = E^{o}_{cell} - \frac{0.0592 V}{1} log \frac{1}{[Ag^{+}][Cl^{-}]}](https://tex.z-dn.net/?f=0.00%20%3D%20E%5E%7Bo%7D_%7Bcell%7D%20-%20%5Cfrac%7B0.0592%20V%7D%7B1%7D%20log%20%5Cfrac%7B1%7D%7B%5BAg%5E%7B%2B%7D%5D%5BCl%5E%7B-%7D%5D%7D) 
    
        
                   = 
                   = 0.577 V
Hence, we will calculate the standard cell potential as follows.
            
        
        
         = 0.222 V
 = 0.222 V
Thus, we can conclude that value of  for the half-cell reaction is 0.222 V.
 for the half-cell reaction is 0.222 V.