The Relative Formula Mass of NaH2PO4 is 120 g/mol
Therefore, the number of moles = 6.6/120
                                                   = 0.055 moles of NaH2PO4 which is also equal to the number of moles of H2PO4.
[H2PO4-] = Number of moles oof H2PO4-/Volume of the solution in L
  = 0.055/ ( 355 ×10^-3)
  = 0.155 M
Na2HPO4 undergoes complete dissociation as follows;
Na2HPO4 (aq)= 2Na+ (aq) + HPO4^2- (aq)
1 mole of Na2HPO4 = 142 g/mol
Therefore; number of moles = 8.0/142
                                             = 0.0563 moles
 [HPO4 ^-2] is given by no of moles HPO4^2- /volume of the solution in L
     = 0.0563/(355×10^-3)
     =  0.1586 M
Both H2PO4^2- and HPO4^2- are weak acids the undergoes partial dissociation 
Ka of H2PO4- = 6.20 × 10^-8
 [H+] =Ka*([H2PO4-]/[HPO4(2-)]
        = (6.20 ×10^-8)×(0.155/0.1586)
        = 6.059 ×10^-8 M
pH = - log[H+]
     = - log (6.059×10^-8)
     = 7.218
        
             
        
        
        
First, we need to get the number of moles:
from the reaction equation when Y4+ takes 4 electrons and became Y, X loses 4 electrons and became X4+  
∴ the number of moles n = 4
we are going to use this formula:
㏑K = n *F *E/RT
when K is the equilibrium constant = 4.98 x 10^-5
and F is Faraday's constant = 96500
and the constant R = 8.314
and T is the temperature in Kelvin = 298 K
and n is number of moles of electrons = 4 
so, by substitution:
㏑4.98 x 10^-5 = 4*96500*E / 8.314*298
∴E = -0.064 V
        
             
        
        
        
The volume of the water in cubic meter is determined as 3.2 x 10⁶ m³ .
<h3>Weight of one gallon of water</h3>
The weight of 1 gal of water is given as 3785 g
Mass of 8.48 x 10⁸ gal = 3785 x 8.48 x 10⁸ = 3.2 x 10¹² g
<h3>Volume of the water in cubic meters</h3>
Volume = mass/density
Volume = 3.2 x 10¹² g/1 gmL
Volume = 3.2 x 10¹² mL x 10⁻⁶ m³/mL = 3.2 x 10⁶ m³ 
Thus, the volume of the water in cubic meter is determined as 3.2 x 10⁶ m³ .
Learn more about volume here: brainly.com/question/1972490
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Hello!
The molarity of the HBr solution is 0,172 M.
Why?
The neutralization reaction between LiOH and HBr is the following:
HBr(aq) + LiOH(aq) → LiBr(aq) + H₂O(l)
To solve this exercise, we are going to apply the common titration equation:


Have a nice day!