D would be correct (i did this Assessment today)<span />
        
                    
             
        
        
        
Answer: 
Explanation:
A double displacement reaction is one in which exchange of ions take place. The salts which are soluble in water are designated by symbol (aq) and those which are insoluble in water and remain in solid form are represented by (s) after their chemical formulas.

The equation can be written in terms of ions as:

Spectator ions are defined as the ions which does not get involved in a chemical equation or they are ions which are found on both the sides of the chemical reaction present in ionic form.
The ions which are present on both the sides of the equation are potassium and nitrate ions and hence are not involved in net ionic equation.
Hence, the net ionic equation is :

 
        
             
        
        
        
Concentration of Solutions is oftenly expressed in Molarity. Molarity is the number of moles of solute dissolved per volume of solution.
                             Molarity  =  Moles / Volume
As,
      Moles  =  Mass / M.mass
So,
                             Molarity  =  Mass / M.mass × Volume   ---- (1)
Data Given;
                   Volume  =  0.750 L
                   Mass  =  52 g
                   M.mass  =  180 g/mol
Putting Values in eq.1,
                     Molarity  =  52 g ÷ (180 g.mol⁻¹ × 0.750 L)
                     Molarity  =  0.385 mol.L⁻¹
        
             
        
        
        
Using off road vehicles does help contribute to the process of erosion.
        
                    
             
        
        
        
Answer:

Explanation:
We know, 
where, R = 0.0821 L.atm/(mol.K), T is temperature in kelvin and 
 is difference in sum of stoichiometric coefficient of products and reactants
Here 
 and T = 311 K
So, ![K_{p}=(0.0111)\times [(0.0821L.atm.mol^{-1}.K^{-1})\times 311K]^{-1}=4.35\times 10^{-4}](https://tex.z-dn.net/?f=K_%7Bp%7D%3D%280.0111%29%5Ctimes%20%5B%280.0821L.atm.mol%5E%7B-1%7D.K%5E%7B-1%7D%29%5Ctimes%20311K%5D%5E%7B-1%7D%3D4.35%5Ctimes%2010%5E%7B-4%7D)
Hence value of equilibrium constant in terms of partial pressure 
 is 