Here we have to calculate the amount of  ion present in the sample.
 ion present in the sample.
In the sample solution 0.122g of  ion is present.
 ion is present.
The reaction happens on addition of excess BaCl₂ in a sample solution of potassium sulfate (K₂SO₄) and sodium sulfate [(Na)₂SO₄] can be written as-
K₂SO₄ = 2K⁺ +  
(Na)₂SO₄=2Na⁺ +  
Thus, BaCl₂+   = BaSO₄↓ + 2Cl⁻ .
 = BaSO₄↓ + 2Cl⁻ .
(Na)₂SO₄ and  K₂SO₄ is highly soluble in water and the precipitation or the filtrate is due to the BaSO₄ only. As a precipitation appears due to addition of excess BaCl₂ thus the total amount of   ion is precipitated in this reaction.
 ion is precipitated in this reaction.  
The precipitate i.e. barium sulfate (BaSO₄)is formed in the reaction which have the mass 0.298g. 
Now the molecular weight of BaSO₄ is 233.3 g/mol. 
We know the molecular weight of sulfate ion ( ) is 96.06 g/mol. Thus in 1 mole of BaSO₄ 96.06 g of
) is 96.06 g/mol. Thus in 1 mole of BaSO₄ 96.06 g of  ion is present.
 ion is present. 
Or. we may write in 233.3 g of BaSO₄ 96.06 g of  ion is present. So in 1 g of BaSO₄
 ion is present. So in 1 g of BaSO₄  g of
 g of  ion is present.
 ion is present. 
Or, in 0.298 g of the filtered mass (0.298×0.411)=0.122g of  ion is present.
 ion is present.        
 
        
             
        
        
        
In the equation,
2Al(s) + 3Cl2(g) —> 2AlCl3(s),
the large number "3" in front of Cl2 indicates the the number of moles of Chlorine molecules needed to balance the equation. 
Hope this will help you.  
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Actually your answer is correct. Your answer that you picked is correct because the basic SI unit for mass is a kilograms, so therefore your answer is 13kg which you picked.
        
             
        
        
        
Answer:
I think B
Explanation:
There are more negative ions than positive ions
 
        
             
        
        
        
A change in position over a certain amount of time