Answer:
 1.87x10⁻³ M SO₄²⁻
Explanation:
The reaction of SO₄²⁻ with Ba²⁺ (From Ba(NO₃)₂) is:
SO₄²⁻(aq) + Ba²⁺(aq) → BaSO₄(s)
<em>Where 1 mole of  SO₄²⁻ reacts per mole of Ba²⁺</em>
<em />
To reach the end point in this titration, we need to add the same moles of Ba²⁺ that the moles that are of SO₄²⁻.
Thus, to find molarity of SO₄²⁻ we need to find first the moles of Ba²⁺ added (That will be the same of SO₄²⁻). And as the volume of the initial sample was 100mL we can find molarity (As ratio of moles of SO₄²⁻ per liter of solution).
<em>Moles Ba²⁺:</em>
7.48mL = 7.48x10⁻³L ₓ (0.0250moles / L) = 1.87x10⁻⁴ moles of Ba²⁺ = Moles of SO₄²⁻
<em>Molarity SO₄²⁻:</em>
As there are 1.87x10⁻⁴ moles of SO₄²⁻ in 100mL = 0.1L, molarity is:
 1.87x10⁻⁴ moles of SO₄²⁻ / 0.1L =
<h3> 1.87x10⁻³ M SO₄²⁻</h3>
 
        
             
        
        
        
I think its 
<span>The Mechanical energy in a mechanical system is determined by adding the potential and kinetic energy together. </span>
 
        
                    
             
        
        
        
The only true answer is A. 
products are on the right side of the reaction
products and reactants don't necessarily have same physical or chemical properties
It is the reactants that <span>are the atoms, molecules, or compounds that participate in the reaction.</span>
        
                    
             
        
        
        
Answer:30 L
Explanation:
Initial Volume
=
V
1
=
60
l
i
t
e
r
Initial Temperature
=
T
1
=
546
K
Final Temperature
=
T
2
=
273
K
Final Vloume
=
V
2
=
?
?
Sol:-
Since the pressure is constant and the question is asking about temperature and volume, i.e,
V
1
T
1
=
V
2
T
2
⇒
V
2
=
V
1
⋅
T
2
T
1
=
60
⋅
273
546
=
60
2
=
30
l
i
t
e
r
⇒
V
2
=
30
l
i
t
e
r
Hence the new volume of the gas is 
30
l
i
t
e
r
 
        
             
        
        
        
Accept a pair of nonbonding electrons,a Lewis acid is an electron-pair acceptor. A Lewis<span> base is any </span>substance, such as the OH-<span> ion, that </span>can<span> donate a pair of nonbonding electrons. </span>A Lewis<span> base is therefore an electron-pair donor.</span>