Answer:
1.667L of a 0.30M BaCl₂ solution
Explanation:
<em>Of a 0.30M barium chloride, contains 500.0mmol of barium chloride.</em>
<em />
Molarity is an unit of concentration used in chemistry defined as the moles of solute present in 1 liter of solution.
In a 0.30M BaCl₂ solution there are 0.30 moles of BaCl₂ in 1 liter of solution.
Now, in your solution you have 500mmol of BaCl₂ = 0.500 moles of BaCl₂ (1000 mmol = 1 mol). Thus, 0.500 moles of BaCl₂ are present in:
0.500 moles * (1L / 0.30 moles) =
<h3>1.667L of a 0.30M BaCl₂ solution</h3>
I think the answer would be A ! Hope this helps
From the calcuation, the percent ionic character of the bond is 70%
<h3>What is percent ionic character?</h3>
The term percent ionic character has to do with the degree of ionic bonding that is contained in a compound. It can be estimated from the electronegativity of each element.
We can use the formula; 100(1 - e^(-ΔEN² / 4))
EN = χB − χA * 100/1
EN = 3.5 - 1.5 = 1
100(1 - e^(-1)^2/4)
= 70%
Learn more about percent ionic character:brainly.com/question/7034446
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<span>2Na + Cl2 => 2NaCl
1 mol Na + 0.5 mol Cl2 => 1 mol NaCl</span>
