Answer:
13.4 (w/w)% of CaCl₂ in the mixture
Explanation:
All the Cl⁻ that comes from CaCl₂ (Calcium chloride) will be precipitate in presence of AgNO₃ as AgCl.
To solve this problem we must find the moles of AgCl = Moles of Cl⁻. As 2 moles of Cl⁻ are in 1 mole of CaCl₂ we can find the moles of CaCl₂ and its mass in order to find mass percent of calcium chloride in the original mixture.
<em>Moles AgCl - Molar mass: 143.32g/mol -:</em>
0.535g * (1mol / 143.32g) = 3.733x10⁻³ moles AgCl = Moles Cl⁻
<em>Moles CaCl₂:</em>
3.733x10⁻³ moles Cl⁻ * (1mol CaCl₂ / 2mol Cl⁻) = 1.866x10⁻³ moles CaCl₂
<em>Mass CaCl₂ -Molar mass: 110.98g/mol-:</em>
1.866x10⁻³ moles CaCl₂ * (110.98g/mol) = 0.207g of CaCl₂ in the mixture
That means mass percent of CaCl₂ is:
0.207g CaCl₂ / 1.55g * 100 =
<h3>13.4 (w/w)% of CaCl₂ in the mixture</h3>
I think the answer is B. static charge. I hope this helps! :)
Ionic, think about the way metals share electrons with other metals, resulting in malleability.
Well if you were asking it was a true or false questions the answer is: True
