Answer:
the answer is a because thermal energy is heat and the question said that is not a factor of thermal energy so A does not refer to heat
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>
M ....6,022*10^22 molecules
2,30g .......3,01*10^22 molecules
or 1 mole has 6,022*10^22 molecules so (2,30/Molar mass) = nr of moles with 3,01*10^22 molecules
which will lead to the same thing.
M= 4,6
Remember the equation for molarity:
![molarity = \frac{moles}{volume}](https://tex.z-dn.net/?f=molarity%20%3D%20%20%5Cfrac%7Bmoles%7D%7Bvolume%7D%20)
*note that volume is in liters*
They give you grams (14.3 g), which can give you moles if you divide it by the molar mass of LiCl (about 42.4 g/mole).
14.3g/42.4= 0.34 moles
They also give you volume, already in liters:
0.45L
Now, it's a "plug and chug" question.
![molarity = \frac{0.34moles}{0.45liters}](https://tex.z-dn.net/?f=molarity%20%3D%20%20%5Cfrac%7B0.34moles%7D%7B0.45liters%7D%20)
Then, you should get your answer.
Hope that helps!
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