Answer:
CaF2 will not precipitate
Explanation:
Given
Volume of Ca(NO3)2
ml
Molar concentration of Ca(NO3)2 
Volume of NaF
ml
Molar concentration of NaF 
Ksp for CaF2 
CaF2 will precipitate if Q for the reaction is greater than ksp of CAF2
Moles of calcium ion

![[Ca2+] = \frac{0.01}{10 + 10} \\= \frac{0.01}{20} \\= 5 * 10^{-4}](https://tex.z-dn.net/?f=%5BCa2%2B%5D%20%3D%20%5Cfrac%7B0.01%7D%7B10%20%2B%2010%7D%20%5C%5C%3D%20%5Cfrac%7B0.01%7D%7B20%7D%20%5C%5C%3D%205%20%2A%2010%5E%7B-4%7D)
Moles of F- ion

![[F-] = \frac{0.001}{10 + 10} \\= \frac{0.001}{20} \\= 5 * 10^{-5}](https://tex.z-dn.net/?f=%5BF-%5D%20%3D%20%5Cfrac%7B0.001%7D%7B10%20%2B%2010%7D%20%5C%5C%3D%20%5Cfrac%7B0.001%7D%7B20%7D%20%5C%5C%3D%205%20%2A%2010%5E%7B-5%7D)
![Q = [Ca2+] [F-]^2\\= (5 * 10^{-4}) * (0.5* 10^-4)\\= 1.25 * 10^{-12}](https://tex.z-dn.net/?f=Q%20%3D%20%5BCa2%2B%5D%20%5BF-%5D%5E2%5C%5C%3D%20%285%20%2A%2010%5E%7B-4%7D%29%20%2A%20%280.5%2A%2010%5E-4%29%5C%5C%3D%201.25%20%2A%2010%5E%7B-12%7D)
Q is lesser than Ksp value of CaF2. Hence it will not precipitate
An intensive property is the physical characteristics that have an independent magnitude. The thermometer can be used to measure the temperature. Thus, option C is correct.
<h3>What is an intensive property?</h3>
An intensive property has been constituted of the parameters that are not dependent on the size and the mass of the sample. Density, pressure, and temperature are some intensive properties.
The first image shows a weighing balance, the second shows a volumetric cylinder, and the fourth shows a ruler used to measure mass, volume, and length respectively, which are extensive properties.
Therefore, option C. thermometer measures temperature, which is an intensive property.
Learn more about the intensive property here:
brainly.com/question/17323212
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Answer:
Part 1: - 1.091 x 10⁴ J/mol.
Part 2: - 1.137 x 10⁴ J/mol.
Explanation:
Part 1: At standard conditions:
At standard conditions Kp= 81.9.
∵ ΔGrxn = -RTlnKp
∴ ΔGrxn = - (8.314 J/mol.K)(298.0 K)(ln(81.9)) = - 1.091 x 10⁴ J/mol.
Part 2: PICl = 2.63 atm; PI₂ = 0.324 atm; PCl₂ = 0.217 atm.
For the reaction:
I₂(g) + Cl₂(g) ⇌ 2ICl(g).
Kp = (PICl)²/(PI₂)(PCl₂) = (2.63 atm)²/(0.324 atm)(0.217 atm) = 98.38.
∵ ΔGrxn = -RTlnKp
∴ ΔGrxn = - (8.314 J/mol.K)(298.0 K)(ln(98.38)) = - 1.137 x 10⁴ J/mol.
There are two valence electrons.