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
1 pm = 10∧-10 cm
Therefore, 230 pm is equivalent to 2.3 ×10∧-8 cm.
Atom is in the shape of a sphere,
The volume of a sphere is given by 4/3πr³
Thus, volume of the atom = 4/3π( 2.3 ×10∧-8)³
= 4/3 (3.142 ×12.167×10∧-24
= 5.096 ×10∧-23 cm³
but 1m³= 1000000cm³
Therefore, the volume of the atom = 5.096 ×10∧-29 m³
Answer:
Freezing point = 1.25
Explanation:
If we increase the concentration of the solution, the concentration of H+ does not change.
Convert 2.5% in to decimal
2.5% = 2.5 ÷100
= 0.025
The freezing point = 0.025 × 50
= 1.25
Explanations:- Part 1: We could count the total number of electrons by looking at the electron configurations. Both of these electrons configurations have 47 electrons. If we look at the periodic table then 47 is the atomic number of silver. So, the name of the element is silver and its represented as Ag.
Part 2: As per the rule, Completely filled and half filled orbitals are more stable. First electron configuration has 9 electrons in 4d and we know that d is more stable if it has 5 electrons(half filled) or it has 10 electrons(full filled).
For stability reasons, one of the electron from 5s goes to 4d and for this reason the second electron configuration is found most often in nature for silver.
Few other examples are Cr and Cu.
Answer:
5.625 moles of oxygen, O₂.
Explanation:
The balanced equation for the reaction is given below:
4Al + 3O₂ —> 2Al₂O₃
From the balanced equation above,
4 moles of Al reacted with 3 moles of O₂.
Finally, we shall determine the number of mole of O₂ required to react with 7.5 moles of aluminum, Al. This can be obtained as illustrated below:
From the balanced equation above,
4 moles of Al reacted with 3 moles of O₂.
Therefore, 7.5 moles of Al will react with = (7.5 × 3)/4 = 5.625 moles of O₂.
Thus, 5.625 moles of O₂ is needed for the reaction.