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
The chemical reaction would be expressed as follows:
HBr + LiOH = LiBr + H2O
We are given the volumes and corresponding concentration to be used for the reaction. We use these values to solve for the concentration of the other reactant. We do as follows:
0.253 mol LiOH / L solution ( 0.01673 L ) ( 1 mol HBr / 1 mol LiOH ) = 0.00423 HBr needed
Concentration of HBr =0.00423mol / .010 L = 0.423 M HBr
An ion is an atom that has lost or gained an electron so the relation ship you would see would be a negative or a positive one depending on if it gained or lost an electron.<span />
It's adenosine triphosphate !
it has Penrose sugar and phosphate as backbone !
and nitrogenous base ... adenine.... in the middle !
Answer:
1 mole of gas = 22.4L
2.5 moles of gas takes up = ( 22.4 L/ 1 mole ) x 2.5 mole
= 56 L
Explanation:
Please mark brainliest and have a great day!
