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
Answer:
Any of the answers given will work
Explanation:
I literally just did it.
Answer:
14.77 mol.
Explanation:
- It is known that every 1.0 mole of compound or element contains Avogadro's number (6.022 x 10²³) of molecules or atoms.
<u><em>Using cross multiplication:</em></u>
1.0 mole of He contains → 6.022 x 10²³ atoms.
??? mole of He contains → 8.84 x 10²⁴ atoms.
<em>∴ The no. of moles of He contains (8.84 x 10²⁴ atoms) </em>= (1.0 mol)(8.84 x 10²⁴ atoms)/(6.022 x 10²³ atoms) =<em> 14.77 mol.</em>
Answer:
True
Explanation:
Significant digits include non-zero digits (unless the zero is between two non-zero numbers)
In 12,785.000, there are 5 non-zero digits:
1 2 7 8 5
<h3><u>Answer;</u></h3>
Directly proportional
<h3><u>Explanation;</u></h3>
- <em><u>Concentration is one of the factors that determine the rate of a reaction. Reaction rates increases with increase in the concentration of the reactants, which means they are directly proportional.</u></em>
- An increase in the concentration of reactants produces more collisions and thus increasing the rate at which the reaction is taking place. Therefore, <u>Increasing the concentration of a reactant increases the frequency of collisions between reactants and will cause an increase in the rate of reaction.</u>