Answer:
its c
Explanation:
solvent is the part of the solution that is in the greater quantity, and solute is the smaller portion of the solution
hope this helps :)
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:
16.46 g.
Explanation:
- It is a stichiometry problem.
- We should write the balance equation of the mentioned chemical reaction:
<em>2Cu + Zn(NO₃)₂ → Zn + 2Cu(NO₃).</em>
- It is clear that 2.0 moles of Cu reacts with 1.0 mole of Zn(NO₃)₂ to produce 1.0 mole of Zn and 2.0 moles of Cu(NO₃).
- We need to calculate the number of moles of the reacted Cu (32.0 g) using the relation:
<em>n = mass / molar mass</em>
- The no. of moles of Cu = mass / atomic mass = (32.0 g) / (63.546 g/mol) = 0.503 mol.
<u><em>Using cross multiplication:</em></u>
2.0 moles of Cu produces → 1.0 mole of Zn, from the stichiometry.
0.503 mole of Cu produces → ??? mole of Zn.
- The no. of moles of Zn produced = (1.0 mol)(0.503 mol) / (2.0 mol) = 0.2517 mol.
∴ The grams of Zn produced = no. of moles x atomic mass of Zn = (0.2517 mol)(65.38 g/mol) = 16.46 g.
Answer:
21883.75 Joules are required to melt the ice!
