Nal is the formula for sodium iodine
Answer:
Explanation:
Hello!
In this case, since the pH of the given metal is 10.15, we can compute the pOH as shown below:
Now, we compute the concentration of hydroxyl ions in solution:
![[OH^-]=10^{-pOH}=10^{-3.95}=1.41x10^{-4}M](https://tex.z-dn.net/?f=%5BOH%5E-%5D%3D10%5E%7B-pOH%7D%3D10%5E%7B-3.95%7D%3D1.41x10%5E%7B-4%7DM)
Now, since this hydroxide has the form MOH, we infer the concentration of OH- equals the concentration of M^+ at equilibrium, assuming the following ionization reaction:
Whose equilibrium expression is:
![Ksp=[M^+][OH^-]](https://tex.z-dn.net/?f=Ksp%3D%5BM%5E%2B%5D%5BOH%5E-%5D)
Therefore, the Ksp for the saturated solution turns out:
Best regards!
W*V i believe because it comes down to this :
I is the current, W is wattage V is volts and the * is the thing that represents the Amperes (i’m not 100% but this is my best)
Answer:
V = 3.1 L
Explanation:
Given data:
Molarity of solution = 0.37 M
Mass of LiF = 29.53 g
Volume of solution = ?
Solution:
Number of moles of LiF:
Number of moles = mass/molar mass
Number of moles = 29.53 g/ 25.94g/mol
Number of moles = 1.14 mol
Volume:
Molarity = number of moles of solute / Volume in L
0.37 M = 1.14 mol / V
V = 1.14 mol / 0.37 M
V = 3.1 L (M = mol/L)
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