For ksp of silver sulfate (ag2so4) at 1. 2 ✕ 10-5, the solubility of silver sulfate is mathematically given as
x = 5.26*10^{-4}
<h3>What is the
solubility of silver
sulfate in 0.15 M AgNO3?</h3>
Generally, the equation for the Chemical reaction is mathematically given as
Ag2SO4 ⇄ 2Ag+ + SO4 2-
Therefore
Ksp = [Ag+ ]2 *[SO4 2-]
1.2*10-5 = ( 0.15 +2x)2*(x)
x = 5.26*10^{-4}
In conclusion, the solubility of silver sulfate is
x = 5.26*10^{-4}
Read more about Chemical reaction
brainly.com/question/16416932
Answer:
8,5 E10 nm
Explanation:
⇒ 0.85 E-2 Km * ( 1000 m / Km ) * ( 1 E9 nm / m ) = 8.5 E10 nm
⇒ 0.85 E-2 Km = 8.5 E10 nm
Alkali metals have only one valence electron and so have low binding energy to the metallic crystal lattice. ... A lower amount of energy needed to break a bond means a lower melting/boiling point.
The molecular weight is the same as molecular mass. You find this by using the periodic table
The molar mass of Mg(NO3)2 is 148.3
It is B, and also for a moment I didn't understand that 4.69 x 10^22. I almost did this whole problem wrong.
