Answer:
a) S = 0.0152 mol/L
b) S' = 4.734 g/L
Explanation:
S 2S S...............in the equilibrium
- Ksp = 1.4 E-5 = [ Ag+ ]² * [ SO42-]
a) molar solubility:
⇒ Ksp = ( 2S) ² * S = 1.4 E-5
⇒ 4S² * S = 1.4 E-5
⇒ S = ∛ ( 1.4 E-5 / 4 )
⇒ S = 0.0152 mol/L
b) solubility ( S' ) in grams per liter:
∴ Mw Ag2SO4 = 311.799 g/mol
⇒ S' = 0.0152 mol/L * ( 311.799 g/mol )
⇒ S' = 4.734 g/L
Answer:
The answer to your question is: 55.84 u. The element is Iron
Explanation:
isotope 1 mass 53.94 u abundance 5.85%
2 mass 55.9349 abundance 91.75%
3 mass 56.9354 abundance 2.12%
4 mass 57.9333 abundance 0.28%
average atomic mass = (53.94 x 0.0585) + (55.9349 x 0.9175) + (56.9354 x 0.0212) + (57.9333 x 0.0028)
average atomic mass = 3.155 + 51.32 + 1.207 + 0.1622
average atomic mass = 55.84 u. The element is Iron
Answer:
Here is the solution..Hope it helps:)
A binary ionic compound is composed of ions of two different elements. One of which is a metal, and the other a non-metal
Fe3N2, also known as Iron (II) nitride, is an ionic compound.
Ionic compounds are compounds that consists of metals and non-metals bonded with ionic bonds. The metal ion gives up electron(s) to the non-metals.
Since iron is a metal and nitrogen is an non-metal, the bond they would form would be an ionic bond. Iron gives up 2 electrons to form iron(II) ion, while nitrogen gains 3 electrons to form nitride ion. Since one iron cannot let a nitrogen gain 3 electrons, so in the compound, there would be 3 iron (ii) ions that has given up 6 electrons in total while 2 nitride ions have gained 6 electrons in total.
