In rubidium oxide - Rb₂O , the ions are Rb⁺ and O²⁻
Rb is a group one element with one valence electron. To become stable it loses its outer electron to gain a complete outer shell.
Electronic configuration of Rb is - 1s² 2s² 2p⁶ 3s² 3p⁶ 3d¹⁰ 4s² 4p⁶ 5s¹
Once it loses its valence electron the configuration is;
- 1s² 2s² 2p⁶ 3s² 3p⁶ 3d¹⁰ 4s² 4p⁶
The noble gas with this configuration is Krypton - Kr
Oxygen electron configuration is 1s² 2s² 2p⁴
Once it gains 2 electrons the configuration is - 1s² 2s² 3p⁶
The noble gas with this configuration is Neon - Ne
Volume= Mass/ Density= 20/184
= 0.108 cubic cm
Answer:
Look on the picture.
Explanation:
He could find only 2 isomers of n-hexane alkenes for this reaction. Other two could be marked from other direction.
Answer is: n<span>o, because the ion product is less than the Ksp of lead iodide. </span>
Chemical dissociation 1: KI(s) → K⁺(aq) + I⁻(aq).
Chemical dissociation 2: Pb(NO₃)₂(s) → Pb²⁺(aq) + 2NO₃⁻(aq).
Chemical reaction: Pb²⁺(aq) + 2I⁻(aq) → PbI₂(s).
Ksp(PbI₂) = 7.1·10⁻⁹.
V = 500 mL ÷ 1000 mL/L = 0.5 L.
c(KI) = c(I⁻) = 0.0025 mol ÷ 0.5 L.
c(I⁻) = 0.005 M.
c(Pb(NO₃)₂) = c(Pb²⁺) = 0.00004 mol ÷ 0.5 L.
c(Pb²⁺) = 0.00008 M.
Q = c(Pb²⁺) · c(I⁻)².
Q = 8·10⁻⁵ M · (5·10⁻³ M)².
Q = 2·10⁻⁹; <span> the ion product.</span>
Answer:
Explanation:
is insufficient to overcome intermolecular forces.
