Answer:
see explaination
Explanation:
Molecular equation;
2Li3PO4(aq) + 3CaCl2(aq) >>>> Ca3(PO4)2(s) + 6LiCl(aq)
Total ionic equation; . Includes all ions ;
6Li^+(aq) + 2PO4^-3(aq) + 3Ca^+2(aq) + 6Cl^-(aq) >>>> Ca3(PO4)2(s) + 6Li^+(aq) + 6Cl^-(aq)
Net ionic equation; remove common ions from total ionic;
2PO4^-3(aq) + 3Ca^+2(aq) >>>> Ca3(PO4)2(s)
Answer is: a. Rubidium (Rb) is more reactive than strontium (Sr) because strontium atoms must lose more electrons.
The ionization energy (Ei) is the minimum amount of energy required to remove the valence electron, when element lose electrons, oxidation number of element grows (oxidation process).
Alkaline metals (group 1), in this example rubidium, have lowest ionizations energy and easy remove valence electrons (one electron), they are most reactive metals.
Earth alkaline metals (group 2), in this example strontium, have higher ionization energy than alkaline metals, because they have two valence electrons, they are less reactive.
Rubidium electron configuration: ₃₇Rb 1s²2s²2p⁶3s²3p⁶3d¹⁰4s²4p⁶5s¹; one valence electron is 5s¹ orbital.
Strontium electron configuration: ₃₈Sr 1s²2s²2p⁶3s²3p⁶3d¹⁰4s²4p⁶5s²; two valence electrons is 5s² orbital.
Solute=potassium chloride, solvent=water
Answer:
First one is 5.0 M ammonia and the Second one ?
Explanation: