Metal (Ca) Polyatomic anion (CO3). You do not need to use the roman numeral because Ca is in group 1. The answer is metal + polyatomic or Calcium Carbonate
Answer:
*result with sodium hydroxide: white ppt of Zn(OH)2
*excess sodium hydroxide: Precipitate dissolves to form a COLOURLESS solution
*Result with Aqueous ammonia: White ppt of Zn(OH)2
*excess aqueous ammonia: Precipitate dissolves to form a COLOURLESS solution
Answer:
0.13 M ( 2 s.f)
Explanation:
2Cl2O5 (g)-->2Cl2(g) +5O2 (g)
rate= (17.4 M -1 .s -1 ) [Cl2O5]2
From the rte above, we can tell that our rate constant (k) = 17.4 M -1 .s -1
The units of k tells us this is a second order reaction.
Initial Concentration [A]o = 1.46M
Final Concentration [A] = ?
Time = 0.400s
The integrated rate law for second order reactions is given as;
1 / [A] = (1 / [A]o) + kt
1 / [A] = [ (1/ 1.46) + (17.4 * 0.4) ]
1 / [A] = 0.6849 + 6.96
1 / [A] = 7.6496
[A] = 1 / 7.6496
[A] = 0.13073 M ≈ 0.13 M ( 2 s.f)
Answer: A, C, E
Explanation: PLATO. all testable questions.
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.
