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.
Answer:
Mg(s) + 2H^+(aq) ----> Mg^2+(aq) + H2(g)
Explanation:
We must first write the molecular equation, then obtain the complete ionic equation and subsequently get the net ionic equation as shown below;
Molecular reaction equation;
Mg(s) + 2HCl(aq) ------> MgCl2(aq) + H2(g)
Complete ionic equation;
Mg(s) + 2H^+(aq) + 2Cl^-(aq) ----> Mg^2+(aq) + 2Cl^-(aq) + H2(g)
Net ionic equation;
Mg(s) + 2H^+(aq) ----> Mg^2+(aq) + H2(g)
Answer: "Yes".
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Answer:
C 108
Explanation:
1 It was found that 4n atoms of metal x
weigh 501.6 g. The relative atomic mass
of X is 209. If n atoms of another metal,
Y, weigh 65.00 g, what is the relative
atomic mass of metal Y?
A 27 B 52 C 108 D 137
if 4n atoms of x weigh 501.6 gm, the n atoms weigh 501.6/4 = 125.4 gms
since the atomic mass is 209, then 6.022 X10^23 atom weigh 209
the ratio of 125.4 to 209 is
125.4/Y =0.6
therefor n is the number of atom in 0.6 moles
if metal Y has the same, number of atoms "n" and weighs 65 gm
then then the relative atomic mass is
65/0.6 =108
so the answer is
C 108
Answer:
IT would not
Explanation:
because the design would be too bulky and would be very unsafe.
