Answer: 44
Explanation: Because its easy
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.
First you calculate how many moles there are in 2.0 grams of hydrogen (H2) atoms.
Hydrogen has a relative atomic mass (RAM) of 1 g/mol, but there are 2 hydrogen atoms: 1 x 2 = 2 g/mol
To work out how many moles there are,
use the formula: n(moles) = mass ÷ molar mass
n(moles) = 2 grams ÷ 2 g/mol = 1 mol
Then use Avogadro's Constant : 6.023 x 10^23
= 1 x 6.023 x 10^23
= 6.023 x 10^23
Final step is to multiply it by the number of atoms, in this case there are 2.
= 6.023 x 10^23 x 2
= 12.046 x 10^23
= 1.205 x 10^24
that ^ should be your final answer
have a great day :)
A and d is physical, b and c is chemical
