We are given an equation 2Mg+O2-> 2MgO and a starting chemical Mg of about 212 g. In order to solve for the amount of O2 needed, we need the molecular weight of Mg and O2.
Molecular weight:
Mg=24.305 g/mol
O2=16(2)=32 g/mol
Note that for every 1 mol of O2, the amount of Mg must be 2 mol.
So,
g O2 = 212 g Mg x1mo Mgl/24.305 g Mg x1mol O2 /2 mol Mg x 32 g O2/mol O2
gO2=139.56 g
Therefore, 139.56 g of O2 is needed for every 212 g Mg.
Answer:
By using magnet.
Explanation:
Put sand and zinc in the bowl.
Take a magnet an bring it close to the bowl.
The zinc from the sand will be attracted towards the magnet.
In this way it will be separated.
La materia no se crea ni se destruye solo se transforma
Explanation:
«En toda reacción química la masa se conserva, es decir, la masa total de los reactivos es igual a la masa total de los productos».
Los átomos ni de crean ni se destruyen,durante una reacción química.Por lo tanto una ecuación química ha de tener el mismo numero de atomos de cada elemento a ambos lados de la flecha. Se dice entonces que la ecuación esta balanceada
Espero y te sirva!!
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.
