Answer:
The common ion will be di-positive ion.
Explanation:
The ionization energy is defined as the amount of energy needed for removal of most loosely bound electron from an isolated atom in gaseous state.
The low ionization energy shows that the atom is able to give electron easily as after losing electron it may attain noble gas configuration or half filled stability.
Here the first and second ionization energy, both are low suggesting that the element is ready to give two electrons easily to form a di-positive ion however the third ionization energy is high which shows that it will not form tri-positive ion commonly.
M=11.20 g
m(H₂)=0.6854 g
M(H₂)=2.016 g/mol
M(Mg)=24.305 g/mol
M(Zn)=65.39 g/mol
w-?
m(Mg)=wm
m(Zn)=(1-w)m
Zn + 2HCl = ZnCl₂ + H₂
m₁(H₂)=M(H₂)m(Zn)/M(Zn)=M(H₂)(1-w)m/M(Zn)
Mg + 2HCl = MgCl₂ + H₂
m₂(H₂)=M(H₂)m(Mg)/M(Mg)=M(H₂)wm/M(Mg)
m(H₂)=m₁(H₂)+m₂(H₂)
m(H₂)=M(H₂)(1-w)m/M(Zn)+M(H₂)wm/M(Mg)=M(H₂)m{(1-w)/M(Zn)+w/M(Mg)}
m(H₂)=M(H₂)m{(1-w)/M(Zn)+w/M(Mg)}
(1-w)/M(Zn)+w/M(Mg)=m(H₂)/{M(H₂)m}
1/M(Zn)-w/M(Zn)+w/M(Mg)=m(H₂)/{M(H₂)m}
w(1/M(Mg)-1/M(Zn))=m(H₂)/{M(H₂)m}-1/M(Zn)
w=[m(H₂)/{M(H₂)m}-1/M(Zn)]/(1/M(Mg)-1/M(Zn))
w=0.583 (58.3%)
The key is a process called <span>pasteurization</span>
Answer: To solve this question, we need to use the Avogadro's Number, which is a constant first discovered by Amadeo Avogadro, an Italian scientist. He discovered that in a mole of a substance, there are 6,02*10²³ molecules. Using this relationship, we apply the following conversion factor:
So, 8,50 * 10²⁴ molecules of Na₂SO₃ represent 14,12 moles of Na₂SO₃
Explanation:
