Since nonmetals gain electrons, the correct question to ask about group 16 elements is; "Will group 16 elements gain electrons to bond with group 2 in an XY format?"
Group 16 elements are divalent and they form divalent negative ions. The periodic table is arranged in groups and periods. The elements in the same group have the same number of valence electrons. All elements in group 2 have six valence electrons.
If a wants to form a compound with the non metals of group 16, the correct question to ask is;"Will group 16 elements gain electrons to bond with group 2 in an XY format?"
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The mass of carbon required to produce iron from 5 67 g iron (iii) oxide is 9.0 g.
<h3>What is the mass of carbon required to produce iron from iron (iii) oxide?</h3>
The mass of carbon required to produce iron from iron (iii) oxide is determined from the equation of the reaction.
From the given equation of reaction, 3 moles of carbon produces 4 moles of iron.
- Moles of a substance = mass/molar mass
Molar mass of iron = 56.0 g
molar mass of carbon = 12.0 g
Moles of iron in 5.67 g of iron = 5.67/56 = 0.1 moles
Moles of carbon required = 3/4 × 0.1 = 0.75 moles
Mass of carbon = 0.75 × 12 = 9.0 g
Therefore, the mass of carbon required to produce iron from 5 67 g iron (iii) oxide is 9.0 g.
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It can be oxygen in only 1 atom. All elements on the periodic table can stand alone in one atom.
(I took a lucky guess, using some of my knowledge about elements )
Explanation:
The given data is as follows.
= 100 mm Hg or 
= 0.13157 atm
= 
= (1080 + 273) K = 1357 K
= 
= (1220 + 273) K = 1493 K
= 600 mm Hg or 
= 0.7895 atm
R = 8.314 J/K mol
According to Clasius-Clapeyron equation,
![log(\frac{0.7895}{0.13157}) = \frac{\Delta H_{vap}}{2.303 \times 8.314 J/mol K}[\frac{1}{1357 K} - \frac{1}{1493 K}]](https://tex.z-dn.net/?f=log%28%5Cfrac%7B0.7895%7D%7B0.13157%7D%29%20%3D%20%5Cfrac%7B%5CDelta%20H_%7Bvap%7D%7D%7B2.303%20%5Ctimes%208.314%20J%2Fmol%20K%7D%5B%5Cfrac%7B1%7D%7B1357%20K%7D%20-%20%5Cfrac%7B1%7D%7B1493%20K%7D%5D)
![log (6) = \frac{\Delta H_{vap}}{19.147}[\frac{(1493 - 1357) K}{1493 K \times 1357 K}]](https://tex.z-dn.net/?f=log%20%286%29%20%3D%20%5Cfrac%7B%5CDelta%20H_%7Bvap%7D%7D%7B19.147%7D%5B%5Cfrac%7B%281493%20-%201357%29%20K%7D%7B1493%20K%20%5Ctimes%201357%20K%7D%5D)
0.77815 = 
= 
J/mol
= 
= 221.9 kJ/mol
Thus, we can conclude that molar heat of vaporization of substance X is 221.9 kJ/mol.
