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KIM [24]
3 years ago
9

n atom “likes” to obtain a full outer shell. The fourth shell (in the atoms that we are studying today) can hold a maximum of 8

electrons. Think: Energetically, would the atom above “prefer” to give away its valence electrons or borrow some from another atom in order to fill its outer shell? Explain your reasoning
Chemistry
1 answer:
fgiga [73]3 years ago
4 0

It depends on how many electrons are already in the outer shell.

If it is a large amount (if it is almost to the maximum), then it will want to borrow from another atom.

But if it is close to empty then it will want to give away so it will go back to the inner ring with will be full.

Hope this helps, good luck!

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One liter of oxygen gas at standard temperature and pressure has a mass of 1.43 g. The same volume of hydrogen gas under these c
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Answer:

Indeed, the two samples should contain about the same number of gas particles. However, the molar mass of \rm O_2\; (g) is larger than that of \rm H_2\; (g) (by a factor of about 16.) Therefore, the mass of the \rm O_2\; (g) sample is significantly larger than that of the \rm H_2\; (g) sample.

Explanation:

The \rm O_2\; (g) and the \rm H_2\; (g) sample here are under the same pressure and temperature, and have the same volume. Indeed, if both gases are ideal, then by Avogadro's Law, the two samples would contain the same number of gas particles (\rm O_2\; (g) and \rm H_2\; (g) molecules, respectively.) That is:

n(\mathrm{O_2}) = n(\mathrm{H}_2).

Note that the mass of a gas m is different from the number of gas particles n in it. In particular, if all particles in this gas have a molar mass of M, then:

m = n \cdot M.

In other words,

  • m(\mathrm{O_2}) = n(\mathrm{O_2}) \cdot M(\mathrm{O_2}).
  • m(\mathrm{H_2}) = n(\mathrm{H_2}) \cdot M(\mathrm{H_2}).

The ratio between the mass of the \rm O_2\; (g) and that of the \rm H_2\; (g) sample would be:

\begin{aligned}& \frac{m(\mathrm{O_2})}{m(\mathrm{H_2})} = \frac{n(\mathrm{O_2})\cdot M(\mathrm{O_2})}{n(\mathrm{H_2})\cdot M(\mathrm{H_2})}\end{aligned}.

Since n(\mathrm{O_2}) = n(\mathrm{H}_2) by Avogadro's Law:

\begin{aligned}& \frac{m(\mathrm{O_2})}{m(\mathrm{H_2})} = \frac{n(\mathrm{O_2})\cdot M(\mathrm{O_2})}{n(\mathrm{H_2})\cdot M(\mathrm{H_2})} = \frac{M(\mathrm{O_2})}{M(\mathrm{H_2})}\end{aligned}.

Look up relative atomic mass data on a modern periodic table:

  • \rm O: 15.999.
  • \rm H: 1.008.

Therefore:

  • M(\mathrm{O_2}) = 2 \times 15.999 \approx 31.998\; \rm g \cdot mol^{-1}.
  • M(\mathrm{H_2}) = 2 \times 1.008 \approx 2.016\; \rm g \cdot mol^{-1}.

Verify whether \begin{aligned}& \frac{m(\mathrm{O_2})}{m(\mathrm{H_2})}= \frac{M(\mathrm{O_2})}{M(\mathrm{H_2})}\end{aligned}:

  • Left-hand side: \displaystyle \frac{m(\mathrm{O_2})}{m(\mathrm{H_2})}= \frac{1.43\; \rm g}{0.089\; \rm g} \approx 16.1.
  • Right-hand side: \displaystyle \frac{M(\mathrm{O_2})}{M(\mathrm{H_2})}= \frac{31.998\; \rm g \cdot mol^{-1}}{2.016\; \rm g \cdot mol^{-1}} \approx 15.9.

Note that the mass of the \rm H_2\; (g) sample comes with only two significant figures. The two sides of this equations would indeed be equal if both values are rounded to two significant figures.

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