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ANEK [815]
3 years ago
5

Question 10 (1 point)

Chemistry
1 answer:
choli [55]3 years ago
4 0
Answer
Manganese II oxide
Thank you
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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
Alchen [17]

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.

7 0
3 years ago
Drag each label to the correct image. Each label can be used more than once.
guajiro [1.7K]

1) Acidic - gastric juices pH 1, lemon juice pH 2.2, tomato juice pH 4.

2) Neutral - salt solution pH 7.

3) Alkaline - baking soda pH 9, bleach pH 13.

pH is a numeric scale used to specify the acidity or basicity of an aqueous solution.

When pH is less than seven (pH<7), solution is acidic.

When is equal seven (pH = 7), solution is neutral.  

When pH is greater than seven (pH > 7), solution is basic.

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An ionic solid is placed in water. which information is described by the solubility product constant?
Illusion [34]

Answer:

\boxed{\text{C. the equilibrium between the solid and its ions in solution}}

Explanation:

\rm MX(s) $\, \rightleftharpoons \,$ M$^{+}$(aq) + $^{-}$(aq); $K_{\text{sp}}$ = [M$^{+}$][X$^{-}$]\\\\\text{$K_{\text{sp}}$ gives us information on}\\\\\boxed{\textbf{ the equilibrium between the solid and its ions in solution}}

It tells us nothing about the amount of precipitate that will form or the temperature at which the equilibrium occurs.

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sound waves travel.faster through materials that are more dense

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Which model represents an element? O A. OB. O c. O D.​
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Answer: B represents an element.
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