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2 years ago

What are the oxides of oxygen

Chemistry

2 answers:

galina1969 [7]2 years ago

6 0

Answer:

Oxides are chemical compounds with one or more oxygen atoms combined with another element (e.g. Li2O). Oxides are binary compounds of oxygen with another element, e.g., CO2, SO2, CaO, CO, ZnO, BaO2, H2O, etc. These are termed as oxides because here, oxygen is in combination with only one element.

Explanation:

Trava [24]2 years ago

5 0

Answer:

<u>Oxides</u> are chemical compounds with one or more oxygen atoms combined with another element (e.g. Li2O). Oxides are binary compounds of oxygen with another element, e.g., CO2, SO2, CaO, CO, ZnO, BaO2, H2O, etc. These are termed oxides because here, oxygen is in combination with only one element.

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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

Consider the balanced chemical reaction when phosphorus and iodine react to produce phosphorus triodide: 2 P(s) + 3 I2(g) → 2 PI

melamori03 [73]

Answer:

Percent yield of PI3 = 95.4%

Explanation:

This is the reaction:

2P (s) + 3I2 (g) > 2PI3 (g)

Let's determine the moles of iodine that has reacted.

58.6 g / 253.8 g/mol = 0.231 mol

Ratio is 3:2. Let's make a rule of three to state the moles produced at 100 % yield reaction.

3 moles of I2 react to make 2 moles of PI3

0.231 moles of I2 would make (0.231 .2) / 3 = 0.154 moles of PI3

As we have produced 0.147 moles let's determine the percent yield.

(Yield produced / Theoretical yield) . 100 > (0.147 / 0.154) . 100 = 95.4%

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2 years ago

Symptoms of a mental disorder are ways of

kkurt [141]

Answer: thinking

Explanation:

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2 years ago

Which of the following will increase the number of collisions without changing the energy of reactant molecules?

natulia [17]

Answer:increasing the concentration of reactants

Explanation:

Collision is the phenomenon in which the reactant molecules come to nearest closness,as a result the reactants are converted into products.

Now the number of effective collision is directly proportional to the number of reactants added..

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2 years ago

The elements in yellow are referred to as ___________.

Lemur [1.5K]

Answer:

Metalloids

Explanation:

hope this helps! :)

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2 years ago

What are the oxides of oxygen​

What are the oxides of oxygen