Describes the process by which gas

Hydrogen, carbon, and gold are all examples of? Chemical Reactions Elements Molecules Compounds

ddd [48]

Elements :)
>>On the periodic table of elements

5 0

3 years ago

a)Nitrous oxide gas (systematic name: dinitrogen monoxide) is used by some dental practitioners as an anesthetic. Nitrous oxide

Mama L [17]

Explanation:

Nitrous oxide or dinitrogen monoxide can be simple be prepared by the decomposition of ammonium nitrate in the following way -

<u>The reaction is as follows -</u>

NH₄NO₃ ----------------> N₂O + H₂O

This is the unbalanced reaction as was asked in the question , without mentioning the states of matter .

The reaction is carried out in an anhydrous condition ,since , ammonium nitrate can explode on heating .

6 0

4 years ago

Read 2 more answers

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

4 years ago

What is the final pressure (expressed in atm) of a 3.05 L system initially at 724 mm Hg and 298 K, that is compressed to a final

maksim [4K]

Hey there!:

V1 = 3.05 L

V2 = 3.00 L

P1 = 724 mmHg

P2 = to be calculated

T1 = 298 K

T2 = 273 K

Therefore:

P1*V1 / T1 = P2*V2 / T2

P2 = ( P1*V1 / T1 ) * T2 / V2

P2 = 724 * 3.05 * 273 / 298 * 3.00

P2 = 602838.6 / 894

P2 = 674.31 mmHg

1 atm ----------- 760 mmHg

atm ------------- 674.31 mHg

= 674.31 * 1 / 760

= 0.887 atm

Hope this helps!

5 0

4 years ago

How many electrons does sulfur have to gain to achieve a noble-gas electron configuration

Tpy6a [65]

Sulfur has to gain 2 electrons to achieve a noble gas electron.

8 0

3 years ago