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3 years ago

All of the following are physical properties of matter EXCEPT ____.

Chemistry

2 answers:

olga_2 [115]3 years ago

7 0

Mass. The mass stays the same no matter what form it is in.

timofeeve [1]3 years ago

7 0

Ability to rust is a chemical property

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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

Chicago is about 2,800 km from Los Angeles. About how long would it take a jet plane flying 800 km/h to fly from Chicago to Los

BaLLatris [955]

<u>Answer:</u> The time taken by jet plane to reach Los Angeles from Chicago is 3.5 hours

<u>Explanation:</u>

Speed is defined as the rate at which an object moves with respect to time.

To calculate the time taken for the given speed, we use the equation:

$s=\frac{d}{t}$

where,

s = speed of the jet plane = 800 km/hr

d = distance traveled = 2800 km

t = time taken by jet plane = ?

Putting values in above equation, we get:

$800km/hr=\frac{2800km}{t}\\\\t=\frac{2800km}{800km/hr}=3.5hr$

Hence, the time taken by jet plane to reach Los Angeles from Chicago is 3.5 hours

4 0

3 years ago

What if you measured 15cc of liquid elemental mercury, how much would that weigh in pounds

PIT_PIT [208]

Answer:

0.447 pounds

Explanation:

The density of Mercury in grams/mL is 13.55 - I multiplied that by the volume and got approximately 203.25 grams, I then turned it into Kilograms and multiplied by 2.2 lbs / kg.

4 0

3 years ago

Calculate the mass of 3 moles of clorine

Vitek1552 [10]

Moles Chlorine to grams = 35.453 grams.
moles Chlorine to grams = 70.906 grams.
moles Chlorine to grams = 106.359 grams.
moles Chlorine to grams = 141.812 grams.
moles Chlorine to grams = 177.265 grams.
moles Chlorine to grams = 212.718 grams.

4 0

3 years ago

How many molecules are there in 3 moles of aluminum chloride ?

Svetradugi [14.3K]

there are 4 atoms in aluminum chloride

6 0

3 years ago