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3 years ago

Which colorless and odorless gas, produced by radioactive decay of Uranium-238, is considered to be a cancer-causing agent?

1 answer:

7 0

B, radon is correct. Interestingly, it often collects in basements from radioactive decay of rocks such as granite that contain uranium. Because it is an unreactive noble gas and because it is denser than air it sits in basements and must be pumped out. It collects in human lungs and is the second leading cause of lung cancer behind smoking.

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Match the mountain range with its formation.

zimovet [89]

Yea it turns to a 35

8 0

4 years ago

Explain the bonding that occurs between sodium and chlorine. Specifically, what kind of bonding takes

Alex787 [66]

Answer:

About bonding

Explanation:

A chemical bond is a lasting attraction between atoms, ions or molecules that enables the formation of chemical compounds. The bond may result from the electrostatic force of attraction between oppositely charged ions as in ionic bonds or through the sharing of electrons as in covalent bond.

7 0

3 years ago

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

The solubility of CaF2 is 0.00021 mole per liter. What is the solubility product constant for CaF2? A) 7.3 x 10-12 B) 3.7 x 10-1

Naddika [18.5K]

Answer:

B) Ksp = 3.7 E-11

Explanation:

CaF2 ↔ Ca2+ + 2F-

S S 2S,,,,,,,,,,,,,,,in the equilibrium

⇒ Ksp = [ Ca2+ ] * [ F- ]²

⇒ Ksp = S * ( 2S )²

⇒ Ksp = 4S³

⇒ Ksp = 4 * ( 0.00021 )³

⇒ Ksp = 4 * 9.261 E-12

⇒ Ksp = 3.704 E-11

3 0

3 years ago

What happens to the size of the an atom when it becomes an negative ion?

Ronch [10]

Answer

The negative ion is larger than the original atom.

Explanation:

This means that with fewer electrons but the same number of positive protons, the size of the ionic radius will decrease.

I hope this helped!

3 0

3 years ago