<span>The correct answer is b. Radon. Oxygen, Hyrdrogen, and Boron, are not radioactivel ike Radon and it regenerates quickly meaning that even though it has a short half-life period, it stays for a long time once released. It also has no taste so it's difficult to notice without proper gear because you can't feel it.</span>
Balanced chemical reaction: 2CH₄(g) ⇄ C₂H₂(g) + 3H₂(g).
1) In a chemical reaction, chemical equilibrium is the state in which both reactants (methane CH₄) and products (ethyne C₂H₂ and hydrogen H₂) are present in concentrations which have no further tendency to change with time.
2) At equilibrium, both the forward and reverse reactions are still occurring.
3) Reaction rates of the forward and backward reactions are equal and there are no changes in the concentrations of the reactants and products.
Answer:
A) 54.04%
B) 13-karat
Explanation:
A) From the problem we have
<em>1)</em> Mg + Ms = 9.40 g
<em>2)</em> Vg + Vs = 0.675 cm³
Where M stands for mass, V stands for volume, and g and s stand for gold and silver respectively.
We can rewrite the first equation using the density values:
<em>3)</em> Vg * 19.3 g/cm³ + Vs * 10.5 g/cm³ = 9.40
So now we have<em> a system of two equations</em> (2 and 3) <em>with two unknowns</em>:
We <u>express Vg in terms of Vs</u>:
We <u>replace the value of Vg in equation 3</u>:
- Vg * 19.3 + Vs * 10.5 = 9.40
- (0.675-Vs) * 19.3 + Vs * 10.5 = 9.40
- 13.0275 - 19.3Vs + 10.5Vs = 9.40
Now we <u>calculate Vg</u>:
- Vg + 0.412 cm³ = 0.675 cm³
We <u>calculate Mg from Vg</u>:
- 0.263 cm³ * 19.3 g/cm³ = 5.08 g
We calculate the mass percentage of gold:
- 5.08 / 9.40 * 100% = 54.04%
B)
We multiply 24 by the percentage fraction:
- 24 * 54.04/100 = 12.97-karat ≅ 13-karat
