The term used to describe the rapid release of bubbles, or rapid release of a gas from a liquid or a solution is called Effervescence. The bubbling of a solution is due to the escape of a gas which may be from a chemical reaction, as in fermenting liquid, or by coming out of a solution after having been under pressure, as in a carbonated drink. For example; soda, champagne among others.
Answer:
OPTION (A) : Testing a rock sample for gold content
Explanation:
For testing a rock sample of gold content you will need a Chemist. To test the material, the sample is rubbed on black stone which will leave a mark on the stone. This mark is tested by applying aqua fortis i.e nitric acid on the mark. If the mark gets dissolve then the material is not gold. If the mark sustain the it is further tested by applying aqua regia i.e nitric acid and hydrochloric acid which will prove the sample is of gold if it gets dissolve on using hydrochloric acid. The purity of the sample can be checked by differing the concentration of the aqua regia and comparing it with the gold material of the known purity.
Answer:
Mass = 785.9 g
Explanation:
Given data:
Atoms of gold = 2.4 × 10²⁴ atoms
Mass of gold = ?
Solution:
First of all we will convert the number of atoms into moles.
2.4 × 10²⁴ atoms × 1 mol/ 6.02 × 10²³ atoms
number of moles = 3.99 mol
Now we will determine the mass of gold.
Mass = number of moles × molar mass
Mass = 3.99 mol × 196.97 g/mol
Mass = 785.9 g
<span>As new discoveries are made, existing theories are revised or replaced.</span>
Answer:
67.8%
Explanation:
La reacción de descomposición del CaCO₃ es:
CaCO₃ → CO₂ + CaO
<em>Donde 1 mol de CaCO₃ al descomponerse produce 1 mol de CO₂ y 1 mol de CaO.</em>
Usando la ley general de los gases, las moles de dioxido de carbono son:
PV = nRT.
<em>Donde P es presión (1atm), V es volumen (20L), n son moles de gas, R es la constante de los gases (0.082atmL/molK) y T es temperatura absoluta (15 + 273.15 = 288.15K). </em>Reemplazando los valores en la ecuación:
PV / RT = n
1atmₓ20L / 0.082atmL/molKₓ288.15K = 0.846 moles
Como 1 mol de CO₂ es producido desde 1 mol de CaCO₃, las moles iniciales de CaCO₃ son 0.846moles.
La masa molar de CaCO₃ es 100.087g/mol. Así, la masa de 0.846moles de CaCO₃ es:
0.846moles ₓ (100.087g / mol) = <em>84.7g de CaCO₃</em>
Así, la pureza del marmol es:
(84.7g de CaCO₃ / 125g) ₓ 100<em> = </em>
<h3>67.8%</h3>
