Answer : The energy removed must be, -67.7 kJ
Solution :
The process involved in this problem are :
The expression used will be:
where,
= heat released by the reaction = ?
m = mass of benzene = 125 g
= specific heat of gaseous benzene =
= specific heat of liquid benzene =
= enthalpy change for vaporization =
Molar mass of benzene = 78.11 g/mole
Now put all the given values in the above expression, we get:
Therefore, the energy removed must be, -67.7 kJ
Answer:
A. It keeps the electrolyte solutions neutral.
Explanation:
The electrons flow from the anode to the cathode and the problem is that the cathode may become too negative. The purpose of the salt bridge is to maintain charge balance because the electrons are moving from one-half cell to the other.
Answer:
Alkaline earth metals
Explanation:
Group 2 has 2 valence electrons that they lose when being ionized, thus making their charge 2+.
Answer:
No matter how well how much you accomplish, or how much good you put into the world, you will always have haters. An unfortunate aspect of life is that some people hate when others do well.It’s just a way of life.some people just are jealous but dont let people brake you down.
Explanation:
hope this helps
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>