Answer:
"2.48 mole" of H₂ are formed. A further explanation is provided below.
Explanation:
The given values are:
Mole of Al,
= 3.22 mole
Mole of HBr,
= 4.96 mole
Now,
(a)
The number of mole of H₂ are:
⇒ 
or,
⇒ 
⇒ 
⇒ 
(b)
The limiting reactant is:
= 
(c)
The excess reactant is:
= 
This is because solids have less energy than liquids do, hence it takes more energy to excite a solid into its gaseous phase than it does a liquid.
When opposed to merely reducing their separation, from solid to liquid, the energy needed to totally separate the molecules as they move from liquid to gas is substantially higher. The latent heat of vaporization is therefore bigger than the latent heat of fusion for this reason.
<h3>
What is heat of sublimation?</h3>
The amount of energy required to change one mole of a substance from its solid to its gaseous state under particular conditions—typically the standard ones—is known as the enthalpy of sublimation or heat of sublimation (STP). A solid's worth is based on its cohesive energy.
<h3>
What is heat of vaporization?</h3>
The term "enthalpy of vaporization," which is often referred to as "heat of vaporization" or "heat of evaporation," refers to the amount of energy that must be applied to a liquid substance in order to cause a part of that substance to transform into a gas. Vaporization's enthalpy varies with the pressure at which the transition takes place.
Learn more about heat of sublimation: brainly.com/question/13200793
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The correct question is:
Why heat of the sublimation of a substance is greater than the heat of vaporization?
Answer:
- <u>Yes, it is 14. g of compound X in 100 ml of solution.</u>
Explanation:
The relevant fact here is:
- the whole amount of solute disolved at 21°C is the same amount of precipitate after washing and drying the remaining liquid solution: the amount of solute before cooling the solution to 21°C is not needed, since it is soluble at 37°C but not soluble at 21°C.
That means that the precipitate that was thrown away, before evaporating the remaining liquid solution under vacuum, does not count; you must only use the amount of solute that was dissolved after cooling the solution to 21°C.
Then, the amount of solute dissolved in the 600 ml solution at 21°C is the weighed precipitate: 0.084 kg = 84 g.
With that, the solubility can be calculated from the followiing proportion:
- 84. g solute / 600 ml solution = y / 100 ml solution
⇒ y = 84. g solute × 100 ml solution / 600 ml solution = 14. g.
The correct number of significant figures is 2, since the mass 0.084 kg contains two significant figures.
<u>The answer is 14. g of solute per 100 ml of solution.</u>
