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.
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The correct question is:
Why heat of the sublimation of a substance is greater than the heat of vaporization?
Answer:
19. Option B. ⁰₋₁B
20. Option D. ²¹⁰₈₄Po
Explanation:
19. ²²⁸₈₈Ra —> ²²⁸₈₉Ac + ʸₓZ
Thus, we can determine ʸₓZ as follow:
228 = 228 + y
Collect like terms
228 – 228 = y
y = 0
88 = 89 + x
Collect like terms
88 – 89 = x
x = –1
Thus,
ʸ ₓZ => ⁰₋₁Z => ⁰₋₁B
²²⁸₈₈Ra —> ²²⁸₈₉Ac + ʸₓZ
²²⁸₈₈Ra —> ²²⁸₈₉Ac + ⁰₋₁B
20. ᵘᵥX —> ²⁰⁶₈₂Pb + ⁴₂He
Thus, we can determine ᵘᵥX as follow:
u = 206 + 4
u = 210
v = 82 + 2
v = 84
Thus,
ᵘᵥX => ²¹⁰₈₄X => ²¹⁰₈₄Po
ᵘᵥX —> ²⁰⁶₈₂Pb + ⁴₂He
²¹⁰₈₄Po —> ²⁰⁶₈₂Pb + ⁴₂He
Answer:
los metales alcalinotérreos: berilio (Be), magnesio (Mg), calcio (Ca), estroncio (Sr), bario (Ba) y radio (Ra).
o simplemente llamado grupo 2A
Explanation:
Answer:
Na₁₁ = 1s² 2s² 2p⁶ 3s¹
Explanation:
Sodium is present in group 1.
It is alkali metal.
It has one valence electron.
The atomic number of sodium is 11.
Its atomic mass is 23 amu.
The longhand notation of electronic configuration of sodium can be written as,
Na₁₁ = 1s² 2s² 2p⁶ 3s¹
The electronic configuration in shorthand notation( noble gas) would be written as,
Na₁₁ = [Ne] 3s¹
Sodium loses its one valence electron to complete the octet and get stable thus form +1 cation.
It react with halogen and form salt. Such as sodium chloride.
2Na + Cl₂ → 2NaCl
Hey there!:
density = 3.51 g/cm³
Volume = 0.0270 cm³
Therefore:
D = m / V
3.51 = m / 0.0270
m = 3.51 * 0.0270
m = 0.09477 g