Answer:
Δ S = 26.2 J/K
Explanation:
The change in entropy can be calculated from the formula -
Δ S = m Cp ln ( T₂ / T₁ )
Where ,
Δ S = change in entropy
m = mass = 2.00 kg
Cp =specific heat of lead is 130 J / (kg ∙ K) .
T₂ = final temperature 10.0°C + 273 = 283 K
T₁ = initial temperature , 40.0°C + 273 = 313 K
Applying the above formula ,
The change in entropy is calculated as ,
ΔS = m Cp ln ( T₂ / T₁ ) = (2.00 )( 130 ) ln( 283 K / 313 K )
ΔS = 26.2 J/K
Answer:
0. 414
Explanation:
Octahedral interstitial lattice sites.
Octahedral interstitial lattice sites are in a plane parallel to the base plane between two compact planes and project to the center of an elementary triangle of the base plane.
The octahedral sites are located halfway between the two planes. They are vertical to the locations of the spheres of a possible plane. There are, therefore, as many octahedral sites as there are atoms in a compact network.
The Octahedral interstitial void ratio range is 0.414 to 0.732. Thus, the minimum cation-to-anion radius ratio for an octahedral interstitial lattice site is 0. 414.
Answer:
The solar wind is a stream of charged particles released from the upper atmosphere of the Sun, called the corona. ... Its particles can escape the Sun's gravity because of their high energy resulting from the high temperature of the corona, which in turn is a result of the coronal magnetic field.
Explanation:
Answer:
Ksp = 1.07x10⁻²¹
Explanation:
Molar solubility is defined as moles of solute can be dissolved in 1L.
Ksp for NiS is defined as:
NiS(s) ⇄ Ni²⁺(aq) + S²⁻(aq)
Ksp = [Ni²⁺] [S²⁻]
As molar solubility is 3.27x10⁻¹¹M, concentration of [Ni²⁺] and [S²⁻] is 3.27x10⁻¹¹M for both.
Replacing:
Ksp = [3.27x10⁻¹¹M] [3.27x10⁻¹¹M]
<em>Ksp = 1.07x10⁻²¹</em>
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Answer:
crust, mantle, inner core, outer core
