Answer:25,06 kJ of energy must be added to a 75 g block of ice.
ΔHfusion(H₂O) = 6,01 kJ/mol.
T(H₂O) = 0°C.
m(H₂O) = 75 g.
n(H₂O) = m(H₂O) ÷ M(H₂O).
n(H₂O) = 75 g ÷ 18 g/mol.
n(H₂O) = 4,17 mol.
Q = ΔHfusion(H₂O) · n(H₂O)
Q = 6,01 kJ/mol · 4,17 mol
Q = 25,06 kJ.
Explanation:
Answer:
![[SO_2Cl_2] = 0.09983 M](https://tex.z-dn.net/?f=%5BSO_2Cl_2%5D%20%3D%200.09983%20M)
Explanation:
Write the balance chemical equation ,

initial concenration of 
lets assume that degree of dissociation=
concenration of each component at equilibrium:
![[SO_2Cl_2] = 0.1-0.1\alpha](https://tex.z-dn.net/?f=%5BSO_2Cl_2%5D%20%3D%200.1-0.1%5Calpha)
![[SO_2] = 0.1\alpha](https://tex.z-dn.net/?f=%5BSO_2%5D%20%3D%200.1%5Calpha)
![[Cl_2] = 0.1\alpha](https://tex.z-dn.net/?f=%5BCl_2%5D%20%3D%200.1%5Calpha)


as
is very small then we can neglect 
therefore ,



Eqilibrium concenration of ![[SO_2Cl_2] = 0.1-0.1\alpha = 0.1-0.1\times 0.00173](https://tex.z-dn.net/?f=%5BSO_2Cl_2%5D%20%3D%200.1-0.1%5Calpha%20%3D%200.1-0.1%5Ctimes%200.00173)
![[SO_2Cl_2] = 0.09983 M](https://tex.z-dn.net/?f=%5BSO_2Cl_2%5D%20%3D%200.09983%20M)
The amount of calories you weight
The density of an object determines whether it will float or sink in another substance. An object will float if it is less dense than the liquid it is placed in. An object will sink if it is more dense than the liquid it is placed in.
Answer:
ΔH₁₂ = -867.2 Kj
Explanation:
Find enthalpy for 3H₂ + O₃ => 3H₂O given ...
2H₂ + O₂ => 2H₂O ΔH₁ = -483.6 Kj
3O₂ => 2O₃ ΔH₂ = + 284.6 Kj
_____________________________
3(2H₂ + O₂ => 2H₂O) => 6H₂ + 3O₂ => 6H₂O (multiply by 3 to cancel O₂)
6H₂ + 3O₂ => 6H₂O ΔH₁ = 3(-483.6 Kj) = -1450.6Kj
2O₃ => 3O₂ ΔH₂ = -284.6Kj (reverse rxn to cancel O₂)
_______________________________
6H₂ + 2O₃ => 6H₂O ΔH₁₂ = -1735.2 Kj (Net Reaction - not reduced)
________________________________
divide by 2 => target equation (Net Reaction - reduced)
3H₂ + O₃ => 3H₂O ΔH₁₂ = (-1735.2/2) Kj = -867.2 Kj