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Metal and Minerals may have survived 85 years of weather.
This uses something called <span>Le Chatelier's principle. It states essentially that any stress put upon a system will be corrected.
In more simple terms, it means that in an equilibrium, such as the equation N2(g) + 3H2(g) <=> 2NH3(g), removing a reactant will cause the system to create more of said reactant to compensate for its loss, or adding excess reactant will cause the system to remove some of the added reactant. For future reference, the same principle applies to products in an equilibrium as well.
In this case, hydrogen gas is a reactant, and hydrogen is being removed. According to </span><span>Le Chatelier's principle, the system will shift to create more hydrogen gas. In essence, it will shift in the direction of the hydrogen gas, so there will be a shift toward the reactants.
To clear something up, Keq will not change, as it is a constant value with constant conditions (such as temperature, pressure, etc.).</span>
Answer:
mass CaI2 = 23.424 Kg
Explanation:
From the periodic table we obtain for CaI2:
⇒ molecular mass CaI2: 40.078 + ((2)(126.90)) = 293.878 g/mol
∴ mol CaI2 = (4.80 E25 units )×(mol/6.022 E23 units) = 79.708 mol CaI2
⇒ mass CaI2 = (79.708 mol CaI2)×(293.878 g/mol) = 23424.43 g
⇒ mass CaI2 = 23.424 Kg
Part (a) :
H₂(g) + I₂(s) → 2 HI(g)
From given table:
G HI = + 1.3 kJ/mol
G H₂ = 0
G I₂ = 0
ΔG = G(products) - G(reactants) = 2 (1.3) = 2.6 kJ/mol
Part (b):
MnO₂(s) + 2 CO(g) → Mn(s) + 2 CO₂(g)
G MnO₂ = - 465.2
G CO = -137.16
G CO₂ = - 394.39
G Mn = 0
ΔG = G(products) - G(reactants) = (1(0) + 2*-394.39) - (-465.2 + 2*-137.16) = - 49.3 kJ/mol
Part (c):
NH₄Cl(s) → NH₃(g) + HCl(g)
ΔG = ΔH - T ΔS
ΔG = (H(products) - H(reactants)) - 298 * (S(products) - S(reactants))
= (-92.31 - 45.94) - (-314.4) - (298 k) * (192.3 + 186.8 - 94.6) J/K
= 176.15 kJ - 84.78 kJ = 91.38 kJ
The volume of the new solution is
calculation
by use of the formula
M1V1= M2V2
M1 (molarity 1) = 2.13 M
V1 (volume 1) = 1.24 l
M2 ( molarity 2) = 1.60 M
V2 (volume 2) = ?
by making V2 the subject of the formula
V2 = M1 V1/ M2
V2=( 1.24 x 2.13)÷ 1.60= 1.651 L
