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2 years ago

C5H12 + 8O2 → 5CO2 + 6H2O Calculate the energy change for the reaction:

1 answer:

8 0

C5H12 (l) + 8O2 (g) ----> 5CO2 (g) + 6H2O (l)
Delta H = -3505.8 kJ/mol

C (s) + O2 (g) -----> CO2 (g)
Delta H = -393.5 kJ/mol

H2 (g) + (1/2)O2 (g) ------> H2O (l)
Delta H = -286 kJ/mol

Possible answers:
a. +35 kJ/mol
b. + 1,073 kJ/mol
c. -4,185 kJ/mol
d. -2,826 kJ/mol
e. -178 kJ/mol

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3 years ago

Barium fluoride (BaF2) has a Ksp = 2. 5 × 10–5 (mol/L)3.

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This problem is asking for the dissolution reaction of barium fluoride, both the equilibrium and Ksp expressions in terms of concentrations and x and its molar solubility in water. Thus, answers shown below:

$BaF_2(s)\rightleftharpoons Ba^{2+}(aq)+2F^-(aq)$
$Ksp=[Ba^{2+}][F^-]^2$
$Ksp=(x)(2x)^2$
0.0184 M.

<h3>Solubility product</h3>

In chemistry, when a solid is dissolved in water, one must take into account the fact that not necessarily its 100 % will be able to break into ions and thus undergo dissolution.

In such a way, and specially for sparingly soluble solids, one ought to write the dissolution reaction at equilibrium as shown below for the given barium fluoride:

$BaF_2(s)\rightleftharpoons Ba^{2+}(aq)+2F^-(aq)$

Next, we can write its equilibrium expression according to the law of mass action, which also demands us to omit any solid and refer it to the solubility product constant (Ksp):

$Ksp=[Ba^{2+}][F^-]^2$

Afterwards, one can insert the reaction extent, x, as it stands for the molar solubility of this solid in water, taking into account the coefficients balancing the reaction:

$Ksp=(x)(2x)^2$

Finally, we solve for the x as the molar solubility of barium fluoride as shown below:

$2.5x10^{-5}=(x)(2x)^2\\\\2.5x10^{-5}=4x^3\\\\x=\sqrt[3]{\frac{2.5x10^{-5}}{4} } \\\\x=0.0184M$

Learn more about chemical equilibrium: brainly.com/question/26453983

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1 year ago