The appropriate answer is option E. Gibbs free energy can be expressed using the equation: ΔG = ΔH - TΔS, where ΔH denotes the change in enthalpy of the reaction, T is the reaction temperature, and ΔS signifies entropy change. For our calculations, we have ΔH = -720.5 kJ/mol which converts to -720500 J/mol (given that 1 kJ = 1000 J), ΔS = -263.7 J/K, and T = 141.0°C, which equals 414.15 K. Consequently, the Gibbs free energy for the specified reaction at 141.0°C is calculated as -611.3 kJ/mol.
The element with atomic number 58 is Cerium, meaning its symbol should be Ce rather than Co, which belongs to Cobalt with atomic number 27. Therefore, the notation for isotopes consists of the element's symbol accompanied by a superscript and a subscript, properly aligned. The superscript indicates the mass number.
Mass number = protons + neutrons = 58 + 33 = 91
The subscript denotes the atomic number, which is 58. This notation is illustrated in the attached image.
<span>4.3065 g
To begin with, consult the atomic masses for each involved element.
Atomic weight of Calcium = 40.078
Atomic weight of Carbon = 12.0107
Atomic weight of Hydrogen = 1.00794
Atomic weight of Oxygen = 15.999
Atomic weight of Sulfur = 32.065
Next, compute the molar masses of both reactants and the product.
Molar mass H2SO4 = 2 * 1.00794 + 32.065 + 4 * 15.999
= 98.07688 g/mol
Molar mass CaCO3 = 40.078 + 12.0107 + 3 * 15.999
= 100.0857 g/mol
Molar mass CaSO4 = 40.078 + 32.065 + 4 * 15.999
= 136.139 g/mol
The balanced equation for the reaction between H2SO4 and CaCO3 is:
CaCO3 + H2SO4 ==> CaSO4 + H2O + CO2
Thus, 1 mole each of CaCO3 and H2SO4 is necessary to generate 1 mole of CaSO4. Let's check the amount of moles we have for CaCO3 and H2SO4.
CaCO3: 3.1660 g / 100.0857 g/mol = 0.031632891 mol
H2SO4: 3.2900 g / 98.07688 g/mol = 0.033545113 mol
H2SO4 is in slight excess, therefore CaCO3 is the limiting reactant, suggesting we can expect 0.031632891 moles of product. To find the mass, multiply the number of moles by the molar mass calculated previously.
0.031632891 mol * 136.139 g/mol = 4.306470148 g
Given that we have 5 significant figures from our data, we round the final result to 5 figures, yielding 4.3065 g</span>
