Although I may not be the smartest, I can definitely answer.
This represents a chemical change because the substances' chemical identities were altered. The fizzing was a clear sign, and the temperature increase was another indicator of the reaction.
1. The total moles of the solution is 0.3079193 mol.
2. The mole fraction for gold is 0.2473212, and for silver, it is 0.7526787.
3. The molar entropy of mixing for gold is 2.87285 j/K, while for silver, it is 1.77804 j/K.
4. The total entropy of mixing sums to 4.65089 j/K.
5. Molar free energy amounts to -2325.445 kJ.
6. Chemical potential for silver is -1750.31129 j/mol and for gold, it is -575.13185 j/mol. To elaborate:
(1) The molar mass of silver stands at 107.8682 g/mol, and gold's at 196.96657 g/mol. Hence, calculating moles leads to mass/molar mass for silver: 25 g/107.8682 g/mol = 0.2317643 mol and for gold: 15 g/196.96657 g/mol = 0.076155 mol, resulting in a total of 0.30791193 mol.
(2) For the mole fractions, silver's fraction is 0.2317643/0.3079193 = 0.7526787, and for gold, it's 0.076155/0.3079193 = 0.2473212.
(3) To find molar entropy mixing ∆Sm, we use the formula ∆Sm = -R * Xi * ln(Xi) where R = 8.3144598. For silver, substituting gives us 1.77804 j/K, while for gold, we get 2.87285 j/K.
(4) Overall entropy of mixing totals 4.65089 j/K thus calculated.
(5) The Gibbs free energy at 500 °C can be derived through G = H - TS, accounting to H = 0 (as T is 500 + 273 = 773 K and S is 4.65089), resulting in G equating to -3595.138 kJ.
(6) The chemical potentials calculated derive from multiplying the Gibbs free energy by their mole fractions.
Answer:
The resulting salt is Cesium Carbonate
Explanation:
2CsOH + H2CO3 → Cs2CO3 + 2H2O
The correct answer is Option A.
The calculation goes as follows:
Number of millimoles of Na3PO4 = 1 × 100 = 100
Number of millimoles of AgNO3 = 1 × 100 = 100
Dissociating 1 mole of Na3PO4 yields 3 moles of sodium ions and 1 mole of phosphate ions, whereas 1 mole of AgNO3 releases 1 mole of Ag+ and 1 mole of NO3-.
The Ag+ ion concentration becomes negligible since it forms a precipitate with the phosphate ion, indicating that the concentration of phosphate ions is also low.
With 100 millimoles of Na3PO4, we get 300 millimoles of Na+ and 100 millimoles of PO43-, and with 100 millimoles of AgNO3 we have 100 millimoles of Ag+ and 100 millimoles of NO3-.
Thus, the order of increasing concentration is: PO43- < NO3- < Na+.
