Answer:
7.00
Explanation:
When the solutions are mixed, the HCl dissociates to form the ions H+ and Cl-. The ion H+ will react with the NH3 to form NH4+. The stoichiometry for this is 1 mol of HCl to 1 mol of H+ to 1 mol of Cl-, and 1 mol of H+ to 1 mol of NH3 to 1 mol of NH4+.
First, let's find the number of moles of each one of them, multiplying the concentration by the volume:
nH+ = 0.15 M * 25 mL = 3.75 mmol
nNH3 = 0.52 M * 25 mL = 13 mmol
So, all the H+ is consumed, and the neutralization is completed, thus pH will be the pH of the solvent (water), pH = 7.00.
Answer: The Answer is 18.7ml.
Explanation: Solved in the attached picture.
Potassium sulfide, also
known as dipotassium monosulfide, consists of two potassium ions bonded to a
sulfide atom, rendering the chemical formula K2S.<span>Rarely
found in nature due to its high reactivity with water, potassium sulfide is
refined from the more common potassium sulfate (K2SO4) and is used in many
industries</span>
Since the compound has 1.38 time that of oxygen gas at the same conditions of temperature and pressure, we have the relationship:
MW/MWoxygen = 1.38
MW = 44.16
Since there is water formed during the reaction, the formula of the compound must be:
XaHb
where a and b are the coefficients of each element.
If the compound reactions with oxygen forming water and an oxide of the element X, the combustion reaction must be:
XaHb + ((2a + (b/2))/2) O2 = a (XO2) + (b/2)(H2O)
Using dimensional analysis:
10 (1/44.16) (b/2 / 1) (18) = 16.3
Solving for b:
b = 8
The compound now is XaH8. Most probably, the compound is C3H8 since it has a molecular formula of 44 and it reacts with O2 to form water and CO2.
Answer:
131.5 kJ
Explanation:
Let's consider the following reaction.
CaCO₃(s) → CaO(s) + CO₂(g)
First, we will calculate the standard enthalpy of the reaction (ΔH°).
ΔH° = 1 mol × ΔH°f(CaO(s)) + 1 mol × ΔH°f(CO₂(g)
) - 1 mol × ΔH°f(CaCO₃(s)
)
ΔH° = 1 mol × (-634.9 kJ/mol) + 1 mol × (-393.5 kJ/mol) - 1 mol × (-1207.6 kJ/mol)
ΔH° = 179.2 kJ
Then, we calculate the standard entropy of the reaction (ΔS°).
ΔS° = 1 mol × S°(CaO(s)) + 1 mol × S°(CO₂(g)
) - 1 mol × S°(CaCO₃(s)
)
ΔS° = 1 mol × (38.1 J/mol.K) + 1 mol × (213.8 J/mol.K) - 1 mol × (91.7 J/mol.K)
ΔS° = 160.2 J/K = 0.1602 kJ/K
Finally, we calculate the standard Gibbs free energy of the reaction at T = 25°C = 298 K.
ΔG° = ΔH° - T × ΔS°
ΔG° = 179.2 kJ - 298 K × 0.1602 kJ/K
ΔG° = 131.5 kJ
