Convert HCl and H2O to moles.
36.0 g of HCl = 0.987 moles HCl
98.0 g of H2O = 5.44 moles H2O
Based on the stoichiometric ratio for HCl,
there are 0.987 moles of H and 0.987 moles of Cl.
For H₂O, according to the stoichiometric ratio, you have 10.88 moles of H and 5.44 moles of O.
Combining them:
11.867 moles H
0.987 moles Cl
5.44 moles O
Revert the moles back to grams, then divide by the total mass and multiply by 100 for the percentage by mass.
11.867 moles H = 11.96 g H
0.987 moles Cl = 34.99 g Cl
5.44 moles O = 87.03 g O
11.96/(36.0+98.0)(100) = 8.93% for H
34.99/(36.0+98.0)(100) = 26.11% for Cl
87.03/(36.0+98.0)(100) = 64.96% for O.
Answer:
Explanation:
In KCl, the two elements that combine to create KCl are potassium (K) and chlorine (Cl).
Potassium, as a Group 1 element, possesses one valence electron in its outermost shell which it readily donates during bonding. Every element aims to achieve a stable electron configuration, typically with 2 or 8 electrons in its outer shell. Potassium is characterized by its lower electronegativity and higher ionization energy, making it more likely to donate its electron than to accept one. On the other hand, chlorine belongs to Group 17 and has 7 electrons in its outer shell, requiring just one additional electron to complete its octet. Chlorine’s higher electronegativity and lower ionization energy facilitate its tendency to accept an electron rather than donate it.
The bond between potassium and chlorine that results in KCl is termed an electrovalent bond.
Reaction equation:
K + Cl → KCl
Answer:
180.56 kilojoules of heat energy is extracted when 1.00 kg of freon-11 evaporates.
Explanation:
The molar mass of freon-11 is 137.35 g/mol
The enthalpy of vaporization for freon-11 is 24.8 kJ/mol at its normal boiling point of 24°C. Given that,
Mass of freon-11 evaporated = 1.00 kg = 1000 g
Moles of freon-11 evaporated can be calculated as
The energy removed in the form of heat when 1.00 kg of freon-11 vaporizes is:
The correct equation is (C) H3O+(aq) + C2H3O2−(aq) -> HC2H3O2(aq) + H2O(l). A buffer system is composed of a weak acid and its corresponding salt, effectively stabilizing the pH levels within a solution. The buffer works by adjusting the concentrations of the conjugate acid and base, maintaining the pH constant.
The formula for a monoprotic acid can be represented as HA, and its reaction with a base is shown as follows: HA + NaOH ---> NaA + H₂O. The stoichiometry between the acid and the base is 1:1. At the point of neutralization, the moles of HA equals the moles of the base. The moles of NaOH that reacted can be calculated as 0.100M / 1000 mL/L x 30.0 mL = 0.003 mol. Consequently, the moles of HA that reacted equal 0.003 mol. The mass of the acid is 0.384 g, yielding a molar mass of 0.384 g / 0.003 mol = 128 g/mol.
