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.
Q is determined to be 12.38. The Nernst equation is expressed as Ecell = E°cell - (2.303RT/nF) log Q, where Q represents the reaction quotient. The reaction quotient Q is calculated by taking the product of the products' concentrations divided by the product of the reactants' concentrations. For an electrochemical cell, Q is the concentration ratio of the solution at the anode compared to that at the cathode. Consequently, Q = [anode]/[cathode], specifically Q = 0.052/0.0042, arriving at a value of Q = 12.38.
For KNO₃, the mass is 346g. The molar mass can be computed as (39.098) + (14) + (15.99*3), which results in 101.068 gmol⁻¹. The volume of the solution is given as 750ml, equivalent to 0.75dm³. The formula for molarity is (mass of solute/molar mass of solute)*(1/volume of solution in dm³). Accordingly, molarity = (346/101.068)*(1/0.75), yielding 4.56 mol dm⁻³.
Answer: Pentane C5H12
Explanation:
The boiling point is defined as the temperature at which a liquid's vapor pressure matches the external pressure, causing the liquid to turn into vapor.
This compound is likely Pentane, represented as C5H12, since its boiling point falls between that of Butane, with the formula C4H10, and Hexane, with the formula C6H14.
