Answer:
The answer to your question is 75%
Explanation:
Data
Theoretical production = 4 moles
Experimental production = 3 moles
Percent yield = ?
Formula
Substitution
Result
Percent yield = 75 %
Answer:
The answer to your question is 1.36 x 10²³ atoms
Explanation:
Data
number of atoms = ?
mass of the sample = 34.2 g
Molecule = Cl₂O₅
Process
1.- Calculate the molar mass of Cl₂O₅
Cl₂O₅ = (35.5 x 2) + (16 x 5) = 71 + 80 = 151 g
2.- Calculate the atoms of Cl₂O₅
151 g of Cl₂O₅ ---------------- 6 .023 x 10²³ atoms
34.2 g of Cl₂O₅ ------------ x
x = (34.2 x 6.023 x 10²³) / 151
x = 1.36 x 10²³ atoms
Answer: C. ethanol
The enthalpy of combustion is the amount of heat produced when one mole of ethanol undergoes complete combustion at 25 ° C and 1 atmosphere pressure, yielding products also at 25 ° C and 1 atm.
<u>The enthalpy of combustion of the unknown compound is</u>
ΔH = - 320 kJ / 0.25 mol = - 1280 kJ / mol
<u>To choose a probable compound according to this combustion enthalpy, we must evaluate the deviation in relation to the values reported in the literature for the three probable compounds</u> (methane, ethylene and ethanol). The deviation (e%) will be calculated according to the following equation,
e% = ( | ΔHx - ΔH | / ΔHx ) x 100%
where ΔHx is the enthalpy of combustion of the probable compound.
The following table shows the combustion enthalpies of the probable compounds and their deviation in relation to the enthalpy of ΔH = - 1280 kJ / mol
Compound Enthalpy of combustion (kJ/mol) Deviation
Methane - 890.7 43.8%
Ehylene -1411.2 9.3%
Ethanol -1368.6 6.5%
According to the previous table, we can say that the most probable compound is ethanol, since it has the smallest deviation in relation to the experimental enthalpy value of combustion.
Atoms are (stable) if it has (valence) electrons in its outer most energy level.
