Answer:
Percentage Yield is given as,
%age Yield = Actual Yield / Theoretical Yield × 100
This shows that the %age yield is directly depending upon the actual yield. And most of the time the percentage yield is less than 100 % because of the following factors.
Impure Starting Materials:
If the starting materials (reactants) are not pure then reaction will not completely form the desired product. Different by products will form which will decrease the %age yield.
Incomplete Reactions:
Not all reactions go to completion. In many reactions the starting material after some time stops forming the product due to different conditions. Some reactions attain equilibrium and stop increasing the amount of product. While, in some reactions a by products (like water) formed often react with the product to give a reverse reactions. Hence, the chemistry of reactions also causes the decrease in %age yield.
Handling:
Another major reason for decrease in yield is handling the product. Always some of the product is lost during the workup of the reaction like, taking TLC, doing solvent extraction, doing column chromatography, taking characterization spectrums. So, we can conclude that the %age yield will always be less than 100%.
When 0.514 g of biphenyl (C12H10) undergoes combustion in a bomb calorimeter, the temperature rises from 25.8 C to 29.4 C. Find ⌂E rxn for the combustion of biphenyl in kJ/mol biphenyl. The heat capacity of the bomb calorimeter, determined in a separate experiment, is 5.86 kJ/ C.
<span>The answer is - 6.30 * 10^3 kJ/mol
</span>
If this is just a general question it seems to vary from about 4.5g to 5g. Is there more data to the question?
Answer:
Pb3O4
Explanation:
According to this question, 3.425g of lead oxide was reduced to form 3.105g of lead in an experiment. Since lead oxide contains both lead (Pb) and oxygen (O) element,
Mass of lead oxide = 3.425g
Mass of lead = 3.105g
Mass of oxygen = (3.425g - 3.105g) = 0.320g
Next, we convert each mass value to mole by dividing by respective molar mass
Pb = 3.105g ÷ 207.2 = 0.0149mol
O = 0.320g ÷ 16 = 0.02mol
Next, we divide each mole value by the smallest (0.0149)
Pb = 0.0149mol ÷ 0.0149mol = 1
O = 0.02mol ÷ 0.0149mol = 1.342
Multiply each ratio value by 3 to get:
Pb = 1 × 3 = 3
O = 1.342 × 3 = 4.026
The whole number ratio, approximately, of Pb and O is 3:4, hence, their empirical formula is Pb3O4.
