Electrons is located outside the nucleus
According to the question, the determined melting point of the compound is 112.5-113.0oC. When the solidified compound was retried, the melting point was found to be 133.6-154.5oC. This greater range higher than 112°C is caused by reusing samples leads to errors.
A pure sample is known by its sharp melting point. A pure sample does not melt over a large range. We can see this in the predetermined melting points of the pure sample(112.5-113.0oC).
However, reusing a sample introduces errors because the pure sample may become contaminated leading to a larger and higher range of melting point (133.6-154.5oC) which is far above 112°C.
Learn more: brainly.com/question/5325004
A general equation for a combustion reaction would be expressed as follows:
CxHy + (x+y/2)O2 = xCO2 + y/2H2O
Propane would obviously would only have carbon and hydrogen in its structure. Assuming a complete combustion, all of the carbon atoms would go to carbon dioxide and all of the hydrogen atoms to water. To determine the empirical, we determine the number of carbon and hydrogen atoms present.
moles C = 2.461 g CO2 ( 1 mol / 44.01 g ) ( 1 mol C / 1 mol CO2 ) = 0.06 mol C
moles H = 1.442 g H2O ( 1 mol / 18.02 g ) ( 2 mol H / 1 mol H ) = 0.16 mol H
Then, we divide the smallest amount to the each mole of the atoms. We do as follows:
C = 0.06 / 0.06 = 1
H = 0.16 / 0.06 = 2.67
Then we multiply a number in order to obtain a whole number ratio between the atoms.
1 CH2.67
2 C2H5.34
3 C3H8 <-------- empirical formula
