Nobody on here is going to write a entire cer for you
Try googleing it that is how i got some of mine
B. Rotten orange is the correct answer. Hope this helps!
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
Answer:
The standard enthalpy of formation of this isomer of is -220.1 kJ/mol.
Explanation:
The given chemical reaction is as follows.
The expression for the entropy change for the reaction is as follows.
Substitute the all values in the entropy change expression.
Therefore, The standard enthalpy of formation of this isomer of is -220.1 kJ/mol.