Answer:
B. an element
Explanation:
An atom is smallest indivisible particle that takes part in a chemical reaction. Different atoms due to the number of their protons called atomic number gives an element. Every element is a singular atom on it's own. Combination of atoms leads to the formation of molecules and compounds.
When compounds mix together without an actual chemical change, a mixture forms.
Elements are distinct substances that cannot be split up into simpler substances. They are usually made up of only one kind of atom.
Answer:
True
Explanation:
Confirmation Bias is the tendency to look for information that supports, rather than rejects, one’s preconceptions, typically by interpreting evidence to confirm existing beliefs while rejecting or ignoring any conflicting data
B is the correct answer to your question
Answer:
Theoretical maximum moles of hydroquinone: 0.2167 mol.
Explanation:
Hello,
In this case, the undergoing chemical reaction is like:

In such a way, since the molar mass of quinone is 108.1 g/mol and it is in a 1:1 molar ratio with hydroquinone, we can easily compute the theoretical maximum moles of hydroquinone by stoichiometry:

Clearly, this is the theoretical yield which in grams is:

Which allows us to compute the percent yield as well since the obtained mass of the product is 13.0 g:

Best regards.