Answer:
HCO₂
Explanation:
From the information given:
The mass of the elements are:
Carbon C = 26.7 g; Hydrogen H = 2.24 g Oxygen O = 71.1 g
To determine the empirical formula;
First thing is to find the numbers of moles of each atom.
For Carbon:

For Hydrogen:

For Oxygen:

Now; we use the smallest no of moles to divide the respective moles from above.
For carbon:

For Hydrogen:

For Oxygen:

Thus, the empirical formula is HCO₂
The substance doesn't have a specific name. We just say that that substance is being reduced. Remember this mnemonic - OILRIG where "Oxidation is Loss, Reduction is Gain" of electrons.
Great question, but I believe you are mixing up atomic number with mass number. Assuming you are, 12.011 amu is the average mass of a carbon atom. For carbon, it can come in three forms: carbon-12, carbon-13, carbon-14. The number following carbon is the mass number of that particular carbon "isotope". The reason the average is so close to 12 is because carbon-12 is by far the most common, so the average should be (and is) very close to 12. Therefore, 12.011 is a weighted average of all carbon molecules, and carbon-14 is a particular carbon molecule that weighs 14 amu.
Answer:
See the explanation
Explanation:
In this case, we have to keep in mind that in the monosubstituted product we only have to replace 1 hydrogen with another group. In this case, we are going to use the methyl group
.
In the axial position, we have a more steric hindrance because we have two hydrogens near to the
group. If we have <u>more steric hindrance</u> the molecule would be <u>more unstable</u>. In the equatorial positions, we don't <u>any interactions</u> because the
group is pointing out. If we don't have <u>any steric hindrance</u> the molecule will be <u>more stable</u>, that's why the molecule will <u>the equatorial position.</u>
See figure 1
I hope it helps!