The oxidation of glucose can result in several products. you perform combustion analysis on a 2.750 g sample that resulted from the oxidation of glucose and obtain 3.702 g co2 and 1.514 g h2o. how many carbon atoms are in the empirical formula of the isolated sample? (assume no other products present from the oxidation of glucose.)]

There is 1 carbon atom in the empirical formula.

Explanation:

First, let's use the given mass of CO₂ to calculate the moles of C, which will be the same amount of C moles present in the sample:

3.702 g CO₂ ÷ 44g/mol = 0.0841 mol CO₂ = 0.0841 mol C

Then with the mass of H₂O we calculate the moles of H:

1.514 g H₂O ÷ 18gH₂O/mol * = 0.1682 mol H

Now we calculate the mass of C, H in the sample, in order to calculate the mass of O, which is calculated by substracting the mass of C and H from the total mass:

0.0841 mol C * 12 g/mol = 1.0092 g C0.1682 mol H * 1 g/mol = 0.1682 g Hg O = 2.750 - (1.0092+0.1682) = 1.5726 g O

Then we convert g O to mol O

1.5726 g O ÷ 16g/mol = 0.0983 mol OIn short, we have:

C ⇒ 0.0841 mol C

H ⇒ 0.1682 mol H

O ⇒ 0.0983 mol O

Now we divide those values by the lowest among them:

C ⇒ 0.0841 / 0.0841 = 1H ⇒ 0.1682 / 0.0841 = 2O ⇒ 0.0983 / 0.0841 = 1.1689 ≅ 1

So the empirical formula is CH₂O

There is 1 carbon atom in the empirical formula.

Most likely a is the answer - because  you generate   38 atp's but also consume an atp.   each atp gives approx. less than 4 kcal so probably a is the closest answer.

Okay well gravitational energy is matter forced by gravity making energy though the kinetic energy then the electric energy produces electricity from the kinetic energy and the kinetic energy gets its energy from the gravitational energy. was that the answer you wanted or something else?