List the following molecules in order from most reducedstate to most oxidizedstate: formaldehyde (ch2o), methane (ch4), carbon dioxide (co2), formic acid (hcooh), and methanol (ch3oh).

CH₄ < CH₃OH < CH₂O < HCOOH < CO₂

Explanation:

In order to know the oxidation number of C in these compounds, we will use the rule that states that the sum of the oxidation numbers of the atoms is equal to the charge of the compound, in these cases, zero because these are neutral molecules.

Formaldehyde (CH₂O)

C + 2.H + O = 0

C + 2.(1) + (-2) = 0

C = 0

Methane (CH₄)

C + 4.H = 0

C + 4.(1) = 0

C = -4

Carbon Dioxide (CO₂)

C + 2.(O-double bond) = 0

C + 2.(-2) = 0

C = +4

Formic Acid (HCOOH)

In this case, we will just consider the H atom linked to the C atom.

C + 1.(H) + 1.(O-double bond) + 1.(O-single bond) = 0

C + 1.(1) + 1.(-2) + 1.(-1) = 0

C = 2

Methanol (CH₃OH)

In this case, we will just consider the H  atoms linked to the C atom.

C + 3.(H) + 1.(O-single bond) = 0

C + 3.(1) + 1.(-1) = 0

C = -2

The order from most reduced to most oxidized is:

CH₄ < CH₃OH < CH₂O < HCOOH < CO₂

your answer would be a. 12.5 kg

in the given question, the partial pressure is calculated by using the ideal gas equation, that is, pv = nrt

here, p is pressure, v is volume = 25 l, n is number of moles = 0.0104 moles, r is gas constant = -0.0821 l.atm / mol.k

t is temperature = 273 k

p = nrt / v

p = (0.0104 mol × 0.0821 l.atm / mol.k × 273 k) / 25 l

= 0.0093 atm