If 11.00 mol of helium gas is at 12.0 ∘c and a gauge pressure of 0.400 atm .calculate the temperature if the gas is compressed to precisely half the volume at a gauge pressure of 1.00 atm.

Assuming ideal behavior of the helium gas for a fixed amount, the value of PV/T is constant, that is,
      P1V1 = nRT1, P2V2 = nRT2
If the gas is compressed to precisely half the volume and the absolute pressure is equal to gauge pressure plus atmospheric pressure, we rearrange the expression to solve for the temperature:
     T2/T1 = (P2/P1)(V2/V1)
     T2 = (P2/P1)(V2/V1)T1
     T2 = (2.00atm/1.350atm)(1/2)(283K)
     T2 = 210 K = -63°C

Hello!

Determine the empirical formula of a compound containing 47.37 grams of carbon, 10.59 grams of hydrogen, and 42.04 grams of oxygen.  

In an experiment, the molar mass of the compound was determined to be 228.276 g/mol. What is the molecular formula of the compound?

We have the following data:

Carbon (C) ≈ 12 a.m.u (g/mol)

Hydrogen (H) ≈ 1 a.m.u (g/mol)

Oxygen (O) ≈ 16 a.m.u (g/mol)

We use the amount in grams (mass ratio) based on the composition of the elements, see: (in 100g solution)

C: 47.37 g

H: 10.59 g

O: 42.04 g

The values ​​(in g) will be converted into quantity of substance (number of mols), dividing by molecular weight (g / mol) each of the values, we will see:We realize that the values ​​found above are not integers, so we divide these values ​​by the smallest of them, so that the proportion does not change, let us see:

convert number of atomic radio into whole number

2 * (1.5 : 4 : 1)  

= 3 : 8 : 2  ← whole number of atomic radio

C = 3

H =  8

O = 2

Thus, the empirical formula found for the compound will be:We are going to find the Molar Mass (MM) of the Empirical Formula (EF), let's see:

if: C3H8O2

C = 3*(12 a.m.u) = 36 a.m.u  

H = 8*(1 a.m.u) = 8 a.m.u  

O = 2*(16 a.m.u) = 32 a.m.u  

Knowing that the Molar Mass of the Molecular Formula is 228.276 (in g/mol) and that the Molar Mass of the Empirical Formula is 76 (in g/mol), then we will find the number of terms (n) for the molecular formula of the compound, let us see:The Molecular Formula is the Empirical Formula times the number of terms (n), then, we have:

Answers:

The empirical formula: C3H8O2

The molecular formula: C9H24O6