First, we have to get how many grams of C & H & O in the compound:
- the mass of C on CO2 = mass of CO2*molar mass of C /molar mass of CO2
= 0.5213 * 12 / 44 = 0.142 g
- the mass of H atom on H2O = mass of H2O*molar mass of H / molar mass of H2O
=0.2835 * 2 / 18 = 0.0315 g
- the mass of O = the total mass - the mass of C atom - the mass of H atom
= 0.3 - 0.142 - 0.0315 = 0.1265 g
Convert the mass to mole by divided by molar mass
C(0.142/12) H(0.0315/2) O(0.1265/16)
C(0.0118) H(0.01575) O(0.0079) by dividing by the smallest value 0.0079
C1.504 H3.99 O1 by rounding to the nearst fraction
C3/2 H4/1 )1/1 multiply by 2
∴ the emprical formula C3H8O2
Answer:
The gas that Dr. Brightguy added was O₂
Explanation:
Ideal Gases Law to solve this:
P . V = n . R . T
Firstly, let's convert 736 Torr in atm
736 Torr is atmospheric pressure = 1 atm
20°C = 273 + 20 = 293 T°K
125 mL = 0.125L
0.125 L . 1 atm = n . 0.082 L.atm / mol.K . 293K
(0.125L .1atm) / (0.082 mol.K /L.atm . 293K) = n
5.20x10⁻³ mol = n
mass / mol = molar mass
0.1727 g / 5.20x10⁻³ mol = 33.2 g/m
This molar mass corresponds nearly to O₂
Answer:
1.635 M
Explanation:
Given:
10 mL of 20 volumes Hydrogen Peroxide
Here,
20 volumes of Hydrogen Peroxide means that on decomposition of 1 mL of H₂O₂ 20 mL of O₂ is obtained
also,
means 1 dm³ of H₂O₂ solution produces 20 dm³ oxygen
Now,
at 298K and 1 atm
20 dm³ oxygen =
moles
or
= 0.817 moles
also,
2H₂O₂ → 2H₂O + O₂
thus,
1 dm³ of solution must contain 2 × moles of O₂ as moles of H₂O₂
thus,
Number of moles of H₂O₂ = 2 × 0.817
or
Number of moles of H₂O₂ = 1.635 moles
Hence,
For 20 volume hydrogen peroxide is 1.635 M
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