1. The empirical formula of the hydrocarbon is CH₃
2. The molecular formula of the hydrocarbon is C₂H₆
<h3>How to determine the mass of Carbon </h3>
- Mass of CO₂ = 1.47 g
- Molar mass of CO₂ = 44 g/mol
- Molar of C = 12 g/mol
- Mass of C =?
Mass of C = (12 / 44) × 1.47
Mass of C = 0.4 g
<h3>How to determine the mass of H</h3>
- Mass of compound = 0.5 g
- Mass of C = 0.4 g
- Mass of H = ?
Mass of H = (mass of compound) – (mass of C)
Mass of H = 0.5 – 0.4
Mass of H =0.1 g
<h3>1. How to determine the empirical formula </h3>
- C = 0.4 g
- H = 0.1 g
- Empirical formula =?
Divide by their molar mass
C = 0.4 / 12 = 0.03
H = 0.1 / 1 = 0.1
Divide by the smallest
C = 0.03 / 0.03 = 1
H = 0.1 / 0.03 = 3
Thus, the empirical formula of the compound is CH₃
<h3>2. How to determine the molecular formula</h3>
- Empirical formula = CH₃
- Molar mass = 30 g/mol
- Molecular formula =?
Molecular formula = empirical × n = mass number
[CH₃]n = 30
[12 + (3×1)]n = 30
15n = 30
Divide both side by 15
n = 30 / 15
n = 2
Molecular formula = [CH₃]n
Molecular formula = [CH₃]₂
Molecular formula = C₂H₆
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Its FeSO3 or iron(iii)sulfite = Fe2(SO3)3
From Ohm's law, we get the general equation that would relate the voltage, current, and resistance,
V = I x R
where V is voltage, R is resistance, and I is current. Deriving the equation for R
R = V / I
R = (1.5 volts) / (0.7 amps) = 2.14 Ohms
... Really wanna go sleep
<u>Answer:</u> The
for the reaction is 54.6 kJ/mol
<u>Explanation:</u>
For the given balanced chemical equation:

We are given:

- To calculate
for the reaction, we use the equation:
![\Delta G^o_{rxn}=\sum [n\times \Delta G_f(product)]-\sum [n\times \Delta G_f(reactant)]](https://tex.z-dn.net/?f=%5CDelta%20G%5Eo_%7Brxn%7D%3D%5Csum%20%5Bn%5Ctimes%20%5CDelta%20G_f%28product%29%5D-%5Csum%20%5Bn%5Ctimes%20%5CDelta%20G_f%28reactant%29%5D)
For the given equation:
![\Delta G^o_{rxn}=[(2\times \Delta G^o_f_{(COCl_2)})]-[(1\times \Delta G^o_f_{(CO_2)})+(1\times \Delta G^o_f_{(CCl_4)})]](https://tex.z-dn.net/?f=%5CDelta%20G%5Eo_%7Brxn%7D%3D%5B%282%5Ctimes%20%5CDelta%20G%5Eo_f_%7B%28COCl_2%29%7D%29%5D-%5B%281%5Ctimes%20%5CDelta%20G%5Eo_f_%7B%28CO_2%29%7D%29%2B%281%5Ctimes%20%5CDelta%20G%5Eo_f_%7B%28CCl_4%29%7D%29%5D)
Putting values in above equation, we get:
![\Delta G^o_{rxn}=[(2\times (-204.9))-((1\times (-394.4))+(1\times (-62.3)))]\\\Delta G^o_{rxn}=46.9kJ=46900J](https://tex.z-dn.net/?f=%5CDelta%20G%5Eo_%7Brxn%7D%3D%5B%282%5Ctimes%20%28-204.9%29%29-%28%281%5Ctimes%20%28-394.4%29%29%2B%281%5Ctimes%20%28-62.3%29%29%29%5D%5C%5C%5CDelta%20G%5Eo_%7Brxn%7D%3D46.9kJ%3D46900J)
Conversion factor used = 1 kJ = 1000 J
- The expression of
for the given reaction:

We are given:

Putting values in above equation, we get:

- To calculate the Gibbs free energy of the reaction, we use the equation:

where,
= Gibbs' free energy of the reaction = ?
= Standard gibbs' free energy change of the reaction = 46900 J
R = Gas constant = 
T = Temperature = ![25^oC=[25+273]K=298K](https://tex.z-dn.net/?f=25%5EoC%3D%5B25%2B273%5DK%3D298K)
= equilibrium constant in terms of partial pressure = 22.92
Putting values in above equation, we get:

Hence, the
for the reaction is 54.6 kJ/mol