A charged rubber rod can pick up a small neutrally charged piece of paper by what process

Combustion analysis is performed on 0.50 g of a hydrocarbon, and 1.47 g of CO2 and 0.902 g of H2O are produced. What is the empi

JulijaS [17]

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

Learn more about empirical formula:

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1 year ago

Iron sulfite has the formula?

Mademuasel [1]

Its FeSO3 or iron(iii)sulfite = Fe2(SO3)3

7 0

3 years ago

You might have a flashlight that runs (2) d-cells of 1.5 volts each in series. the bulb is rated for 0.7 amps. what’s its resist

sweet [91]

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

7 0

2 years ago

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MArishka [77]

... Really wanna go sleep

3 0

2 years ago

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Consider the following reaction: CO2(g)+CCl4(g)⇌2COCl2(g) Calculate ΔG for this reaction at25 ∘C under these conditions: PCO2PCC

Salsk061 [2.6K]

<u>Answer:</u> The $\Delta G$ for the reaction is 54.6 kJ/mol

<u>Explanation:</u>

For the given balanced chemical equation:

$CO_2(g)+CCl_4(g)\rightleftharpoons 2COCl_2(g)$

We are given:

$\Delta G^o_f_{CO_2}=-394.4kJ/mol\\\Delta G^o_f_{CCl_4}=-62.3kJ/mol\\\Delta G^o_f_{COCl_2}=-204.9kJ/mol$

To calculate $\Delta G^o_{rxn}$ for the reaction, we use the equation:

$\Delta G^o_{rxn}=\sum [n\times \Delta G_f(product)]-\sum [n\times \Delta G_f(reactant)]$

For the given equation:

$\Delta G^o_{rxn}=[(2\times \Delta G^o_f_{(COCl_2)})]-[(1\times \Delta G^o_f_{(CO_2)})+(1\times \Delta G^o_f_{(CCl_4)})]$

Putting values in above equation, we get:

$\Delta G^o_{rxn}=[(2\times (-204.9))-((1\times (-394.4))+(1\times (-62.3)))]\\\Delta G^o_{rxn}=46.9kJ=46900J$

Conversion factor used = 1 kJ = 1000 J

The expression of $K_p$ for the given reaction:

$K_p=\frac{(p_{COCl_2})^2}{p_{CO_2}\times p_{CCl_4}}$

We are given:

$p_{COCl_2}=0.760atm\\p_{CO_2}=0.140atm\\p_{CCl_4}=0.180atm$

Putting values in above equation, we get:

$K_p=\frac{(0.760)^2}{0.140\times 0.180}\\\\K_p=22.92$

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

$\Delta G=\Delta G^o+RT\ln K_p$

where,

$\Delta G$ = Gibbs' free energy of the reaction = ?

$\Delta G^o$ = Standard gibbs' free energy change of the reaction = 46900 J

R = Gas constant = $8.314J/K mol$

T = Temperature = $25^oC=[25+273]K=298K$

$K_p$ = equilibrium constant in terms of partial pressure = 22.92

Putting values in above equation, we get:

$\Delta G=46900J+(8.314J/K.mol\times 298K\times \ln(22.92))\\\\\Delta G=54659.78J/mol=54.6kJ/mol$

Hence, the $\Delta G$ for the reaction is 54.6 kJ/mol

7 0

2 years ago