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Svetlanka [38]
3 years ago
6

A gaseous compound is 30.4 % N and 69.6% OF. A 5.25 g sample of the gas occupies a volume of 1.00 L and exerts a pressure of 958

mm of Hg at –4.0 C. Which of the following is its molecular formula
Chemistry
1 answer:
Harman [31]3 years ago
8 0

Answer:

The molecular formula = N2O4

Explanation:

<u>Step 1</u>: Data given

A gaseous compound is 30.4 % N and 69.6%

Mass of the compound = 5.25 grams

Volume of the gas = 1.00 L

Pressure of the gas = 958 mmHg = 1.26 atm

Temperature of the gas = -4 °C = 273 -4°C = 269 Kelvin

Molar mass of N = 14 g/mol

Molar mass of O = 16 g/mol

<u>Step 2</u>: Calculate mass of N

Mass of Nitrogen = 5.25 grams * 0.304 = 1.596 grams

<u>Step 3:</u> Calculate mass of O

Mass of Oxygen = 5.25 grams * 0.696 = 3.654 grams

<u>Step 4:</u> Calculate number of moles N

Number of moles N = Mass of N/ Molar mass of N

Moles of N = 1.596 grams / 14g/mol

Moles of N = 0.114 moles N

<u>Step 5:</u> Calculate moles of O

Moles O = 3.654 grams / 16 g/mol

Moles 0 = 0.2884 moles

<u>Step 6:</u> Calculate empirical formule

We calculate the empirical formule by dividing number of moles by the smallest number of mol

N : 0.114 / 0.114 = 1

O: 0.2284 / 0.114 = 2

Empirical formule = NO2

<u>Step 7: </u>Calculate number of moles of 5.25 g sample via gas law:

p*V = nRT

⇒ with p = the pressure = 1.26 atm

⇒ with v = 1.00 L

⇒ with n = the number of moles = TO BE DETERMINED

⇒ with R = the gas constant = 0.08206 L*atm/ K*mol

⇒ with T = the temperature = 269 K

number of moles n = (p*V)/(R*T)

n = (1.26*1L)/(0.08206*269)

n = 0.057 mol  

<u>Step 8:</u> Calculate molar mass of the compound

This means 5.25 grams of the gas = 0.057 moles

So 1 mol of the compound has a molar mass of: 5.25 / 0.057 = 92.11 g/mol

<u>Step 9</u>: Calculate molar mass of the empirical formula NO2

N = 14 g/mol

O = 16 g/mol

NO2 = 14 + 16 + 16 = 46 g/mol

The empirical formule NO2 has a molar mass of 46 g/mol

<u>Step 10</u>: Calculate molecular formula

92.11 / 46 = 2

This means the empirical formula should be multiplied by 2

2*(NO2) = N2O4

The molecular formula = N2O4

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Answer: The enthalpy of combustion, per mole, of butane is -2657.4 kJ

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2C_4H_{10}(g)+13O_2(g)\rightarrow 8CO_2(g)+10H_2O(g)

The expression for enthalpy change is,

\Delta H=[n\times H_f_{products}]-[n\times H_f_{reactants}]

Putting the values we get :

\Delta H=[8\times H_f_{CO_2}+10\times H_f_{H_2O}]-[2\times H_f_{C_4H_{10}+13\times H_f_{O_2}}]

\Delta H=[(8\times -393.5)+(10\times -241.82)]-[(2\times -125.7)+(13\times 0)]

\Delta H=-5314.8kJ

2 moles of butane releases heat = 5314.8 kJ

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Arsine, AsH3, is a highly toxic compound used in the electronics industry for the production of semiconductors. Its vapor pressu
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Answer:

-79.8 × 10⁴ J/mol

Explanation:

<em>Arsine, AsH₃, is a highly toxic compound used in the electronics industry for the production of semiconductors. Its vapor pressure is 35 Torr at 111.95 °C and 253 Torr at 83.6 °C.</em>

<em />

Then,

P₁ = 35 torr

T₁ = 111.95 + 273.15 = 385.10 K

P₂ = 253 torr

T₂ = 83.6 + 273.15 = 356.8 K

We can calculate the standard enthalpy of vaporization (ΔH°vap) using the two-point Clausius-Clapeyron equation.

ln(\frac{P_{2}}{P_{1}} )=\frac{-\Delta H\°_{vap}}{R} .(\frac{1}{T_{2}} -\frac{1}{T_{1}} )

where,

R is the ideal gas constant

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