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3 years ago

The enthalpies of formation of the compounds in the combustion of methane,

Chemistry

2 answers:

Naya [18.7K]3 years ago

5 0

Answer:

The amount of heat released by the combustion of 2 moles of methane = 1605.1 kJ

Option 3 is correct.

ΔH°(combustion of 2 moles of methane)

= -1,605.1 kJ

Explanation:

The balanced equation for the combustion of methane is presented as

CH₄ + 2O₂ → CO₂ + 2H₂O

The heat of formation for the reactants and products are

CH₄ (g): ΔHf = –74.6 kJ/mol;

CO₂ (g): ΔHf = –393.5 kJ/mol;

and H₂O(g): ΔHf = –241.82 kJ/mol.

ΔHf for O₂ = 0 kJ/mol

ΔH°rxn for the combustion of methane is given as

ΔH°rxn = ΣnH°(products) - ΣnH°(reactants)

ΣnH°(products) = (1×-393.5) + (2×-241.82)

= -877.14 kJ/mol

ΣnH°(reactants) = (1×-74.6) + (2×0)

= -74.6 kJ/mol

ΔH°rxn = -877.14 - (-74.6) = -802.54 kJ/mol

For 2 moles of methane, the heat of combustion = 2 moles × -802.54 kJ/mol

= -1,605.08 kJ = -1,605.1 kJ

Hope this Helps!!!

inn [45]3 years ago

4 0

Answer:

C, -1,605.1 kJ

Explanation:

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A sample of ethanol (C2H6O) has a mass of 0.2301 g. Complete combustion of this sample causes the temperature of a bomb calorime

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Answer: $4.994\times 10^{-3}$ moles

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$\text{Number of moles}=\frac{\text{Given mass}}{\text {Molar mass}}$

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4 years ago

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A 30.0 L sample of nitrogen inside a rigid, metal container at 20.0 °C is placed inside an oven whose temperature is 50.0 °C. Th

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Answer:

3.31 atm.

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∵ P α T.

P₁ = 3.00 atm, T₁ = 20.0 °C + 273.15 = 293.15 K.

P₂ = ??? atm, T₂ = 50.0 °C + 273.15 = 323.15 K.

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3 years ago

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4 years ago

How many molecules are in 3.6 grams of NaCl? Question options:

Levart [38]

Answer:

$\boxed {\boxed {\sf 3.7 * 10^{22} \ molecules \ NaCl}}$

Explanation:

We are asked to find how many molecules are in 3.6 grams of sodium chloride.

<h3>1. Convert Grams to Moles </h3>

First, we convert grams to moles using the molar mass. These values are equivalent to atomic masses on the Periodic Table, but the units are grams per moles instead of atomic mass units. Look up the molar masses of the individual elements: sodium and chlorine.

Na: 22.9897693 g/mol
Cl: 35.45 g/mol

There are no subscripts in the chemical formula (NaCl), so we simply add the 2 molar masses.

NaCl: 22.9897693 + 35.45 = 58.4397693 g/mol

Now we will convert using dimensional analysis. First, set up a ratio using the molar mass.

$\frac {58.4397693 \ g \ NaCl}{ 1 \ mol \ NaCl}$

We are converting 3.6 grams to moles, so we must multiply the ratio by this value.

$3.6 \ g \ NaCl *\frac {58.4397693 \ g \ NaCl}{ 1 \ mol \ NaCl}$

Flip the ratio so the units of grams of sodium chloride cancel.

$3.6 \ g \ NaCl *\frac { 1 \ mol \ NaCl}{58.4397693 \ g \ NaCl}$

$3.6 *\frac { 1 \ mol \ NaCl}{58.4397693}$

$\frac { 3.6}{58.4397693} \ mol \ NaCl$

$0.06160188589 \ mol \ NaCl$

<h3>2. Convert Moles to Molecules </h3>

Next, we convert moles to molecules using Avogadro's Number. This is 6.022 × 10²³ and it tells us the number of particles (atoms, molecules, formula units, etc). In this case, the particles are molecules of sodium chloride. Let's set up another ratio.

$\frac {6.022 \times 10^{23} \ molecules \ NaCl}{ 1 \ mol \ NaCl}$

Multiply by the number of moles we calculated.

$0.06160188589 \ mol \ NaCl * \frac{6.022 \times 10^{23} \ molecules \ NaCl}{1 \ mol \ NaCl}$

The units of moles of sodium chloride cancel.

$0.06160188589 * \frac{6.022 \times 10^{23} \ molecules \ NaCl}{1 }$

$3.70966557*10^{22} \ molecules \ NaCl$

<h3>3. Round </h3>

The original measurement of grams (3.6) has 2 significant figures, so our answer must have the same. For the number we found, that is the tenths place. The 0 in the hundredth place tells us to leave the 7 in the tenth place.

$3.7 * 10^{22} \ molecules \ NaCl$

There are $3.7 * 10^{22} \ molecules \ NaCl$ in 3.6 grams and the correct answer is choice D.

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3 years ago