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Veseljchak [2.6K]
3 years ago
11

Gas stoichiometry a sample of methane gas having a volume of 2.80 L at 25c and 1.65 atm was mixed with a sample oxygen gas havin

g a volume 35.0L at 31c and 1.25 atm . The mixture was then ignited to form carbon dioxide and water . Calculate the volume of co2 formed at a pressure of 2.50 atm and a temperature of 125 c
Chemistry
1 answer:
gladu [14]3 years ago
8 0

Answer:

2.47 L. CO2

Explanation:

First, let's establish what we need to find. The volume of CO2 formed at a pressure of 2.50 atm and a temperature of 125 C. Now, don't let this overwhelm you. To find the volume of CO2 formed, we must first find the amount of CO2 formed.

But to find the amount of CO2 formed, we have to write out the equation for this reaction. The type of reaction depicted here is combustion, where oxygen gas reacts with a substance to form carbon dioxide and water. And we also know that this equation must be balanced, according to the Law of Conservation of Matter. It is as follows:

CH4 + 2 O2 => CO2 + 2 H2O.

Now that we have the equation of the reaction, we can now find which of the two reactants is the limiting reactant. To do this, we need to convert the volumes of methane and oxygen gas to moles. Simply use the Ideal Gas Law to do this, and you get:

Methane Gas: .189 mol     |     Oxygen Gas: 1.75 mol

Using the equation, we can then see that methane gas is the limiting reactant.

We know, also due to the equation that 1 mol of CO2 is produced for every mol of methane gas. This gives us the answer of .189 mol of CO2 produced. But remember, the question asks us for the <em>volume</em> of CO2 produced, so we must apply the Ideal Gas Law once again, getting the answer of 2.47 L. CO2.

<u>Note</u>: Please remember to use significant figures during your calculations, and if you have any questions, feel free to post them down below.

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Answer:

I think the answer is……

O B.H2S

Explanation:

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The combustion of ethene in the presence of excess oxygen yields carbon dioxide and water: c2h4 (g) + 3o2 (g) → 2co2 (g) + 2h2o
meriva

Answer:

\boxed{-267.5}

Explanation:

You can calculate the entropy change of a reaction by using the standard molar entropies of reactants and products.

The formula is

\Delta_{r} S^{\circ} = \sum_n {nS_{\text{products}}^{\circ} - \sum_{m} {mS_{\text{reactants}}^{\circ}}}

The equation for the reaction is

                        C₂H₄(g) + 3O₂(g) ⟶ 2CO₂(g) + 2H₂O(ℓ)

ΔS°/J·K⁻¹mol⁻¹   219.5      205.0         213.6         69.9

\Delta_{r} S^{\circ} = (2\times213.6 + 2\times69.9) - (1\times219.5 + 3\times205.0)\\\\= 567.0 - 834.5 = \boxed{-267.5 \text{ J}\cdot\text{K}^{-1} \text{mol}^{-1}}

3 0
3 years ago
The density of solid Cr is 7.15 g/cm3. How many atoms are present per cubic centimeter of Cr?
Masja [62]
To calculate the number of atoms of Cr, we first find the number of moles per unit of cubic centimeter of Cr. Then, use avogadros number for the number of atoms. Calculations are as follows:

1 cm^3 (7.15 g/cm^3) (1 mol / 51.996 g Cr) = 0.14 mol Cr

0.14 mol Cr ( 6.022 x 10^23 atoms Cr / 1 mol Cr ) = 8.28 x 10^22 atoms Cr
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Which of the following pairs contains a conjugate acid-base pair?
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An example of a conjugate acid-base pair would be NH₃ and NH₄⁺NH₃ + H₂O --> NH₄⁺ + OH⁻NH3 is the base, and NH₄⁺ is the conjugate acid
5 0
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If Kc = 4.0×10−2 for PCl3(g)+Cl2(g)⇌PCl5(g) at 520 K , what is the value of Kp for this reaction at this temperature?
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Here we have to get the K_{p} of the reaction at 520 K temperature.

The K_{p} of the reaction is 1.705 atm

We know the relation between K_{p} and K_{c} is K_{p}=K_{c}(RT)^{N}, where  K_{p} = The equilibrium constant of the reaction in terms of partial pressure, K_{c}  = The equilibrium constant of the reaction in terms of concentration and N = number of moles of gaseous products - Number of moles of gaseous reactants.

Now in this reaction, PCl₃ + Cl₂ ⇄ PCl₅

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The given value of  K_{c} is 4.0×10⁻²

The molar gas constant, R = 0.082 L. Atm. mol⁻¹. K⁻¹ and temperature, T = 520 K.

On plugging the values in the equation we get,

K_{p} = 4.0 X 10^{-2}(0.082X520)^{1}

Or, K_{p} = 1.705 atm

Thus, the K_{p} of the reaction is 1.705 atm

7 0
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