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OleMash [197]
3 years ago
9

Uppose you have two 1-L flasks, one containing a gas with a molar mass of 30 g and the other a gas with a molar mass of 60 g. Bo

th flasks are at the same temperature, both gas samples have a mass of 1.2 g, but the pressure in Flask X is 1.0 atm and in Flask Y is 0.5 atm. Which flask contains the gas molar mass 30 g?
flask X
flask Y
not enough information problem.
Chemistry
1 answer:
kobusy [5.1K]3 years ago
3 0

Answer:

The answer to your question is: Flask X

Explanation:

Data

                            Flask X              Flask B

Molar mass            30 g                    60 g

mass                       1.2g                     1.2 g

Pressure                  1 atm                   0.5 atm

Formula                  PV = nRT

                         In the formula, we can notice that the number of moles (n)

                         is directly proportional to the pressure.

Then, let's calculate the number of moles

               flask X                                flask Y

  30 g  ---------------  1 mol           60 g -------------- 1 mol

  1.2 g ----------------   x                 1.2 g -------------   x

   x = (1.2 x 1) / 30                        x = (1.2 x 1) / 60

   x = 0.04 mol                             x = 0.02 mol

From the results, we conclude that the flask with the gas of molar mass 30g is the flask with pressure of 1 atm, because the higher the number of moles, the higher the pressure.

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Explanation:

Calories in food such as candy bars refer to the amount of energy that the food has. Calories are also called Kilocalories because they are the equivalent of 1,000 calories.

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C. Use Hess's law and the following equations to calculate ΔH for the reaction 4NH3 (g) + 5O2 (g) 4NO(g) + 6H2 O(g). Show your w
Monica [59]

Considering the Hess's Law, the enthalpy change for the reaction is -906.4 kJ/mol.

<h3>Hess's Law</h3>

Hess's Law indicates that the enthalpy change in a chemical reaction will be the same whether it occurs in a single stage or in several stages. That is, the sum of the ∆H of each stage of the reaction will give us a value equal to the ∆H of the reaction when it occurs in a single stage.

<h3>ΔH in this case</h3>

In this case you want to calculate the enthalpy change of:

4 NH₃ + 5 O₂ → 4 NO + 6 H₂O

which occurs in three stages.

You know the following reactions, with their corresponding enthalpies:

Equation 1: 2 N₂ + 6 H₂ → 4 NH₃   ΔH = –183.6 kJ/mol

Equation 2:  2 N₂ + 2 O₂ → 4 NO     ΔH = 361.1 kJ/mol

Equation 3: 2 H₂ + O₂→ 2 H₂O     ΔH = -483.7 kJ/mol

Because of the way formation reactions are defined, any chemical reaction can be written as a combination of formation reactions, some going forward and some going back.

In this case, first, to obtain the enthalpy of the desired chemical reaction you need 4 moles of NH₃ on reactant side and it is present in first equation on product side. So you need to invert the reaction, and when an equation is inverted, the sign of delta H also changes.

Now, 4 moles of NO must be a product and is present in the second equation, so let's write this as such.

Finally, you need 6 moles of H₂O on the product side, so you need to multiply by 3 the third equation to obtain the amount of water that you need. Since enthalpy is an extensive property, that is, it depends on the amount of matter present, since the equation is multiply by 3, the variation of enthalpy also.

In summary, you know that three equations with their corresponding enthalpies are:

Equation 1: 2 4 NH₃ → N₂ + 6 H₂  ΔH = 183.6 kJ/mol

Equation 2:  2 N₂ + 2 O₂ → 4 NO     ΔH = 361.1 kJ/mol

Equation 3: 6 H₂ + 3 O₂→ 6 H₂O     ΔH = -1,451.1 kJ/mol

Adding or canceling the reactants and products as appropriate, and adding the enthalpies algebraically, you obtain:

4 NH₃ + 5 O₂ → 4 NO + 6 H₂O ΔH= -906.4 kJ/mol

Finally, the enthalpy change for the reaction is -906.4 kJ/mol.

Learn more about Hess's law:

brainly.com/question/5976752

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#SPJ1

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Answer:

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From the question given above, the following data were obtained:

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Final volume (V2) =?

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P × 1.04 = 2P × V2

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V2 = 0.52 L

Thus, the new volume of the gas is 0.52 L.

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