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Ket [755]
3 years ago
13

A sample of gas has a volume of 135 ml at 0.600 atm. what would be the volume if the pressure is decreased to 0.200 atm while te

mperature is held constant? 405 ml 101 ml 180 ml 135 ml 45.0 ml
Chemistry
1 answer:
Orlov [11]3 years ago
5 0

Answer: -

405 mL

Explanation: -

Initial Pressure P 1 = 0.6 atm

Final Pressure P 2 = 0.2 atm

Initial volume V 1 = 135 mL

Using Boyle's Law,

At constant temperature,

P 1 x V 1 = P 2 x V 2

0.6 atm x 135 mL = 0.2 atm x V 2

Final volume = V 2 = \frac{0.6 atm x 135 mL}{0.2 atm}

= 405 mL

Thus if a sample of gas has a volume of 135 ml at 0.600 atm. 405 mL would be the volume if the pressure is decreased to 0.200 atm while temperature is held constant.

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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:

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How many grams of hydrogen chloride can be produced from 1g of hydrogen and 55g of chlorine? What is the limiting reactant?
vova2212 [387]

Answer:

The limiting reactant is hydrogen, and the grams HCl produced is 36.175 g.

Explanation:

Balanced equation is 2 H + Cl2 = 2 HCl.

First thing, convert grams to moles via using molar mass.

Molar mass for hydrogen is 1.0079 g/mol. 1g x 1 mol / 1.0079 g = 0.99216 mol.

Molar mass for chlorine is 70.906 g/mol. 55g x 1 mol / 70.906 g = 0.7756748 mol.

Next, determine which is the limiting reactant - probably the fastest way to do it is just to take one of the reactants, say it's the limiting one, and calculate how much of the other reactant would be needed if that really was the limiting reactant, and then compare it to the actual moles of reactant available.

If hydrogen was the limiting reactant at 0.992 mol, you'd need .496 mol of Cl2 to complete the reaction.

If chloride was the limiting reactant at 0.776 mol, you'd need 1.55 mol of H to complete the reaction.

Comparing these numbers to the amounts we actually have available, the limiting reactant is hydrogen.

Once you've determined that, just plug in the amounts to the balanced equation to get the number of moles of HCL produced, which in this case, is just 0.992 mol.

Now, reverse the process that you took to get the moles of reactant, and you have the grams of product produced.

0.992 mol x 36.4609 g / 1 mol = 36.175 g.

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