Give two examples of an ionic compound

Ainat [17]

The answer is D or 4,3,2!

3 0

3 years ago

At this point Ron is slightly confused, this isn’t surprising. However, Hermione is doing rather well with them. This also isn’t

Zigmanuir [339]

Answer:

$\boxed{\text{0.780 atm}}$

Explanation:

Hermione is pretty smart. She realizes that, according to Dalton's Law of Partial Pressures, each gas exerts its pressure independently of the others, as if the others weren't even there.

She shows Ron how to use the Ideal Gas Law to solve the problem.

pV = nRT

She collects the data:

V = 1.00 L; n = 0.0319 mol; T = 25.0 °C

She reminds him to convert the temperature to kelvins

T = (25.0 +273.15) K = 298.15 K

Then she shows him how to do the calculation.

$p \times \text{1.00 L} = \text{0.0319 mol} \times \text{L}\cdot\text{atm}\cdot\text{0.082 06 K}^{-1}\text{mol}^{-1} \times \text{298.15 K}\\\\1.00p = \text{0.7805 atm}\\\\p = \textbf{0.780 atm}\\\\\text{The partial pressure of the nitrogen is } \boxed{\textbf{0.780 atm}}$

Isn't she smart?

4 0

3 years ago

What of the following could be classified as matter

Sholpan [36]

Answer:

Desk

Water

Cloud

Helium

Explanation:

7 0

3 years ago

According to the concept of entropy, what will likely happen to most molecules over time?.

Vsevolod [243]

They're going to fall apart.

<h3>What most molecules will likely experience over time?</h3>

A system's entropy cannot decrease unless heat is being lost from it. This indicates that entropy cannot decrease in a closed system (i.e., nothing enters or leaves, and that means nothing at all), so it must either remain constant or rise. Entropy can decrease, rise, or stay constant in an open system when movement between the system and its surroundings is conceivable. Both of these outcomes leave open the option of a constant state.

Entropy:-

The quantity of atom-alignment combinations that can exist in a system is known as entropy. The quantity of energy that cannot be used for work can also be calculated from an object's entropy.

Learn more about Entropy here:-

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#SPJ4

5 0

2 years ago

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.

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.

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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