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ArbitrLikvidat [17]
4 years ago
15

The change in the standard Gibbs free energy (ΔGº) for the dissociation of nitrous acid (HNO2) at 298 K is 19.09 kJ. If the pH o

f the solution after equilibrium is reached is 1.30 and the NO2 – concentration at equilibrium is 0.00060 M. What is the equilibrium concentration of HNO2? Hint: Use pH to determine the equilibrium concentration of H+ .
Chemistry
1 answer:
Fynjy0 [20]4 years ago
4 0
The change in the standard Gibbs free energy (ΔGº) for the dissociation of nitrous acid (HNO2) at 298 K is 19.09 kJ. If the pH of the solution after equilibrium is reached is 1.30 and the NO2 – concentration at equilibrium is 0.00060 M. What is the equilibrium concentration of HNO2? Hint: Use pH to determine the equilibrium concentration of H+ .
HNO2(aq) H + (aq) + NO2 – (aq)
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Calcium would be more reactive because it is closer to francium on the periodic table, francium is the MOST reactive element.  
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What compoponents of the atom are found outside of nucleus
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Electrons are the smallest of the three particles that make up atoms. Electrons are found in shells or orbitals that surround the nucleus of an atom. Protons and neutrons are found in the nucleus. They group together in the center of the atom.

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3 years ago
How much energy is required to move the electron of the hydrogen atom from the 1s to the 2s orbital?
Alja [10]

Answer:

1.63425 × 10^- 18 Joules.

Explanation:

We are able to solve this kind of problem, all thanks to Bohr's Model atom. With the model we can calculate the energy required to move the electron of the hydrogen atom from the 1s to the 2s orbital.

We will be using the formula in the equation (1) below;

Energy, E(n) = - Z^2 × R(H) × [1/n^2]. -------------------------------------------------(1).

Where R(H) is the Rydberg's constant having a value of 2.179 × 10^-18 Joules and Z is the atomic number= 1 for hydrogen.

Since the Electrons moved in the hydrogen atom from the 1s to the 2s orbital,then we have;

∆E= - R(H) × [1/nf^2 - 1/ni^2 ].

Where nf = 2 = final level= higher orbital, ni= initial level= lower orbital.

Therefore, ∆E= - 2.179 × 10^-18 Joules× [ 1/2^2 - 1/1^2].

= -2.179 × 10^-18 Joules × (0.25 - 1).

= - 2.179 × 10^-18 × (- 0.75).

= 1.63425 × 10^- 18 Joules.

7 0
3 years ago
To improve air quality and obtain a useful product, chemists often remove sulfur from coal and natural gas by treating the conta
Norma-Jean [14]

Answer:

1) Increase

2) Decreases

3) increases

4) Increase

Explanation:

These questions can only be answered by considering the principle which states that, "When a constraint such as a change in concentration, pressure or volume is imposed on a reaction system in equilibrium. The system will readjust itself in order to annul the constraint."  

Now, if more reactants are added, the equilibrium position will shift towards the right, If more products are added, the equilibrium position will shift to the left.

Similarly, the removal of H2S causes the O2 concentration to increase since the equilibrium position now shifts to the left.

Also, addition of O2 causes H2S to be removed as the equilibrium moves to the right.

8 0
3 years ago
Hydrogen, a potential future fuel, can be produced from carbon (from coal) and steam by the following reaction: C(s)+2H2O(g)→2H2
madam [21]

The question is incomplete , complete question is:

Hydrogen, a potential future fuel, can be produced from carbon (from coal) and steam by the following reaction:

C(s)+ 2 H_2O(g)\rightarrow 2H_2(g)+CO_2(g).\Delta H=?

Note that the average bond energy for the breaking of a bond in CO2 is 799 kJ/mol. Use average bond energies to calculate ΔH of reaction for this reaction.

Answer:

The ΔH of the reaction is -626 kJ/mol.

Explanation:

C(s)+ 2 H_2O(g)\rightarrow 2H_2(g)+CO_2(g).\Delta H=?

We are given with:

\Delta H_{H-O}=459 kJ/mol

\Delta H_{H-H}=432 kJ/mol

\Delta H_{C=O}=799 kJ/mol

ΔH =  (Energies required to break bonds on reactant side) - (Energies released on formation of bonds on product side)

\Delta H=(4\times \Delta H_{O-H})-(2\times \Delta H_{H-H}+2\times\Delta H_{C=O})

=(4\times 459 kJ/mol)-(2\times 432 kJ/mol+2\times 799 kJ/mol

\Delta H=-626 kJ/mol

The ΔH of the reaction is -626 kJ/mol.

5 0
4 years ago
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