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Igoryamba
3 years ago
14

Select the classification for the following reaction. 2Na(s) + 2H2O(l) → 2NaOH(aq) + H2(g) a. precipitation b. acid-base c. redo

x d. combination e. none of the above
Chemistry
1 answer:
Alexeev081 [22]3 years ago
8 0

Answer:

The answer to your question: I think is letter c

Explanation:

                       2Na(s) + 2H2O(l) → 2NaOH(aq) + H2(g)

a.- Precipitation: is incorrect because for this kind of reactions one of the products  must be a precipitate and in the reaction we see that NAOH is soluble in water (aq) and H2 is a gas, none of them precipitate.

b. acid - base: the reactants must be an acid and a base, and the reactants of this reaction has a base but it's in the products.

c. Redox: This reaction is a redox reaction because for this reaction the oxidation numbers of the reactants must change and this happen in this reaction (H changes from +1 to 0) and Na changes from 0 to +1.

d. Combination. In this reaction the reactants combine to form a new compound and this doesn't happen here.

e. None of the above. is false because letter C is correct.

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3 years ago
The overall reaction and equilibrium constant value for a hydrogen-oxygen fuel cell at 298 K is given below. 2 H2(g) + O2(g) → 2
DENIUS [597]

Answer:

(a) ΔG° = -474 kJ/mol; E° = 1.23 V

(b) ΔH° negative; ΔS° negative

(c) Since ΔS is negative, as T increases, ΔG becomes more positive. Therefore, the maximum work obtained will decrease as T increases.

Explanation:

Let's consider the following reaction.

2 H₂(g) + O₂(g) → 2 H₂O(l)

with an equilibrium constant K = 1.34 × 10⁸³

<em>(a) Calculate E° and ΔG° at 298 K for the fuel-cell reaction.</em>

We can calculate the standard Gibbs free energy (ΔG°) using the following expression:

ΔG° = - R × T × lnK

ΔG° = - 8.314 × 10⁻³ kJ . mol⁻¹.K⁻¹ × 298 K × ln 1.34 × 10⁸³ = -474 kJ/mol

To calculate the standard cell potential (E°) we need to write oxidation and reduction half-reactions.

Oxidation: 2 H₂ ⇒ 4 H⁺ + 4 e⁻

Reduction: O₂ + 4 e⁻ ⇒ 2 O²⁻

The moles of electrons (n) involved are 4.

We can calculate E° using the following expression:

E\°=\frac{0.0591V}{n} .logK\\E\°=\frac{0.0591V}{4} .log1.34 \times 10^{83}=1.23V

<em>(b) Predict the signs of ΔH° and ΔS° for the fuel-cell reaction. ΔH°: positive negative ΔS°: positive negative</em>

The standard Gibbs free energy is related to the standard enthalpy (ΔH°) and standard entropy (ΔS°) through the following expression:

ΔG° = ΔH° - T.ΔS°

Usually, the major contribution to ΔG° is ΔH°. So, if ΔG° is negative (exergonic), ΔH° is expected to be negative (exothermic).

The entropy is related to the number of moles of gases. There are 3 gaseous moles in the reactants and 0 in the products, so the final state is predicted to be more ordered than the initial state, resulting in a negative ΔS°.

<em>(c) As temperature increases, does the maximum amount of work obtained from the fuel-cell reaction increase, decrease, or remain the same?</em>

The maximum amount of work obtained depends on the standard Gibbs free energy.

wmax = ΔG° = ΔH° - T.ΔS°

Since ΔS is negative, as T increases, ΔG becomes more positive. Therefore, the maximum work obtained will decrease as T increases.

5 0
4 years ago
What will change more solid NiCl2 is added
hodyreva [135]
The Molarity will increase.
5 0
3 years ago
Which substance is a base? HCOOH RbOH H2CO3 NaNO3
Alex Ar [27]

Answer:

RbOH

Explanation:

For this question, we have to remember what is the definition of a base. A base is a compound that has the <u>ability to produce hydroxyl ions</u> OH^-, so:

AOH~->~A^+~+~OH^-

With this in mind we can write the <u>reaction for each substance:</u>

HCOOH~->~HCOO^-~+~H^+

RbOH~->~Rb^+~+~OH^-

H_2CO_3~->~CO_3^-^2~+~2H^+

NaNO_3~->~Na^+~+~NO_3^-

The only compound that fits with the definition is RbOH, so this is our <u>base</u>.

I hope it helps!

6 0
3 years ago
Read 2 more answers
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