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Sonja [21]
3 years ago
10

When a magnesium wire is dipped into a solution of lead (II) nitrate, a black deposit forms on the wire. Which of the following

can be concluded from this observation?
(A) The standard reduction potential, Eo , for Pb2+(aq) is greater than that for Mg2+(aq).
(B) Mg(s) is less easily oxidized than Pb(s).
(C) An external source of potential must have been supplied.
(D) The magnesium wire will be the cathode of a Mg/Pb cell.
(E) Pb(s) can spontaneously displace Mg2+(aq) from solution.
...

Chemistry
1 answer:
bogdanovich [222]3 years ago
4 0

Answer:The standard reduction potential, Eo , for Pb2+(aq) is greater than that for Mg2+(aq).

Explanation:

Metals are usually arranged in an order of reactivity called activity series. Metals that are high up in the series are good reducing agents with very low (very negative) reduction potentials. Metals with greater (less negative) reduction potentials are found lower in the series. In the image attached, elements were arranged according to their reducing ability. Magnesium is very electro positive hence it is a better reducing agent with a lesser standard reduction potential than lead(refer to the image for numerical values of standard reduction potentials). Hence it displaces lead from solution and the elemental lead deposits on the wire.

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A sample of ammonia ^NH3h gas is completely decomposed to nitrogen and hydrogen gases over heated iron wool. If the total pressu
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Explanation :

According to the Dalton's Law, the partial pressure exerted by component 'i' in a gas mixture is equal to the product of the mole fraction of the component and the total pressure.

Formula used :

p_i=X_i\times p_T

X_i=\frac{n_i}{n_T}

So,

p_i=\frac{n_i}{n_T}\times p_T

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p_i = partial pressure of gas

X_i = mole fraction of gas

p_T = total pressure of gas

n_i = moles of gas

n_T = total moles of gas

The balanced decomposition of ammonia reaction will be:

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Now we have to determine the partial pressure of N_2 and H_2

p_{N_2}=\frac{n_{N_2}}{n_T}\times p_T

Given:

n_{N_2}=1\\\\n_{H_2}=3\\\\n_{T}=4\\\\p_T=866mmHg

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p_{H_2}=\frac{n_{H_2}}{n_T}\times p_T

Given:

n_{H_2}=1\\\\n_{H_2}=3\\\\n_{T}=4\\\\p_T=866mmHg

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