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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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2 years ago
At 2°C, the vapor pressure of pure water is 23.76 mmHg and that of a certain seawater sample is 23.09 mmHg. Assuming that seawat
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Answer:

0.808  M

Explanation:

Using Raoult's Law

\frac{P_s}{Pi}= x_i

where:

P_s = vapor pressure of sea water( solution) = 23.09 mmHg

P_i = vapor pressure of pure water (solute) = 23.76 mmHg

x_i = mole fraction of water

∴

\frac{23.09}{23.76}= x_i

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x_2 = 1- x_i

x_2 = 1- 0.9718

x_2 = 0.0282

x_i = \frac{n_i}{n_i+n_2}  ------ equation (1)

x_2 = \frac{n_2}{n_i+n_2}  ------ equation (2)

where; (n_2) = number of moles of sea water

(n_i) = number of moles of pure water

equating above equation 1 and 2; we have :

\frac{n_2}{n_i}= \frac{0.0282}{0.9178}

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Answer:

32.23 to 4 significant figures.

Explanation:

The molar mass of the element is the mass of 6.022 * 10^23 atoms (Avogadro's number).

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The density of benzene is 0.879g/mL. The volume (with 1 decimal) of 131.9g sample of benzene is how many mL?
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Answer:

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Explanation:

Step 1: Given data

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Step 2: Calculate the volume of the sample of benzene

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