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Karolina [17]
3 years ago
14

Cathodic protection of iron involves using another more reactive metal as a sacrificial anode. Classify each of the following me

tals by whether they would or would not act as a sacrificial anode to iron under standard conditions.
a. Ag
b. Mg
c. Cu
d. Pb
e. Sn
f. Zn
g. Au
Chemistry
1 answer:
Anika [276]3 years ago
6 0

Answer:

a. Ag ---> cannot serve as a sacrificial anode for iron because it is lower than iron in the reactivity series. Hence, it is less reactive than iron.

b. Mg ---> can serve as a sacrificial anode for iron because it is higher than iron in the reactivity series. Hence, it is more reactive than iron.

c. Cu ---> cannot serve as a sacrificial anode for iron because it is lower than iron in the reactivity series. Hence, it is less reactive than iron.

d. Pb ---> cannot serve as a sacrificial anode for iron because it is lower than iron in the reactivity series. Hence, it is less reactive than iron.

e. Sn ---> cannot serve as a sacrificial anode for iron because it is lower than iron in the reactivity series. Hence, it is less reactive than iron.

f. Zn ---> can serve as a sacrificial anode for iron because it is higher than iron in the reactivity series. Hence, it is more reactive than iron.

g. Au ---> cannot serve as a sacrificial anode for iron because it is lower than iron in the reactivity series. Hence, it is less reactive than iron.

Explanation:

Cathodic protection of iron involves using another more reactive metal as a sacrificial anode. The reactivity series of metals arranges metals based on decreasing order of reactivity. The more reactive metals are found higher up in the series while the least reactive metals are found at the lower ends of the series. Thus, metals above iron in the reactivity series can serve as sacrificial anodes by protecting against corrosion, while those lower than iron cannot.

Based on the reactivity series, the following metals can be classified as either a sacrificial anode for iron or not:

a. Ag ---> cannot serve as a sacrificial anode for iron because it is lower than iron in the reactivity series. Hence, it is less reactive than iron.

b. Mg ---> can serve as a sacrificial anode for iron because it is higher than iron in the reactivity series. Hence, it is more reactive than iron.

c. Cu ---> cannot serve as a sacrificial anode for iron because it is lower than iron in the reactivity series. Hence, it is less reactive than iron.

d. Pb ---> cannot serve as a sacrificial anode for iron because it is lower than iron in the reactivity series. Hence, it is less reactive than iron.

e. Sn ---> cannot serve as a sacrificial anode for iron because it is lower than iron in the reactivity series. Hence, it is less reactive than iron.

f. Zn ---> can serve as a sacrificial anode for iron because it is higher than iron in the reactivity series. Hence, it is more reactive than iron.

g. Au ---> cannot serve as a sacrificial anode for iron because it is lower than iron in the reactivity series. Hence, it is less reactive than iron.

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The question is incomplete. Here is the complete question.

When 2.10 g of a certain molecular compound X are dissolved in 65.0 g of benzene (C₆H₆), the freezing point of the solution is measured to be 3.5°C. Calculate the molar mass of X. If you need any additional information on benzene, use only what you find in the ALEKS Data resource. Also, be sure your answer has a unit symbol, and is rounded to 2 significant digits.

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<u>Freezing</u> <u>Point</u> <u>Depression</u> occurs when a solute is added to a solvent: the freezing point of the solvent decreases when a non-volatile solute is incremented.

<u>Molality</u> or <u>molal</u> <u>concentration</u> is a quantity of solute dissolved in a certain mass, in kg, of solvent. Its symbol is m and it's defined as

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Freezing point depression and molal are related as the following:

\Delta T_{f}=K_{f}.m

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\Delta T_{f} is freezing point depression of solution

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Now, to determine molar mass, first, find molality of the mixture:

\Delta T_{f}=K_{f}.m

m=\frac{\Delta T_{f}}{K_{f}}

For benzene, constant is 5.12°C/molal. Then

m=\frac{3.5}{5.12}

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Second, knowing the relationship between molal and moles of solute, determine the last one:

m=\frac{moles(solute)}{kg(solvent)}

mol(solute)=m.kg(solvent)

mol(solute) = 0.683(0.065)

mol(solute) = 0.044 mol

The definition for <u>Molar</u> <u>mass</u> is the mass in grams of 1 mol of substance:

n(moles)=\frac{m(g)}{MM(g/mol)}

MM=\frac{m}{n}

In the mixture, there are 0.044 moles of X, so its molecular mass is

MM=\frac{2.1}{0.044}

MM = 47.30 g/mol

The molecular compound X has molecular mass of 47.30 g/mol.

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Explanation:

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2. Convert grams to decigrams

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