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Anit [1.1K]
4 years ago
14

8. The best way to obtain pure, solid household salt from a solid mixture of household salt and sand is to:

Chemistry
1 answer:
dedylja [7]4 years ago
8 0

The best way to obtain pure, solid household salt from a solid mixture of household salt and sand is to "add water, stir, filter and evaporate the filtrate".

<u>Option: C</u>

<u>Explanation:</u>

The salt and sand can be separated on the basis of solubility, as we know the salt is chemically known as sodium chloride, which have good solubility in water. The most general method is the mixture is taken into a flask or beaker and water is added with stirring, where salt get dissolved and sand remain as it is, because NaCl is capable enough to form hydrogen bonding with water, while sand have absence of such property. Then this solution containing insoluble sand is filtered by using filter paper. The sand is received in filter paper while filtrate in beaker is evaporated by boiling it in order to receive salt as residue.

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The solubility of acetanilide is 12.8 g in 100 mL of ethanol at 0 ∘C, and 46.4 g in 100 mL of ethanol at 60 ∘C. What is the maxi
weqwewe [10]

Answer: 72.41% and 26.90% respectively.

Explanation:

At 60°C, you can dissolve 46.4g of acetanilide in 100mL of ethanol. If you lower the temperature, at 0°C, you can dissolve just 12.8g, which means (46.4g-12.8g)=33.6g of acetanilide must have precipitated from the solution.

We can calculate recovery as:

\%R=\frac{crystalized\ mass}{initial\ mass}*100 =\frac{33.6\ g}{46.4\ g}*100=72.41\%

So the answer to the first question is 72.41%.

For the second part just use the same formula, the mass of the precipitate is the final mass minus the initial mass, (171mg-125mg)=46mg.

\%R=\frac{crystalized\ mass}{initial\ mass}*100 =\frac{46\ mg}{171\ mg}*100=26.90\%

So the answer to the second question is 26.90%.

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3 years ago
Use the ruler to determine the length of this object. Record your answer to the nearest tenth. The object is ____ long.
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The object is 2.7 cm long
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3 years ago
An electrochemical cell has the following standard cell notation: Al(s) | Al3+(aq) || Mg2+(aq) | Mg(s)
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Answer:

a. Al(s) ⇄ Al⁺³(aq) + 3e⁻ (oxidation)

Mg²⁺(aq) + 2e⁻ ⇄ Mg(s) (reduction)

b. ΔE° = + 0.715 V

c.  It's an electrolytic cell, because it's a nonspontaneous reaction.

d. 2Al(s) + 3Mg²⁺(aq) ⇄ 2Al⁺³(aq) + 3Mg(s)

Explanation:

a. By the notation given, first is represented the oxidation reaction and then the reduction reaction, so they are:

Al(s) ⇄ Al⁺³(aq) + 3e⁻ (oxidation)

Mg²⁺(aq) + 2e⁻ ⇄ Mg(s) (reduction)

b. The standard potential of the cell (ΔE°) is the reduction potential of the oxidation less the reduction potential of the reduction. The reduction potentials are:

Al(s) = -1.66 V

Mg(s) = -2.375 V

ΔE° = -1.66 - (-2.375)

ΔE° = + 0.715 V

c. It's an electrolytic cell.

A galvanic cell is spontaneous, so the cathode (reduction) has a higher E° than the cathode (oxidation). In this case, the oxidation reaction has a higher E°, so the reaction is nonspontaneous and it's necessary an external force to it happen, so it's an electrolytic cell.

d. 2Al(s) + 3Mg²⁺(aq) ⇄ 2Al⁺³(aq) + 3Mg(s)

The number of electrons must be the same, so the oxidation reaction is multiplied by 2, and the reduction reaction by 3.

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3 years ago
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