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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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3 years ago
Liquid hexane will react with gaseous oxygen to produce gaseous carbon dioxide and gaseous water . Suppose 6.9 g of hexane is mi
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Answer:

There, there are no leftover for C6H14 instead, an additional mass of 4g of C6H14 is needed to completely react with 38.4g of O2.

Explanation:

Step 1:

We'll begin by writing the balanced equation for the reaction. This is shown below:

2C6H14 + 19O2 —> 12CO2 + 14H2O

Step 2:

Let us calculate the masses of C6H14 and O2 that reacted from the balanced equation. This is illustrated below:

2C6H14 + 19O2 —> 12CO2 + 14H2O

Molar Mass of C6H14 = (12x6) + (14x1) = 72 + 14 = 86g/mol

Mass of C6H14 from the balanced equation = 2 x 86 = 172g

Molar Mass of O2 = 16x2 =32g/mol

Mass of O2 from the balanced equation = 19 x 32 = 608g

From the balanced equation above, 172g of C6H14 reacted with 608g of O2.

Step 3.

Now, let us determine the mass of C6H14 that will react with 38.4 g of oxygen. This is illustrated below:

From the balanced equation above, 172g of C6H14 reacted with 608g of O2.

Therefore, Xg of C6H14 will react with 38.4g of O2 i.e

Xg of C6H14 = (172 x 38.4) /608

Xg of C6H14 = 10.9g

From the calculations made above, we can see clearly that the mass of C6H14 is limited as the reaction requires 10.9g of C6H14 and only 6.9g was given. There, there are no leftover for C6H14 instead, an additional mass ( 10.9 - 6.9 = 4g) of 4g of C6H14 is needed to completely react with 38.4g of O2.

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