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3 years ago

31. A mixture with sand, pink sugar and water would be best separated using

Chemistry

1 answer:

ozzi3 years ago

3 0

Answer:

Filtration and evaporation

Explanation:

A mixture of sand, pink sugar and water can be separated by filtration and evaporation.

The pink sugar is already dissolved in the water but the sand remains undissolved in the water hence it can be filtered off. The filtrate now contains the sugar.

The sugar is obtained from the filtrate by evaporation of the solution to yield the solid pink sugar and steam which can be condensed again to give liquid water.

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Is the formation of sulfur tetrafluoride monoxide endothermic or exothermic?

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A geochemist in the field takes a 46.0 mL sample of water from a rock pool lined with crystals of a certain mineral compound X.

inysia [295]

Answer:

The solubility of the mineral compound X in the water sample is 0.0189 g/mL.

Explanation:

Step 1: Given data

The volume of water sample = 46.0 mL.

The weight of the mineral compound X after evaporation, drying, and washing = 0.87 g.

Step 2: Calculate the solubility of X in water

46.00 mL of water sample contains 0.87 g of the mineral compound X.

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3 years ago

Lithium diisopropylamide (LDA) is used as a strong base in organic synthesis. LDA is itself prepared by an acid-base reaction be

kakasveta [241]

Explanation:

Lithium diisopropylamide (LDA) is used in many organic synthesis and is a strong base. It is prepared by the acid base reaction of N,N-diisopropylamine ( [(CH₃)₂CH]₂NH ) and butyllithium ( Li⁺⁻CH₂CH₂CH₂CH₃ ).

The equation is show below as:

[(CH₃)₂CH]₂NH + Li⁺⁻CH₂CH₂CH₂CH₃ ⇒ [(CH₃)₂CH]₂N⁻Li⁺ + CH₃CH₂CH₂CH₃

N,N-diisopropylamine ( [(CH₃)₂CH]₂NH ) is a weaker acid and hence, LDA ( [(CH₃)₂CH]₂N⁻Li⁺ ) is stronger base. (Weaker acid has stronger conjugate base)

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3 years ago

658 mL of 0.250 M HCl solution is mixed with 325 mL of 0.600 M HCl solution. What is the molarity of the resulting solution

Tju [1.3M]

The molarity of the resulting solution obtained by mixing 658 mL of 0.250 M HCl solution with 325 mL of 0.600 M HCl solution is 0.366 M

We'll begin by calculating the number of mole of HCl in each solution. This can be obtained as follow:

<h3>For solution 1:</h3>

Volume = 658 mL = 658 / 1000 = 0.658 L

Molarity = 0.250 M

Mole = Molarity x Volume

Mole of HCl = 0.250 × 0.658

<h3>For solution 2:</h3>

Volume = 325 mL = 325 / 1000 = 0.325 L

Molarity = 0.6 M

Mole = Molarity x Volume

Mole of HCl = 0.6 × 0.325

Next, we shall determine the total mole of HCl in the final solution. This can be obtained as follow:

Mole of HCl in solution 1 = 0.1645 mole

Mole of HCl in solution 2 = 0.195 mole

Total mole = 0.1645 + 0.195

<h3>Total mole = 0.3595 mole</h3>

Next, we shall determine the total volume of the final solution.

Volume of solution 1 = 0.658 L

Volume of solution 2 = 0.325 L

Total Volume = 0.658 + 0.325

<h3>Total Volume = 0.983 L</h3>

Finally, we shall determine the molarity of the resulting solution.

Total mole = 0.3595 mole

Total Volume = 0.983 L

<h3>Molarity =?</h3>

Molarity = mole / Volume

Molarity = 0.3595 / 0.983

<h3>Molarity = 0.366 M</h3>

Therefore, the molarity of the resulting solution is 0.366 M

Learn more: brainly.com/question/25342554

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3 years ago