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Vikki [24]
2 years ago
7

A student wants to prepare a 1.0-liter solution of a specific Molarity. The student determines that the mass of the solute needs

to be 30. grams. What is the proper procedure to follow?
(1) Add 30. g of solute to 1.0 L of solvent.
(2) Add 30. g of solute to 970. mL of solvent to make 1.0 L of solution.
(3) Add 1000. g of solvent to 30. g of solute.
(4) Add enough solvent to 30. g of solute to make 1.0 L of solution.
Chemistry
1 answer:
Mars2501 [29]2 years ago
4 0
The answer is (4) Add enough solvent to 30.0 g of solute to make 1.0 L solution. The molarity is calculated using volume of the solution. When solute dissolving, the total volume will change. So the final volume of solution need to be 1.0 L.
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Answer:

The answer is 0.046 mol.

Explanation:

By looking at the balanced equation, you can form a ratio of lithium chloride and lithium bromide using the coefficient value :

ratio of lithium bromide <u>(</u>2LiBr)

= 2

ratio of lithium chloride (2LiCl)

= 2

So the ratio is 2 : 2 then simplify into 1 : 1 . Which means that 1 mol of lithium bromide is equal to 1 mole of lithium chloride.

In this case, 0.046 mol of lithium bromide will form <u>0</u><u>.</u><u>0</u><u>4</u><u>6</u><u> </u><u>m</u><u>o</u><u>l</u> of lithium chloride.

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Alika [10]

Answer:

1. Hot water

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Explanation:

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4 0
2 years ago
How many g i2 should be added to 750g ccl4 to prepare a 0.200m solution?
vesna_86 [32]
<span>There are a number of ways to express concentration of a solution. This includes molality. Molality is expressed as the number of moles of solute per mass of the solvent. We calculate as follows:

0.200 mol I2 / kg CCl4 ( .750 kg CCl4 ) ( 253.809 g I2 / mol I2) = 38.07 g I2 needed

Hope this helps.

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6 0
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Mademuasel [1]

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