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2 years ago

In a simple distillation setup, what is the sequence of equipment from the bench top to the round bottom flask

Chemistry

1 answer:

Julli [10]2 years ago

3 0

In a simple distillation setup, the sequence of equipment from the bench top to the round bottom flask is:

Thermometer
Distillation flask

Liebig condenser

Round bottom flask
Bunsen burner

<h3>What is Distillation?</h3>

This is the process in which a mixture is separated through selective boiling and condensation.

The distillation flask and liebig condenser are usually located above the round bottom flask in the set up.

Read more about Distillation here brainly.com/question/24553469

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What is the concentration of the barium hydroxide solution if 50.0 mL of a 0.425 M HNO3 solution is required to neutralize a 36.

fomenos

The molarity of Barium Hydroxide is 0.289 M.

<u>Explanation:</u>

We have to write the balanced equation as,

Ba(OH)₂ + 2 HNO₃ → Ba(NO₃)₂ + 2 H₂O

We need 2 moles of nitric acid to react with a mole of Barium hydroxide, so we can write the law of volumetric analysis as,

V1M1 = 2 V2M2

Here V1 and M1 are the volume and molarity of nitric acid

V2 and M2 are the volume and molarity of Barium hydroxide.

So the molarity of Ba(OH)₂, can be found as,

$$ M2 = \frac{V1 \times M1}{2 \times V2}$

$$M2 = \frac{50 \times 0.425 }{2 \times 36.8}$

= 0.289 M

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3 years ago

1 mol Fe2(CrO4)3=____g

IgorLugansk [536]

111.6g of Fe + 156g of Cr + 192g of O = 459.6g of Fe2(CrO4)3

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3 years ago

Question - Complete and balance the following chemical equations:

Mekhanik [1.2K]

Answer is because
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2 years ago

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vichka [17]

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3 years ago

5.36 liters of nitrogen gas are at STP. What would be the new volume if we increased the moles from 3.5 moles to 6.0 moles?

Aleksandr [31]

Answer:

$V_2=9.20L$

Explanation:

Hello there!

In this case, according to the given STP (standard pressure and temperature), it is possible for us to realize that the equation to use here is the Avogadro's law as a directly proportional relationship between moles and volume:

$\frac{V_2}{n_2}= \frac{V_1}{n_1}$

In such a way, given the initial volume and both initial and final moles, we can easily compute the final volume as shown below:

$V_2= \frac{V_1n_2}{n_1} \\\\V_2=\frac{5.36L*6.0mol}{3.5mol}\\\\V_2=9.20L$

Best regards!

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3 years ago