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Ainat [17]
1 year ago
8

The solutions that are used in this experiment have alcohols as both the solvent and the reactant. How is the excess alcohol rem

oved?.
Chemistry
1 answer:
ElenaW [278]1 year ago
4 0

Excess alcohol is removed by Vacuum filtration.

<h3>What is vacuum filtration?</h3>
  • Vacuum filtration is a methodology where a tension differential is kept up with across the channel medium by clearing the air beneath the channel paper.
  • Vacuum filtration gives a power on the arrangement notwithstanding that of gravity and builds the pace of filtration.
  • Vacuum filtration is the standard strategy utilized for isolating a strong fluid blend when the objective is to hold the strong (for instance in crystallization).
<h3>How do you dry a product after vacuum filtration?</h3>
  • The strong can be dried rapidly by washing the strong with an unpredictable fluid (methanol, CH3)2CO, or ether are normal) that the strong won't break up in. In the event that the filtrate is to be gathered, change the getting jar prior to washing.

To learn more about vacuum filtration from the given link

brainly.com/question/24082009

#SPJ4

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Express the answer in scientific notation and with the correct number of significant figures: (6.32 x 10-4) ÷ 12.64
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4 0
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Read 2 more answers
4) What volume will the gas in the balloon at right occupy at 250k?<br><br> balloon: 4.3L 350K
swat32

Answer:

2.87 liter.

Explanation:

Given:

Initially volume of balloon = 4.3 liter

Initially temperature of balloon = 350 K

Question asked:

What volume will the gas in the balloon occupy at 250 K ?

Solution:

By using:

Pv =nRT

Assuming pressure as constant,

V∝ T

Now, let  K is the constant.

V = KT

Let initial volume of balloon , V_{1} = 4.3 liter

1000 liter = 1 meter cube

1 liter = \frac{1}{1000} m^{3} = 10^{-3} m^{3

4.3 liter = 4.3\times10^{-3}=4.3\times10^{-3} m^{3}

And initial temperature of balloon, T_{1} = 350 K

Let the final volume of balloon is V_{2}

And as given, final temperature of balloon, T_{2} is 250 K

Now, V_{1} = KT_{1}

4.3\times10^{-3}=K\times350\ (equation\ 1 )

V_{2} = KT_{2}

=K\times250\ (equation 2)

Dividing equation 1 and 2,

\frac{4.3\times10^{-3}}{V_{2} } =\frac{K\times350}{K\times250}

K cancelled by K.

By cross multiplication:

350V_{2} =4.3\times10^{-3} \times250\\V_{2} =\frac{ 4.3\times10^{-3} \times250\\}{350} \\          = \frac{1075\times10^{-3}}{350} \\          =2.87\times10^{-3}m^{3}

Now, convert it into liter with the help of calculation done above,

2.87\times10^{-3} \times1000\\2.87\times10^{-3} \times10^{3} \\2.87\ liter

Therefore, volume of the gas in the balloon at 250 K will be  2.87 liter.

4 0
3 years ago
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