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Drupady [299]
3 years ago
12

A student carried out a simple distillation on a compound known to boil at 124 oc and they reported a boiling point of 116-117 o

analysis of the compound showed it was pure and calibration of the thermometer indicated that it was accurate. What procedural error might the student have made in setting up the distillation?

Chemistry
1 answer:
chubhunter [2.5K]3 years ago
4 0

Answer :

The correct answer is due to incorrect position of thermometer .

Distillation  is process of separating two volatile liquids on the basis of their boiling points .In involves the following set up ( shown in image )

Since the Boiling point is found   is 116-117 C which is near to 124 C , also the purity was checked and thermometer was calibrated .

So the only error could be possible is position of thermometer .

The position of  bulb of thermometer is very important to get an accurate reading of boiling point .The vapor so formed should touch the side walls of bulb as soon as they are formed .

May be  the bulb of thermometer was   kept far from the vapors formed by liquid so,  s<u>ome of  the vapor before reaching to  bulb start condensing and their Temperature decreases </u>.  

Hence , the  temperature shown by thermometer was decrease than actual boiling point  due to incorrect position of thermometer.


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The enthalpy of a pure liquid at 75oC is 100 J/mol. The enthalpy of the pure vapor of that substance at 75oC is 1000 J/mol. What
pentagon [3]

Answer:

900 J/mol

Explanation:

Data provided:

Enthalpy of the pure liquid at 75° C = 100 J/mol

Enthalpy of the pure vapor at 75° C = 1000 J/mol

Now,

the heat of vaporization is the the change in enthalpy from the liquid state to the vapor stage.

Thus, mathematically,

The heat of vaporization at 75° C

=  Enthalpy of the pure vapor at 75° C - Enthalpy of the pure liquid at 75° C

on substituting the values, we get

The heat of vaporization at 75° C = 1000 J/mol - 100 J/mol

or

The heat of vaporization at 75° C = 900 J/mol

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3 years ago
(Extra points+Brainliest Answer, :) happy holidays)
grigory [225]

the answer is 360 g H2O with a 90.3 yeild percentage

7 0
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What is the molecular weight mass for KOH
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The weight is 56.1056 g/Mol hoped this helped
5 0
3 years ago
How many liters of oxygen gas, at standard
Karo-lina-s [1.5K]

Answer:

Explanation:

  • For the balanced reaction:

<em>4Fe(s) + 3O₂(g) → 2Fe₂O₃(s)​.</em>

It is clear that 4 mol of Fe react with 3 mol of O₂ to produce 2 mol of Fe₂O₃.

  • Firstly, we need to calculate the no. of moles of 35.8 grams of Fe metal:

no. of moles of Fe = mass/molar mass = (35.8 g)/(55.845 g/mol) = 0.64 mol.

  • Now, we can find the no. of moles of O₂ is needed to react with the proposed amount of Fe:

<em><u>Using cross multiplication:</u></em>

4 mol of Fe is needed to react with → 3 mol of O₂, from stichiometry.

0.64 mol of Fe is needed to react with → ??? mol of O₂.

∴ The no. of moles of O₂ needed = (3 mol)(0.64 mol)/(4 mol) = 0.48 mol.

  • Finally, we can get the volume of oxygen using the information:

<em>It is known that 1 mole of any gas occupies 22.4 L at standard P and T (STP).</em>

<em></em>

<em><u>Using cross multiplication:</u></em>

1 mol of O₂ occupies → 22.4 L, at STP conditions.

0.48 mol of O₂ occupies → ??? L.

∴ The no. of liters of O₂ = (0.48 mol)(22.4 L)/(1 mol) = 10.752 L.

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Answer:

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Compound of sodium carbonate and sodium bicarbonate was the name of the resulting hydrate that formed.

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