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NikAS [45]
4 years ago
6

Eugenol is to be isolated using steam distillation. At 100 oC and 744.67 torr calculate the vapor pressure of pure eugenol at 10

0oC when 16.48% of the distillate collected is eugenol. The vapor pressure of water at 100oC is 760 torr. Put your answer in 5 significant figures.
Chemistry
1 answer:
STALIN [3.7K]4 years ago
4 0

Answer:

Vapor pressure of Eugenol = 667.04torr

Explanation:

By applying dalton's law of partial pressure;

at 100degree celsius, total pressure = 744.67torr

vapor pressure of water at 100 degree celsius = 760torr

mole fraction of eugenol = 0.1648

mole fraction of water = 1 - 0.1648 = 0.8352

Total pressure = vapor P(water) + vapor P( Eugenol)

for water; vapour pressure = mole fraction x total pressure

for eugenol; vapor pressure =  mole fraction x total pressure

substituting into the above gives the vapor pressure of eugenol = 667.04torr

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Answer:

   adding silver ions , adding chloride ions, removing chloride ions and removing silver chloride.

Explanation:

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We can differentiate barium ions and sulfate ions away from the surface if we add a small amount of solid barium sulfate is mixed with water and shaken.

  The action would shift this reaction away from solid barium sulfate and toward the dissolved ions are adding silver ions , adding chloride ions, removing chloride ions and removing silver chloride.

An ionic and covalent quality has in every molecular bond.

In barium sulfate barium has lowest electronegativities.

Oxygen has highest electronegativities.

so the bond between both barium and sulfate is ionic in character.

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To test the effectiveness of a gunpowder mixture, 1 gram was exploded under controlled STP conditions, and the reaction chamber
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Answer:

The volume of the gas at given temperature is 1300 cm^3

Explanation:

P_1 = initial pressure of gas = 1 atm

T_1 = initial temperature of gas = 0^oC=273.15 K

V_1 = initial volume of gas = 310 cm^3 = 310 mL = 0.310 L

(1 cm^3= 1mL , 1 mL = 0.001 L)

P_1V_1=nRT_1..[1]

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T_2 = final temperature of gas = 2,200^oC=273+2200 K=2473.15 K

V_2 = final volume of gas = ?

P_2V_2=nRT_2..[2]

By dividing [1] and [2] we get combined gas equation :,

\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}

V_2=\frac{P_1V_1\times T_2}{T_1\times P_2}

Now put all the given values in the above equation, we get:

V_2=\frac{1 atm\times 0.310 L\times 2473.15 K}{273.15 K\times 2.1 atm}

V_2=1.3 L=1300 mL= 1300 cm^3

The volume of the gas at given temperature is 1300 cm^3

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3 years ago
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3 years ago
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The escape of molecules from the surface of a liquid is known as _____.
RUDIKE [14]
Its c. evaporation :)

8 0
3 years ago
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2AgNO3 + CaCl2 → 2AgCl + Ca(NO3)2
Harrizon [31]

It can be found that 337.5 g of AgCl formed from 100 g of silver nitrate and 258.4 g of AgCl from 100 g of CaCl₂.

<u>Explanation:</u>

2AgNO₃ + CaCl₂ → 2 AgCl + Ca(NO₃)₂

We have to find the amount of AgCl formed from 100 g of Silver nitrate by writing the expression.

100 g \text { of } A g N O_{3} \times \frac{2 \text { mol } A g N O_{3}}{169.87 g A g N O_{3}} \times \frac{2 \text { mol } A g C l}{1 \text { mol } A g N O_{3}} \times \frac{143.32 g A g C l}{1 \text { mol } A g C l}

= 337.5 g AgCl

In the same way, we can find the amount of silver chloride produced from 100 g of Calcium chloride.

It can be found as 258.4 g of AgCl produced from 100 g of Calcium chloride.

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