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Nesterboy [21]
3 years ago
11

Why do the deprotonated organic acid (RCO2-) and protonated organic base (RNH3+) go into the aqueous layer?

Chemistry
1 answer:
rodikova [14]3 years ago
4 0

Answer:

The deprotonated organic acid (RCO2-) and protonated organic base (RNH3+) go into the aqueous layer because Organic compounds that are neither acids or bases do not react with either NaOH or HCl and, therefore remain more soluble in the organic solvent and are not extracted.

Explanation:

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Solids, liquids, and gases are three common states of matter. Which of these states of matter can flow?
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Suppose that 0.48 g of water at 25∘C∘C condenses on the surface of a 55-gg block of aluminum that is initially at 25∘C∘C. If the
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Answer : The final temperature of the metal block is, 25^oC

Explanation :

heat_{absorbed}=heat_{released}

As we know that,  

Q=m\times c\times \Delta T=m\times c\times (T_{final}-T_{initial})

m_1\times c_1\times (T_{final}-T_1)=-[m_2\times c_2\times (T_{final}-T_2)]         .................(1)

where,

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T_2 = temperature of water = 25^oC

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c_2 = specific heat of water= 4.184J/g^oC

Now put all the given values in equation (1), we get

55g\times 0.900J/g^oC\times (T_{final}-25)^oC=-[0.48g\times 4.184J/g^oC\times (T_{final}-25)^oC]

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A sample of a gas at 25°C has a volume of 150 mL when its pressure is 0.947 atm. What will the temperature of the gas be at a pr
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Answer: The temperature of the gas at a pressure of  0.987 atm and volume of 144mL is 25.16^0C

Explanation:

The combined gas equation is,

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

where,

P_1 = initial pressure of gas = 0.947 atm

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T_1 = initial temperature of gas = 25^0C=(25+273.15)K=298.15K

T_2 = final temperature of gas = ?

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

\frac{0.947\times 150}{298.15}=\frac{0.987\times 144}{T_2}

T_2=298.31K=(298.31-273.15)^0C=25.16^0C

The temperature of the gas at a pressure of  0.987 atm and volume of 144mL is 25.16^0C

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