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N76 [4]
3 years ago
5

Classify which compounds will dissolve in water and which ones will not dissolve in water.

Chemistry
1 answer:
vredina [299]3 years ago
6 0

Answer:

Dissolves in water: Acetone, 1-propanol, Methanol

Do not dissolve in water: Hexane, Decane

Explanation:

From the principle that like dissolves like,polar substances will dissolve in polar solutions, while non-polar substances will dissolve in non-polar substances.

Water is a polar solution, therefore, polar substances will dissolve in it.From the given options:

Acetone is a polar molecule, therefore, it will dissolve in water.

1-propanol is a polar molecule, therefore, it will dissolve in water.

Hexane is a non-polar molecule, therefore, it will not dissolve in water.

Methanol is a polar molecule, therefore, it will dissolve in water.

Decane is a non-polar molecule, therefore, it will not dissolve in water

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Answer : The volume of CO_2 will be, 514.11 ml

Explanation :

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HCO_3^-+HCl\rightarrow Cl^-+H_2O+CO_2

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Molar mass of HCO_3^- = 1 + 12 + 3(16) = 61 g/mole

\text{Moles of }HCO_3^-=\frac{\text{Mass of }HCO_3^-}{\text{Molar mass of }HCO_3^-}=\frac{1.23175g}{61g/mole}=0.0202moles

Now we have to calculate the moles of CO_2.

From the balanced chemical reaction, we conclude that

As, 1 mole of HCO_3^- react to give 1 mole of CO_2

So, 0.0202 mole of HCO_3^- react to give 0.0202 mole of CO_2

The moles of CO_2 = 0.0202 mole

Now we have to calculate the volume of CO_2 by using ideal gas equation.

PV=nRT

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P = pressure of gas = 1.00 atm

V = volume of gas = ?

T = temperature of gas = 37^oC=273+37=310K

n = number of moles of gas = 0.0202 mole

R = gas constant = 0.0821 L.atm/mole.K

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

(1.00atm)\times V=0.0202 mole\times (0.0821L.atm/mole.K)\times (310K)

V=0.51411L=514.11ml

Therefore, the volume of CO_2 will be, 514.11 ml

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