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skelet666 [1.2K]
3 years ago
5

The high polarity of the oxygen-carbon bond in alcohols is what allows them to be soluble in water.

Chemistry
1 answer:
Anna11 [10]3 years ago
8 0

This is false. An alcohol does indeed have a polar C-O single bond, but what we should really be focusing on is the extraordinarily polar O-H single bond. When oxygen, fluorine, or nitrogen is bound to a hydrogen atom, there is a small (but not negligible) charge separation, where the eletronegative N, O, or F has a partial negative charge, and the H has a partial positive charge. Water has two O-H single bonds in it (structure is H-O-H). The partially negative charge on the O of the water molecule (specifically around the lone pair) can become attracted either a neighboring water molecule's partially positive H atom, or an alcohol's partially positive H atom. This is weak (and partially covalent) attraction is called a hydrogen bond. This is stronger than a typical dipole-dipole attraction (as would be seen between neighboring C-O single bonds), and much stronger than dispersion forces (between any two atoms). When the solvent (water) and the solute (the alcohol) both exhibit similar intermolecular forces (hydrogen bonding being the most important in this case), they can mix completely in all proportions (i.e. they are miscible) in water.

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<u>Answer:</u>

<u>For a:</u> The number of moles of air present in the RV is 0.047 moles

<u>For b:</u> The number of molecules of gas is 2.83\times 10^{22}

<u>Explanation:</u>

  • <u>For a:</u>

To calculate the number of moles, we use the equation given by ideal gas follows:

PV=nRT

where,

P = pressure of the air = 1.00 atm

V = Volume of the air = 1200 mL = 1.2 L    (Conversion factor:  1 L = 1000 mL)

T = Temperature of the air = 37^oC=[37+273]K=310K

R = Gas constant = 0.0821\text{ L. atm }mol^{-1}K^{-1}

n = number of moles of air = ?

Putting values in above equation, we get:

1.00atm\times 1.2L=n_{air}\times 0.0821\text{ L atm }mol^{-1}K^{-1}\times 310K\\n_{air}=\frac{1.00\times 1.2}{0.0821\times 310}=0.047mol

Hence, the number of moles of air present in the RV is 0.047 moles

  • <u>For b:</u>

According to mole concept:

1 mole of a compound contains 6.022\times 10^{23} number of molecules.

So, 0.047 moles of air will contain (0.047\times 6.022\times 10^{23})=2.83\times 10^{22} number of gas molecules.

Hence, the number of molecules of gas is 2.83\times 10^{22}

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What volume of a 1.5 M KOH solution is needed to provide 3.0 moles of KOH?
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Answer:

  • Volume = <u>2.0 liter</u> of 1.5 M solution of KOH

Explanation:

<u>1) Data:</u>

a) Solution: KOH

b) M = 1.5 M

c) n = 3.0 mol

d) V = ?

<u>2) Formula:</u>

Molarity is a unit of concentration, defined as number of moles of solute per liter of solution:

  • M = n / V in liter

<u>3) Calculations:</u>

  • Solve for n: M = n / V ⇒ V = n / M

  • Substitute values: V = 3.0 mol / 1.5 M = 2.0 liter

You must use 2 significant figures in your answer: <u>2.0 liter.</u>

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Answer: 4.1 g of barium precipitated.

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To calculate the moles, we use the equation:

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Given : moles of barium = 0.030

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0.030=\frac{x}{137}

x= 4.1 g

Thus there are 4.1 g of barium that precipitated.

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