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Nostrana [21]
3 years ago
6

Complete the sentences to best explain the ranking.

Chemistry
1 answer:
Jobisdone [24]3 years ago
4 0

Answer:

dipole-dipole forces, ion-dipole forces, higher molar mass, hydrogen bonding, stronger intermolecular forces

Explanation:

<em>1. H₂S and H₂Se exhibit the following intermolecular forces: </em><em>dipole-dipole forces </em><em>and </em><em>ion-dipole forces</em><em>.</em>  These molecules have a bent geometry, thus, a dipolar moment which makes them dipoles. When they are in the aqueous form they are weak electrolytes whose ions interact with the water dipoles

<em>2. Therefore, when comparing H₂S and H₂Se the one with a </em><em>higher molar mass</em><em> has a higher boiling point.</em>  In this case, H₂Se has a higher boiling point than H₂S due to its higher molar mass.

<em>3. The strongest intermolecular force exhibited by H₂O is </em><em>hydrogen bonding</em><em>.  </em>This is a specially strong dipole-dipole interaction in which the positive density charge on the hydrogens is attracted to the negative density charge on the oxygen.

<em>4. Therefore, when comparing H₂Se and H₂O the one with </em><em>stronger intermolecular forces</em><em> has a higher boiling point. </em>That's why the boiling point of H₂O is much higher than the boiling point of H₂Se.

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Explanation:

Formula to calculate osmotic pressure is as follows.

 Osmotic pressure = concentration × gas constant × temperature( in K)

Temperature = 25^{o} C

                      = (25 + 273) K

                      = 298.15 K  

Osmotic pressure = 531 mm Hg or 0.698 atm     (as 1 mm Hg = 0.00131)

Putting the given values into the above formula as follows.

       0.698 = C \times 0.082 \times 298.15 K&#10;

               C = 0.0285

This also means that,

  \frac{\text{moles}}{\text{volume (in L)}} = 0.0285

So,     moles = 0.0285 × volume (in L)

                      = 0.0285 × 0.100

                     = 2.85 \times 10^{-3&#10;}

Now, let us assume that mass of C_{12}H_{23}O_{5}N = x grams

And, mass of C_{12}H{22}O_{11} = (1.00 - x)

So, moles of C_{12}H_{23}O_{5}N = \frac{mass}{\text{molar mass}}

                              = \frac{x}{369}

Now, moles of C_{12}H_{22}O_{11} = \frac{(1.00 - x)}{342}

                   = \frac{x}{369} + \frac{(1.00 - x)}{342}

                  = 2.85 \times 10^{-3}

             = x = 0.346

Therefore, we can conclude that amount of C_{12}H_{23}O_{5}N present is 0.346 g  and amount of C_{12}H_{22}O_{11} present is (1 - 0.346) g = 0.654 g.

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3 years ago
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