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kramer
3 years ago
10

Classify the substances according to the strongest solute-solvent interaction that will occur between the given substances and w

ater during dissolution.
a. CuCl2
b. CuSO4
c. NH3
d. CH3OH
Chemistry
1 answer:
-BARSIC- [3]3 years ago
5 0

Explanation:

Ion-dipole interactions are defined as the interactions that occur when an ion interacts with the dipole of a molecule (polar molecule).

For example, CuCl_{2}, CuSO_{4}, CaCl_{2}, NaCl etc are all the molecules that will show ion-dipole interactions.

Dipole-dipole interactions are defined as the interactions that occur when partial positive charge on an atom is attracted by partial negative charge on another atom.

When a polar molecules produces a dipole on a non-polar molecule through distribution of electrons then it is known as dipole-induced forces.

For example, NH_{3}, CH_{3}OH etc are the molecules which show dipole-dipole interactions.

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3 years ago
Which of the following was not a big ideal of chemistry
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The Michaelis constant for pancreatic lipase is 5 mM. At 60 C, lipase is subject to deactivation with a half-life of 8 min. Fat
Dmitriy789 [7]

Explanation:

According to the given data, we will calculate the following.

 Half life of lipase t_{1/2} = 8 min x 60 s/min

                                       = 480 s

Rate constant for first order reaction is as follows.

         k_{d} = \frac{0.6932}{480}

                        = 1.44 \times 10^{-3}s^{-1}


Initial fat concentration S_{o} = 45 mol/m^{3}

                                                = 45 mmol/L

Rate of hydrolysis V_m_{o} = 0.07 mmol/L/s

Conversion X = 0.80

Final concentration (S) = S_{o} (1 - X)

                                      = 45 (1 - 0.80)

                                      = 9 mol/m^{3}

or,                                  = 9 mmol/L

It is given that K_{m} = 5mmol/L

Therefore, time taken will be calculated as follows.

                    t = -\frac{1}{K_{d}}ln[1 - \frac{K_{d}}{V}{K_{M} ln (\frac{S_{o}}{S}) + (S_{o} - S)]

Now, putting the given values into the above formula as follows.

            t = -\frac{1}{K_{d}}ln[1 - \frac{K_{d}}{V}{K_{M} ln (\frac{S_{o}}{S}) + (S_{o} - S)]  

             = -\frac{1}{1.44 \times 10^{-3}s^{-1}}ln[1 - \frac{1.44 \times 10^{-3}s^{-1}}{0.07 mmol/L/s
}{K_{M} ln (\frac{45 mmol/L
}{9 mmol/L
}) + (45 mmol/L - 9 mmol/L
)]

              = 1642.83 s \times \frac{1 min}{60 sec}

              = 27.38 min

Therefore, we can conclude that time taken by the enzyme to hydrolyse 80% of the fat present is 27.38 min.

6 0
3 years ago
Calculate the net change in enthalpy for the formation of one mole of nitric acid from nitrogen, hydrogen and oxygen from these
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The net enthalpy of formation of nitric acid is calculated by,
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Since all the reactants are in their elemental forms, the simplified equation would be,

    Hrxn = Hproduct 

Substituting,

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<em>Answer: -372.42 kJ</em>

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4 years ago
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