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77julia77 [94]
3 years ago
12

acetone and ethanol Choose one or more: A. London dispersion B. dipole–dipole C. hydrogen bonding D. ion-induced dipole

Chemistry
1 answer:
andreyandreev [35.5K]3 years ago
3 0

Explanation:

The weak intermolecular forces which can arise either between nucleus and electrons or between electron-electron are known as dispersion forces. These forces are also known as London dispersion forces and these are temporary in nature.

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.

Hydrogen bonding is defined as a bonding which exists between a hydrogen atom and an electronegative atom like O, N and F.

Chemical formula of acetone is CH_{3}COCH_{3}. Due to the presence of oxygen atom there will be partial positive charge on carbon and a partial negative charge on oxygen atom. Hence, dipole-dipole forces will exist in a molecule of acetone.

Whereas hydrogen bonding will exist in a molecule of ethanol (CH_{3}CH_{2}OH). Since, hydrogen atom is attached with electronegative oxygen atom.

Whereas London dispersion forces will also exist in both acetone and ethanol molecule.

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Upon dissolving in , undergoes a disproportionation reaction according to the following unbalanced equation: This disproportiona
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Answer:

The question in the narrative seems not to be complete because the unbalanced equation was not given and the values of the second(s) and  chemical dissolved was also not given.

Kindly find the complete question below and if you feel the question is still correct, the solution provided is still implies but without the values inserted.

Correct Question:

Upon dissolving InCl(s) in HCl , In(aq)  undergoes a disproportionation reaction according to the following unbalanced equation:

In ⁺ (aq) → In(s) → In³⁺ (aq)

This disproportionation follows first-order kinetics with a half-life of 667s. What is the concentration of In⁺ (aq) after 1.25 h if the initial solution of In⁺ (aq) was prepared by dissolving 2.38 g InCl(s) in dilute HCl to make 5.00 x 10² mL of solution? What mass of In(s) is formed after 1.25h?

Solution / Explanation:

Given half life of In⁺ at 66.7 s,

We recall the formula used in claculating the rate of constant for the first oreder reactin as :

K = 0.693 / t₁÷2,

Noting that:

t₁÷2 = half life

and K= rate constant

Therefore, if we replace the value of  t₁÷2  in the formular above,

We have,

K = 0.693 / 667s

K = 0.00 /s

Now, if we recall the mass of InCl(s) as 2.38g,

Volume of dilute HCl = 500 mL,

and the molar mass of  InCl(s) - 150.271 g/mol,

The number of moles is then calculated using the formular:

Number of moles: = Given Mass/Molar Mass

Now replacing the given values of given mass and the molar mass in the above formular,

= 2.38g / 150.271g/mol

= 0.0158 mol

Volume of diluted 500 mL.

Recalling also that we need to convert from mL into Liters

Therefore,  1mL = 10⁻³L

Therefore,

500mL = (500 X 10⁻³)L

0.5 L

Now, the molarrity of In⁺ (aq) is calculated using

morality of In⁺ (aq) = moles of In⁺ (aq)/volume of solution

= 0.0158/0.5L

=0.0316M (This is the initial concentration of In⁺ (aq))

The time of the reaction is 1.25h

There is 3600s in one hour

1.25h = 1.25 x 3600

= 4500s

7 0
3 years ago
Determine whether each of the following salts will form a solution that is acidic, basic, or pH-neutral.
storchak [24]

Answer:

a. Neutral

b. Basic

Explanation:

To determine which of the salts are acidic, neutral or basci we should dissociate them and determine if the ions, can make hydrolysis to water.

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We need to know, where do the ions come from. In this case, K⁺ comes from the KOH which is a strong base and Cl⁻ comes from the HCl, a strong acid. In conclussion, both are the conjugate weak acid and base, respectively. They do not make hydrolysis, so this salt is neutral. No protons or hydroxides are given.

NaClO → Na⁺  + ClO⁻

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We are giving hydroxides to medium, so the salt is basic.

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