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11111nata11111 [884]
3 years ago
12

The dipole moment of hf is 1.91d. what is the dipole moment of hf in c⋅m? express your answer in coulomb-meters to three signifi

cant figures.
Chemistry
1 answer:
Drupady [299]3 years ago
3 0
<span>6.37x10^-30 coulomb-meters The unit for dipole moments is the Debye which is 1x10^-21 C*m^2/s divided by the speed of light (299792458 m/s). So the conversion factor is 1x10^-21 C*m^2/s / 299792458 m/s = 3.33564095E-30 C*m And converting 1.91 d to Cm gives 1.91 * 3.33564095E-30 C*m = 6.37x10^-30 coulomb-meters</span>
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A. The force of gravity.

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a prescriber has ordered childrens motrin 400 mg po q6h for a child who weighs 60 kg. how many mg/kg is the child receiving
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2 years ago
identify the reagents you would use to convert each of the following compounds into pentanoic acid: (a) 1-pentene (b) 1-bromobut
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a)BH3.THF is used to convert 1-pentane to pentanoic acid and b)NaCN is used to convert Bromobutane to pentanoic acid.

a) The conversion of 1-pentane to pentanoic acid using BH3, also known as hydroboration-oxidation, is a two-step reaction involving the reaction of 1-pentane with borane (BH3), followed by oxidation of the resulting 1-pentylborane with hydrogen peroxide or other oxidizing agents.

In the first step, 1-pentane reacts with borane (BH3) to form 1-pentylborane, through a process known as hydroboration. This reaction is catalyzed by a Lewis acid, such as aluminum chloride, and proceeds via a hydride transfer from the borane to the 1-pentane.

In the second step, the 1-pentylborane is oxidized to pentanoic acid using hydrogen peroxide (H₂O₂) or other suitable oxidizing agents. The oxidation is catalyzed by an acid, such as hydrochloric acid (HCl), and proceeds via a proton transfer from the 1-pentylborane to the hydrogen peroxide. The end result is the conversion of 1-pentane to pentanoic acid.

The overall chemical reaction for the conversion of 1-pentane to pentanoic acid using borane (BH₃) and hydrogen peroxide (H₂O₂) is as follows:

1-pentane + BH₃ + H₂O₂ → pentanoic acid + H₂O + BH₂

b)The conversion of 1-Bromo butane to pentanoic acid using sodium cyanide (NaCN) proceeds via a nucleophilic substitution reaction. The reaction mechanism involves the following steps:

1. Attack of the nucleophile, NaCN, on the carbon atom of 1-Bromo butane to form a tetrahedral intermediate.

2. Loss of a proton from the tetrahedral intermediate to form a carbanion.

3. Protonation of the carbanion by water (or another proton source) to form pentanoic acid.

The overall reaction can be represented as follows:

1-Bromo butane + NaCN → Pentanoic Acid + NaBr

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1 year ago
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Permanganate (VII) ion is a strong oxidizing agent and gets easily reduced to manganese ion in presence of the hydrogen ion in an acidic solution.

The balanced half-reaction for reduction is shown as,

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