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gavmur [86]
3 years ago
12

Match the words in the left column to the appropriate blanks in the sentences on the right. Make certain each sentence is comple

te before submitting your answer. ResetHelp 1. NaCl ; ionic bonds{\rm NaCl} ; blank are stronger than the blank in {\rm HCl}. are stronger than the dispersion forces{\rm NaCl} ; blank are stronger than the blank in {\rm HCl} . in HCl. 2. H2O ; hydrogen bonds{\rm H_2O} ; blank are stronger than the blank in {\rm H_2Se}. are stronger than the dispersion forces{\rm H_2O} ; blank are stronger than the blank in {\rm H_2Se}. in H2Se. 3. NH3 ; hydrogen bonds{\rm NH_3} ; blank are stronger than the blank in {\rm PH_3}. are stronger than the dipole-dipole attractions{\rm NH_3} ; blank are stronger than the blank in {\rm PH_3}. in PH3. 4. HF ; hydrogen bonds{\rm HF} ; blank are stronger than the blank in {\rm F_2}. are stronger than the dispersion forces{\rm HF} ; blank are stronger than the blank in {\rm F_2}. in F2.
Chemistry
1 answer:
Kazeer [188]3 years ago
7 0

Answer:

The ionic bond in NaCl are stronger than the stronger than the dispersion forces in HCl.

The hydrogen bonds in H2O are stronger than the dispersion forces in H2Se

Hydrogen bonds in NH3 are stronger than the dipole-dipole attractions in PH3.

Hydrogen bonds in HF are stronger than the dispersion forces in F2

Explanation:

Ionic bonds occur in molecules with high differences in their electronegative value where there are actual transfer of electrons. HCl has a bond which is involved in the sharing of electrons.

Hydrogen bonds are present in H2O which is stronger than the dispersion forces.

PH3 is a larger molecule with greater dispersion forces than ammonia, NH3 has very polar N-H bonds leading to strong hydrogen bonding. This dominant intermolecular force results in a greater attraction between NH3 molecules than there is between PH3 molecules.

F2 is a non-polar molecule, therefore they have London dispersion forces between molecules while HF has a hydrogen bond because F is highly electronegative.

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8 0
3 years ago
When bsia is assembled as an octamer, what is most likely to be true regarding l76, l77, and l79?
telo118 [61]

The option that is most likely to be true regarding L76, L77, and L79  is that option B: The assembly would incur an entropic penalty if they occupied a solvent-exposed site.

<h3>What is entropic penalty?</h3>

The entropic penalty in regards to ordered water is known to be one that tells or account  for any form of  weaker binding of the antibiotic novobiocin to what we call resistant mutant of DNA gyrase.

Note that in regards to the scenario above, the Entropic penalty is seen as the thermodynamically disfavored needs that is required in forming a cage of polar solvent molecules that is known to be seen around surface that has exposed hydrophobic potion of a molecule.

Hence, The option that is most likely to be true regarding L76, L77, and L79  is that option B: The assembly would incur an entropic penalty if they occupied a solvent-exposed site.

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brainly.com/question/13198179

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See full question below

When Bs1A is assembled as an octamer, what is most likely to be true regarding L76, L77, and L79?

A. They are oriented toward the solvent-exposed exterior of the protein assembly.

B. The assembly would incur an entropic penalty if they occupied a solvent-exposed site.

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D. Their physiochemical properties are not substantially dependent on their hydrophobicity

5 0
2 years ago
If 2.00g of p-aminophenol ( 109.1 g/mol) reacts with 5.00 ml of acetic anhydride (102.1 g/mol and density = 1.08 g/ml), what mas
inna [77]
The complete reaction is as,

      4-Aminophenol + Acetic Anhydride → <span>Acetaminophen + Acetic Acid

First of all convert the ml of Acetic anhydrite to grams,
As,
                                Density  =  mass / volume
Solving for mass,
                                mass  =  Density </span>× Volume
<span>Putting values,
                                mass  =  1.08 g/ml </span>× 5ml
<span>
                                mass =  5.4 g of acetic anhydride

First Find amount of  acetic anhydride required to react completely with 2 g of p-Aminophenol,
As,
       109.1 g of p-aminophenol required  =  102.1 g of acetic anhydride
so,     2 g of p-aminophenol will require  =  X g of Acetic Anhydride

Solving for X,
                                X  =  (2 g </span>× 102.1 g) ÷ 109.1 g
 
                                X  =  1.87 g of acetic anhydride is required to be reacted.

But, we are provided with 5.4 g of Acetic Anhydride, means p-aminophenol is the limiting reactant and it controls the formation of product. Now Let's calculate for product,
As,
       109.1 g of p-aminophenol produced  =  180.2 g of <span>Acetaminophen
So    2.00 g of p-aminophenol will produce  =  X g of Acetaminophen

Solving for X,
                               X  =  (2.00 g </span>× 180.2 g) ÷ 109.1 g
 
                               X  =  3.30 g of Acetaminophen

Result:
          <span>If 2.00g of p-aminophenol reacts with 5.00 ml of acetic anhydride 3.30 g of acetaminophen is made.</span>
4 0
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