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IceJOKER [234]
3 years ago
5

The relative strengths of covalent bonds and van der Waals interactions remain the same when tested in a vacuum or in water. How

ever, this is not true of hydrogen bonds or ionic bonds, whose bond strength is lowered considerably in the presence of water in comparison with the bond strength observed in a vacuum. A logical explanation to these observations is that the estimated bond strengths measure the amount of energy needed to break them. Covalent and Van der Waals attractions have intrinsic value independent of the environment while hydrogen bonds depend on any charged or polar molecule with reducing strength of the interaction they would otherwise have in the absence of water (in a vacuum).
Chemistry
1 answer:
trasher [3.6K]3 years ago
7 0

Answer and Explanation:

The explanation given in the problem is correct but not totally encompassing.

Van der waals interactions are a type of hydrophobic interaction, in which they do not interact with the polar water molecule. Covalent bonds involve the sharing of electrons between atoms of relatively similar electronegativities, and are most often too strong to disrupt by polar molecules of water. Therefore, covalent bonds and van der waals forces have an Intrinsic bond strength value that is independent of the environment.

However, either the partial negative oxygen atom or the partial positive hydrogen atoms in water molecules disrupt hydrogen or ionic bonds. Water is known to form hydrogen bonds with other polar or charged molecules, thus reducing the strength of interaction these molecules would normally have in the absence of water. Basically, these compounds with Hydrogen or Ionic bonds ionize, whether partially or fully in water, thereby leading to a decrease in bond strength in water.

QED!

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What is true about the electrolysis of water? Use the picture to choose 2 correct answers.
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Answer:

oxygen is produced at the anode and hydrogen gas is produced at the cathode

Explanation:

7 0
4 years ago
47.0ml of a HBr solution were titrated with 37.5ml of a 0.215M LiOH solution to reach the equivalence point. what is the molarit
faltersainse [42]

Hello!

The molarity of the HBr solution is 0,172 M.

Why?

The neutralization reaction between LiOH and HBr is the following:

HBr(aq) + LiOH(aq) → LiBr(aq) + H₂O(l)

To solve this exercise, we are going to apply the common titration equation:

M1*V1=M2*V2

M1=\frac{M2*V2}{V1}= \frac{0,215 M * 37,5 mL}{47 mL}=0,172 M

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4 0
4 years ago
28. How many formula units are found in 0.250 moles of potassium nitrate?
Soloha48 [4]

Answer:

\boxed {\boxed {\sf B. \ 1.5 *10^{23} \ formula \ units}}

Explanation:

1 mole of any substance contains the same number of particles. The particles can vary (atoms, molecules, formula units), but there are always 6.022*10²³ particles. In this case, the particles are formula units of potassium nitrate or KNO₃.

Let's create a ratio.

\frac {6.022*10^{23} \ formula \ units  \ KNO_3}{1 \ mol \ KNO_3}

Since we are trying to find the formula units in 0.250 moles, we multiply by that number.

0.250 \ mol \ KNO_3 *\frac {6.022*10^{23} \ formula \ units  \ KNO_3}{1 \ mol \ KNO_3}

The units of moles of potassium nitrate cancel.

0.250  *\frac {6.022*10^{23} \ formula \ units  \ KNO_3}{1 }

The denominator of 1 can be ignored, so we can make a simple multiplication problem.

0.250  *{6.022*10^{23} \ formula \ units  \ KNO_3}

1.5055 * 10^{23} \ formula \ units \ KNO_3

If we round to the nearest tenth, the 0 in the hundredth place tells us to leave the 5 in the tenth place.

1.5 *10^{23} \ formula \ units \ KNO_3

0.250 moles of potassium nitrate is approximately equal to 1.5*10²³ formula units of potassium nitrate and choice B is correct.

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3 years ago
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Answer:

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Explanation:

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