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seraphim [82]
3 years ago
11

All of the statements describe hydrogen bonds EXCEPT: a. Hydrogen bonds account for the anomalously high boiling point of water.

b. In liquid water, the average water molecule forms hydrogen bonds with two to three other water molecules. c. Individual hydrogen bonds are much weaker than covalent bonds. d. Individual hydrogen bonds in liquid water exist for many seconds and sometimes for minutes. e. The strength of a hydrogen bond depends on the linearity of the atoms involved in the bond.
Chemistry
1 answer:
Mamont248 [21]3 years ago
3 0

Answer:

The correct option is e

Explanation:

Hydrogen bond is an intermolecular interaction/bonding that are formed between an electronegative atom (such as nitrogen, oxygen and fluorine) and a hydrogen atom. They are weak intermolecular bonds compared to covalent bonds but account for the high boiling point of water because of the strong hydrogen bond presence between the water molecules. Water molecules form hydrogen bonds between each other; since an oxygen atom (in a water molecule) has two lone pairs on it's outermost shell, it forms an hydrogen bond with two hydrogen atoms of other water molecule. Due to the fluidity of liquid water molecules, hydrogen bonds keep getting broken (although recreated/formed almost immediately), hence, individual hydrogen bonds in liquid water does not exist for long.

In the explanation above, it was stated that the strength of the hydrogen bond in water is the reason for it's high boiling point. The atoms in a water molecule are bent NOT linear hence the strength of hydrogen bond does not depend on the linearity of the atoms involved in the bond.

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A certain reaction is exothermic in the forward direction. The reaction has more moles of gas on the product side. Which of the
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(c) Assume you have an equilibrium mixture of [A], [B], and [C] at 298K and that the
djyliett [7]

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P

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7. Add N2; add H2; decrease the container volume; heat the mixture.

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CO

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13. (a)  

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CO

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[

H

2

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[

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O

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3 0
3 years ago
Cal is titrating 50.8 mL of 0.319 M HBr with 0.337 M Ba(OH)2. How many mL of Ba(OH)2 does Cal need to add to reach the equivalen
alexgriva [62]

Answer:

V_{base}=24.04mL

Explanation:

Hello!

In this case, according to the chemical reaction by which HBr reacts with Ba(OH)2:

2HBr+Ba(OH)_2\rightarrow BaBr_2+2H_2O

We can see there is a 2:1 mole ratio between the acid and the base; thus, at the equivalent point we can write:

2M_{base}V_{base}=M_{acid}V_{acid}

Therefore, for is to compute the volume of the used base, we proceed as shown below:

V_{base}=\frac{M_{acid}V_{acid}}{2M_{base}}

And we plug in to obtain:

V_{base}=\frac{0.319M*50.8mL}{2*0.337M}\\\\V_{base}=24.04mL

Best regards!

8 0
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