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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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Determine molar mass! help please
bezimeni [28]

Answer:

the molar mass is the mass of a given chemical element or chemical compound divided by the amount of substance

8 0
3 years ago
Three chemistry students measured the length of a copper bar. The recorded lengths were 5.05 cm, 5 cm , and 5.1 cm, What is the
Salsk061 [2.6K]
5.05 + 5 + 5.1 = 15.15cm Then you just divide it by the amount of measurements you had like this:15.15 ÷ 3 = 5.04999971cm Then you can just round it to the 3rd figure: 5.05cm < And that's the mean/average length of the bar. :) (Or the one above if you want all of the decimals too) 
3 0
4 years ago
A gas with a volume of 4.0 L at a pressure of 2.02 atm is allowed to expand to a volume of 12.0
Maurinko [17]
<h3>Answer:</h3>

P₂ = 0.67 atm

<h3>General Formulas and Concepts:</h3>

<u>Math</u>

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

  1. Brackets
  2. Parenthesis
  3. Exponents
  4. Multiplication
  5. Division
  6. Addition
  7. Subtraction
  • Left to Right  

Equality Properties

  • Multiplication Property of Equality
  • Division Property of Equality
  • Addition Property of Equality
  • Subtraction Property of Equality<u> </u>

<u>Chemistry</u>

<u>Gas Laws</u>

Boyle's Law: P₁V₁ = P₂V₂

  • P₁ is pressure 1
  • V₁ is volume 1
  • P₂ is pressure 2
  • V₂ is volume 2
<h3>Explanation:</h3>

<u>Step 1: Define</u>

[Given] P₁ = 2.02 atm

[Given] V₁ = 4.0 L

[Given] V₂ = 12.0 L

[Solve] P₂

<u>Step 2: Solve</u>

  1. Substitute in variables [Boyle's Law]:                                                              (2.02 atm)(4.0 L) = P₂(12.0 L)
  2. [Pressure] Multiply:                                                                                           8.08 atm · L = P₂(12.0 L)
  3. [Pressure] [Division Property of Equality] Isolate unknown:                          0.673333 atm = P₂
  4. [Pressure] Rewrite:                                                                                           P₂ = 0.673333 atm

<u>Step 3: Check</u>

<em>Follow sig fig rules and round. We are given 2 sig figs as our smallest.</em>

0.673333 atm ≈ 0.67 atm

4 0
3 years ago
Water is poured into a conical container at the rate of 10 cm3/sec. The cone points directly down, and it has a height of 20 cm
8090 [49]

Answer:

\frac{dh}{dt}_{h=2cm} =\frac{40}{9\pi}\frac{cm}{2}

Explanation:

Hello,

The suitable differential equation for this case is:

\frac{dV}{dt}=10\frac{cm^3}{s}

As we're looking for the change in height with respect to the time, we need a relationship to achieve such as:

\frac{dh}{dt} = ?*\frac{dV}{dt}

Of course, ?=\frac{dh}{dV}.

Now, since the volume of a cone is V=\pi r^2h/3 and the ratio r/h=15/20=3/4 or r=3/4h, the volume becomes:

V=\pi (\frac{3}{4} h)^2h/3= \frac{3}{16}\pi h^3

We proceed to its differentiation:

\frac{dV}{dh} =\frac{9}{16} \pi h^2\\\frac{dh}{dV} =\frac{16}{9 \pi h^2}

Then, we compute \frac{dh}{dt}

\frac{dh}{dt} = \frac{16}{9 \pi h^2}*\frac{dV}{dt}\\\frac{dh}{dt} = \frac{16}{9\pi h^2}*10\frac{cm^3}{s} =\frac{160}{9 \pi h^2}

Finally, at h=2:

\frac{dh}{dt}_{h=2cm} =\frac{160}{9\pi 2^2}\\\frac{dh}{dt}_{h=2cm} =\frac{40}{9\pi}\frac{cm}{s}

Best regards.

4 0
3 years ago
As temperature increase and the volume stays constant, how will the pressure change
crimeas [40]

Answer:

There will be more collisions and so a greater pressure. The number of particles is proportional to pressure, if the volume of the container and the temperature remain constant. ... Volume is inversely proportional to pressure, if the number of particles and the temperature are constant.

5 0
2 years ago
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