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GaryK [48]
2 years ago
13

a) (1 point) Build anthracene, optimize its geometry and examine its structure. Describe its shape. b) (1 point) Measure the C-C

bond lengths in anthracene. How do the bond lengths compare to typical C−C or C=C bond distances? c) (1 point) Build N-maleimide, optimize the geometry and examine the structure. Is the molecule flat?
Chemistry
1 answer:
oksano4ka [1.4K]2 years ago
8 0

Answer:

a) The structure of anthracene is planar with all the pi electrons delocalized in the structure to maintain aromaticity.

b) The C-C bond length in anthracene is about 140 pm with all the bond lengths being similar to each other.

The standard C-C bond length is 154 pm while standard C=C bond is about 134 pm. Therefore the bond length in anthracene is smaller than standard C-C bond length and longer than standard C=C bond length. This can be explained from the fact that the C-C bonds in anthracene has be mixed characteristics of single and double bond because of the delocalization of pi electrons over the whole structure. As a result, they are neither fully single nor fully double bond in nature. Hence the observed bond lengths.

c) This molecule is not flat. The N-atom is sp3 hybridized here and the H-atom attached to N will remain out of plane.

Explanation:

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A 20.0 -\,L volume of an ideal gas in a cylinder with a piston is at a pressure of 3.2 atm. Enough weight is suddenly removed fr
zzz [600]

Answer:

1. ΔE = 0 J

2. ΔH = 0 J

3. q = 3.2 × 10³ J

4. w = -3.2 × 10³ J

Explanation:

The change in the internal energy (ΔE) and the change in the enthalpy (ΔH) are functions of the temperature. If the temperature is constant, ΔE = 0 and ΔH = 0.

The gas initially occupies a volume V₁ = 20.0 L at P₁ = 3.2 atm. When the pressure changes to P₂ = 1.6 atm, we can find the volume V₂ using Boyle's law.

P₁ × V₁ = P₂ × V₂

3.2 atm × 20.0 L = 1.6 atm × V₂

V₂ = 40 L

The work (w) can be calculated using the following expression.

w = - P . ΔV

where,

P is the external pressure for which the process happened

ΔV is the change in the volume

w = -1.6 atm × (40L - 20.0L) = -32 atm.L × (101.325 J/1atm.L) = -3.2 × 10³ J

The change in the internal energy is:

ΔE = q + w

0 = q + w

q = - w = 3.2 × 10³ J

6 0
2 years ago
A sixth grade teacher takes students on a field trip to the beach. One student finds several pebbles that have a rounded shape a
Fittoniya [83]

Answer:

C: The shape of the pebbles is a result of weathering and deposition

Explanation:

For the several pebbles to have a rounded shape and smooth to the touch, it will undergo weathering and deposition. This is because weathering involves breaking down of rocks and creating new sediments. This weathering could be either chemical weathering or physical weathering where Chemical weathering is the decomposition of rocks which are caused by chemical reactions and which result in formation of new compound while physical weathering is the breakdown of rocks into smaller pieces. On the other hand, deposition occurs when the agents of erosion such as wind or water deposit sediments from one spot to another which in turn changes the shape of the land.

Thus, the shape of the pebbles are as a result weathering of the parent rocks and from deposition.

7 0
2 years ago
2.1. The lithium ion in a 250,00 mL sample of mineral water was
juin [17]

Answer:

11482 ppt of Li

Explanation:

The lithium is extracted by precipitation with B(C₆H₄)₄. That means moles of Lithium = Moles B(C₆H₄)₄. Now, 1 mole of B(C₆H₄)₄ produce the liberation of 4 moles of EDTA. The reaction of EDTA with Mg²⁺ is 1:1. Thus, mass of lithium ion is:

<em>Moles Mg²⁺:</em>

0.02964L * (0.05581mol / L) = 0.00165 moles Mg²⁺ = moles EDTA

<em>Moles B(C₆H₄)₄ = Moles Lithium:</em>

0.00165 moles EDTA * (1mol B(C₆H₄)₄ / 4mol EDTA) = 4.1355x10⁻⁴ mol B(C₆H₄)₄ = Moles Lithium

That means mass of lithium is (Molar mass Li=6.941g/mol):

4.1355x10⁻⁴ moles Lithium * (6.941g/mol) = 0.00287g. In μg:

0.00287g * (1000000μg / g) = 2870μg of Li

As ppt is μg of solute / Liter of solution, ppt of the solution is:

2870μg of Li / 0.250L =

<h3>11482 ppt of Li</h3>

4 0
2 years ago
What is the kinetic energy acquired by the electron in hydrogen atom, if it absorbs a light radiation of energy 1.08x101 J. (A)
Delvig [45]

Explanation:

The given data is as follows.

            Energy of radiation absorbed by the electron in hydrogen atom = 1.08 \times 10^{-17} J

As energy is absorbed as a photon. Hence, frequency will be calculated will be as follows.

                                    E = h \nu

               1.08 \times 10^{-17} J = 6.626 \times 10^{-34} Js \times \nu

               \nu = 0.163 \times 10^{17} s^{-1}

or,                \nu = 1.63 \times 10^{16} s^{-1}    

It is known that,        \nu = \frac{c}{\lambda}

                1.63 \times 10^{16} s^{-1} = \frac{3 \times 10^{8} m/s}{\lambda}                  

                   \lambda = 1.84 \times 10^{-8} m

And, according to De-Broglie equation \lambda = \frac{h}{p}

as,        p = m \times \nu

So,          \lambda = \frac{h}{m \times \nu}

            m \times \nu = \frac{6.626 \times 10^{-34} Js}{1.84 \times 10^{-8} m}          

                             = 3.6 \times 10^{-26} J/m

Now, on squaring both the sides we get the following.

           (m \times \nu)^{2} = (3.6 \times 10^{-26} J/m)^{2}    

                              = 12.96 \times 10^{-52}  

               m \times \nu^{2} = \frac{12.96 \times 10^{-52}}{m}

where,   m = mass of electron

So,           m \times \nu^{2} = \frac{12.96 \times 10^{-52}}{m}

                             = \frac{12.96 \times 10^{-52}}{9.1 \times 10^{-31}}

                                   = 1.42 \times 10^{-21} J

Since,  K.E = \frac{1}{2}m \nu^{2}

                 = \frac{1.42 \times 10^{-21} J}{2}

                 = 0.71 \times 10^{-21} J

Thus, we can conclude that kinetic energy acquired by the electron in hydrogen atom is 7.1 \times 10^{-22} J.

4 0
2 years ago
A tanker truck carrying 6.05×103 kg of concentrated sulfuric acid solution tips over and spills its load. The sulfuric acid solu
irinina [24]

Answer:

6,216.684 kilograms of sodium carbonate must be added to neutralize 6.05\times 10^3 kg of sulfuric acid solution.

Explanation:

Mass of sulfuric acid solution = 6.05\times 10^3 kg=6.05\times 10^6 g

1 kg = 10^3 g

Percentage mass of sulfuric acid = 95.0%

Mass of sulfuric acid = \frac{95.0}{100}\times 6.05\times 10^6 g

=5,747,500 g

Moles of sulfuric acid = \frac{5,747,500 g}{98 g/mol}=58,647.96 mol

H_2SO_4+Na_2CO_3\rightarrow Na_2SO_4+CO_2+H_2O

According to reaction , 1 mole of sulfuric acid is neutralized by 1 mole of sodium carbonate.

Then 58,647.96 moles of sulfuric acisd will be neutralized by :

\frac{1}{1}\times 58,647.96 mol=58,647.96 mol of sodium carbonate

Mass of 58,647.96 moles of sodium carbonate :

106 g/mol\times 58,647.96 mol=6,216,683.76 g

6,216,683.76 g = 6,216,683.76 × 0.001 kg = 6,216.684 kg

6,216.684 kilograms of sodium carbonate must be added to neutralize 6.05\times 10^3 kg of sulfuric acid solution.

3 0
2 years ago
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