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Nat2105 [25]
3 years ago
9

Estimate the surface-to-volume ratio of a C60 fullerene by treating the molecule as a hollow sphere and using 77pm for the atomi

c radius of carbon.
Chemistry
1 answer:
AleksAgata [21]3 years ago
7 0

Answer:

The surface-to-volume ratio of a C-60 fullerene is 3:77.

Explanation:

Surface area of sphere = S=4\pi r^2

Volume of the sphere = V=\frac{4}{3}\pi r^3

where : r  = radius of the sphere

Radius of the C-60 fullerene sphere = r = 77 pm

Surface area of the C-60 fullerene = S=4\pi (77 pm)^2...[1]

Volume area of the C-60 fullerene = V=\frac{4}{3}\pi (77 pm)^3..[2]

Dividing [1] by [2]:

\frac{S}{V}=\frac{4\pi (77 pm)^2}{\frac{4}{3}\pi (77 pm)^3}

=\frac{3}{77}

The surface-to-volume ratio of a C-60 fullerene is 3:77.

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In Haber’s process, 30 moles of hydrogen and 30 moles of nitrogen react to make ammonia. If the yield of the product is 50%, wha
devlian [24]

Answer:

Therewill be produced 170.6 grams NH3, there will remain 25 moles of N2, this is 700 grams

Explanation:

<u>Step 1:</u> Data given

Number of moles hydrogen = 30 moles

Number of moles nitrogen = 30 moles

Yield = 50 %

Molar mass of N2 = 28 g/mol

Molar mass of H2 = 2.02 g/mol

Molar mass of NH3 = 17.03 g/mol

<u>Step 2:</u> The balanced equation

N2 + 3H2 → 2NH3

<u>Step 3:</u> Calculate limiting reactant

For 1 mol of N2, we need 3 moles of H2 to produce 2 moles of NH3

Hydrogen is the limiting reactant.

The 30 moles will be completely be consumed.

N2 is in excess. There will react 30/3 =10 moles

There will remain 30 -10 = 20 moles (this in the case of a 100% yield)

In a 50 % yield, there will remain 20 + 0,5*10 = 25 moles. there will react 5 moles.

<u>Step 4:</u> Calculate moles of NH3

There will be produced, 30/ (3/2) = 20 moles of NH3 (In case of 100% yield)

For a 50% yield there will be produced, 10 moles of NH3

<u>Step 5</u>: Calculate the mass of NH3

Mass of NH3 = mol NH3 * Molar mass NH3

Mass of NH3 = 20 moles * 17.03

Mass of NH3 = 340.6 grams = theoretical yield ( 100% yield)

<u>Step 6: </u>Calculate actual mass

50% yield = actual mass / theoretical mass

actual mass = 0.5 * 340.6

actual mass = 170.3 grams

<u>Step 7:</u> The mass of nitrogen remaining

There remain 20 moles of nitrogen + 50% of 10 moles = 25 moles remain

Mass of nitrogen = 25 moles * 28 g/mol

Mass of nitrogen = 700 grams

6 0
4 years ago
Read 2 more answers
Examining the equations or equilibrium constants related to a base, salt, or an acid is an indirect way to determine strength of
Gwar [14]

Answer:

1 - Weak electrolyte

2- Non electrolyte

3- Weak electrolyte

4- Strong electrolyte

Explanation:

A strong electrolyte refers to an electrolyte that decomposes completely in solution. This means that there are more charge carriers in solution when a strong electrolyte is dissolved in water. A strong electrolyte produces a strong glow. LiOH is a strong electrolyte.

A weak electrolyte is not completely dissociated in water. Only a small amount dissociates in water. HF is a weak electrolyte. A weak electrolyte does not produce a bright light.

A non-electrolyte does not dissociate in solution at all hence it does not power a bulb E.g  C12H22O11.

7 0
3 years ago
How much water needs to be added to 100 mL of a 20%w solution of copper (II) sulfate to prepare a 0.01-M solution?
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Answer:

12430mL of water must be added

Explanation:

To solve this question we need to convert the 20%w of CuSO₄ to molarity. Then, using the <em>diulution factor </em>we can find the amount of water required:

20g CuSO₄ / 100mL * (1mol / 159.609g CuSO₄) = 0.1253 moles / 100mL = 0.1253mol / 0.1L =

1.25M is the concentration of CuSO₄. To dilute this concentration to 0.01M, the dilution factor must be of:

1.25M / 0.01M = 125 times must be diluted the solution.

As the volume of the concentrated solution is 100mL, the total volume of the solution to have a 0.01M solution must be of:

100mL * 125 times = 12530mL is the final volume of the solution. That means the amount of water added must be of:

12530mL - 100mL =

<h3>12430mL of water must be added</h3>
8 0
3 years ago
Triphenylmethane can be prepared by reaction of benzene and chloroform in the presence of AlCl3. Draw curved arrows to show the
never [62]

Answer:

Reaction follows frieldel-craft alkylation mechanism

Explanation:

  • Reaction between benzene and chloroform in the presence of AlCl_{3} follows friedel-craft alkylation of benzene
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  • Remaining two equivalent of Cl in chloroform are similarly replaced by two equivalent of benzene to produce triphenylmethane
  • Reaction mechanism has been shown below

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3 years ago
Heat that flows by conduction is the transfer of thermal energy between substances in contact. What must occur for this to happe
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Answer:

c. one must have a higher kinetic energy than the other system

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