What is the reaction that joins thousands of small, identical molecules to form one very long molecule?

An unknown liquid is composed of 34.31% c, 5.28% h, and 60.41% i. The molecular weight is 210.06 amu. What is the molecular form

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In an unknown liquid, the percentage composition with respect to carbon, hydrogen and iodine is 34.31%, 5.28% and 60.41% respectively.

Let the mass of liquid be 100 g thus, mass of carbon, hydrogen and oxygen will be 34.31 g, 5.28 g and 60.41 g respectively.

To calculate molecular formula of compound, convert mass into number of moles as follows:

$n=\frac{m}{M}$

Molar mass of carbon, hydrogen and iodine is 12 g/mol, 1 g/mol and 126.90 g/mol.

Taking the ratio:

$C:H:I=n_{C}:n_{H}:n_{I}$

Putting the values,

$C:H:I=\frac{34.31 g}{12 g/mol}:\frac{5.28 g}{1 g/mol}:\frac{60.41 g}{126.90 g/mol}=6:11:1$

Thus, molecular formula of compound will be $C_{6}H_{11}I$ .

4 0

3 years ago

Question 4 of 10

Luba_88 [7]

Answer:

C. P = nRT

Explanation:

PV = nRT, where n is a number of moles and R is the universal gas constant, R = 8.31 J/mol ⋅ K.

Hope this helps :)

7 0

3 years ago

What observations can you make about this group of people? Name one inference you can make from your observations.

Stells [14]

Answer:

no no no who are these some look good but are black what is this

7 0

3 years ago

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Fill in the blanks. People can identify the _____ of an element by looking at its _____ on the periodic table

Monica [59]

Molecule, formulaaaaaaaaaa

7 0

3 years ago

If a single gold atom has a diameter of 2.9×x10−8 cm, how many atoms thick was rutherford's foil

ale4655 [162]

27,586

<h3>Further explanation</h3>

<u>Given:</u>

A single gold atom has a diameter of $\boxed{ \ 2.9 \times 10^{-8} \ cm. \ }$

From a reference, the Rutherford gold foil used in his scattering experiment had a thickness of approximately $\boxed{ \ 8 \times 10^{-3} \ mm. \ }$

<u>Question:</u>

How many atoms thick were Rutherford's foil?

<u>The Process:</u>

Convert thickness from mm to cm.

$\boxed{ \ 8 \times 10^{-3} \ mm = 8 \times 10^{-3} \times 10{-1} \ cm \ } \rightarrow \boxed{ \ 8 \times 10^{-4} \ cm \}$

The number of atoms is calculated from gold foil thickness divided by the atomic diameter.

$\boxed{ \ = \frac{8 \times 10^{-4} \ cm}{2.9 \times 10^{-8} \ cm} \ }$

$\boxed{ \ =2.7586 \times 10^4 \ atoms \ }$

Therefore, we get an atomic thickness of 27,586 atoms.

<u>Notes:</u>

In 1909-1910, Ernest Rutherford with two of his assistants, namely Hans Geiger and Ernest Marsden, conducted a series of experiments to find out more about the arrangement of atoms. They fired at a very thin gold plate with high-energy alpha particles.
One of their observations is that a small portion of alpha particles are reflected. This greatly surprised Rutherford. The reflected alpha particle must have hit something very dense in the atom. This fact is incompatible with the atomic model proposed by J.J. Thomson where the atoms are described as homogeneous in all parts with electrons and protons evenly distributed.
In 1911, Rutherford was able to explain the scattering of alpha rays by proposing ideas about atomic nuclei. According to him, most of the mass and positive charge of atoms are concentrated at the center of the atom, hereinafter referred to as the nucleus.

<h3>Learn more</h3>

The energy density of the stored energy brainly.com/question/9617400
The theoretical density of platinum which has the FCC crystal structure. brainly.com/question/5048216
Compound microscope brainly.com/question/4000241

Keywords: if a single gold atom, has a diameter of 2.9 x 10⁻⁸ cm, how many, atoms thick, Rutherford's foil, his scattering experiment

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