The classification of the alcohols gives;
- Compound 1 - Primary alcohol
- Compound 2 - Tertiary alcohol
- Compound 3 - Secondary alcohol
- Compound 4 - Secondary alcohol
<h3>What are alcohols?</h3>
Organic compounds occurs in families. The family of compounds is called a homologous series. The homologous series always have a functional group. The functional group is the atom, group of atoms or bond that is responsible for the chemical reactivity of the members of a given homologous series.
Now we know that the alcohols are those organic compounds that contains the -OH group. The could be aliphatic or alicyclic compounds. We shall now proceed to name the kind of alcohols that each of the compounds shown are;
- Compound 1 - Primary alcohol
- Compound 2 - Tertiary alcohol
- Compound 3 - Secondary alcohol
- Compound 4 - Secondary alcohol
Learn ore about alcohols:brainly.com/question/4698220
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Answer
is: 0.375 moles are present in 8.4 liters of nitrous oxide at stp.
V(N₂O) = 8.4 L.
V(N₂O) =
n(N₂O) · Vm.
Vm = 22,4 L/mol.<span>
n</span>(N₂O) = V(N₂O) ÷ Vm.
n(N₂O) = 8.4 L ÷ 22.4 L/mol.
n(N₂O) = 0.375 mol.<span>
Vm - molare volume on STP.</span>
Answer:
The percentage composition of a given compound is defined as <u>the ratio of the amount of each element to the total amount of individual elements present in the compound multiplied by 100. </u>Here, the quantity is measured in terms of grams of the elements present.
please give me brainliest
27,586
<h3>
Further explanation</h3>
<u>Given:</u>
A single gold atom has a diameter of 
From a reference, the Rutherford gold foil used in his scattering experiment had a thickness of approximately 
<u>Question:</u>
How many atoms thick were Rutherford's foil?
<u>The Process:</u>
Convert thickness from mm to cm.
The number of atoms is calculated from gold foil thickness divided by the atomic diameter.
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
