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iragen [17]
3 years ago
8

Describe a way in which you use organic compounds in your home or work environment. choose a compound that is classified as an e

ther, aldehyde, ketone, ester, or alcohol. let's see how many different examples can be described.
Chemistry
2 answers:
Tatiana [17]3 years ago
5 0
<span>I use several different organic compounds at home. An example of a ketone would be acetone, which I use as nail polish remover. An example of an ester would be octyl acetate, which lends an orange fragrance to my perfume. An example of an alcohol would be isopropyl alcohol, commonly known as "rubbing alcohol", which I use to clean around my home.</span>
fgiga [73]3 years ago
3 0
(1) Alcohol is used all the time in our life:
   a) It may be used in body lotion industry.
   b) It is used in lipstick industry as glycerol is an alcohol and used in other cosmetics.
   c) It is used for fuel in some cars as the ethanol is an alcohol.
   d) The alcohol that you drink in usually ethyl alcohol CH3CH2OH
(2) Esters are used in more things in our life:
   a) Easters that are have scented odors are used in perfumes industry.
   b) It is used to make surfactants like detergents and soap.
   c) Phosphodiesters form the backbone of DNA molecules.
   d) The fatty acid esters of glycerol naturally occurring fats and oils.
(3) Aldehyde and ketone:
   a)  aldehyde is used in preserving and tanning and insecticide for plants and vegetables. 
   - The reaction between formaldehyde and the aromatic compound phenol form the plastic Bakelite.
   b) Formaldehydes used to form strong adhesives as it involved in some reactions (like used in plywood).
   c) Other aldehydes ketones are used as perfumes, plastics industry, pharmaceuticals, and dyes.
   d) Methadone is a ketone used to cure addiction to opiates.

You might be interested in
10. How many g of Cu(OH)2 can be made from 9.1 x 1025 atoms of O?
il63 [147K]
Molar mass Cu(OH)₂ = 97.561 g/mol

97.561 g Cu(OH)₂ --------------- 6.02x10²³ atoms
  ? g Cu(OH)₂ -------------------- 9.1x10²⁵ atoms

mass = 9.1x10²⁵ * 97.561 / 6.02x10²³

mass = 8.87x10²⁷ / 6.02x10²³

mass = 14734.2 g

hope this helps!
8 0
3 years ago
The standard free energy of formation, ΔG∘f, of a substance is the free energy change for the formation of one mole of the subst
OLEGan [10]

Answer:

B. 2 Na(s) + O₂(g) → Na₂O₂(s); ΔG∘f=−451.0 kJ/mol

D. 2 SO(g) + O₂(g) → 2 SO₂(g); ΔG°f=−600.4 kJ/mol

Explanation:

The spontaneity of a reaction  is given by the value of the standard Gibbs free energy of the reaction (ΔG°rxn). The more negative is the ΔG°rxn, the more spontaneous is a reaction.

The ΔG°rxn can be calculated using the following expression:

ΔG°rxn = ∑np × ΔG°f(products) − ∑nr × ΔG°f(reactants)

By definition, the standard Gibbs free energy of formation of simple substances in their most stable state is zero. That is why, in the reaction of formation of a compound ΔG°rxn = ΔG°f(product).

<em>Based on the standard free energies of formation, which of the following reactions represent a feasible way to synthesize the product? </em>

<em>     A. N₂(g) + H₂(g) → N₂H₄(g); ΔG°f=159.3 kJ/mol. </em>

<em>     </em>Not feasible. ΔG°rxn = ΔG°f(product) > 0.

    <em>B. 2 Na(s) + O₂(g) → Na₂O₂(s); ΔG°f=−451.0 kJ/mol</em>

    Feasible. ΔG°rxn = ΔG°f(product) < 0.

    <em>C. 2 C(s) + 2 H₂(g) → C₂H₄(g); ΔG°f=68.20 kJ/mol</em>

    Not feasible. ΔG°rxn = ΔG°f(product) > 0.

    <em>D. 2 SO(g) + O₂(g) → 2 SO₂(g); ΔG°f=−600.4 kJ/mol</em>

    Feasible. ΔG°rxn = ΔG°f(product) < 0.

3 0
3 years ago
H<img src="https://tex.z-dn.net/?f=H_%7B2%7D%20%2BO_%7B2%7D" id="TexFormula1" title="H_{2} +O_{2}" alt="H_{2} +O_{2}" align="abs
Snowcat [4.5K]

Explanation:

<h3 /><h2><em><u>H2 </u></em><em><u>+</u></em><em><u> </u></em><em><u>O2 </u></em><em><u>=</u></em><em><u> </u></em><em><u>H2O</u></em></h2>

<h2><em><u>Hydrogen</u></em><em><u> </u></em><em><u>+</u></em><em><u> </u></em><em><u>Oxygen</u></em><em><u> </u></em><em><u>=</u></em><em><u> </u></em><em><u>Water</u></em><em><u>.</u></em><em><u>.</u></em><em><u>.</u></em><em><u>.</u></em></h2>

<em><u>(~‾▿‾)~</u></em><em><u>(~‾▿‾)~</u></em><em><u>(~‾▿‾)~</u></em><em><u>(~‾▿‾)~</u></em><em><u>(~‾▿‾)~</u></em><em><u>(~‾▿‾)~</u></em><em><u>(~‾▿‾)~</u></em>

4 0
2 years ago
Which best describes the effect of j.j. thompsons theory
dangina [55]
 The accepted model of the atom was changed.
8 0
3 years ago
A sample of calcium oxide (CaO) has a mass of 2.80 g. The molar mass of CaO is 56.08 g/mol. How many moles of CaO does this samp
Molodets [167]

Answer is "0.05 mol".

<em>Explanation;</em>

We can do calculation by using a simple formula as

n = m/M

Where, n is the number of moles of the substance (mol), m is the mass of the substance (g) and M is the molar mass of the substance (g/mol).

Here,

n = ?

m = 2.80 g

M = 56.08 g/mol

By substitution,

n = 2.80 g /56.08 g/mol

n = 0.0499 mol ≈ 0.05 mol

4 0
3 years ago
Read 2 more answers
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