Answer:
Coconut oil, Olive oil and Sunflower oil
Explanation:
Fatty acids are carboxylic acids with a long unbranched chain of carbon and hydrogen atoms.
There are three main classes of fatty acids which are explained as under:
1. Saturated Fatty acids: These fatty acids have long carbon chain with two hydrogen atoms bonded to each carbon atom. This saturation of fatty acids make the fatty acids more stable towards high temperature. These fatty acids becomes solid at room temperature. Coconut oil and butter are the examples of saturated fatty acids.
2. Monounsaturated Fatty Acids: In a long carbon chain, if there is a carbon atom which is double bonded with another carbon atom and rest is saturated with hydrogen atoms, because of this single double-bond, the fatty acid is termed as monounsaturated fatty acids. These fatty acids are liquid at room temperature but solidify in refrigerator. Olive oil is an example of such fatty acids.
3. Polyunsaturated Fatty Acids: In a long carbon chain, if there are two or more than two carbon atoms which are double bonded with each other and rest is saturated with hydrogen atoms, because of multiple double bonds, such fatty acids are termed as polyunsaturated fatty acids. Because of higher unsaturation, these fatty acids are liquid in both normal room temperature and in refrigerator. Such unsaturation also make them unfit for cooking purposes. Sunflower oil, Soyabean oil and Flaxseed oil are examples of polyunsaturated fatty acids.
Answer:
C
Explanation:
garbage being reduced is a far more long term other than the other stupid answer choices
Explanation:
An enzyme inhibitor is a molecule that binds to an enzyme and decreases its activity. ... Since blocking an enzyme's activity can kill a pathogen or correct a metabolic imbalance, many drugs are enzyme inhibitors. They are also used in pesticides.
Answer:
Coefficient = 1.58
Exponent = - 5
Explanation:
pH = 2.95
Molar concentration = 0.0796M
Ka = [H+]^2 / [HA]
Ka = [H+]^2 / 0.0796
Therefore ;
[H+] = 10^-2.95
[H+] = 0.0011220 = 1.122 × 10^-3
Ka = [H+] / molar concentration
Ka = [1.122 × 10^-3]^2 / 0.0796
Ka = (1.258884 × 10^-6) / 0.0796
Ka = 15.815 × 10^-6
Ka = 1.58 × 10^-5
Coefficient = 1.58
Exponent = - 5
Explanation:
1.
Cu(NO3)2 + 2NaCl(aq) --> CuCl2(aq) + 2NaNO3(aq)
2.
Cu(NO3)2 + 2NaOH(aq) --> Cu(OH)2(s) + 2NaNO3(aq)
A light blue precipitate of Cu(OH)2 is formed and NaNO3 in solution.
3.
Cu(NO3)2(aq) --> Cu2+(aq) + 2NO3^-2(aq)
2NaOH(aq) --> 2Na+(aq) + 2OH-(aq)
Cu2+(aq) + 2OH-(aq) --> Cu(OH)2(aq)
2Na+(aq) + 2NO3^-2(aq) --> 2NaNO3(aq)
4.
The reaction in both Questions 1 and 2 is called Double displacement reaction. A double-replacement reaction exchanges the cations and/or or the anions of two ionic compounds. A precipitation reaction is a double-replacement reaction in which one product is a solid precipitate (precipitated) while the other in solution.
Since the cation and anions in Qustion 1 were exchanged, the same was done for Question 2, hence the identity of the precipitate in Question 2 was got.
