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zysi [14]
3 years ago
12

Fatty acids are carboxylic acids with a long, unbranched hydrocarbon chain. There are three main classes of fatty acids. Classif

y the fatty acids as saturated, monounsaturated, or polyunsaturated.
Chemistry
1 answer:
LuckyWell [14K]3 years ago
8 0

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.

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How to solve x² in differential​
g100num [7]

Answer:

x² = mutiphy by them self

Explanation:

4 0
3 years ago
1. Pewien tlenek azotu o masie cząsteczkowej 108 u zawiera 74,07% tlenu. Wykonaj stosowne obliczenia i napisz wzór sumaryczny te
love history [14]

Answer:

1. Stąd empiryczny wzór substancji to N₂O₅

2. W związku z tym ilość w gramach chlorku sodu NaCl is 114,4 g.

Explanation:

1. Mamy tutaj;

Masa molowa tlenku azotu = 108u

Masa azotu = 14,0067u

Masa tlenu = 15,999 u

74,07% masy tlenku azotu to tlen

Dlatego masa obecnego tlenu = 108 × 74,07 / 100 = 79,9956 u

Masa obecnego azotu = 108 - 79,9956 = 28,0044u

Liczba moli tlenu = 79,9956 / 15,999 = 5,00003 ≈ 5

Liczba moli azotu = 28,0044 / 14,0067 = 1,99935 ≈ 2

Stąd empiryczny wzór substancji to N₂O₅.

2. Kiedy sód reaguje z chlorem, mamy;

2Na (s) + Cl₂ (g) → 2NaCl (s)

Dlatego 2 mole sodu Na reaguje z 1 molem chloru gazowego Cl₂, z wytworzeniem 2 moli chlorku sodu NaCl

W związku z tym 1 mol sodu Na reaguje z 1/2 molem chloru gazowego Cl₂ z wytworzeniem 1 mola chlorku sodu NaCl

Masa Na obecnego w reakcji = 45 g

Masa molowa sodu = 22,989769u

Liczbę moli sodu w 45 g sodu podano w następujący sposób;

Liczba \, \, moli \, \, Na= \frac{Mass \, of \, Na}{Molowy \, masa \, z \, Na} = \frac{45}{22.989769} = 1.96 \, mole

Z czego 1,96 moli sodu Na reaguje z 1/2 × 1,96 mola chloru gazowego Cl₂ z wytworzeniem 1,96 mola chlorku sodu NaCl

Masa molowa NaCl = 58,44 g / mol

Dlatego masa NaCl = liczba moli NaCl × masa molowa NaCl

Masa NaCl = 1,96 × 58,44 = 114,39001 g ≈ 114,4 g

W związku z tym ilość w gramach chlorku sodu NaCl = 114,4 g.

4 0
3 years ago
Does mint actually cool things down
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The cooling of a mint is just a sensation if you were to put mints into warm water the water would remain warm because it is just a cooling sensation. I hope this helps:)
3 0
3 years ago
Enthalpy of combustion of ethyl alcohol, C2 H5 OH, is – 950 kJ mol–1. How much heat is evolved when one gram of ethyl alcohol bu
melomori [17]
<h3>Answer:</h3>

20.62 Kilo-joules

<h3>Explanation:</h3>
  • The Enthalpy of combustion of ethyl alcohol is -950 kJ/mol.
  • This means that 1 mole of ethyl alcohol evolves a quantity of heat of 950 Joules when burned.

Molar mass of ethyl ethanol = 46.08 g/mol

Therefore;

46.08 g of  C₂H₅OH evolves heat equivalent to 950 kilojoules

We can calculate the amount of heat evolved by 1 g of C₂H₅OH

Heat evolved by 1 g of C₂H₅OH  = Molar enthalpy of combustion ÷ Molar mass

                                      = 950 kJ/mol ÷ 46.08 g/mol

                                      = 20.62 Kj/g

Therefore, a gram of C₂H₅OH  will evolve 20.62 kilo-joules of heat

7 0
3 years ago
For this reaction: 4 Al + 3O2 = 2 Al2O3
MariettaO [177]
0.347 mols, working out shown on photo

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