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zysi [14]
4 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]4 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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A chunk of zinc is added to a solution of gold (III) nitrate to extract the gold. The reaction forms,
alina1380 [7]

Answer:

it forms :

1. Gold ( Au )

2. Zinc nitrate ( Zn(NO3)2 )

Explanation:

When a chunk of zinc is added to a solution of gold (III) nitrate to extract the gold. The reaction forms Gold and Zinc nitrate .

it's a single displacement reaction,

here's the balanced equation for above reaction :

3 Zn + 2 Au(NO3)3 =》3 Zn(NO3)2 + 2 Au

7 0
3 years ago
Read 2 more answers
A balloon occupies 1.50 L with 0.205 mol of carbon dioxide. How many moles would be required to increase the size of the balloon
Gekata [30.6K]

Answer:

0.683 moles of the gas are required

Explanation:

Avogadro's law relates the moles of a gas with its volume. The volume of a gas is directely proportional to its moles when temperature and pressure of the gas remains constant. The law is:

V₁n₂ = V₂n₁

<em>Where V is volume and n are moles of 1, initial state and 2, final state of the gas.</em>

<em />

Computing the values of the problem:

1.50Ln₂ = 5L*0.205mol

n₂ = 0.683 moles of the gas are required

<em />

8 0
3 years ago
Is Sr3(PO4)2 ionic or covalent
lana66690 [7]
Sr3(PO4)2 is definitely Ionic
6 0
3 years ago
How many parts per million of lead is found in 250 ml of water if there is 1.30g of lead in the water? 1ml=1g
SSSSS [86.1K]

Answer:

5200 ppm

Explanation:

As per the definition, parts per million of a contaminant is a measure of the amount of mass of contaminant present per million amount of the solution. It is denoted by ppm.

Given in the question,

Water = 250 ml = 250 g

Lead = 1.30 g

So,

ppm of Lead = \frac{Lead}{Water} \times 10^6 = \frac{1.30}{250} \times 10^6 = 5200 ppm

So, as calculated above, there is 5200 ppm of lead present in 250 ml of water.

6 0
3 years ago
A sample of argon gas has a volume of 795 mL at a pres-sure of 1.20 atm and a temperature of 116 ∘C. What is the final volume of
jek_recluse [69]

<u>Answer:</u> The volume when the pressure and temperature has changed is 1.6\times 10^2mL

<u>Explanation:</u>

To calculate the volume when temperature and pressure has changed, we use the equation given by combined gas law.

The equation follows:

\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}

where,

P_1,V_1\text{ and }T_1 are the initial pressure, volume and temperature of the gas

P_2,V_2\text{ and }T_2 are the final pressure, volume and temperature of the gas

Let us assume:

P_1=1.20atm\\V_1=795mL\\T_1=116^oC=[116+273]K=389K\\P_2=0.55atm\\V_2=?mL\\T_2=75^oC=[75+273]K=348K

Putting values in above equation, we get:

\frac{1.20atm\times 795mL}{389K}=\frac{0.55atm\times V_2}{348K}\\\\V_2=\frac{1.20\times 795\times 348}{0.55\times 389}=1.6\times 10^3mL

Hence, the volume when the pressure and temperature has changed is 1.6\times 10^2mL

5 0
3 years ago
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