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Butoxors [25]
2 years ago
7

Draw a triacylglycerol (triglyceride) made from glycerol, myristic acid [CH3(CH2)12COOH], palmitic acid [CH3(CH2)14COOH], and ol

eic acid [CH3(CH2)7CH=CH(CH2)7COOH(cis)]. Place palmitic acid at the middle position. Cis/trans isometry is graded

Chemistry
1 answer:
Sergio039 [100]2 years ago
7 0

Answer:

The triglyceride made from glycerol, myristic acid, palmitic acid and cis-oleic acid is shown in the attached pictures.

Explanation:

A triglyceride or triacylglyceride is an ester derived from glycerol and three fatty acids. The fatty acids of this exercise are myristic acid, palmitic acid and oleic acid.

A fatty acid is a biomolecule of lipid nature formed by a long linear hydrocarbon chain, of different length or number of carbon atoms, at the end of which is a carboxyl group.

The myristic acid has a carbon chain of 14 carbon atoms with a carboxyl group at one end.

Palmitic acid has a carbon chain of 16 carbon atoms with a carboxyl group at one end.

Oleic acid has a carbon chain of 18 carbon atoms with a carboxyl group at one end and a double bond at position 9.

Triglyceride is formed by the esterification of the glycerol oxidrile group and the carboxyl acid group.

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The density of phosphorus vapor at 310 degrees celcius and 775 mmHg is 2.64g/L. what is the molecular formula of the phosphorus
lidiya [134]

Answer:

The molecular formula of the phosphorus is P4

Explanation:

<u>Step 1:</u> Data given

Density of phosphorus vapor at 310 °C and 775 mmHg = 2.64g /L

<u>Step 2: </u>Calculate the molecular weight

We assume phosphorus to be an ideal gas

So p*V = n*R*T

 ⇒ with p = the pressure of phosphorus = 775 mmHg

⇒ with V = the Volume

⇒ with n = the number of moles = mass/molecular weight

⇒ with R = ideal gas constant  = 0.08206 L*atm/K*mol

⇒ with T = the absolute temperature

p*V = m/MW *R*T

MW = mRT/PV

 ⇒ Since the volume is unknown but can be written as density = mass/volume

MW = dRT/P

MW = (2.64g/L * 0.08206 L*atm/K*mol * 583 Kelvin)/1.0197 atm

MW = 123.86 g/mol

<u>Step 3</u>: Calculate molecular formula of phosphorus

The relative atomic mass of phosphorus = 30.97 u

123.86 / 30.97 = 4

The molecular formula of the phosphorus is P4

5 0
2 years ago
The question is on the pic, thanks :)
Inessa05 [86]
It’s the BOA not the dog or kangaroo
8 0
1 year ago
Calculate the cell potential E at 25°C for the reaction 2 Al(s) + 3 Fe2+(aq) → 2 Al3+(aq) + 3 Fe(s) given that [Fe 2+] = 0.020 M
Elodia [21]

Answer:

1.18 V

Explanation:

The given cell is:

Al(s)/Al^{3+}(0.10M)||Fe^{2+}(0.020M)/Fe(s)

Half reactions for the given cell follows:

Oxidation half reaction: Al(s)\rightarrow Al^{3+}(0.10M)+2e^-;E^o_{Al^{3+}/Al}=-1.66V

Reduction half reaction: Fe^{2+}(0.020M)+2e^-\rightarrow Fe(s);E^o_{Fe^{2+}/Fe}=-0.45V

Multiply Oxidation half reaction by 2 and Reduction half reaction by 3

Net reaction: 2Al(s)+3Fe^{2+}(0.020M)\rightarrow 2Al^{3+}(0.10M)+3Fe(s)

Oxidation reaction occurs at anode and reduction reaction occurs at cathode.

To calculate the E^o_{cell} of the reaction, we use the equation:

E^o_{cell}=E^o_{cathode}-E^o_{anode}

Putting values in above equation, we get:

E^o_{cell}=-0.45-(-1.66)=1.21V

To calculate the EMF of the cell, we use the Nernst equation, which is:

E_{cell}=E^o_{cell}-\frac{0.059}{n}\log \frac{[Al^{3+}]^2}{[Fe^{2+}]^3}

where,

E_{cell} = electrode potential of the cell = ?V

E^o_{cell} = standard electrode potential of the cell = +1.21 V

n = number of electrons exchanged = 6

Putting values in above equation, we get:

E_{cell}=1.21-\frac{0.059}{6}\times \log(\frac{0.10^2}{0.020^3})\\\\E_{cell}=1.18V

5 0
2 years ago
A student is given a sample of CuSO4(s) that contains a solid impurity that is soluble and colorless. The student wants to deter
DochEvi [55]

Answer:

The impurity which is present in the solution of copper sulphate (CuSO4) is determined by the an instrument known as spectrophotometer.

Explanation:

Spectrophotometer is a device or an instrument which is used to determine the concentration of a chemical by measuring the detection of light intensity that is coming from the solution. If the solution of copper sulphate is checked through spectrophotometer, we can can determined or measure the amount of copper sulphate and the impurity in the solution.

3 0
2 years ago
Element X is a radioactive isotope such that every 82 years, its mass decreases by half. Given that the initial mass of a sample
lesantik [10]

Answer: 17 years

Explanation:

Expression for rate law for first order kinetics  for radioactive substance is given by:

t=\frac{2.303}{k}\log\frac{a}{a-x}

where,

k = rate constant

t = age of sample

a = let initial amount of the reactant

a - x = amount left after decay process  

a) for completion of half life:

Half life is the amount of time taken by a radioactive material to decay to half of its original value.

t_{\frac{1}{2}}=\frac{0.693}{k}

k=\frac{0.693}{82years}=8.4\times 10^{-3}years^{-1}

b) for 8900 g of the mass of the sample to reach  7700 grams

t=\frac{2.303}{8.4\times 10^{-3}}\log\frac{8900}{7700}

t=17years

Thus it will take 17 years

4 0
2 years ago
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