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vovangra [49]
4 years ago
10

) palmitic acid is a 16 carbon acid. in a balanced equation, the products of the saponification of glyceryl tripalmitate (tripal

mitin) are
Chemistry
1 answer:
lesya [120]4 years ago
8 0

Answer: The products of the saponification of glyceryl tripalmitate (tripalmitin) are one molecule of glycerol and three molecules of sodium salt of palmitic acid.

Explanation:

A chemical reaction in which triglycerides react with sodium hydroxide and leads to the formation of one molecule of glycerol and three molecules of a salt of fatty acid is known as saponification.

For example, when tripalmitin reacts with sodium hydroxide then it leads to the formation of one molecule of glycerol and three molecules of sodium salt of palmitic acid.

The reaction equation is as follows.

   C_{51}H_{98}O_{6} + NaOH \rightarrow C_{3}H_{8}O_{3} + 3C_{16}H_{31}O_{2}Na^{+}

Thus, we can conclude that the products of the saponification of glyceryl tripalmitate (tripalmitin) are one molecule of glycerol and three molecules of sodium salt of palmitic acid.

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A 1.3 g sample of a substance is heated from 0°C to 45°C and is found to have absorbed 45 j of heat. What is the specific heat o
wlad13 [49]

Q=mcat

45=(1.3)(c)(45)

45=58.5c

.769=c

~.77

The answer is .77

5 0
3 years ago
Determine the mole fractions and partial pressures of CO2, CH4, and He in a sample of gas that contains 1.20 moles of CO2, 1.79
BARSIC [14]

Answer :  The mole fraction and partial pressure of CH_4,CO_2 and He gases are, 0.267, 0.179, 0.554 and 1.54, 1.03 and 3.20 atm respectively.

Explanation : Given,

Moles of CH_4 = 1.79 mole

Moles of CO_2 = 1.20 mole

Moles of He = 3.71 mole

Now we have to calculate the mole fraction of CH_4,CO_2 and He gases.

\text{Mole fraction of }CH_4=\frac{\text{Moles of }CH_4}{\text{Moles of }CH_4+\text{Moles of }CO_2+\text{Moles of }He}

\text{Mole fraction of }CH_4=\frac{1.79}{1.79+1.20+3.71}=0.267

and,

\text{Mole fraction of }CO_2=\frac{\text{Moles of }CO_2}{\text{Moles of }CH_4+\text{Moles of }CO_2+\text{Moles of }He}

\text{Mole fraction of }CO_2=\frac{1.20}{1.79+1.20+3.71}=0.179

and,

\text{Mole fraction of }He=\frac{\text{Moles of }He}{\text{Moles of }CH_4+\text{Moles of }CO_2+\text{Moles of }He}

\text{Mole fraction of }He=\frac{3.71}{1.79+1.20+3.71}=0.554

Thus, the mole fraction of CH_4,CO_2 and He gases are, 0.267, 0.179 and 0.554 respectively.

Now we have to calculate the partial pressure of CH_4,CO_2 and He gases.

According to the Raoult's law,

p_i=X_i\times p_T

where,

p_i = partial pressure of gas

p_T = total pressure of gas  = 5.78 atm

X_i = mole fraction of gas

p_{CH_4}=X_{CH_4}\times p_T

p_{CH_4}=0.267\times 5.78atm=1.54atm

and,

p_{CO_2}=X_{CO_2}\times p_T

p_{CO_2}=0.179\times 5.78atm=1.03atm

and,

p_{He}=X_{He}\times p_T

p_{He}=0.554\times 5.78atm=3.20atm

Thus, the partial pressure of CH_4,CO_2 and He gases are, 1.54, 1.03 and 3.20 atm respectively.

4 0
3 years ago
Science plz help due soon 6th grade Major grade
pentagon [3]

Answer:

the answer is d.

Explanation:

because the core is deepest then is the mantle and the lithpsphere is last .

I put a picture incase you need a visual example.

5 0
3 years ago
What type of mixture is a sandstorm​
sveta [45]

Answer:homogeneous mixture

A sandstorm is a homogeneous mixture.

Explanation:

6 0
3 years ago
Read 2 more answers
Convert 0.95 kilograms into centigrams. (1 g = 100 cg; 1 kg = 1000 g)
svet-max [94.6K]

Answer:

95,000 centigrams

Explanation:

There is 1000 CG in 0.01 kilograms

so you do 1000*95 which equals 95,000 centigrams.

6 0
3 years ago
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