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zepelin [54]
3 years ago
13

A common structural feature of membrane lipids is their amphipathic nature. For example, in phosphatidylcholine, the two fatty a

cid chains are hydrophobic and the phosphocholine head group is hydrophilic. For each of the following membrane lipids, name the components that serve as the hydrophobic and hydrophilic units: (a) phosphatidylethanolamine; (b) sphingomyelin; (c) galactosylcerebroside; (d) ganglioside; (e) cholestrol.
Engineering
1 answer:
BabaBlast [244]3 years ago
6 0

Answer:

A. Phosphatidylethanolamine:

phosphoethanolamine <em>(hydrophilic)</em> 2 fatty acids <em>(hydrophobic )</em>

<em />

B. Sphingomyelin:

phosphocholine <em>(hydrophilic)</em> ceramide (sphingosine + 1 fatty acid) <em>(hydrophobic )</em>

<em />

C. Galactosylcerebroside:

D-galactose <em>(hydrophilic)</em> ceramide <em>(hydrophobic )</em>

<em />

D. Ganglioside:

oligosaccharide <em>(hydrophilic)</em> ceramide <em>(hydrophobic )</em>

<em />

E. Cholesterol:

-OH group <em>(hydrophilic)</em> hydrocarbon ring skeleton <em>(hydrophobic )</em>

<em />

Explanation:

The amphipathic nature of lipids shows  that all lipids have a <em>hydrophilic </em>(water-loving or polar end) and a <em>hydrophobic</em><em>  </em>(water fearing or non-polar end.)

<em />

The various lipids in question, and their ampiphatic groups are detailed above.

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3 years ago
Identify the true statements about the relationship between the distillate and the sample mixture in a simple distillation.
fenix001 [56]

Answer:

The answer is "Choice b,c, and d".

Explanation:

The distillation would be the separation phase of working liquids that boil close to each other. Whenever the formulation is long-term, a much more fraction of a compound collected will have a lower boiling point than it would otherwise be. In other phrases, it'll be pulverized with both the lower level, also with the boiling point. Its structure of both the diluted volatile component may therefore include another cell's contaminants.

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When you look at a drawing, how do you know if you are looking at US
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is it glizzy bread

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When comparing solids to fluids, the following is true: for elastic solids, the stress must be normal. For Newtonian fluids, the
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Answer: D

Find the answer in the explanation

Explanation:

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4 0
4 years ago
Water flows at a rate of 0.011 m3/s in a horizontal pipe whose diameter increases from 6 to 11 cm by an enlargement section. If
lakkis [162]

Answer:

Pressure change in pipe is 766.96 N/m²

Explanation:

The continuity equation is stated below,

AV = Q

A1V1 = A2V2

Where A is the cross-sectional area, V is the velocity and Q is the fluid flow rate.

To calculate the inlet velocity of the pipe,

V1 = Q/A1

V1 = Q/(π x d1²)

d1 is the inlet diameter of the pipe

Substituting values,

V1 = 0.011/(π x 1/4 x 0.06²)

V1 = 3.89 ms-¹

To determine the outlet velocity,

V2 = Q/A2

d2 is the outlet diameter of pipe

V2 = 0.011/(π x 1/4 x 0.11²)

V2 = 1.157 ms-²

Applying Bernoulli's equation for steady flow between the points,

P1/pg + a1(V1²/2g) + z1 + hp = P2/pg + a2(V2²/2g) + z2 + ht + hL

Collecting like terms,

The kinetic energy correction factor, a = a1 = a2

a((V1² - V2²)/2g) - hL = (P2 - P1)/pg

apg((V1² - V2²)/2g) - hLpg = P2 - P1

ap((V1² - V2²)/2) - hLpg = ∆P

p - density of water, g is the acceleration due to gravity and hL is the head loss due to friction in pipe.

Substituting values,

a = 1.05, p = 1000kg/m³, g = 9.81m/s², hL = 0.66m

∆P = (1000)(1.05)((3.89² - 1.157²)/2) - (0.66 x 1000 x 9.81)

∆P = 7241.56 - 6474.6

∆P = 766.96 N/m²

4 0
3 years ago
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