1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
Fed [463]
3 years ago
7

In a phospholipid, which components constitute the hydrophilic head? Select the correct answer below: Question 1 options: amino-

alcohol modified phosphate, ester groups, and glycerol fatty acid and glycerol phosphate and fatty acid only the amino alcohol that modifies the phosphate group
Chemistry
1 answer:
KIM [24]3 years ago
4 0

Answer: Option (a) is the correct answer.

Explanation:

A lipid that contains a phosphate group in its molecule is known as a phospholipid.

These are the primary molecules present in a plasma membrane. These tend to contain two hydrophobic fatty acid tails and one hydrophilic head.

Hydrophilic means water loving in nature.

As phosphate group is a polar group due to this the hydrophilic region present in a phospholipid is able to interact with water.

In an amino-alcohol modified phosphate, ester groups, and glycerol, the glycerol is hydrophobic in nature. It also contains two fatty acids and a phosphoric acid which are present as ester. The polar head that mainly consists of phosphate group is hydrophilic and thus, it helps in combining to a water molecules.

Hence, we can conclude that in a phospholipid, amino-alcohol modified phosphate, ester groups, and glycerol are the components which constitute the hydrophilic head.

 

You might be interested in
Set up a Hess’s law cycle, and use the following information to calculate ΔH∘f for aqueous nitric acid, HNO3(aq). You will need
Assoli18 [71]

Answer:

\Delta H_{f}^{\circ } (HNO_{3}, aq) = (-207.39) kJ/mol

Explanation:

Given:

\Delta H_{f}^{\circ } (NH_{3}, g)= -46.1 kJ/mol\\ \Delta H_{f}^{\circ } (H_{2}O, l)= -285.8 kJ/mol\\ \Delta H_{f}^{\circ } (HNO_{3}, aq)= ? kJ/mol

3NO₂(g)+H₂O(l)→ 2HNO₃(aq)+NO(g);  ΔH° = -137.3 kJ            ....equation 1

2NO(g)+O₂(g)→ 2NO₂(g)                   ; ΔH° = -116.2 kJ             ....equation 2

4NH₃(g)+5O₂(g)→ 4NO(g)+6H₂O(l)   ;ΔH° = -1165.2 kJ            ....equation 3

Multiplying equation 1 with (2/3), we get

2NO₂(g) + 2/3 H₂O(l) → 4/3 HNO₃(aq) + (2/3) NO(g)              ....equation 4

Adding equation 2 and 4, we get

2NO₂(g) + 2/3 H₂O(l) + 2NO(g) + O₂(g) → 4/3 HNO₃(aq) + (2/3) NO(g) + 2NO₂(g)  

⇒ (4/3) NO(g) + O₂(g) + 2/3 H₂O(l) → 4/3 HNO₃(aq)              ....equation 5

Multiplying equation 3 with (1/3), we get  

(4/3) NH₃(g) + (5/3) O₂(g) → (4/3) NO(g)+ 2 H₂O(l)                  ....equation 6

Now adding equation 5 and 6, we get

(4/3) NH₃(g) + (5/3) O₂(g) + (4/3) NO(g) + O₂(g) + 2/3 H₂O(l) → (4/3) NO(g)+ 2 H₂O(l) + 4/3 HNO₃(aq)

⇒ (4/3) NH₃(g) + (8/3) O₂(g)  → (4/3) HNO₃(aq) + (4/3) H₂O(l)  ....equation 7

Now multiplying equation 7 with (3/4), we get

NH₃(g) + 2 O₂(g) → HNO₃(aq) + H₂O(l)                                       ...equation 8

Therefore, by Hess's law the standard enthalpy of formation is:

ΔH° = (3/4) [ (-137.3 kJ) × (2/3) + (-116.2 kJ) + (-1165.2 kJ) × (1/3)]

ΔH° = (3/4) [ - 91.53 - 116.2 - 388.4]

ΔH° =  (3/4) [-596.13] = -447.09 kJ

Since the change in enthalpy of a reaction:

ΔH° = [\Delta H_{f}^{\circ } (HNO_{3}, aq) + \Delta H_{f}^{\circ } (H_{2}O, l)] - [\Delta H_{f}^{\circ } (NH_{3}, g) + 2 \Delta H_{f}^{\circ } (O_{2}, g)]

(-447.09 kJ) = [\Delta H_{f}^{\circ } (HNO_{3}, aq) + (-285.8 kJ)] - [-46.1 kJ + 2 (0 kJ)]

⇒ \Delta H_{f}^{\circ } (HNO_{3}, aq) = (-447.09) + 285.8 kJ - 46.1 kJ

⇒ \Delta H_{f}^{\circ } (HNO_{3}, aq) = (-207.39) kJ/mol

6 0
3 years ago
In a constant‑pressure calorimeter, 60.0 mL of 0.300 M Ba ( OH ) 2 was added to 60.0 mL of 0.600 M HCl . The reaction caused the
densk [106]

Answer:

Q sln = 75.165 J

Explanation:

a constant pressure calorimeter:

  • Q sln = mCΔT

∴ m sln = m Ba(OH)2 + m HCl

∴ molar mass Ba(OH)2 = 171.34 g/mol

∴ mol Ba(OH)2 = (0.06 L)(0.3 mol/L) = 0.018 mol

⇒ mass Ba(OH)2 = (0.018 mol)(171.34 g/mol) = 3.084 g

∴ molar mass HCl = 36.46 g/mol

∴ mol HCl = (0.06 L)(0.60 mol/L) = 0.036 mol

⇒ mass HCl = (0.036 mol)(36.46 g/mol) = 1.313 g

⇒ m sln = 3.084 g + 1.313 g = 4.3966 g

specific heat (C):

∴ C sln = C H2O = 4.18 J/g°C

∴ ΔT = 26.83°C - 22.74°C = 4.09°C

heat absorbed (Q):

⇒ Q sln = (4.3966 g)(4.18 J/g°C)(4.09°C)

⇒ Q sln = 75.165 J

8 0
4 years ago
Question 7 (1 point
Anna35 [415]

Answer:

the elements to the right including hydrogen

Explanation:

as you know non-metal tend to gain eletrons

basically group 5, 6, and 7 are nonmetals

group 1, 2, and 3 are metals and group 4 are transitional metals so be careful with group 4

7 0
3 years ago
Read 2 more answers
Which two statements are true about a system
Y_Kistochka [10]
What are the two statements?
8 0
3 years ago
Delta H for each reaction ty!
Lubov Fominskaja [6]

Answer:

Reaction 1 = -462.5 kJ/mol

Reaction 2 = -572 kJ/mol

Magnesium combustion = -601.83 kJ/mol

5 0
3 years ago
Other questions:
  • Which equation explains where the energy that is released from stars comes from?
    11·2 answers
  • 30 POINTS
    5·1 answer
  • 2. Which balanced equation represents an oxidation-reduction reaction?
    5·1 answer
  • What’s the percent by mass of carbon in c10h14n2
    6·1 answer
  • Organisms that are producers
    15·1 answer
  • When CaCl2 and Zn react together, zinc (Zn) can replace chlorine (Cl) in the compound because zinc is higher on the periodic tab
    5·2 answers
  • Why are free radicals unstable<br> atoms and how can they harm our bodies?
    15·2 answers
  • Help ASAP !! <br> will give brainiest for correct answer !! <br> &lt;3
    5·1 answer
  • A river depositing sediment in a delta <br><br>A. weathering<br>B. erosion<br>C. deposition​
    7·2 answers
  • Help asap please, What does Newton's law of gravitation state that two masses will do?
    5·2 answers
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!