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Margaret [11]
4 years ago
13

What functional feature(s) does the phosphate group contribute to the structure of a phospholipid? select all that apply. select

all that apply. nonpolar group that avoids water negative charge to interact with water place to attach fatty acids place where bonds can form between adjoining phospholipids place to attach another small charged molecule?
Chemistry
1 answer:
sergejj [24]4 years ago
4 0
The phosphate group contribute the following functional features to the structure of phospholipids:
1. Negative charge to interact with water.
2. Place to attach another small molecule. 
The phosphate group is made up of four atoms of oxygen which are attached to one atom of phosphorus. This molecule has a net negative charge of -3. In the phospholipid molecule, the phosphate enhanced the polarity of the phospholipid head by mean of its negative charge which react with water. Phosphate group also provide a point where other small molecules such as alcohol, serine, etc can be attached.
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A 8.249 gram sample of copper is heated in the presence of excess fluorine. A metal fluoride is formed with a mass of 13.18 g. D
levacccp [35]

Answer:

CuF_2 the empirical formula of the metal fluoride.

Explanation:

Mass of copper heated = 8.249 g

Mass of copper fluoride formed = 13.18 g

Mass of fluorine gas in copper fluoride = x

13.18 g = 8.249 g + x\\x= 13.18 - 8.249 g = 4.931 g

Moles of copper :

= \frac{8.249 g}{63.546 g/mol}=0.1298 mol

Moles of fluorine:

= \frac{4.931 g}{18.998 g/mol}=0.2596 mol

For the empirical formula divide the smallest mole of an element with all the moles of elements present in the compound.

Copper= \frac{0.1298 mol}{0.1298 mol}=1\\Fluorine = \frac{0.2596 mol}{0.1298 mol}=2

The empirical formula of the copper fluoride = CuF_2

CuF_2 the empirical formula of the metal fluoride.

6 0
3 years ago
Which of the following is an example of ionization of an acid?
Amiraneli [1.4K]

Answer:

NH3(g) + H2O(1) → NH4+(aq) + OH (aq)

HF(aq) + H2O(1) → H3O+(aq) + F (aq)

Explanation:

Acid-base reactions are chemical reactions involving acids and bases. Acids tend to ionize/dissociate in water, a property which determines their strength. Ionization of an acid refers to the acid losing its hydrogen ion (H+) in water solution. An acid ionizes or dissociates to form a conjugate base.

A strong acid is so because it ionizes completely in water i.e. loses all its hydrogen ion (H+) while a weak acid partially ionizes in water.

In the chemical reactions;

1) NH3(g) + H2O(1) → NH4+(aq) + OH (aq)

H20 loses its hydrogen ion (H+) in this reaction to form an anion (OH-). Hence, water (H20) is an acid in this case which ionizes to form a conjugate base (OH-). This is an example of ionization of acid.

2) HF(aq) + H2O(1) → H3O+(aq) + F (aq)

Hydrogen fluoride (HF) loses its hydrogen ion (H+) in the presence of water to form anion (F-). The HF is the acid while F- is it's conjugate base. Thus, an example of ionization of acid

4 0
3 years ago
Give the ground-state electron configuration for each of the following elements. After each atom is its atomic number in parenth
Anni [7]

Answer:

(a) ₁₉K: 1s² 2s² 2p⁶ 3s² 3p⁶ 4s¹

(b) ₁₀Ne: 1s² 2s² 2p⁶

---

(a) 3

(b) 6

(c) 7

Explanation:

We can state the ground-state electron configuration for each element following Aufbau's principle.

(a) ₁₉K: 1s² 2s² 2p⁶ 3s² 3p⁶ 4s¹

(b) ₁₀Ne: 1s² 2s² 2p⁶

Second part

(a) Al belongs to Group 13 in the Periodic Table. It has 13-10=3 electrons in the valence shell.

(b) O belongs to Group 16 in the Periodic Table. It has 16-10=6 electrons in the valence shell.

(c) F belongs to Group 17 in the Periodic Table. It has 17-10=7 electrons in the valence shell.

3 0
3 years ago
For the following reaction, 9.60 grams of butane (C4H10) are allowed to react with 17.0 grams of oxygen gas. butane (C4H10) (g)
dangina [55]

Answer:

14.4 g of CO_{2} can be produced.

Explanation:

Balanced equation: 2C_{4}H_{10}+13O_{2}\rightarrow 8CO_{2}+10H_{2}O

                                        Molar mass (g/mol)

                  C_{4}H_{10}                    58.12

                    O_{2}                         32

                  CO_{2}                       44.01

So, 9.60 g of C_{4}H_{10} = \frac{9.60}{58.12}mol=0.165mol

      17.0 g of O_{2} = \frac{17.0}{32}mol=0.531mol

According to balanced equation-

2 moles of C_{4}H_{10} produce 8 moles of CO_{2}

So, 0.165 moles of C_{4}H_{10} produce (\frac{8}{2}\times 0.165)mol of CO_{2}  or 0.660 moles of CO_{2}

13 moles of O_{2} produce 8 moles of CO_{2}

So, 0.531 moles of O_{2} produce (\frac{8}{13}\times 0.531)moles of CO_{2} or 0.327 moles of CO_{2}

As least number of moles of CO_{2} are produced from O_{2} therefore O_{2} is the limiting reagent.

So, maximum amount of CO_{2} that can be formed = 0.327 moles

                                                                              = (0.327\times 44.01)g

                                                                              = 14.4 g

3 0
3 years ago
7. The smallest representation of an lonic compound is
BlackZzzverrR [31]

Answer:

c Formula unit

hope this helped

3 0
3 years ago
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