Plllllzzz help :(

Consider an experimental setup with two compartments separated by a phospholipid bilayer membrane containing ion channels select

Schach [20]

This question is incomplete, the complete question is;

Consider an experimental setup with two compartments separated by a phospholipid bilayer membrane containing ion channels selectively permeable only to chloride ions. The left compartment (also called inside) contains 1 mM Cl- and the right compartment (also called outside) contains 100 mM Cl-. What will the electrical potential be when the system attains equilibrium? [ assume body temperature; log 100 = 2, log 10 = 1, log 1 = 0, log 0.1 = -1, log 0.01 = -2]

Options;

a) -62 mV

b) -124 mV

c) +62 mV

d) 0 mV

e) +124 mV

Answer:

the electrical potential be when the system attains equilibrium is –124mV

Option b) –124mV is the correct answer

Explanation:

Given the data in the question;

Two compartments are divided by lipid bilayer;

In inside compartment Cl- ion concentration- 1mM and out side of the cell concentration is 100mM

now we apply the Nernst equilibrium potential equation;

Chlorine ion valency is z = –1

So

$E_{eq}$ = 62/z × log(ion outside/ ion inside) [for Cl‐ ions]

$E_{eq}$ = (62 / –1) x log( 100 / 1 )

$E_{eq}$ = -62 x 2 =

$E_{eq}$ = –124mV

Therefore, the electrical potential be when the system attains equilibrium is –124mV

Option b) –124mV is the correct answer

4 0

2 years ago

BH+ClO4- is a salt formed from the base B (Kb = 1.00e-4) and perchloric acid. It dissociates into BH+, a weak acid, and ClO4-, w

Len [333]

Answer:

The pH of 0.1 M BH⁺ClO₄⁻ solution is 5.44

Explanation:

Given: The base dissociation constant: $K_{b}$ = 1 × 10⁻⁴, Concentration of salt: BH⁺ClO₄⁻ = 0.1 M

Also, water dissociation constant: $K_{w}$ = 1 × 10⁻¹⁴

The acid dissociation constant ( $K_{a}$ ) for the weak acid (BH⁺) can be calculated by the equation:

$K_{a}. K_{b} = K_{w}$

$\Rightarrow K_{a} = \frac{K_{w}}{K_{b}}$

$\Rightarrow K_{a} = \frac{1\times 10^{-14}}{1\times 10^{-4}} = 1\times 10^{-10}$

Now, the acid dissociation reaction for the weak acid (BH⁺) and the initial concentration and concentration at equilibrium is given as:

Reaction involved: BH⁺ + H₂O ⇌ B + H₃O+

Initial: 0.1 M x x

Change: -x +x +x

Equilibrium: 0.1 - x x x

The acid dissociation constant: $K_{a} = \frac{\left [B \right ] \left [H_{3}O^{+}\right ]}{\left [BH^{+} \right ]} = \frac{(x)(x)}{(0.1 - x)} = \frac{x^{2}}{0.1 - x}$

$\Rightarrow K_{a} = \frac{x^{2}}{0.1 - x}$

$\Rightarrow 1\times 10^{-10} = \frac{x^{2}}{0.1 - x}$

$As, x$

$\Rightarrow 0.1 - x = 0.1$

$\therefore 1\times 10^{-10} = \frac{x^{2}}{0.1 }$

$\Rightarrow x^{2} = (1\times 10^{-10})\times 0.1 = 1\times 10^{-11}$

$\Rightarrow x = \sqrt{1\times 10^{-11}} = 3.16 \times 10^{-6}$

Therefore, the concentration of hydrogen ion: x = 3.6 × 10⁻⁶ M

Now, pH = - ㏒ [H⁺] = - ㏒ (3.6 × 10⁻⁶ M) = 5.44

Therefore, the pH of 0.1 M BH⁺ClO₄⁻ solution is 5.44

5 0

2 years ago

Repost: What is weathering?

Stella [2.4K]

Your drawing looks so good :p

4 0

2 years ago

Read 2 more answers

Equal amounts of N2 and O2 are added, under certain conditions, to a closed container. Which changes occur in the reverse reacti

geniusboy [140]

1) The forward reaction is N2 (g) + O2 (g) → 2NO

(that reaction requires special contitions because at normal pressures and temperatures N2 and O2 do not react to form another compound.

2) The equiblibrium equation is

N2 (g) + O2 (g) ⇄ 2NO

3) Then, the reverse reaction is

2NO → N2(g) + O2(g)

Answer: 2NO → N2(g) + O2(g)

4 0

3 years ago

Wht are the trends for electronegativity and ionzation energy similar

FinnZ [79.3K]

Answer:

Along period electronegativity and ionization energy increases.

Along group electronegativity and ionization energy decreases.

Explanation:

Along period:

As we move from left to right across the periodic table the number of valance electrons in an atom increase. The atomic size tend to decrease in same period of periodic table because the electrons are added with in the same shell. When the electron are added, at the same time protons are also added in the nucleus. The positive charge is going to increase and this charge is greater in effect than the charge of electrons. This effect lead to the greater nuclear attraction. Thus the attraction of the atoms for valance electrons increases. The electrons are pull towards the nucleus and valance shell get closer to the nucleus. As a result of this greater nuclear attraction atomic radius decreases and ionization energy increases because it is very difficult to remove the electron from atom and more energy is required, and electronegativity also increases.

Along group:

As we move from top to bottom in periodic table the atomic sizes increases.The electrons are added in next energy level in every next element. Thus the valance electrons farther away from the nucleus and hold of nucleus becomes weaker, because of weak nuclear attraction atomic radii increases and electronegativity and ionization energy decreases.

3 0

2 years ago