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Talja [164]
3 years ago
11

Let's consider a scenario in which the resting membrane potential changes from −70 mV to +70 mV, but the concentrations of all i

ons in the intracellular and extracellular fluids are unchanged. Predict how this change in membrane potential affects the movement of Na+. The electrical gradient for Na+ would tend to move Na+ __________ while the chemical gradient for Na+ would tend to move Na+ __________.
Chemistry
1 answer:
Inessa [10]3 years ago
6 0

Answer: The correct answer is 1. outside and 2. inside.

Explanation:

During the resting potential (-70 mV), the concentration of Na+ is bigger outside than inside, so this situation generates a trend for moving inside. If we consider a scenario where the resting potential changes from -70 mV to +70 mV, we would realize that the electrical gradient would stop the entrance of Na+ ions and would make them to move outside.

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sashaice [31]
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8 0
3 years ago
Consider 2 NOCl(g) \Longleftrightarrow⟺ 2 NO(g) + Cl2 (g) At 25oC under conditions other than equilibrium, there are 1.20 moles
nekit [7.7K]

Answer:

a) Q = 6.1875x10⁻³

b) The direction of the reaction is to form the products.

c) [Cl₂]e = 0.094 M

[NO]e =  0.190 M

[NOCl]e = 0.140 M

Explanation:

a) Q is the reaction quotient, and for a generic reaction aA + bB ⇄ cC + dD it is

Q = \frac{[C]^cx[D]^d}{[A]^ax[B]^b}

Which is the same equation for Kc, but in Kc expressions, the concentrations are in the equilibrium. Q is calculated at any time. So, for the reaction given

2NOCl(g) ⇄ 2NO(g) + Cl2(g)

Q = \frac{[Cl2]x[[NO]^2}{[NOCl]^2}

[Cl₂] = 0.220/5.00 = 0.044 M

[NO] = 0.450/5.00 = 0.090 M

[NOCl] = 1.20/5.00 = 0.240 M

Q = (0.044)x(0.090)²/(0.240)²

Q = 6.1875x10⁻³

b) Q < Kc, which means that there are fewer products to what are needed to the equilibrium. So the direction of the reaction is to form the products.

c)

2NOCl(g) ⇄ 2NO(g) + Cl2(g)

0.240           0.090      0.044         <em>Initial</em>

-2x                 +2x          +x             <em> Reacts</em> (stoichiometry is 2:2:1)

0.240-2x   0.090+2x   0.044+x    <em> Equilibrium</em>

Kc = \frac{(0.044+x)x(0.090+2x)^2}{(0.240 - 2x)^2}

1.86x10^{-1} = \frac{(0.044+x)*(8.1x10^{-3} +0.36x + 4x^2)}{(0.0576 - 0.96x +4x^2)}

3.564x10⁻⁴+0.01584x+0.176x²+8.1x10⁻³x+0.36x²+4x³ = 0.010714-0.17856x+0.744x²

4x³ - 0.208x² + 0.2025x - 0.01036 = 0

Solving this third grade equation in a computer program:

x = 0.05 M

So:

[Cl₂]e = 0.044 + 0.05 = 0.094 M

[NO]e = 0.090 + 2x0.05 = 0.190 M

[NOCl]e = 0.240 - 2x0.05 = 0.140 M

7 0
3 years ago
CARDS
valina [46]

Answer:d

Explanation:

8 0
3 years ago
What type of reaction is...
Alexxandr [17]

Answer:

d: double replacement

Explanation:

7 0
3 years ago
How many grams of naoh are needed to prepare 500 ml of 0.125 m naoh?
Romashka-Z-Leto [24]

First let us calculate for the number of moles needed:

moles NaOH = 0.125 M * 0.500 L = 0.0625 mol

 

The molar mass of NaOH is 40 g/mol, hence the mass is:

mass NaOH = 0.0625 mol * 40 g/mol

<span>mass NaOH = 2.5 grams</span>

6 0
3 years ago
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