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Marina86 [1]
3 years ago
5

It is usually assumed that an action potential begins immediately at the cathode. If this were true, both methods for calculatin

g conduction velocity would provide the same answer. However, when a strong stimulus intensity is used, the action potential may begin some distance away from the cathode. Under these conditions, the difference method would be more accurate.Did you observe any important difference between the conduction velocity values?
Chemistry
1 answer:
jarptica [38.1K]3 years ago
7 0

Answer: Provided in the explanation section

Explanation:

Our questions says that:

It is usually assumed that an action potential begins immediately at the cathode. If this were true, both methods for calculating conduction velocity would provide the same answer. However, when a strong stimulus intensity is used, the action potential may begin some distance away from the cathode. Under these conditions, the difference method would be more accurate.Did you observe any important difference between the conduction velocity values?

Answer to this :

By using the difference method, you subtract out any "uncertainties" involved in the measurement of latencies. Say for example, saw we are uncertain as to where the AP's are actually originating within the vicinity of the stimulating electrodes, this "error" will be introduced into both latency measurements, and therefore subtracted out when performing a difference method calculation.

However, the difference method is only experimentally sound when one is dealing with the same population of nerve fibres over the recording electrodes used, which is not the case with the sciatic nerve, as it is a short nerve, and thin at one end.

    The non-uniformity of the nerve, and the difficulty in making accurate measurements of very small distances and latencies are principal points to consider when making conduction velocity measurements. Naturally if the nerve studied were longer and more uniform, we would improve the accuracy of our calculations.

cheers i hope this helped !!!!

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Calculate the pH of each of the following solutions: (a) 0.1000M Propanoic acid (HC3H5O2, Ka= 1.3x10-5 ) (b) 0.1000M sodium prop
jek_recluse [69]

(a) The pH of 0.1000 M propanoic acid (HC3H5O2) is 2.9.

(b) The pH of 0.1000 M sodium propanoate (NaC3H5O2) is 8.9.

(c) The pH of 0.1000 M propanoic acid (HC3H5O2) and 0.1000 M sodium propanoate (NaC3H5O2) is 4.9.

<h3>Further explanation:</h3>

(a)

Given information:

The value of acid ionization constant for propanoic acid is  1.3 x 10^{-5} .

The initial concentration of propanoic acid is  .

To calculate:

The pH of 0.1000 M propanoic acid solution.

Solution:

Propanoic acid  is a weak acid. It ionizes partially in water as follows:

 

The expression for acid dissociation constant is,

                                                            …… (1)

Here,

 is ionization constant of propanoic acid.

is the equilibrium concentration of propanoate ion.

is the equilibrium concentration of hydronium ion.

 is the equilibrium concentration of propanoic acid.

ICE table (1):

 

Refer ICE table (1),

 

Substitute the values form the ICE table (1) in equation (1).

 

The approximation x is very small is valid. Therefore, the value of x can be neglected. Above equation can be modified as,

 

Rearrange above equation for x.

                                                                                                           …… (2)

Substitute   for   in equation (2) to calculate the value of x.

 

Therefore, from the ICE table (1) the concentration of hydronium ion is,

 .

The negative logarithm of hydronium ion concentration is defined as the pH of the solution. Mathematically,

                                                                                                               …… (3)

Substitute    for    in equation (3) to calculate the pH of the solution.

 

(b)

Given information:

The value of acid ionization constant for propanoic acid is  .

The initial concentration of sodium propanoate is  .

To calculate:

The pH of 0.1000 M sodium propanoate solution.

Solution:

Sodium propanoate  is conjugate base of weak propanoic acid. It undergoes hydrolysis in water to yield hydroxide ion in the solution as follows:

                                                        …… (4)

Propanoic acid  is a weak acid. It ionizes partially in water as follows:

                                                       …… (5)

Dissociation reaction for water is written as follows:

                                                                                       …… (6)

From equation (4), (5), and (6) the relationship between   and   is,

                                                                                                                              …… (7)

Substitute   for   and   for   in equation (7).

 

ICE table (2):

 

The expression for base dissociation constant is,

                                                                                                     …… (8)

Here,

is base ionization constant.

is the equilibrium concentration of propanoate ion.

is the equilibrium concentration of hydroxide ion.

 is the equilibrium concentration of propanoic acid.

From the ICE table (2),

 

Substitute the values form the ICE table (2) in equation (8).

 

The approximation y is very small is valid. Therefore, the value of y can be neglected. Above equation can be modified as,

 

Rearrange above equation for y.

                                                                                                           …… (9)

Substitute   for   in equation (9) to calculate the value of y.

 

Therefore, from the ICE table (2) the concentration of hydroxide ion is,

 

The negative logarithm of hydroxide ion concentration is defined as pOH of the solution. Mathematically,

                                                                                                           …… (10)

Substitute    for    in equation (10) to calculate pOH of the solution.

 

The relation between pH and pOH is as follows:

                                                                                                                   …… (11)

Substitute 5.057 for pOH in equation (11) to calculate the pH of the solution.

 

(c)

Given information:

The value of acid ionization constant for propanoic acid is  .

The initial concentration of sodium propanoate is  .

The initial concentration of sodium propanoate is  .

To calculate:

The pH of 0.1000 M sodium propanoate and 0.1000 M propanoic acid solution.

Solution:

Propanoic acid is a weak acid, and sodium propanoate is salt of the conjugate base of propanoic acid. Thus, propanoic acid and sodium propanoate will form a buffer system.

The pH of the buffer solution can be determined with the help of the Henderson-Hasselbalch equation. Mathematically,

 

For propanoic acid and sodium propanoate buffer system, the Henderson-Hasselbalch equation can be modified as,

                                                                                               …… (12)

The negative logarithm of acid ionization constant is equal to  .

                                                                                                                …… (13)

Substitute   for  in equation (13).

 

Substitute    for  ,   for   and 4.9 for    in equation (12).

 

Learn more:

1. About Henderson-Hasselbalch equation brainly.com/question/12999557

2. Learn more about how to calculate moles of the base in given volume brainly.com/question/4283309

Answer details:

Grade: Senior School

Subject: Chemistry

Chapter: Ionic equilibria

Keywords: ionic equilibrium, propanoic acid, sodium propanoate, ionization constant, weak acid, conjugate base, equilibrium concentration, hydronium ion, hydroxide ion, pH, pOH, ICE table, negative logarithm, buffer solution, Henderson-Hasselbalch equation, 0.1000 M, 4.9, 8.9, 2.9.

3 0
4 years ago
Read 2 more answers
A flask with a volume of 125.0mL contains air with a density of 1.269 g/L. What is the mass of the air contained in the flask?
Aliun [14]
<span>In order to solve this problem you must first make sure all your numbers are in like terms. From the density value you can see that it is grams per liter. The first conversion you must do in convert the 125.0 mL value to Liters which you would do by dividing by 1000 because 1 liter is equal to 1000 mL. 125.0 divided by 1000 is 0.125 Liter. Now you will use the density equation to solve. The density equation is density is equal to mass divided by volume. Plug in your known numbers for density and volume. Then solve for mass. So Density (1.269 g/l is equal to mass divided by volume (.125 Liter) You must rearrange the equation to multiple density by volume which is 1.269 times 0.125 which will give you 0.1586. Because the Liters cancel each other out, the answer's unit will be grams. Your final answer is 0.1586 grams.</span>
4 0
3 years ago
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