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serg [7]
4 years ago
12

Imagine you created a toxin such that when a neuron fired an action potential, the toxin would bind immediately to the sodium-po

tassium pump at the top of the action potential, but the sodium and potassium channels would still function. What process(es) would be affected
Chemistry
1 answer:
loris [4]4 years ago
4 0

Answer:Maintaining resting potential and returning to resting potential after the hyperpolarization phase of an action potential

Explanation:TOXINS are chemical substances which are known to be POISONOUS produced with living organisms that causes harm to other organisms, examples include Venom from snakes which when a person is bitten by a Snake it will possibly lead to death if not adequate treated.

HYPERPOLARIZATION is a term that explains the change in membrane potential due to toxin,it make the membrane more electronegative. When the toxin has hyped the level of Sodium-Potassium level returning to a rest state will be most affected.

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What are the four important steps needed to obtain accurate melting points on a mel-temp
AveGali [126]

Explanation:

1) refine the specimen into fine powder 2) place the smallest amount you can see in the capillary tube 3) set the voltage to increase exponentially to 200 below the predicted temperature, then adjust so that the temperature rises to 20 per minute 4) report the temperature at which the liquid first appears and the temperature at which the last crystal disappears.

3 0
4 years ago
Determine the molar concentration of na+ and po4 3- in a 2.25 M Na3 PO4 solution
muminat

Answer:

A. The concentration of Na^+ in the solution is 6.75 M.

B. The concentration of PO4^3- in the solution is 2.25 M.

Explanation:

We'll begin by writing the balanced dissociation equation for Na3PO4.

This is illustrated below:

Na3PO4 will dissociate in solution as follow:

Na3PO4(aq) —> 3Na^+(aq) + PO4^3-(aq)

Thus, from the balanced equation above,

1 mole of Na3PO4 produce 3 moles of Na^+ and 1 mole of PO4^3-

A. Determination of the concentration of Na+ in 2.25 M Na3PO4 solution.

This can be obtained as follow:

From the balanced equation above,

1 mole of Na3PO4 produce 3 moles of Na^+.

Therefore, 2.25 M Na3PO4 solution will produce = (2.25 x 3) /1 = 6.75 M Na^+.

Therefore, the concentration of Na^+ in the solution is 6.75 M

B. Determination of the concentration of PO4^3- in 2.25 M Na3PO4 solution.

This can be obtained as follow:

From the balanced equation above,

1 mole of Na3PO4 produce 1 mole of PO4^3-

Therefore, 2.25 M Na3PO4 solution will also produce 2.25 M PO4^3-.

Therefore, the concentration of PO4^3- in the solution is 2.25 M.

7 0
3 years ago
How do you measure the volume of a rubber duck?
EleoNora [17]

Answer:

Im sorry i though this was a joke i don't know I'm sorry:

8 0
3 years ago
Read 2 more answers
For a particular reaction, Δ=−111.4 kJ and Δ=−25.0 J/K.
vichka [17]

Answer:

\Delta G =-103.95kJ

Explanation:

Hello there!

In this case, since the thermodynamic definition of the Gibbs free energy for a change process is:

\Delta G =\Delta H-T\Delta S

It is possible to plug in the given H, T and S with consistent units, to obtain the correct G as shown below:

\Delta G =-111.4kJ-(298K)(-25.0\frac{J}{K}*\frac{1kJ}{1000J} )\\\\\Delta G =-103.95kJ

Best regards!

6 0
3 years ago
Vinegar is a solution of acetic acid, CH3COOH, dissolved in water. A 5.54-g sample of vinegar was neutralized by 30.10 mL of 0.1
tekilochka [14]

Answer:

3.26 % of vinegar is acetic acid

Explanation:

Step 1: Data given

Mass of the sample = 5.54 grams

Volume of NaOH = 30.10 mL

Molarity of NaOH = 0.100M

Step 2: The balanced equation

CH3COOH + NaOH → CH3COONa + H2O

Step 3: Calculate moles NaOH

Moles NaOH = volume * molarity

Moles NaOH = 0.03010 L * 0.100M

Moles NaOH = 0.00301 moles NaOH

Step 4: Calculate moles CH3COOH

For 1 mol NaOH we need 1 mol CH3COOH

For 0.00301 moles NaOH we nee 0.00301 moles CH3COOH

Step 5: Calculate mass CH3COOH

Mass CH3COOH = moles CH3COOH * molar mass CH3COOH

Mass CH3COOH = 0.00301 moles * 60.05 g/mol

Mass CH3COOH = 0.1808 grams

Step 6: Calculate percent by weight of acetic acid

Mass % = ( 0.1808 / 5.54 ) *100%

Mass % = 3.26 %

3.26 % of vinegar is acetic acid

7 0
3 years ago
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