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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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At 35°C, K = 1.6 × 10^-5 for the reaction
TEA [102]

Answer:

a) [NOCl] = 0.968 M

[NO] = 0.032M

[Cl²] = 0.016M

b) [NOCl] = 1.992M

[NO] = 0.008 M

[Cl2]  = 1.004 M

Explanation:

Step 1: Data given

Temperature = 35°C = 308K

K = 1.6 × 10^-5

Step 2: The reaction

2 NOCl(g) ⇌ 2 NO(g) + Cl2(g)

For 2 moles NOCl we'll have 2 moles NO and 1 mol Cl2

Step 3

a. 2.0 mol pure NOCl in a 2.0 L flask

Concentration at the start:

Concentration = mol / volume

[NOCl] = mol / volume

[NOCl] = 2.0 / 2.0 L

[NOCl] = 1.0 M

[NO] = 0 M

[Cl] = 0M

Concentration at the equillibrium

[NOCl] = 1.0M - 2x

[NO] = 2x

[Cl2]= x

K = [Cl2][NO]² / [NOCl]² = 1.6*10^-5

1.6*10^-5 = ((2x)² * x) / (1.0-2x)²

x = 0.016

[NOCl] = 1.0 -  2*0.016 = 0.968 M

[NO] = 2*0.016 = 0.032M

[Cl²] = 0.016M

b. 2.0 mol NOCl and 1.0 mol Cl2 in a 1.0 L flask

Concentration at the equillibrium

[NOCl] = 2.0 mol / 1.0 L = 2.0 M

[NO] = 0 M

[Cl2]= 1.0 mol / 1.0 L = 1.0 M

Concentration at the equillibrium

[NOCl] = 2.0M - 2x

[NO] = 2x

[Cl2]= 1.0 + x

K = [Cl2][NO]² / [NOCl]² = 1.6*10^-5

1.6 *10^-5 = (2x)²*(1.0+x) / ((2.0-2x)²)

1.6 *10^-5= (2x)² * 1 )/2.0²

1.6 *10^-5= 4x² / 4 = x²

x = \sqrt{1.6 *10^-5} = 4.0*10^-3

[NOCl] = 2.0 - 2*0.004 = 1.992M

[NO] = 2*0.004 = 0.008 M

[Cl2] = 1+ 0.004M = 1.004 M

5 0
3 years ago
BRAINLIEST PLEASE HELP!!!
Natalka [10]
Answer : first opinion and also last
Expiation : Note that these last two reactions, and 2H + 2H → 4He + γ, .Nuclear fusion is a reaction in which two nuclei are combined to form a larger nucleus. Nuclear fusion is a reaction in which two nuclei are combined, or fused, to form a larger nucleus. We know that all nuclei have less mass than the sum of the masses of the protons and neutrons that form them. The missing mass times c2 equals the binding energy of the nucleus—the greater the binding energy, the greater the missing mass.
5 0
3 years ago
How many moles of NO2 form when<br> 63.25 g N2O5 decompose?<br><br> 2N₂O54NO₂ + O₂
KonstantinChe [14]

Answer:

1.1713 moles

Explanation:

RFM of N2O5=108

Moles of N2O5= mass/RFM= 63.25/108=0.5856 moles

Mole ratio of N2O5:NO2= 2:4

Therefore moles of NO2= 4/2*0.5856= 1.1713 moles

4 0
2 years ago
The element boron exists in nature as two isotopes: 10B has a mass of 10.0129 u, and 11B has a mass of 11.0093 u. The average at
Shalnov [3]

Answer:

Percentage abundance of B 10 is = 20 %

Percentage abundance of B 11 is = 80 %

Explanation:

The formula for the calculation of the average atomic mass is:

Average\ atomic\ mass=(\frac {\%\ of\ the\ first\ isotope}{100}\times {Mass\ of\ the\ first\ isotope})+(\frac {\%\ of\ the\ second\ isotope}{100}\times {Mass\ of\ the\ second\ isotope})

Given that:

Since the element has only 2 isotopes, so the let the percentage of first be x and the second is 100 -x.

For first isotope, B 10:

% = x %

Mass = 10.0129 u

For second isotope, B 11:

% = 100  - x  

Mass = 11.0093 u

Given, Average Mass = 10.81 u

Thus,  

10.81=\frac{x}{100}\times {10.0129}+\frac{100-x}{100}\times {11.0093}

10.0129x+11.0093\left(100-x\right)=1081

Solving for x, we get that:

x = 20 %

Thus percentage abundance of B 10 is = 20 %

Percentage abundance of B 11 is = 100 - 20 %  = 80 %

8 0
3 years ago
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Identify impacts of air pollution
Valentin [98]
Heart disease, lung cancer and respiratory diseases
7 0
2 years ago
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