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irina [24]
4 years ago
7

According to the all-or-none principle, what would happen if a cell were depolarized twice the normal amount needed to produce a

n action potential?
Chemistry
1 answer:
mixas84 [53]4 years ago
5 0
According to the all-or-none principle, the thing that would happen if a cell were depolarized twice the amount needed to produce an action potential would be : Nothing would happen because the cell must be polarized to fire.

all-or-none principle stated that the strength by which muscle fiber respond to a stimulus is independent to  the strength of the stimulus.
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1.
tester [92]

Answer: New moon!

Explanation: So if your saying lunar eclips it would be new moon becuase we have full moons all the time that means lunar eclipses all the time soo it would be new moon

5 0
3 years ago
Read 2 more answers
How many moles of nitrogen gas will occupy a volume of 5L at 3.85 atm and 27c?
Olegator [25]

Answer:

n=0.78mol

Explanation:

Hello there!

In this case, according to the given information, it possible for us to realize that this problem is solved via the ideal gas equation:

PV=nRT

By which we solve for n (moles) as shown below:

n=\frac{PV}{RT}

Then, we plug in the given pressure, volume and temperature (in K) to obtain:

n=\frac{3.85atm*5L}{0.08206\frac{atm*L}{mol*K}*(273+27)K} \\\\n=0.78mol

Regards!

7 0
3 years ago
A 0.65 M solution of an unknown amine base is found to be 2.2% ionized at equilibrium. Calculate the value for Kb for this amine
mihalych1998 [28]

The value of Kb of the unknown amine is 3.22 x 10^-4.

Weak electrolytes:

The compound which do not dissociate completely when added to the aqueous solution are called weak electrolytes. The electrolyte and its ions exists in equilibrium. In case of weak base, the equilibrium constant is expressed as Kb.

Calculations:

Step 1:

The percentage of amine ionized at equilibrium is 2.2%.

NH3   +     H2O ------>    NH4+   +    OH-

The concentration of conjugate acid at equilibrium is calculated as:

2.2 = (x/0.65) x 100

x = 0.0143 M

Step 2:

The equilibrium constant (Kb) is calculated as:

Kb = x^2/0.65 - x

= (0.0143)^2/(0.65 - x)

= 3.22 x 10^-4

Learn more about weak electrolytes here:

brainly.com/question/19340043

#SPJ4

6 0
2 years ago
A certain first-order reaction a → b is 25% complete in 42 min at 25°c. What is the half-life of the reaction?.
Anika [276]

The half-life of the reaction is 101.9 min

<h3>First order reaction </h3>

When a reaction's pace and reactant concentration are inversely correlated, the process is known as a first-order reaction. To put it another way, the reaction rate doubles when the concentration does. One or two reactants can be present in a first-order reaction, as in the case of the decomposition process.

<h3>The half-life of first-order reaction:-</h3>

The amount of time it takes for the initial concentration of the reactant(s) to decrease by half is known as the half-life of a chemical reaction (abbreviated as "t1/2").

<h3 /><h3>Calculation:-</h3>

a→b

25% reacted means 75% remains

t=42 min

Rate constant

k=(2.303/t)(log a/a-x)

k=(2.303/42)(log 100/100-25 )

k=(0.054) (log 100/75)

k=(0.054)(0.1249)

k=0.0068 per min

half life

t1/2=(0.693/k)

     =(0.693/0.0068)

     =101.9 min

Learn more about first order reaction here :-

brainly.com/question/27754430

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5 0
2 years ago
HELPP MEEE PLEASEEEEEEEEE TT<br>asapppp<br>​
andriy [413]

Answer:

B then A

Explanation:

5 0
3 years ago
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