The temperature of a sample of lead increased by 22.9 °C when 265 J of heat was applied. What is the mass of the sample?

90.41 g

Explanation:

Data provided in the question:

Increase in temperature of lead sample, ΔT = 22.9 °C

Heat supplied to the sample , Q = 265 J

Now,

we know that

Q = mCΔT

here,

m = mass of the sample

C = Specific heat of lead = 0.128 J/g.°C

Thus,  

265 = m × 0.128 × 22.9

or

265 = m × 2.9312  

or  

m = 90.41 g  

the answer that best describes the relative refractory period in the neuron is the resting membrane potential is restored, but concentrations of sodium and potassium are not restored to their original state.

the refractory period is the time during which another stimulus given to the neuron will not lead to a second potential. thus, because na+ channels are inactivated during this time, additional depolarizing stimuli do not lead to new action potentials.  

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