<h3><u>Answer and explanation;</u></h3>
- <u><em>A neutral sodium atom contains 11 protons, 12 neutrons and 11 electrons. By removing an electron from this atom we get a positively charged Na+ ion that has a net charge of +1.</em></u> Thus; Sodium atom is neutral whereas sodium ion is a charged species with a charge of +1.
- <u><em>Sodium ion is positively charged. In sodium ion, there are 11 protons and 12 neutrons but 10 electrons</em></u>, i.e., sodium ion contains lesser number of electrons. Additionally,the size of sodium ion is smaller than that of sodium atom.
The capacitive reactance is reduced by a factor of 2.
<h3>Calculation:</h3>
We know the capacitive reactance is given as,
where,
= capacitive reactance
f = frequency
C = capacitance
It is given that frequency is doubled, i.e.,
f' = 2f
To find,
=?
Therefore, the capacitive reactance is reduced by a factor of 2.
I understand the question you are looking for is this:
A capacitor is connected across an AC source. Suppose the frequency of the source is doubled. What happens to the capacitive reactant of the inductor?
- The capacitive reactance is doubled.
- The capacitive reactance is traduced by a factor of 4.
- The capacitive reactance remains constant.
- The capacitive reactance is quadrupled.
- The capacitive reactance is reduced by a factor of 2.
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The answer is b that what I think and feel like it is
B. evaporation ................