If the current is composed of Na+ ions (q = +e), how many such ions flow in 0.7 s?
You have not provided required information, such as the current in Amperes.
To determine the number of Na+ ions that flow in 0.7 seconds, we need to use the formula:
\( Q = I \cdot t \),
where:
- Q is the charge (in Coulombs),
- I is the current (in Amperes),
- t is the time (in seconds).
In this case, we are given the current I in the form of Na+ ions, with a charge of +e (elementary charge).
The elementary charge, denoted as \( e \), is the magnitude of the electric charge carried by a single electron or proton. Its value is approximately \( 1.6 \times 10^{-19} \) Coulombs.
Thus, we can rewrite the formula as:
\( Q = (q \cdot n) \cdot t \),
where:
- q is the charge of a single Na+ ion, which is +e (\(1.6 \times 10^{-19}\) Coulombs),
- n is the number of Na+ ions,
- t is 0.7 seconds.
Since we are looking for the number of Na+ ions (n), we can rearrange the equation:
\( n = \frac{Q}{q \cdot t} \).
Now we substitute the given values into the equation:
\( n = \frac{0.7 \cdot q}{q \cdot t} \).
Notice that the q and q values cancel out, leaving:
\( n = \frac{0.7}{t} \).
Substituting the given time value t = 0.7 seconds:
\( n = \frac{0.7}{0.7} = 1 \).
Therefore, in 0.7 seconds, there would be one Na+ ion that flows.