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Sagot :
To find the residence time of sodium in sea water, we use the formula:
[tex]\[ T = \frac{m}{f} \][/tex]
where:
- [tex]\( m \)[/tex] is the mass of sodium in tons,
- [tex]\( f \)[/tex] is the flow rate of sodium in tons per year.
Given:
- Mass of sodium, [tex]\( m = 2.8 \times 10^{13} \)[/tex] tons,
- Flow rate, [tex]\( f = 3.5 \times 10^5 \)[/tex] tons/year.
First, we calculate the residence time [tex]\( T \)[/tex]:
[tex]\[ T = \frac{2.8 \times 10^{13}}{3.5 \times 10^5} \][/tex]
Simplifying this fraction:
[tex]\[ T = \frac{2.8}{3.5} \times \frac{10^{13}}{10^5} \][/tex]
[tex]\[ T = 0.8 \times 10^{8} \][/tex]
So, the residence time of sodium is:
[tex]\[ T = 0.8 \times 10^{8} \text{ years} \][/tex]
Therefore, the residence time of sodium written in scientific notation is [tex]\( 0.8 \times 10^8 \)[/tex] years.
In the given placeholders:
- The first blank should be filled with [tex]\( 0.8 \)[/tex],
- The second blank should be filled with [tex]\( 8 \)[/tex].
Thus, the complete answer is [tex]\( \mathbf{0.8} \times 10^{\mathbf{8}} \)[/tex] years.
[tex]\[ T = \frac{m}{f} \][/tex]
where:
- [tex]\( m \)[/tex] is the mass of sodium in tons,
- [tex]\( f \)[/tex] is the flow rate of sodium in tons per year.
Given:
- Mass of sodium, [tex]\( m = 2.8 \times 10^{13} \)[/tex] tons,
- Flow rate, [tex]\( f = 3.5 \times 10^5 \)[/tex] tons/year.
First, we calculate the residence time [tex]\( T \)[/tex]:
[tex]\[ T = \frac{2.8 \times 10^{13}}{3.5 \times 10^5} \][/tex]
Simplifying this fraction:
[tex]\[ T = \frac{2.8}{3.5} \times \frac{10^{13}}{10^5} \][/tex]
[tex]\[ T = 0.8 \times 10^{8} \][/tex]
So, the residence time of sodium is:
[tex]\[ T = 0.8 \times 10^{8} \text{ years} \][/tex]
Therefore, the residence time of sodium written in scientific notation is [tex]\( 0.8 \times 10^8 \)[/tex] years.
In the given placeholders:
- The first blank should be filled with [tex]\( 0.8 \)[/tex],
- The second blank should be filled with [tex]\( 8 \)[/tex].
Thus, the complete answer is [tex]\( \mathbf{0.8} \times 10^{\mathbf{8}} \)[/tex] years.
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