Question

there is an average of 3.5×10^(-2) kilograms of dissolved salt in each liter of seawater. the pacific ocean contains approximately 6.6×10^(20) liters of seawater. how many kilograms of dissolved salt are in the pacific ocean? enter your answer, in scientific notation, by filling in the boxes.

Explanation:

Step1: Identify the formula

To find the total mass of dissolved salt, multiply the mass of salt per - liter by the total volume of seawater. The formula is $m = s\times v$, where $m$ is the total mass of salt, $s$ is the mass of salt per liter, and $v$ is the volume of seawater.

Step2: Substitute the given values

We have $s = 3.5\times10^{-2}$ kg/L and $v = 6.6\times10^{20}$ L. So $m=(3.5\times10^{-2})\times(6.6\times10^{20})$.

Step3: Use the rule of exponents for multiplication

When multiplying numbers in scientific - notation $a\times10^{n}$ and $b\times10^{m}$, we use the formula $(a\times10^{n})\times(b\times10^{m})=(a\times b)\times10^{n + m}$. Here, $a = 3.5$, $b = 6.6$, $n=-2$, and $m = 20$. First, calculate $a\times b$: $3.5\times6.6 = 23.1$. Then, calculate $n + m$: $-2+20 = 18$. So $m = 23.1\times10^{18}$.

Step4: Convert to proper scientific notation

In scientific notation, the coefficient $a$ should satisfy $1\leq a<10$. We can rewrite $23.1\times10^{18}$ as $2.31\times10^{19}$ (since $23.1=2.31\times10^{1}$ and $2.31\times10^{1}\times10^{18}=2.31\times10^{1 + 18}=2.31\times10^{19}$).

Answer:

$2.31\times10^{19}$ kg