Question

the diameter of a water molecule is 0.000000000033 m. which is the best estimate of the diameter of a water molecule expressed as the product a single digit and a power of 10? \\(\boldsymbol{\circ}\\) \\(3 \times 10^{-10}\\) m \\(\boldsymbol{\odot}\\) \\(3 \times 10^{-11}\\) m \\(\boldsymbol{\circ}\\) \\(30 \times 10^{-9}\\) m \\(\boldsymbol{\circ}\\) \\(3 \times 10^{10}\\) m

Explanation:

Step1: Recall scientific notation rules

Scientific notation is in the form \( a \times 10^n \), where \( 1 \leq |a| < 10 \) and \( n \) is an integer. For a number less than 1, we move the decimal point to the right until we get a number between 1 and 10, and the exponent \( n \) is negative (equal to the number of places we moved the decimal).

The given number is \( 0.000000000033 \) m. Let's move the decimal point to the right.

Step2: Move the decimal point

Start with \( 0.000000000033 \). Move the decimal point 11 places to the right to get \( 3.3 \) (since we need a single digit estimate, we can use 3). The number of places moved is 11, so the exponent of 10 is \( - 11 \).

So, \( 0.000000000033 \approx 3 \times 10^{-11} \) m.

Let's check the other options:

• \( 3 \times 10^{-10} \) would be \( 0.0000000003 \), which is larger than \( 0.000000000033 \), so not a good estimate.
• \( 30 \times 10^{-9} = 3 \times 10^{-8} \) (since \( 30 = 3 \times 10^1 \), so \( 3 \times 10^1 \times 10^{-9}=3 \times 10^{-8} \)), which is much larger.
• \( 3 \times 10^{10} \) is a very large positive number, not matching the small decimal.

Answer:

\( 3 \times 10^{-11} \) m (the option with \( 3 \times 10^{-11} \) m)