Question

1. what is the value of n in the equation 1.9×10^7=(1×10^5)(1.9×10^n)?
2. apply math models a scientist is determining the mass of 30,000 molecules of oxygen.

a. how many significant digits should be in the solution?
mass of one molecule of oxygen = 5.3×10^(-23) gram
b. express the mass of 30,000 molecules of oxygen in scientific notation.

1. find (7.2×10^(-8))/(-3×10^(-2)). use what you know about scientific notation to express your answer with a power of 10.

Explanation:

Step1: Simplify the right - hand side of the first equation

We know that \((a\times10^{m})(b\times 10^{n})=(a\times b)\times10^{m + n}\). For \((1\times10^{5})(1.9\times10^{n})\), we get \((1\times1.9)\times10^{5 + n}=1.9\times10^{5 + n}\).
The equation \(1.9\times10^{7}=(1\times10^{5})(1.9\times10^{n})\) becomes \(1.9\times10^{7}=1.9\times10^{5 + n}\).
Since the non - exponential parts are equal (\(1.9 = 1.9\)), we can set the exponents equal to each other: \(7=5 + n\).

Step2: Solve for \(n\)

Subtract 5 from both sides of the equation \(7=5 + n\). We have \(n=7 - 5=2\).

Step3: Solve part a of the second question

The number 30000 has 1 significant digit (the non - zero digit 3). When multiplying or dividing, the number of significant digits in the result is determined by the number with the fewest significant digits in the values used in the calculation. The mass of one molecule of oxygen \(5.3\times10^{-23}\) has 2 significant digits. But since 30000 has 1 significant digit, the solution should have 1 significant digit.

Step4: Solve part b of the second question

The mass of one molecule of oxygen is \(m = 5.3\times10^{-23}\) gram. The mass of 30000 molecules of oxygen is \(M=30000\times(5.3\times10^{-23})\).
First, write 30000 as \(3\times10^{4}\). Then \(M=(3\times10^{4})\times(5.3\times10^{-23})\).
Using the rule \((a\times10^{m})(b\times10^{n})=(a\times b)\times10^{m + n}\), we get \(M=(3\times5.3)\times10^{4+( - 23)}=15.9\times10^{-19}\).
In scientific notation, we rewrite it as \(M = 1.59\times10^{-18}\) gram.

Step5: Solve the third question

We have \(\frac{7.2\times10^{-8}}{- 3\times10^{-2}}\).
Using the rule \(\frac{a\times10^{m}}{b\times10^{n}}=\frac{a}{b}\times10^{m - n}\), we get \(\frac{7.2}{-3}\times10^{-8-( - 2)}\).
\(\frac{7.2}{-3}=-2.4\) and \(-8-( - 2)=-8 + 2=-6\).
So the result is \(-2.4\times10^{-6}\).

Answer:

For the first question, \(n = 2\).
For the second question:
a. 1 significant digit.
b. \(1.59\times10^{-18}\) gram.
For the third question: \(-2.4\times10^{-6}\)