Question

practice & problem solving
leveled practice in 7 and 8, perform the operation and express your answer in scientific notation.

1. (7×10^(-6))(7×10^(-6))

(□·□)×(10^□·10^□)
□×10^□
4.9×10^□

1. (3.76×10^5)+(7.44×10^5)

(□ + □)×(10^□)
□×□
1.12×10^□

1. what is the value of n in the equation 1.9×10^7=(1×10^5)(1.9×10^n)?
2. find (5.3×10^3)-(8×10^2). express your answer in scientific notation.
3. what is the mass of 30,000 molecules? express your answer in scientific notation.

mass of one molecule of oxygen = 5.3×10^(-23) gram

1. critique reasoning your friend says that the product of 4.8×10^8 and 2×10^(-3) is 9.6×10^(-5). is this answer correct? explain.
2. a certain star is 4.3×10^2 light - years from earth. one light - year is about 5.9×10^12 miles. how far from earth (in miles) is the star? express your answer in scientific notation.

Explanation:

Step1: Multiply the coefficients and add the exponents of 10

For $(7\times 10^{-6})(7\times 10^{-6})$, we multiply the coefficients $7\times7 = 49$ and add the exponents of 10: $-6+( - 6)=-12$. So we have $(7\times7)\times(10^{-6}\times10^{-6})=49\times10^{-12}$.

Step2: Convert to proper scientific - notation

Since $49 = 4.9\times10^{1}$, then $49\times10^{-12}=4.9\times10^{1}\times10^{-12}$. Using the rule $a^{m}\times a^{n}=a^{m + n}$, we get $4.9\times10^{1+( - 12)}=4.9\times10^{-11}$.

Step3: For problem 8

$(3.76\times 10^{5})+(7.44\times 10^{5})=(3.76 + 7.44)\times10^{5}=11.2\times10^{5}$. Convert to proper scientific - notation: $11.2\times10^{5}=1.12\times10^{1}\times10^{5}=1.12\times10^{6}$.

Step4: For problem 9

$(1\times 10^{5})(1.9\times 10^{n})=(1\times1.9)\times(10^{5}\times10^{n})=1.9\times10^{5 + n}$. Since $1.9\times10^{7}=1.9\times10^{5 + n}$, then $7=5 + n$, and $n = 2$.

Step5: For problem 10

Rewrite the numbers with the same exponent of 10. $5.3\times10^{3}=53\times10^{2}$. Then $(5.3\times10^{3})-(8\times10^{2})=(53\times10^{2})-(8\times10^{2})=(53 - 8)\times10^{2}=45\times10^{2}$. Convert to proper scientific - notation: $45\times10^{2}=4.5\times10^{1}\times10^{2}=4.5\times10^{3}$.

Step6: For problem 11

The mass of one oxygen molecule is $5.3\times10^{-23}$ grams. The mass of 30000 molecules (where $30000 = 3\times10^{4}$) is $(3\times10^{4})\times(5.3\times10^{-23})=(3\times5.3)\times(10^{4}\times10^{-23})=15.9\times10^{-19}=1.59\times10^{-18}$ grams.

Step7: For problem 12

$(4.8\times10^{8})\times(2\times10^{-3})=(4.8\times2)\times(10^{8}\times10^{-3})=9.6\times10^{8+( - 3)}=9.6\times10^{5}$, not $9.6\times10^{-5}$. So the answer is incorrect.

Step8: For problem 14

The star is $4.3\times10^{2}$ light - years away. One light - year is about $5.9\times10^{12}$ miles. So the distance in miles is $(4.3\times10^{2})\times(5.9\times10^{12})=(4.3\times5.9)\times(10^{2}\times10^{12})=25.37\times10^{14}=2.537\times10^{15}$ miles.

Answer:

1. $4.9\times10^{-11}$
2. $1.12\times10^{6}$
3. $n = 2$
4. $4.5\times10^{3}$
5. $1.59\times10^{-18}$ grams
6. Incorrect. The correct product is $9.6\times10^{5}$
7. $2.537\times10^{15}$ miles